lang
stringclasses 1
value | prompt
stringlengths 1.38k
11.3k
| eval_prompt
stringlengths 37
8.09k
| ground_truth
stringlengths 1
328
| unit_tests
stringclasses 145
values | task_id
stringlengths 23
25
| split
stringclasses 2
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|---|---|---|---|---|---|---|
python
|
Complete the code in python to solve this programming problem:
Description: Eric has an array $$$b$$$ of length $$$m$$$, then he generates $$$n$$$ additional arrays $$$c_1, c_2, \dots, c_n$$$, each of length $$$m$$$, from the array $$$b$$$, by the following way:Initially, $$$c_i = b$$$ for every $$$1 \le i \le n$$$. Eric secretly chooses an integer $$$k$$$ $$$(1 \le k \le n)$$$ and chooses $$$c_k$$$ to be the special array.There are two operations that Eric can perform on an array $$$c_t$$$: Operation 1: Choose two integers $$$i$$$ and $$$j$$$ ($$$2 \leq i < j \leq m-1$$$), subtract $$$1$$$ from both $$$c_t[i]$$$ and $$$c_t[j]$$$, and add $$$1$$$ to both $$$c_t[i-1]$$$ and $$$c_t[j+1]$$$. That operation can only be used on a non-special array, that is when $$$t \neq k$$$.; Operation 2: Choose two integers $$$i$$$ and $$$j$$$ ($$$2 \leq i < j \leq m-2$$$), subtract $$$1$$$ from both $$$c_t[i]$$$ and $$$c_t[j]$$$, and add $$$1$$$ to both $$$c_t[i-1]$$$ and $$$c_t[j+2]$$$. That operation can only be used on a special array, that is when $$$t = k$$$.Note that Eric can't perform an operation if any element of the array will become less than $$$0$$$ after that operation.Now, Eric does the following: For every non-special array $$$c_i$$$ ($$$i \neq k$$$), Eric uses only operation 1 on it at least once. For the special array $$$c_k$$$, Eric uses only operation 2 on it at least once.Lastly, Eric discards the array $$$b$$$.For given arrays $$$c_1, c_2, \dots, c_n$$$, your task is to find out the special array, i.e. the value $$$k$$$. Also, you need to find the number of times of operation $$$2$$$ was used on it.
Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Description of test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$3 \leq n \leq 10^5$$$, $$$7 \leq m \leq 3 \cdot 10^5$$$) — the number of arrays given to you, and the length of each array. The next $$$n$$$ lines contains $$$m$$$ integers each, $$$c_{i,1}, c_{i,2}, \dots , c_{i,m}$$$. It is guaranteed that each element of the discarded array $$$b$$$ is in the range $$$[0,10^6]$$$, and therefore $$$0 \leq c_{i,j} \leq 3 \cdot 10^{11}$$$ for all possible pairs of $$$(i,j)$$$. It is guaranteed that the sum of $$$n \cdot m$$$ over all test cases does not exceed $$$10^6$$$. It is guaranteed that the input is generated according to the procedure above.
Output Specification: For each test case, output one line containing two integers — the index of the special array, and the number of times that Operation 2 was performed on it. It can be shown that under the constraints given in the problem, this value is unique and won't exceed $$$10^{18}$$$, so you can represent it as a $$$64$$$-bit integer. It can also be shown that the index of the special array is uniquely determined. In this problem, hacks are disabled.
Notes: NoteIn the first test case, the secret array $$$b$$$ is $$$[0, 1, 1, 1, 1, 1, 1, 1, 0]$$$. Array $$$c_1$$$ and array $$$c_2$$$ are generated by using operation 1. Array $$$c_3$$$ is generated by using operation 2.For Array $$$c_1$$$,you can choose $$$i=4$$$ and $$$j=5$$$ perform Operation 1 one time to generate it. For Array $$$c_2$$$, you can choose $$$i=6$$$ and $$$j=7$$$ perform Operation 1 one time to generate it. For Array $$$c_3$$$,you can choose $$$i=4$$$ and $$$j=5$$$ perform Operation 2 one time to generate it.In the second test case, the secret array $$$b$$$ is $$$[20, 20, 20, 20, 20, 20, 20]$$$. You can also find that array $$$c_1$$$ and array $$$c_2$$$ are generated by using Operation 1. Array $$$c_3$$$ is generated by using Operation 2.In the third test case, the secret array $$$b$$$ is $$$[20, 20, 20, 20, 20, 20, 20, 20, 20]$$$. You can also find that array $$$c_1$$$ and array $$$c_2$$$ are generated by using Operation 1. Array $$$c_3$$$ is generated by using Operation 2.
Code:
for t in range(int(input())):
N, M = map(int, input().split())
minV = int(1e20)
maxV = -1
ret = 0
for index in range(N):
mlist = list(map(int, input().split()))
cur = sum([index*val for index, val in enumerate(mlist)])
minV = min(minV, cur)
if maxV < cur:
# TODO: Your code here
maxV = max(maxV, cur)
print(f"{ret} {maxV-minV}")
|
for t in range(int(input())):
N, M = map(int, input().split())
minV = int(1e20)
maxV = -1
ret = 0
for index in range(N):
mlist = list(map(int, input().split()))
cur = sum([index*val for index, val in enumerate(mlist)])
minV = min(minV, cur)
if maxV < cur:
{{completion}}
maxV = max(maxV, cur)
print(f"{ret} {maxV-minV}")
|
ret = index+1
|
[{"input": "7\n3 9\n0 1 2 0 0 2 1 1 0\n0 1 1 1 2 0 0 2 0\n0 1 2 0 0 1 2 1 0\n3 7\n25 15 20 15 25 20 20\n26 14 20 14 26 20 20\n25 15 20 15 20 20 25\n3 9\n25 15 20 15 25 20 20 20 20\n26 14 20 14 26 20 20 20 20\n25 15 20 15 25 15 20 20 25\n3 11\n25 15 20 15 25 20 20 20 20 20 20\n26 14 20 14 26 20 20 20 20 20 20\n25 15 20 15 25 20 15 20 20 20 25\n3 13\n25 15 20 15 25 20 20 20 20 20 20 20 20\n26 14 20 14 26 20 20 20 20 20 20 20 20\n25 15 20 15 25 20 20 15 20 20 20 20 25\n3 15\n25 15 20 15 25 20 20 20 20 20 20 20 20 20 20\n26 14 20 14 26 20 20 20 20 20 20 20 20 20 20\n25 15 20 15 25 20 20 20 15 20 20 20 20 20 25\n3 9\n909459 479492 676924 224197 162866 164495 193268 742456 728277\n948845 455424 731850 327890 304150 237351 251763 225845 798316\n975446 401170 792914 272263 300770 242037 236619 334316 725899", "output": ["3 1\n3 10\n3 15\n3 20\n3 25\n3 30\n1 1378716"]}]
|
block_completion_002633
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a string $$$s$$$, consisting of lowercase Latin letters.You are asked $$$q$$$ queries about it: given another string $$$t$$$, consisting of lowercase Latin letters, perform the following steps: concatenate $$$s$$$ and $$$t$$$; calculate the prefix function of the resulting string $$$s+t$$$; print the values of the prefix function on positions $$$|s|+1, |s|+2, \dots, |s|+|t|$$$ ($$$|s|$$$ and $$$|t|$$$ denote the lengths of strings $$$s$$$ and $$$t$$$, respectively); revert the string back to $$$s$$$. The prefix function of a string $$$a$$$ is a sequence $$$p_1, p_2, \dots, p_{|a|}$$$, where $$$p_i$$$ is the maximum value of $$$k$$$ such that $$$k < i$$$ and $$$a[1..k]=a[i-k+1..i]$$$ ($$$a[l..r]$$$ denotes a contiguous substring of a string $$$a$$$ from a position $$$l$$$ to a position $$$r$$$, inclusive). In other words, it's the longest proper prefix of the string $$$a[1..i]$$$ that is equal to its suffix of the same length.
Input Specification: The first line contains a non-empty string $$$s$$$ ($$$1 \le |s| \le 10^6$$$), consisting of lowercase Latin letters. The second line contains a single integer $$$q$$$ ($$$1 \le q \le 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contains a query: a non-empty string $$$t$$$ ($$$1 \le |t| \le 10$$$), consisting of lowercase Latin letters.
Output Specification: For each query, print the values of the prefix function of a string $$$s+t$$$ on positions $$$|s|+1, |s|+2, \dots, |s|+|t|$$$.
Code:
import sys
input = sys.stdin.readline
s = input().strip()
l = len(s)
d, p = [], [0]
d.append([1 if s[0] == chr(x+97) else 0 for x in range(26)])
for idx, i in enumerate(s):
if idx == 0: # TODO: Your code here
x = ord(i) - 97
d.append(d[p[idx-1]][:])
d[idx][x] = idx+1
p.append(d[p[idx-1]][x])
for i in range(10): p.append(0); d.append([])
for i in range(int(input())):
for idx, t in enumerate(input().strip()):
x, pos = ord(t) - 97, idx+l
d[pos] = d[p[pos-1]][:]
d[pos][x] = pos+1
p[pos] = d[p[pos-1]][x]
print(*p[l:l+idx+1])
|
import sys
input = sys.stdin.readline
s = input().strip()
l = len(s)
d, p = [], [0]
d.append([1 if s[0] == chr(x+97) else 0 for x in range(26)])
for idx, i in enumerate(s):
if idx == 0: {{completion}}
x = ord(i) - 97
d.append(d[p[idx-1]][:])
d[idx][x] = idx+1
p.append(d[p[idx-1]][x])
for i in range(10): p.append(0); d.append([])
for i in range(int(input())):
for idx, t in enumerate(input().strip()):
x, pos = ord(t) - 97, idx+l
d[pos] = d[p[pos-1]][:]
d[pos][x] = pos+1
p[pos] = d[p[pos-1]][x]
print(*p[l:l+idx+1])
|
continue
|
[{"input": "aba\n6\ncaba\naba\nbababa\naaaa\nb\nforces", "output": ["0 1 2 3 \n1 2 3 \n2 3 4 5 6 7 \n1 1 1 1 \n2 \n0 0 0 0 0 0"]}, {"input": "aacba\n4\naaca\ncbbb\naab\nccaca", "output": ["2 2 3 1 \n0 0 0 0 \n2 2 0 \n0 0 1 0 1"]}]
|
block_completion_002696
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a string $$$s$$$, consisting of lowercase Latin letters.You are asked $$$q$$$ queries about it: given another string $$$t$$$, consisting of lowercase Latin letters, perform the following steps: concatenate $$$s$$$ and $$$t$$$; calculate the prefix function of the resulting string $$$s+t$$$; print the values of the prefix function on positions $$$|s|+1, |s|+2, \dots, |s|+|t|$$$ ($$$|s|$$$ and $$$|t|$$$ denote the lengths of strings $$$s$$$ and $$$t$$$, respectively); revert the string back to $$$s$$$. The prefix function of a string $$$a$$$ is a sequence $$$p_1, p_2, \dots, p_{|a|}$$$, where $$$p_i$$$ is the maximum value of $$$k$$$ such that $$$k < i$$$ and $$$a[1..k]=a[i-k+1..i]$$$ ($$$a[l..r]$$$ denotes a contiguous substring of a string $$$a$$$ from a position $$$l$$$ to a position $$$r$$$, inclusive). In other words, it's the longest proper prefix of the string $$$a[1..i]$$$ that is equal to its suffix of the same length.
Input Specification: The first line contains a non-empty string $$$s$$$ ($$$1 \le |s| \le 10^6$$$), consisting of lowercase Latin letters. The second line contains a single integer $$$q$$$ ($$$1 \le q \le 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contains a query: a non-empty string $$$t$$$ ($$$1 \le |t| \le 10$$$), consisting of lowercase Latin letters.
Output Specification: For each query, print the values of the prefix function of a string $$$s+t$$$ on positions $$$|s|+1, |s|+2, \dots, |s|+|t|$$$.
Code:
import sys
input = sys.stdin.readline
s = input().strip()
l = len(s)
d, p = [], [0]
d.append([1 if s[0] == chr(x+97) else 0 for x in range(26)])
for idx, i in enumerate(s):
if idx == 0: continue
x = ord(i) - 97
d.append(d[p[idx-1]][:])
d[idx][x] = idx+1
p.append(d[p[idx-1]][x])
for i in range(10): p.append(0); d.append([])
for i in range(int(input())):
for idx, t in enumerate(input().strip()):
# TODO: Your code here
print(*p[l:l+idx+1])
|
import sys
input = sys.stdin.readline
s = input().strip()
l = len(s)
d, p = [], [0]
d.append([1 if s[0] == chr(x+97) else 0 for x in range(26)])
for idx, i in enumerate(s):
if idx == 0: continue
x = ord(i) - 97
d.append(d[p[idx-1]][:])
d[idx][x] = idx+1
p.append(d[p[idx-1]][x])
for i in range(10): p.append(0); d.append([])
for i in range(int(input())):
for idx, t in enumerate(input().strip()):
{{completion}}
print(*p[l:l+idx+1])
|
x, pos = ord(t) - 97, idx+l
d[pos] = d[p[pos-1]][:]
d[pos][x] = pos+1
p[pos] = d[p[pos-1]][x]
|
[{"input": "aba\n6\ncaba\naba\nbababa\naaaa\nb\nforces", "output": ["0 1 2 3 \n1 2 3 \n2 3 4 5 6 7 \n1 1 1 1 \n2 \n0 0 0 0 0 0"]}, {"input": "aacba\n4\naaca\ncbbb\naab\nccaca", "output": ["2 2 3 1 \n0 0 0 0 \n2 2 0 \n0 0 1 0 1"]}]
|
block_completion_002697
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a string $$$s$$$, consisting of lowercase Latin letters.You are asked $$$q$$$ queries about it: given another string $$$t$$$, consisting of lowercase Latin letters, perform the following steps: concatenate $$$s$$$ and $$$t$$$; calculate the prefix function of the resulting string $$$s+t$$$; print the values of the prefix function on positions $$$|s|+1, |s|+2, \dots, |s|+|t|$$$ ($$$|s|$$$ and $$$|t|$$$ denote the lengths of strings $$$s$$$ and $$$t$$$, respectively); revert the string back to $$$s$$$. The prefix function of a string $$$a$$$ is a sequence $$$p_1, p_2, \dots, p_{|a|}$$$, where $$$p_i$$$ is the maximum value of $$$k$$$ such that $$$k < i$$$ and $$$a[1..k]=a[i-k+1..i]$$$ ($$$a[l..r]$$$ denotes a contiguous substring of a string $$$a$$$ from a position $$$l$$$ to a position $$$r$$$, inclusive). In other words, it's the longest proper prefix of the string $$$a[1..i]$$$ that is equal to its suffix of the same length.
Input Specification: The first line contains a non-empty string $$$s$$$ ($$$1 \le |s| \le 10^6$$$), consisting of lowercase Latin letters. The second line contains a single integer $$$q$$$ ($$$1 \le q \le 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contains a query: a non-empty string $$$t$$$ ($$$1 \le |t| \le 10$$$), consisting of lowercase Latin letters.
Output Specification: For each query, print the values of the prefix function of a string $$$s+t$$$ on positions $$$|s|+1, |s|+2, \dots, |s|+|t|$$$.
Code:
def get_next(j, k, nxt, p):
while p[j] != '$':
if k == -1 or p[j] == p[k]:
j += 1
k += 1
if p[j] == p[k]:
nxt[j] = nxt[k]
else:
# TODO: Your code here
else:
k = nxt[k]
return j, k, nxt
def solve():
s = input().strip()
len_s = len(s)
ns = [ch for ch in s]
for i in range(11):
ns.append('$')
# print(ns)
j, k, nxt = get_next(0, -1, [-1 for i in range(len(ns))], ns)
q = int(input().strip())
for _ in range(q):
t = input().strip()
ans = []
for i in range(10):
ns[i + len_s] = '$'
for i in range(len(t)):
ns[i + len_s] = t[i]
# print(ns)
nj, nk, n_nxt = get_next(j, k, nxt, ns)
# print(n_nxt)
ans.append(n_nxt[len_s + i + 1])
print(' '.join(map(str, ans)))
if __name__ == '__main__':
# t = int(input().strip())
# for _ in range(t):
solve()
|
def get_next(j, k, nxt, p):
while p[j] != '$':
if k == -1 or p[j] == p[k]:
j += 1
k += 1
if p[j] == p[k]:
nxt[j] = nxt[k]
else:
{{completion}}
else:
k = nxt[k]
return j, k, nxt
def solve():
s = input().strip()
len_s = len(s)
ns = [ch for ch in s]
for i in range(11):
ns.append('$')
# print(ns)
j, k, nxt = get_next(0, -1, [-1 for i in range(len(ns))], ns)
q = int(input().strip())
for _ in range(q):
t = input().strip()
ans = []
for i in range(10):
ns[i + len_s] = '$'
for i in range(len(t)):
ns[i + len_s] = t[i]
# print(ns)
nj, nk, n_nxt = get_next(j, k, nxt, ns)
# print(n_nxt)
ans.append(n_nxt[len_s + i + 1])
print(' '.join(map(str, ans)))
if __name__ == '__main__':
# t = int(input().strip())
# for _ in range(t):
solve()
|
nxt[j] = k
|
[{"input": "aba\n6\ncaba\naba\nbababa\naaaa\nb\nforces", "output": ["0 1 2 3 \n1 2 3 \n2 3 4 5 6 7 \n1 1 1 1 \n2 \n0 0 0 0 0 0"]}, {"input": "aacba\n4\naaca\ncbbb\naab\nccaca", "output": ["2 2 3 1 \n0 0 0 0 \n2 2 0 \n0 0 1 0 1"]}]
|
block_completion_002698
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You have an image file of size $$$2 \times 2$$$, consisting of $$$4$$$ pixels. Each pixel can have one of $$$26$$$ different colors, denoted by lowercase Latin letters.You want to recolor some of the pixels of the image so that all $$$4$$$ pixels have the same color. In one move, you can choose no more than two pixels of the same color and paint them into some other color (if you choose two pixels, both should be painted into the same color).What is the minimum number of moves you have to make in order to fulfill your goal?
Input Specification: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case consists of two lines. Each of these lines contains two lowercase letters of Latin alphabet without any separators, denoting a row of pixels in the image.
Output Specification: For each test case, print one integer — the minimum number of moves you have to make so that all $$$4$$$ pixels of the image have the same color.
Notes: NoteLet's analyze the test cases of the example.In the first test case, you can paint the bottom left pixel and the top right pixel (which share the same color) into the color r, so all pixels have this color.In the second test case, two moves are enough: paint both top pixels, which have the same color c, into the color b; paint the bottom left pixel into the color b. In the third test case, all pixels already have the same color.In the fourth test case, you may leave any of the pixels unchanged, and paint all three other pixels into the color of that pixel in three moves.In the fifth test case, you can paint both top pixels into the color x.
Code:
import sys
if __name__ == "__main__":
input_arr = sys.stdin.read().split()
tc = int(input_arr[0])
pos = 1
for case in range(tc):
a1, a2 = input_arr[pos:pos + 2]
char = []
for i in a1:
char.append(i)
for j in a2:
char.append(j)
l = len(set(char))
if l == 4:
print(3)
elif l == 3:
# TODO: Your code here
elif l == 2:
print(1)
elif l == 1:
print(0)
pos += 2
|
import sys
if __name__ == "__main__":
input_arr = sys.stdin.read().split()
tc = int(input_arr[0])
pos = 1
for case in range(tc):
a1, a2 = input_arr[pos:pos + 2]
char = []
for i in a1:
char.append(i)
for j in a2:
char.append(j)
l = len(set(char))
if l == 4:
print(3)
elif l == 3:
{{completion}}
elif l == 2:
print(1)
elif l == 1:
print(0)
pos += 2
|
print(2)
|
[{"input": "5\n\nrb\n\nbr\n\ncc\n\nwb\n\naa\n\naa\n\nab\n\ncd\n\nyy\n\nxx", "output": ["1\n2\n0\n3\n1"]}]
|
block_completion_002714
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You have an image file of size $$$2 \times 2$$$, consisting of $$$4$$$ pixels. Each pixel can have one of $$$26$$$ different colors, denoted by lowercase Latin letters.You want to recolor some of the pixels of the image so that all $$$4$$$ pixels have the same color. In one move, you can choose no more than two pixels of the same color and paint them into some other color (if you choose two pixels, both should be painted into the same color).What is the minimum number of moves you have to make in order to fulfill your goal?
Input Specification: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case consists of two lines. Each of these lines contains two lowercase letters of Latin alphabet without any separators, denoting a row of pixels in the image.
Output Specification: For each test case, print one integer — the minimum number of moves you have to make so that all $$$4$$$ pixels of the image have the same color.
Notes: NoteLet's analyze the test cases of the example.In the first test case, you can paint the bottom left pixel and the top right pixel (which share the same color) into the color r, so all pixels have this color.In the second test case, two moves are enough: paint both top pixels, which have the same color c, into the color b; paint the bottom left pixel into the color b. In the third test case, all pixels already have the same color.In the fourth test case, you may leave any of the pixels unchanged, and paint all three other pixels into the color of that pixel in three moves.In the fifth test case, you can paint both top pixels into the color x.
Code:
import sys
if __name__ == "__main__":
input_arr = sys.stdin.read().split()
tc = int(input_arr[0])
pos = 1
for case in range(tc):
a1, a2 = input_arr[pos:pos + 2]
char = []
for i in a1:
char.append(i)
for j in a2:
char.append(j)
l = len(set(char))
if l == 4:
print(3)
elif l == 3:
print(2)
elif l == 2:
# TODO: Your code here
elif l == 1:
print(0)
pos += 2
|
import sys
if __name__ == "__main__":
input_arr = sys.stdin.read().split()
tc = int(input_arr[0])
pos = 1
for case in range(tc):
a1, a2 = input_arr[pos:pos + 2]
char = []
for i in a1:
char.append(i)
for j in a2:
char.append(j)
l = len(set(char))
if l == 4:
print(3)
elif l == 3:
print(2)
elif l == 2:
{{completion}}
elif l == 1:
print(0)
pos += 2
|
print(1)
|
[{"input": "5\n\nrb\n\nbr\n\ncc\n\nwb\n\naa\n\naa\n\nab\n\ncd\n\nyy\n\nxx", "output": ["1\n2\n0\n3\n1"]}]
|
block_completion_002715
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given two arrays $$$a$$$ and $$$b$$$, consisting of $$$n$$$ integers each.Let's define a function $$$f(a, b)$$$ as follows: let's define an array $$$c$$$ of size $$$n$$$, where $$$c_i = a_i \oplus b_i$$$ ($$$\oplus$$$ denotes bitwise XOR); the value of the function is $$$c_1 \mathbin{\&} c_2 \mathbin{\&} \cdots \mathbin{\&} c_n$$$ (i.e. bitwise AND of the entire array $$$c$$$). Find the maximum value of the function $$$f(a, b)$$$ if you can reorder the array $$$b$$$ in an arbitrary way (leaving the initial order is also an option).
Input Specification: The first line contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the size of arrays $$$a$$$ and $$$b$$$. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i < 2^{30}$$$). The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$0 \le b_i < 2^{30}$$$). The sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: For each test case print one integer — the maximum value of the function $$$f(a, b)$$$ if you can reorder the array $$$b$$$ in an arbitrary way.
Code:
import sys
from itertools import permutations, combinations
ls = []
for l in sys.stdin:
lst = l.rstrip('\n')
if len(lst) > 0:
ls.append(lst)
def solve(n, a, b):
ps = [((list(range(n))), (list(range(n))))]
res = (1<<30) - 1
for k in range(30, -1, -1):
next_ps = []
for (pa, pb) in ps:
a0, a1, b0, b1 = [], [], [], []
for pai in pa:
if a[pai] & (1<<k) == 0: a0.append(pai)
else: # TODO: Your code here
for pbi in pb:
if b[pbi] & (1<<k) == 0: b0.append(pbi)
else: b1.append(pbi)
if len(a0) == len(b1):
res = res & (res | (1 << k))
if len(a0) > 0 and len(b1) > 0: next_ps.append((a0, b1))
if len(a1) > 0 and len(b0) > 0: next_ps.append((a1, b0))
else:
res = res & ~(1 << k)
next_ps.append((pa, pb))
ps = next_ps if int(res & (1<<k)) != 0 else ps
return res
#return rec(a, b, max_order-1)
for i in range(1, len(ls)-1, 3):
n = int(ls[i])
a = [int(x) for x in ls[i+1].split(' ')]
b = [int(x) for x in ls[i+2].split(' ')]
print(solve(n, a, b))
|
import sys
from itertools import permutations, combinations
ls = []
for l in sys.stdin:
lst = l.rstrip('\n')
if len(lst) > 0:
ls.append(lst)
def solve(n, a, b):
ps = [((list(range(n))), (list(range(n))))]
res = (1<<30) - 1
for k in range(30, -1, -1):
next_ps = []
for (pa, pb) in ps:
a0, a1, b0, b1 = [], [], [], []
for pai in pa:
if a[pai] & (1<<k) == 0: a0.append(pai)
else: {{completion}}
for pbi in pb:
if b[pbi] & (1<<k) == 0: b0.append(pbi)
else: b1.append(pbi)
if len(a0) == len(b1):
res = res & (res | (1 << k))
if len(a0) > 0 and len(b1) > 0: next_ps.append((a0, b1))
if len(a1) > 0 and len(b0) > 0: next_ps.append((a1, b0))
else:
res = res & ~(1 << k)
next_ps.append((pa, pb))
ps = next_ps if int(res & (1<<k)) != 0 else ps
return res
#return rec(a, b, max_order-1)
for i in range(1, len(ls)-1, 3):
n = int(ls[i])
a = [int(x) for x in ls[i+1].split(' ')]
b = [int(x) for x in ls[i+2].split(' ')]
print(solve(n, a, b))
|
a1.append(pai)
|
[{"input": "3\n\n5\n\n1 0 0 3 3\n\n2 3 2 1 0\n\n3\n\n1 1 1\n\n0 0 3\n\n8\n\n0 1 2 3 4 5 6 7\n\n7 6 5 4 3 2 1 0", "output": ["2\n0\n7"]}]
|
block_completion_002743
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given two arrays $$$a$$$ and $$$b$$$, consisting of $$$n$$$ integers each.Let's define a function $$$f(a, b)$$$ as follows: let's define an array $$$c$$$ of size $$$n$$$, where $$$c_i = a_i \oplus b_i$$$ ($$$\oplus$$$ denotes bitwise XOR); the value of the function is $$$c_1 \mathbin{\&} c_2 \mathbin{\&} \cdots \mathbin{\&} c_n$$$ (i.e. bitwise AND of the entire array $$$c$$$). Find the maximum value of the function $$$f(a, b)$$$ if you can reorder the array $$$b$$$ in an arbitrary way (leaving the initial order is also an option).
Input Specification: The first line contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the size of arrays $$$a$$$ and $$$b$$$. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i < 2^{30}$$$). The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$0 \le b_i < 2^{30}$$$). The sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: For each test case print one integer — the maximum value of the function $$$f(a, b)$$$ if you can reorder the array $$$b$$$ in an arbitrary way.
Code:
import sys
from itertools import permutations, combinations
ls = []
for l in sys.stdin:
lst = l.rstrip('\n')
if len(lst) > 0:
ls.append(lst)
def solve(n, a, b):
ps = [((list(range(n))), (list(range(n))))]
res = (1<<30) - 1
for k in range(30, -1, -1):
next_ps = []
for (pa, pb) in ps:
a0, a1, b0, b1 = [], [], [], []
for pai in pa:
if a[pai] & (1<<k) == 0: a0.append(pai)
else: a1.append(pai)
for pbi in pb:
if b[pbi] & (1<<k) == 0: b0.append(pbi)
else: # TODO: Your code here
if len(a0) == len(b1):
res = res & (res | (1 << k))
if len(a0) > 0 and len(b1) > 0: next_ps.append((a0, b1))
if len(a1) > 0 and len(b0) > 0: next_ps.append((a1, b0))
else:
res = res & ~(1 << k)
next_ps.append((pa, pb))
ps = next_ps if int(res & (1<<k)) != 0 else ps
return res
#return rec(a, b, max_order-1)
for i in range(1, len(ls)-1, 3):
n = int(ls[i])
a = [int(x) for x in ls[i+1].split(' ')]
b = [int(x) for x in ls[i+2].split(' ')]
print(solve(n, a, b))
|
import sys
from itertools import permutations, combinations
ls = []
for l in sys.stdin:
lst = l.rstrip('\n')
if len(lst) > 0:
ls.append(lst)
def solve(n, a, b):
ps = [((list(range(n))), (list(range(n))))]
res = (1<<30) - 1
for k in range(30, -1, -1):
next_ps = []
for (pa, pb) in ps:
a0, a1, b0, b1 = [], [], [], []
for pai in pa:
if a[pai] & (1<<k) == 0: a0.append(pai)
else: a1.append(pai)
for pbi in pb:
if b[pbi] & (1<<k) == 0: b0.append(pbi)
else: {{completion}}
if len(a0) == len(b1):
res = res & (res | (1 << k))
if len(a0) > 0 and len(b1) > 0: next_ps.append((a0, b1))
if len(a1) > 0 and len(b0) > 0: next_ps.append((a1, b0))
else:
res = res & ~(1 << k)
next_ps.append((pa, pb))
ps = next_ps if int(res & (1<<k)) != 0 else ps
return res
#return rec(a, b, max_order-1)
for i in range(1, len(ls)-1, 3):
n = int(ls[i])
a = [int(x) for x in ls[i+1].split(' ')]
b = [int(x) for x in ls[i+2].split(' ')]
print(solve(n, a, b))
|
b1.append(pbi)
|
[{"input": "3\n\n5\n\n1 0 0 3 3\n\n2 3 2 1 0\n\n3\n\n1 1 1\n\n0 0 3\n\n8\n\n0 1 2 3 4 5 6 7\n\n7 6 5 4 3 2 1 0", "output": ["2\n0\n7"]}]
|
block_completion_002744
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given an array $$$a_1, a_2, \dots, a_n$$$, which is sorted in non-descending order. You decided to perform the following steps to create array $$$b_1, b_2, \dots, b_n$$$: Create an array $$$d$$$ consisting of $$$n$$$ arbitrary non-negative integers. Set $$$b_i = a_i + d_i$$$ for each $$$b_i$$$. Sort the array $$$b$$$ in non-descending order. You are given the resulting array $$$b$$$. For each index $$$i$$$, calculate what is the minimum and maximum possible value of $$$d_i$$$ you can choose in order to get the given array $$$b$$$.Note that the minimum (maximum) $$$d_i$$$-s are independent of each other, i. e. they can be obtained from different possible arrays $$$d$$$.
Input Specification: The first line contains the single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of arrays $$$a$$$, $$$b$$$ and $$$d$$$. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$; $$$a_i \le a_{i+1}$$$) — the array $$$a$$$ in non-descending order. The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$; $$$b_i \le b_{i+1}$$$) — the array $$$b$$$ in non-descending order. Additional constraints on the input: there is at least one way to obtain the array $$$b$$$ from the $$$a$$$ by choosing an array $$$d$$$ consisting of non-negative integers; the sum of $$$n$$$ doesn't exceed $$$2 \cdot 10^5$$$.
Output Specification: For each test case, print two lines. In the first line, print $$$n$$$ integers $$$d_1^{min}, d_2^{min}, \dots, d_n^{min}$$$, where $$$d_i^{min}$$$ is the minimum possible value you can add to $$$a_i$$$. Secondly, print $$$n$$$ integers $$$d_1^{max}, d_2^{max}, \dots, d_n^{max}$$$, where $$$d_i^{max}$$$ is the maximum possible value you can add to $$$a_i$$$. All $$$d_i^{min}$$$ and $$$d_i^{max}$$$ values are independent of each other. In other words, for each $$$i$$$, $$$d_i^{min}$$$ is just the minimum value among all possible values of $$$d_i$$$.
Notes: NoteIn the first test case, in order to get $$$d_1^{min} = 5$$$, we can choose, for example, $$$d = [5, 10, 6]$$$. Then $$$b$$$ $$$=$$$ $$$[2+5,3+10,5+6]$$$ $$$=$$$ $$$[7,13,11]$$$ $$$=$$$ $$$[7,11,13]$$$.For $$$d_2^{min} = 4$$$, we can choose $$$d$$$ $$$=$$$ $$$[9, 4, 8]$$$. Then $$$b$$$ $$$=$$$ $$$[2+9,3+4,5+8]$$$ $$$=$$$ $$$[11,7,13]$$$ $$$=$$$ $$$[7,11,13]$$$.
Code:
import sys
from math import ceil
ls = []
for l in sys.stdin:
lst = l.rstrip('\n')
if len(lst) > 0:
ls.append(lst)
for l in ls:
sys.stderr.write(l + ('\n'))
sys.stderr.write("-- output --\n")
def at_least(L, x):
for i in range(len(L)):
if L[i] >= x:
return i
return len(L)-1
def at_most(L, x):
for i in range(len(L)-1, -1, -1):
if L[i] <= x:
return i
return 0
def solve(n, a, b):
dmin, dmax = [-1]*n, [-1]*n
imin, imax = 0, n-1
for i in range(n):
while b[imin] < a[i]:
imin += 1
dmin[i] = b[imin] - a[i]
dmax[n-1-i] = b[imax] - a[n-1-i]
if i < n-1:
if b[n-i-2] < a[n-i-1]:
# TODO: Your code here
return dmin, dmax
for i in range(1, len(ls)-1, 3):
n = int(ls[i])
a = [int(x) for x in ls[i+1].split(' ')]
b = [int(x) for x in ls[i+2].split(' ')]
dmin, dmax = solve(n, a, b)
print(*dmin)
print(*dmax)
|
import sys
from math import ceil
ls = []
for l in sys.stdin:
lst = l.rstrip('\n')
if len(lst) > 0:
ls.append(lst)
for l in ls:
sys.stderr.write(l + ('\n'))
sys.stderr.write("-- output --\n")
def at_least(L, x):
for i in range(len(L)):
if L[i] >= x:
return i
return len(L)-1
def at_most(L, x):
for i in range(len(L)-1, -1, -1):
if L[i] <= x:
return i
return 0
def solve(n, a, b):
dmin, dmax = [-1]*n, [-1]*n
imin, imax = 0, n-1
for i in range(n):
while b[imin] < a[i]:
imin += 1
dmin[i] = b[imin] - a[i]
dmax[n-1-i] = b[imax] - a[n-1-i]
if i < n-1:
if b[n-i-2] < a[n-i-1]:
{{completion}}
return dmin, dmax
for i in range(1, len(ls)-1, 3):
n = int(ls[i])
a = [int(x) for x in ls[i+1].split(' ')]
b = [int(x) for x in ls[i+2].split(' ')]
dmin, dmax = solve(n, a, b)
print(*dmin)
print(*dmax)
|
imax = n-i-2
|
[{"input": "4\n\n3\n\n2 3 5\n\n7 11 13\n\n1\n\n1000\n\n5000\n\n4\n\n1 2 3 4\n\n1 2 3 4\n\n4\n\n10 20 30 40\n\n22 33 33 55", "output": ["5 4 2\n11 10 8\n4000\n4000\n0 0 0 0\n0 0 0 0\n12 2 3 15\n23 13 3 15"]}]
|
block_completion_002756
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The robot is placed in the top left corner of a grid, consisting of $$$n$$$ rows and $$$m$$$ columns, in a cell $$$(1, 1)$$$.In one step, it can move into a cell, adjacent by a side to the current one: $$$(x, y) \rightarrow (x, y + 1)$$$; $$$(x, y) \rightarrow (x + 1, y)$$$; $$$(x, y) \rightarrow (x, y - 1)$$$; $$$(x, y) \rightarrow (x - 1, y)$$$. The robot can't move outside the grid.The cell $$$(s_x, s_y)$$$ contains a deadly laser. If the robot comes into some cell that has distance less than or equal to $$$d$$$ to the laser, it gets evaporated. The distance between two cells $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ is $$$|x_1 - x_2| + |y_1 - y_2|$$$.Print the smallest number of steps that the robot can take to reach the cell $$$(n, m)$$$ without getting evaporated or moving outside the grid. If it's not possible to reach the cell $$$(n, m)$$$, print -1.The laser is neither in the starting cell, nor in the ending cell. The starting cell always has distance greater than $$$d$$$ to the laser.
Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of testcases. The only line of each testcase contains five integers $$$n, m, s_x, s_y, d$$$ ($$$2 \le n, m \le 1000$$$; $$$1 \le s_x \le n$$$; $$$1 \le s_y \le m$$$; $$$0 \le d \le n + m$$$) — the size of the grid, the cell that contains the laser and the evaporating distance of the laser. The laser is neither in the starting cell, nor in the ending cell ($$$(s_x, s_y) \neq (1, 1)$$$ and $$$(s_x, s_y) \neq (n, m)$$$). The starting cell $$$(1, 1)$$$ always has distance greater than $$$d$$$ to the laser ($$$|s_x - 1| + |s_y - 1| > d$$$).
Output Specification: For each testcase, print a single integer. If it's possible to reach the cell $$$(n, m)$$$ from $$$(1, 1)$$$ without getting evaporated or moving outside the grid, then print the smallest amount of steps it can take the robot to reach it. Otherwise, print -1.
Code:
import sys
ls = []
for l in sys.stdin:
lst = l.rstrip('\n')
if len(lst) > 0:
# TODO: Your code here
for l in ls:
sys.stderr.write(l + ('\n'))
sys.stderr.write("-- output --\n")
def ins(u, mn, mx):
return u[0] >= mn[0] and u[1] >= mn[1] and u[0] <= mx[0] and u[1] <= mx[1]
def clmp(x, n):
if x < 0: return 0
if x >= n: return n-1
return x
def clp(u, n, m):
return (clmp(u[0], n), clmp(u[1], m))
def d(a, b):
return abs(a[0] - b[0]) + abs(a[1] - b[1])
def solve(n, m, sx, sy, d):
if d ==0: return n+m-2
smin = clp((sx-d, sy-d), n, m)
smax = clp((sx+d, sy+d), n, m)
if abs(smax[0]-smin[0]) >= n-1 or abs(smax[1]-smin[1]) >= m-1: return -1
if ins((0, 0), smin, smax) or ins((n-1, m-1), smin, smax) : return -1
return n+m-2
for l in ls[1:]:
n, m, sx, sy, d = [int(x) for x in l.split(' ')]
print(solve(n, m, sx-1, sy-1, d))
|
import sys
ls = []
for l in sys.stdin:
lst = l.rstrip('\n')
if len(lst) > 0:
{{completion}}
for l in ls:
sys.stderr.write(l + ('\n'))
sys.stderr.write("-- output --\n")
def ins(u, mn, mx):
return u[0] >= mn[0] and u[1] >= mn[1] and u[0] <= mx[0] and u[1] <= mx[1]
def clmp(x, n):
if x < 0: return 0
if x >= n: return n-1
return x
def clp(u, n, m):
return (clmp(u[0], n), clmp(u[1], m))
def d(a, b):
return abs(a[0] - b[0]) + abs(a[1] - b[1])
def solve(n, m, sx, sy, d):
if d ==0: return n+m-2
smin = clp((sx-d, sy-d), n, m)
smax = clp((sx+d, sy+d), n, m)
if abs(smax[0]-smin[0]) >= n-1 or abs(smax[1]-smin[1]) >= m-1: return -1
if ins((0, 0), smin, smax) or ins((n-1, m-1), smin, smax) : return -1
return n+m-2
for l in ls[1:]:
n, m, sx, sy, d = [int(x) for x in l.split(' ')]
print(solve(n, m, sx-1, sy-1, d))
|
ls.append(lst)
|
[{"input": "3\n\n2 3 1 3 0\n\n2 3 1 3 1\n\n5 5 3 4 1", "output": ["3\n-1\n8"]}]
|
block_completion_002786
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: The robot is placed in the top left corner of a grid, consisting of $$$n$$$ rows and $$$m$$$ columns, in a cell $$$(1, 1)$$$.In one step, it can move into a cell, adjacent by a side to the current one: $$$(x, y) \rightarrow (x, y + 1)$$$; $$$(x, y) \rightarrow (x + 1, y)$$$; $$$(x, y) \rightarrow (x, y - 1)$$$; $$$(x, y) \rightarrow (x - 1, y)$$$. The robot can't move outside the grid.The cell $$$(s_x, s_y)$$$ contains a deadly laser. If the robot comes into some cell that has distance less than or equal to $$$d$$$ to the laser, it gets evaporated. The distance between two cells $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ is $$$|x_1 - x_2| + |y_1 - y_2|$$$.Print the smallest number of steps that the robot can take to reach the cell $$$(n, m)$$$ without getting evaporated or moving outside the grid. If it's not possible to reach the cell $$$(n, m)$$$, print -1.The laser is neither in the starting cell, nor in the ending cell. The starting cell always has distance greater than $$$d$$$ to the laser.
Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of testcases. The only line of each testcase contains five integers $$$n, m, s_x, s_y, d$$$ ($$$2 \le n, m \le 1000$$$; $$$1 \le s_x \le n$$$; $$$1 \le s_y \le m$$$; $$$0 \le d \le n + m$$$) — the size of the grid, the cell that contains the laser and the evaporating distance of the laser. The laser is neither in the starting cell, nor in the ending cell ($$$(s_x, s_y) \neq (1, 1)$$$ and $$$(s_x, s_y) \neq (n, m)$$$). The starting cell $$$(1, 1)$$$ always has distance greater than $$$d$$$ to the laser ($$$|s_x - 1| + |s_y - 1| > d$$$).
Output Specification: For each testcase, print a single integer. If it's possible to reach the cell $$$(n, m)$$$ from $$$(1, 1)$$$ without getting evaporated or moving outside the grid, then print the smallest amount of steps it can take the robot to reach it. Otherwise, print -1.
Code:
import sys
ls = []
for l in sys.stdin:
lst = l.rstrip('\n')
if len(lst) > 0:
ls.append(lst)
for l in ls:
sys.stderr.write(l + ('\n'))
sys.stderr.write("-- output --\n")
def ins(u, mn, mx):
return u[0] >= mn[0] and u[1] >= mn[1] and u[0] <= mx[0] and u[1] <= mx[1]
def clmp(x, n):
if x < 0: # TODO: Your code here
if x >= n: return n-1
return x
def clp(u, n, m):
return (clmp(u[0], n), clmp(u[1], m))
def d(a, b):
return abs(a[0] - b[0]) + abs(a[1] - b[1])
def solve(n, m, sx, sy, d):
if d ==0: return n+m-2
smin = clp((sx-d, sy-d), n, m)
smax = clp((sx+d, sy+d), n, m)
if abs(smax[0]-smin[0]) >= n-1 or abs(smax[1]-smin[1]) >= m-1: return -1
if ins((0, 0), smin, smax) or ins((n-1, m-1), smin, smax) : return -1
return n+m-2
for l in ls[1:]:
n, m, sx, sy, d = [int(x) for x in l.split(' ')]
print(solve(n, m, sx-1, sy-1, d))
|
import sys
ls = []
for l in sys.stdin:
lst = l.rstrip('\n')
if len(lst) > 0:
ls.append(lst)
for l in ls:
sys.stderr.write(l + ('\n'))
sys.stderr.write("-- output --\n")
def ins(u, mn, mx):
return u[0] >= mn[0] and u[1] >= mn[1] and u[0] <= mx[0] and u[1] <= mx[1]
def clmp(x, n):
if x < 0: {{completion}}
if x >= n: return n-1
return x
def clp(u, n, m):
return (clmp(u[0], n), clmp(u[1], m))
def d(a, b):
return abs(a[0] - b[0]) + abs(a[1] - b[1])
def solve(n, m, sx, sy, d):
if d ==0: return n+m-2
smin = clp((sx-d, sy-d), n, m)
smax = clp((sx+d, sy+d), n, m)
if abs(smax[0]-smin[0]) >= n-1 or abs(smax[1]-smin[1]) >= m-1: return -1
if ins((0, 0), smin, smax) or ins((n-1, m-1), smin, smax) : return -1
return n+m-2
for l in ls[1:]:
n, m, sx, sy, d = [int(x) for x in l.split(' ')]
print(solve(n, m, sx-1, sy-1, d))
|
return 0
|
[{"input": "3\n\n2 3 1 3 0\n\n2 3 1 3 1\n\n5 5 3 4 1", "output": ["3\n-1\n8"]}]
|
block_completion_002787
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
_,(*a,),*r=(map(int,s.split())for s in open(0))
b=[[0],[0]]
for x in b:
for u,v in zip([0]+a,a):# TODO: Your code here
max=min
for s,t in r:l=b[s>t];print(l[t]-l[s])
|
_,(*a,),*r=(map(int,s.split())for s in open(0))
b=[[0],[0]]
for x in b:
for u,v in zip([0]+a,a):{{completion}}
max=min
for s,t in r:l=b[s>t];print(l[t]-l[s])
|
x+=x[-1]+max(0,u-v),
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
block_completion_002941
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
n,m=map(int,input().split())
c=[int(i) for i in input().split()]
f=[0]*n
g=[0]*n
for i in range(1,n):
f[i]=f[i-1]+max(0,c[i-1]-c[i])
g[-i-1]=g[-i]+max(0,c[-i]-c[-i-1])
for i in range(m):
x,y=map(int,input().split())
if x<y:
print(f[y-1]-f[x-1])
else:
# TODO: Your code here
|
n,m=map(int,input().split())
c=[int(i) for i in input().split()]
f=[0]*n
g=[0]*n
for i in range(1,n):
f[i]=f[i-1]+max(0,c[i-1]-c[i])
g[-i-1]=g[-i]+max(0,c[-i]-c[-i-1])
for i in range(m):
x,y=map(int,input().split())
if x<y:
print(f[y-1]-f[x-1])
else:
{{completion}}
|
print(g[y-1]-g[x-1])
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
block_completion_002942
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
R=lambda:map(int,input().split())
n,m=R()
*a,=R()
b=[[0],[0]]
f=max
for x in b:
for u,v in zip([0]+a,a):# TODO: Your code here
f=min
for _ in[0]*m:s,t=R();l=b[s>t];print(l[t]-l[s])
|
R=lambda:map(int,input().split())
n,m=R()
*a,=R()
b=[[0],[0]]
f=max
for x in b:
for u,v in zip([0]+a,a):{{completion}}
f=min
for _ in[0]*m:s,t=R();l=b[s>t];print(l[t]-l[s])
|
x+=x[-1]+f(0,u-v),
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
block_completion_002943
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
n,m=map(int,input().split())
world=['x']+list(map(int,input().split()))
L1=[0]
L2=[0]
for i in range(1,n):
L1.append(L1[i-1]+max(world[i]-world[i+1],0))
L2.append(L2[i-1]+max(world[i+1]-world[i],0))
for i in range(m):
s,t=map(int,input().split())
if s<t:
print(L1[t-1]-L1[s-1])
else:
# TODO: Your code here
|
n,m=map(int,input().split())
world=['x']+list(map(int,input().split()))
L1=[0]
L2=[0]
for i in range(1,n):
L1.append(L1[i-1]+max(world[i]-world[i+1],0))
L2.append(L2[i-1]+max(world[i+1]-world[i],0))
for i in range(m):
s,t=map(int,input().split())
if s<t:
print(L1[t-1]-L1[s-1])
else:
{{completion}}
|
print(L2[s-1]-L2[t-1])
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
block_completion_002944
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
n,m=[int(x) for x in input().split()]
a=[int(x) for x in input().split()]
ltr,rtl=[0],[0]
for i in range(1,n):
ltr.append(max(0,a[i-1]-a[i])+ltr[-1])
rtl.append(max(0,a[i]-a[i-1])+rtl[-1])
for i in range(m):
s,t=[int(x) for x in input().split()]
if s<=t:
print(ltr[t-1]-ltr[s-1])
else:
# TODO: Your code here
|
n,m=[int(x) for x in input().split()]
a=[int(x) for x in input().split()]
ltr,rtl=[0],[0]
for i in range(1,n):
ltr.append(max(0,a[i-1]-a[i])+ltr[-1])
rtl.append(max(0,a[i]-a[i-1])+rtl[-1])
for i in range(m):
s,t=[int(x) for x in input().split()]
if s<=t:
print(ltr[t-1]-ltr[s-1])
else:
{{completion}}
|
print(rtl[s-1]-rtl[t-1])
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
block_completion_002945
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
(n,m),(*a,),*r=(map(int,s.split())for s in open(0))
b=[[0],[0]]
for x in b:
for u,v in zip([0]+a,a):# TODO: Your code here
max=min
for s,t in r:l=b[s>t];print(l[t]-l[s])
|
(n,m),(*a,),*r=(map(int,s.split())for s in open(0))
b=[[0],[0]]
for x in b:
for u,v in zip([0]+a,a):{{completion}}
max=min
for s,t in r:l=b[s>t];print(l[t]-l[s])
|
x+=x[-1]+max(0,u-v),
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
block_completion_002946
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
n, m = map(int, input().split())
a = list(map(int, input().split()))
inc = [0]
dec = [0]
for i in range(n-1):
inc.append(inc[i] + max(0, a[i]-a[i+1]))
dec.append(dec[i] + max(0, a[i+1] - a[i]))
for i in range(m):
x, y = map(int, input().split())
#print(x, y)
#ans = 0
if x < y:
ans = inc[y-1] - inc[x-1]
else:
# TODO: Your code here
print(ans)
|
n, m = map(int, input().split())
a = list(map(int, input().split()))
inc = [0]
dec = [0]
for i in range(n-1):
inc.append(inc[i] + max(0, a[i]-a[i+1]))
dec.append(dec[i] + max(0, a[i+1] - a[i]))
for i in range(m):
x, y = map(int, input().split())
#print(x, y)
#ans = 0
if x < y:
ans = inc[y-1] - inc[x-1]
else:
{{completion}}
print(ans)
|
ans = dec[x-1] - dec[y-1]
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
block_completion_002947
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
R=lambda:map(int,input().split())
n,m=R()
*a,=R()
b=[[0],[0]]
for x in b:
for u,v in zip([0]+a,a):# TODO: Your code here
a=a[::-1]
b[1]=[0]+b[1][::-1]
for _ in[0]*m:s,t=R();l=b[s>t];print(abs(l[s]-l[t]))
|
R=lambda:map(int,input().split())
n,m=R()
*a,=R()
b=[[0],[0]]
for x in b:
for u,v in zip([0]+a,a):{{completion}}
a=a[::-1]
b[1]=[0]+b[1][::-1]
for _ in[0]*m:s,t=R();l=b[s>t];print(abs(l[s]-l[t]))
|
x+=x[-1]+max(0,u-v),
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
block_completion_002948
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
n, m = map(int, input().split());a = list(map(int, input().split()));l = [0] + [max(0, a[i] - a[i + 1]) for i in range(n - 1)];r = [0] + [max(0, a[i] - a[i - 1]) for i in range(1, n)]
for i in range(n - 1): # TODO: Your code here
for _ in range(m): s, t = map(int, input().split());print(l[t - 1] - l[s - 1]) if(s < t) else print(r[s - 1] - r[t - 1])
|
n, m = map(int, input().split());a = list(map(int, input().split()));l = [0] + [max(0, a[i] - a[i + 1]) for i in range(n - 1)];r = [0] + [max(0, a[i] - a[i - 1]) for i in range(1, n)]
for i in range(n - 1): {{completion}}
for _ in range(m): s, t = map(int, input().split());print(l[t - 1] - l[s - 1]) if(s < t) else print(r[s - 1] - r[t - 1])
|
l[i + 1] += l[i];r[i + 1] += r[i]
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
block_completion_002949
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
n, m = map(int, input().split());a = list(map(int, input().split()));l = [0] + [max(0, a[i] - a[i + 1]) for i in range(n - 1)];r = [0] + [max(0, a[i] - a[i - 1]) for i in range(1, n)]
for i in range(n - 1): l[i + 1] += l[i];r[i + 1] += r[i]
for _ in range(m): # TODO: Your code here
|
n, m = map(int, input().split());a = list(map(int, input().split()));l = [0] + [max(0, a[i] - a[i + 1]) for i in range(n - 1)];r = [0] + [max(0, a[i] - a[i - 1]) for i in range(1, n)]
for i in range(n - 1): l[i + 1] += l[i];r[i + 1] += r[i]
for _ in range(m): {{completion}}
|
s, t = map(int, input().split());print(l[t - 1] - l[s - 1]) if(s < t) else print(r[s - 1] - r[t - 1])
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
block_completion_002950
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
n,m=(map(int,input().split()))
l=list(map(int,input().split()))
f=[0]*n
b=[0]*n
d=0
for j in range(1,n):
d=d+max(0,l[j-1]-l[j])
f[j]=d
l=l[::-1]
d=0
for k in range(1,n):
d=d+max(0,l[k-1]-l[k])
b[k]=d
b=b[::-1]
for i in range(m):
s,t=(map(int,input().split()))
if s<t:
print(f[t-1]-f[s-1])
else:
# TODO: Your code here
|
n,m=(map(int,input().split()))
l=list(map(int,input().split()))
f=[0]*n
b=[0]*n
d=0
for j in range(1,n):
d=d+max(0,l[j-1]-l[j])
f[j]=d
l=l[::-1]
d=0
for k in range(1,n):
d=d+max(0,l[k-1]-l[k])
b[k]=d
b=b[::-1]
for i in range(m):
s,t=(map(int,input().split()))
if s<t:
print(f[t-1]-f[s-1])
else:
{{completion}}
|
print(b[t-1]-b[s-1])
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
block_completion_002951
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
import sys, collections, math, itertools
input = lambda: sys.stdin.readline().strip()
ints = lambda: list(map(int, input().split()))
Int = lambda: int(input())
n, m = ints()
a = ints()
s = int(m ** .5 + 1)
maxs = [max(a[i:i + s]) for i in range(0, m, s)]
for i in range(Int()):
ys, xs, yf, xf, k = ints()
yes = (xs - xf) % k == 0 and (ys - yf) % k == 0
if not yes:
# TODO: Your code here
mi, ma = min(xs, xf), max(xs, xf)
high = max([0] + a[mi:min(ma, (mi // s + 1) * s)] + a[max(mi, ma // s * s):ma])
for j in range(min(xs, xf) // s + 1, max(xs, xf) // s):
high = max(high, maxs[j])
if high < ys:
print('yes')
continue
print('yes' if ((high - ys) // k + 1) * k + ys <= n else 'no')
|
import sys, collections, math, itertools
input = lambda: sys.stdin.readline().strip()
ints = lambda: list(map(int, input().split()))
Int = lambda: int(input())
n, m = ints()
a = ints()
s = int(m ** .5 + 1)
maxs = [max(a[i:i + s]) for i in range(0, m, s)]
for i in range(Int()):
ys, xs, yf, xf, k = ints()
yes = (xs - xf) % k == 0 and (ys - yf) % k == 0
if not yes:
{{completion}}
mi, ma = min(xs, xf), max(xs, xf)
high = max([0] + a[mi:min(ma, (mi // s + 1) * s)] + a[max(mi, ma // s * s):ma])
for j in range(min(xs, xf) // s + 1, max(xs, xf) // s):
high = max(high, maxs[j])
if high < ys:
print('yes')
continue
print('yes' if ((high - ys) // k + 1) * k + ys <= n else 'no')
|
print('no')
continue
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
block_completion_002989
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
import sys, collections, math, itertools
input = lambda: sys.stdin.readline().strip()
ints = lambda: list(map(int, input().split()))
Int = lambda: int(input())
n, m = ints()
a = ints()
s = int(m ** .5 + 1)
maxs = [max(a[i:i + s]) for i in range(0, m, s)]
for i in range(Int()):
ys, xs, yf, xf, k = ints()
yes = (xs - xf) % k == 0 and (ys - yf) % k == 0
if not yes:
print('no')
continue
mi, ma = min(xs, xf), max(xs, xf)
high = max([0] + a[mi:min(ma, (mi // s + 1) * s)] + a[max(mi, ma // s * s):ma])
for j in range(min(xs, xf) // s + 1, max(xs, xf) // s):
# TODO: Your code here
if high < ys:
print('yes')
continue
print('yes' if ((high - ys) // k + 1) * k + ys <= n else 'no')
|
import sys, collections, math, itertools
input = lambda: sys.stdin.readline().strip()
ints = lambda: list(map(int, input().split()))
Int = lambda: int(input())
n, m = ints()
a = ints()
s = int(m ** .5 + 1)
maxs = [max(a[i:i + s]) for i in range(0, m, s)]
for i in range(Int()):
ys, xs, yf, xf, k = ints()
yes = (xs - xf) % k == 0 and (ys - yf) % k == 0
if not yes:
print('no')
continue
mi, ma = min(xs, xf), max(xs, xf)
high = max([0] + a[mi:min(ma, (mi // s + 1) * s)] + a[max(mi, ma // s * s):ma])
for j in range(min(xs, xf) // s + 1, max(xs, xf) // s):
{{completion}}
if high < ys:
print('yes')
continue
print('yes' if ((high - ys) // k + 1) * k + ys <= n else 'no')
|
high = max(high, maxs[j])
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
block_completion_002990
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
import math
import sys
input = sys.stdin.buffer.readline
y, x_ = input().split()
block = [int(x) for x in input().split()]
list2 = [[0] *20 for i in range( len(block))]
for a in range(len(block)):
list2[a][0] = block[a]
z, x = 0, 1
while (1 << x) <= len(block):
z = 0
while (z + (1 << x) - 1) < len(block):
list2[z][x] = (max(list2[z][x - 1], list2[z + (1 << (x - 1))][x - 1]))
z += 1
x += 1
for i in range(int(input())):
s = [int(x) for x in input().split()]
if abs(s[0] - s[2]) % s[-1] != 0 or abs(s[1] - s[3]) % s[-1] != 0:
print("NO")
else:
smaller = min(s[1], s[3])
bigger = max(s[1], s[3])
k = 0
while (1 << (k + 1)) <= bigger - smaller + 1:
k += 1
highest = max(list2[smaller - 1][k], list2[bigger - (1 << k)][k])
if highest < s[-1] * ((int(y) - s[0]) // s[-1]) + s[0]:
print("YES")
else:
# TODO: Your code here
|
import math
import sys
input = sys.stdin.buffer.readline
y, x_ = input().split()
block = [int(x) for x in input().split()]
list2 = [[0] *20 for i in range( len(block))]
for a in range(len(block)):
list2[a][0] = block[a]
z, x = 0, 1
while (1 << x) <= len(block):
z = 0
while (z + (1 << x) - 1) < len(block):
list2[z][x] = (max(list2[z][x - 1], list2[z + (1 << (x - 1))][x - 1]))
z += 1
x += 1
for i in range(int(input())):
s = [int(x) for x in input().split()]
if abs(s[0] - s[2]) % s[-1] != 0 or abs(s[1] - s[3]) % s[-1] != 0:
print("NO")
else:
smaller = min(s[1], s[3])
bigger = max(s[1], s[3])
k = 0
while (1 << (k + 1)) <= bigger - smaller + 1:
k += 1
highest = max(list2[smaller - 1][k], list2[bigger - (1 << k)][k])
if highest < s[-1] * ((int(y) - s[0]) // s[-1]) + s[0]:
print("YES")
else:
{{completion}}
|
print("NO")
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
block_completion_002991
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
from sys import stdin, stdout
from math import floor, ceil, log
input, print = stdin.readline, stdout.write
def main():
n, m = map(int, input().split())
arr = [0] + [int(x) for x in input().split()]
st = construct(arr, m)
for _ in range(int(input())):
x1, y1, x2, y2, k = map(int, input().split())
if ((y2 - y1) % k != 0 or (x2 - x1) % k != 0):
# TODO: Your code here
if (x1 <= arr[y1] or x2 <= arr[y2]):
print("NO\n")
continue
max_x = ((n-x1)//k)*k + x1
if (max_x <= getMax(st, m, min(y1, y2), max(y1, y2))):
print("NO\n")
continue
print("YES\n")
def construct(arr, n):
x = ceil(log(n, 2))
max_size = 2 * pow(2, x) - 1
st = [0]*max_size
construct2(arr, 0, n-1, st, 0)
return st
def construct2(arr, ss, se, st, si):
if (ss == se):
st[si] = arr[ss]
return arr[ss]
mid = getMid(ss, se)
st[si] = max(construct2(arr, ss, mid, st, si * 2 + 1), construct2(arr, mid + 1, se, st, si * 2 + 2))
return st[si]
def getMid(s, e):
return s + (e - s) // 2
def getMax(st, n, l, r):
return MaxUtil(st, 0, n - 1, l, r, 0)
def MaxUtil(st, ss, se, l, r, node):
if (l <= ss and r >= se):
return st[node]
if (se < l or ss > r):
return -1
mid = getMid(ss, se)
return max(MaxUtil(st, ss, mid, l, r, 2 * node + 1), MaxUtil(st, mid + 1, se, l, r, 2 * node + 2))
main()
|
from sys import stdin, stdout
from math import floor, ceil, log
input, print = stdin.readline, stdout.write
def main():
n, m = map(int, input().split())
arr = [0] + [int(x) for x in input().split()]
st = construct(arr, m)
for _ in range(int(input())):
x1, y1, x2, y2, k = map(int, input().split())
if ((y2 - y1) % k != 0 or (x2 - x1) % k != 0):
{{completion}}
if (x1 <= arr[y1] or x2 <= arr[y2]):
print("NO\n")
continue
max_x = ((n-x1)//k)*k + x1
if (max_x <= getMax(st, m, min(y1, y2), max(y1, y2))):
print("NO\n")
continue
print("YES\n")
def construct(arr, n):
x = ceil(log(n, 2))
max_size = 2 * pow(2, x) - 1
st = [0]*max_size
construct2(arr, 0, n-1, st, 0)
return st
def construct2(arr, ss, se, st, si):
if (ss == se):
st[si] = arr[ss]
return arr[ss]
mid = getMid(ss, se)
st[si] = max(construct2(arr, ss, mid, st, si * 2 + 1), construct2(arr, mid + 1, se, st, si * 2 + 2))
return st[si]
def getMid(s, e):
return s + (e - s) // 2
def getMax(st, n, l, r):
return MaxUtil(st, 0, n - 1, l, r, 0)
def MaxUtil(st, ss, se, l, r, node):
if (l <= ss and r >= se):
return st[node]
if (se < l or ss > r):
return -1
mid = getMid(ss, se)
return max(MaxUtil(st, ss, mid, l, r, 2 * node + 1), MaxUtil(st, mid + 1, se, l, r, 2 * node + 2))
main()
|
print("NO\n")
continue
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
block_completion_002992
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
from sys import stdin, stdout
from math import floor, ceil, log
input, print = stdin.readline, stdout.write
def main():
n, m = map(int, input().split())
arr = [0] + [int(x) for x in input().split()]
st = construct(arr, m)
for _ in range(int(input())):
x1, y1, x2, y2, k = map(int, input().split())
if ((y2 - y1) % k != 0 or (x2 - x1) % k != 0):
print("NO\n")
continue
if (x1 <= arr[y1] or x2 <= arr[y2]):
# TODO: Your code here
max_x = ((n-x1)//k)*k + x1
if (max_x <= getMax(st, m, min(y1, y2), max(y1, y2))):
print("NO\n")
continue
print("YES\n")
def construct(arr, n):
x = ceil(log(n, 2))
max_size = 2 * pow(2, x) - 1
st = [0]*max_size
construct2(arr, 0, n-1, st, 0)
return st
def construct2(arr, ss, se, st, si):
if (ss == se):
st[si] = arr[ss]
return arr[ss]
mid = getMid(ss, se)
st[si] = max(construct2(arr, ss, mid, st, si * 2 + 1), construct2(arr, mid + 1, se, st, si * 2 + 2))
return st[si]
def getMid(s, e):
return s + (e - s) // 2
def getMax(st, n, l, r):
return MaxUtil(st, 0, n - 1, l, r, 0)
def MaxUtil(st, ss, se, l, r, node):
if (l <= ss and r >= se):
return st[node]
if (se < l or ss > r):
return -1
mid = getMid(ss, se)
return max(MaxUtil(st, ss, mid, l, r, 2 * node + 1), MaxUtil(st, mid + 1, se, l, r, 2 * node + 2))
main()
|
from sys import stdin, stdout
from math import floor, ceil, log
input, print = stdin.readline, stdout.write
def main():
n, m = map(int, input().split())
arr = [0] + [int(x) for x in input().split()]
st = construct(arr, m)
for _ in range(int(input())):
x1, y1, x2, y2, k = map(int, input().split())
if ((y2 - y1) % k != 0 or (x2 - x1) % k != 0):
print("NO\n")
continue
if (x1 <= arr[y1] or x2 <= arr[y2]):
{{completion}}
max_x = ((n-x1)//k)*k + x1
if (max_x <= getMax(st, m, min(y1, y2), max(y1, y2))):
print("NO\n")
continue
print("YES\n")
def construct(arr, n):
x = ceil(log(n, 2))
max_size = 2 * pow(2, x) - 1
st = [0]*max_size
construct2(arr, 0, n-1, st, 0)
return st
def construct2(arr, ss, se, st, si):
if (ss == se):
st[si] = arr[ss]
return arr[ss]
mid = getMid(ss, se)
st[si] = max(construct2(arr, ss, mid, st, si * 2 + 1), construct2(arr, mid + 1, se, st, si * 2 + 2))
return st[si]
def getMid(s, e):
return s + (e - s) // 2
def getMax(st, n, l, r):
return MaxUtil(st, 0, n - 1, l, r, 0)
def MaxUtil(st, ss, se, l, r, node):
if (l <= ss and r >= se):
return st[node]
if (se < l or ss > r):
return -1
mid = getMid(ss, se)
return max(MaxUtil(st, ss, mid, l, r, 2 * node + 1), MaxUtil(st, mid + 1, se, l, r, 2 * node + 2))
main()
|
print("NO\n")
continue
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
block_completion_002993
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
import sys
from array import array
input = lambda: sys.stdin.buffer.readline().decode().strip()
inp = lambda dtype: [dtype(x) for x in input().split()]
inp_2d = lambda dtype, n: [dtype(input()) for _ in range(n)]
inp_2ds = lambda dtype, n: [inp(dtype) for _ in range(n)]
ceil1 = lambda a, b: (a + b - 1) // b
debug = lambda *x: print(*x, file=sys.stderr)
out = []
class sparse_table:
def __init__(self, n, a, default=0, func=max):
self.n, self.lg = n, 20
self.func = func
self.table = [array('i', [default] * n) for _ in range(self.lg)]
self.table[0] = a
self.preprocess()
def preprocess(self):
for j in range(1, self.lg):
i = 0
while i + (1 << j) - 1 < self.n:
# TODO: Your code here
def quary(self, l, r):
'''[l,r]'''
l1 = (r - l + 1).bit_length() - 1
r1 = r - (1 << l1) + 1
return self.func(self.table[l1][l], self.table[l1][r1])
n, m = inp(int)
a = array('i', inp(int))
mem = sparse_table(m, a)
for _ in range(int(input())):
xs, ys, xf, yf, k = inp(int)
div, mod = divmod(n - xs, k)
xma = n - mod
qur = mem.quary(min(ys, yf) - 1, max(ys, yf) - 1)
if qur >= xma or abs(xma - xf) % k or abs(ys - yf) % k:
out.append('no')
else:
out.append('yes')
print('\n'.join(out))
|
import sys
from array import array
input = lambda: sys.stdin.buffer.readline().decode().strip()
inp = lambda dtype: [dtype(x) for x in input().split()]
inp_2d = lambda dtype, n: [dtype(input()) for _ in range(n)]
inp_2ds = lambda dtype, n: [inp(dtype) for _ in range(n)]
ceil1 = lambda a, b: (a + b - 1) // b
debug = lambda *x: print(*x, file=sys.stderr)
out = []
class sparse_table:
def __init__(self, n, a, default=0, func=max):
self.n, self.lg = n, 20
self.func = func
self.table = [array('i', [default] * n) for _ in range(self.lg)]
self.table[0] = a
self.preprocess()
def preprocess(self):
for j in range(1, self.lg):
i = 0
while i + (1 << j) - 1 < self.n:
{{completion}}
def quary(self, l, r):
'''[l,r]'''
l1 = (r - l + 1).bit_length() - 1
r1 = r - (1 << l1) + 1
return self.func(self.table[l1][l], self.table[l1][r1])
n, m = inp(int)
a = array('i', inp(int))
mem = sparse_table(m, a)
for _ in range(int(input())):
xs, ys, xf, yf, k = inp(int)
div, mod = divmod(n - xs, k)
xma = n - mod
qur = mem.quary(min(ys, yf) - 1, max(ys, yf) - 1)
if qur >= xma or abs(xma - xf) % k or abs(ys - yf) % k:
out.append('no')
else:
out.append('yes')
print('\n'.join(out))
|
ix = i + (1 << (j - 1))
self.table[j][i] = self.func(self.table[j - 1][i], self.table[j - 1][ix])
i += 1
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
block_completion_002994
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
from sys import setrecursionlimit, stdin
readline = stdin.readline
# def I(): return int(readline())
# def ST(): return readline()[:-1]
# def LI(): return list(map(int, readline().split()))
def solve():
N, m = map(int, readline().split())
A = [0] + list(map(int, readline().split()))
n = len(A)
num = 1 << (n - 1).bit_length()
tree = [0] * 2 * num
for i in range(n):
tree[num + i] = A[i]
for i in range(num - 1, 0, -1):
tree[i] = max(tree[2 * i], tree[2 * i + 1])
def query(l, r):
ret = 0
l += num
r += num
while l < r:
if l & 1:
# TODO: Your code here
if r & 1:
ret = max(ret, tree[r - 1])
l >>= 1
r >>= 1
return ret
for _ in range(int(input())):
sx, sy, gx, gy, k = map(int, readline().split())
dx = abs(sx - gx)
dy = abs(sy - gy)
if dx % k or dy % k:
print("NO")
continue
top = sx + k * ((N - sx) // k)
if sy > gy: sy, gy = gy, sy
if query(sy, gy + 1) < top: print("YES")
else: print("NO")
solve()
|
from sys import setrecursionlimit, stdin
readline = stdin.readline
# def I(): return int(readline())
# def ST(): return readline()[:-1]
# def LI(): return list(map(int, readline().split()))
def solve():
N, m = map(int, readline().split())
A = [0] + list(map(int, readline().split()))
n = len(A)
num = 1 << (n - 1).bit_length()
tree = [0] * 2 * num
for i in range(n):
tree[num + i] = A[i]
for i in range(num - 1, 0, -1):
tree[i] = max(tree[2 * i], tree[2 * i + 1])
def query(l, r):
ret = 0
l += num
r += num
while l < r:
if l & 1:
{{completion}}
if r & 1:
ret = max(ret, tree[r - 1])
l >>= 1
r >>= 1
return ret
for _ in range(int(input())):
sx, sy, gx, gy, k = map(int, readline().split())
dx = abs(sx - gx)
dy = abs(sy - gy)
if dx % k or dy % k:
print("NO")
continue
top = sx + k * ((N - sx) // k)
if sy > gy: sy, gy = gy, sy
if query(sy, gy + 1) < top: print("YES")
else: print("NO")
solve()
|
ret = max(ret, tree[l])
l += 1
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
block_completion_002995
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
from sys import setrecursionlimit, stdin
readline = stdin.readline
# def I(): return int(readline())
# def ST(): return readline()[:-1]
# def LI(): return list(map(int, readline().split()))
def solve():
N, m = map(int, readline().split())
A = [0] + list(map(int, readline().split()))
n = len(A)
num = 1 << (n - 1).bit_length()
tree = [0] * 2 * num
for i in range(n):
tree[num + i] = A[i]
for i in range(num - 1, 0, -1):
tree[i] = max(tree[2 * i], tree[2 * i + 1])
def query(l, r):
ret = 0
l += num
r += num
while l < r:
if l & 1:
ret = max(ret, tree[l])
l += 1
if r & 1:
# TODO: Your code here
l >>= 1
r >>= 1
return ret
for _ in range(int(input())):
sx, sy, gx, gy, k = map(int, readline().split())
dx = abs(sx - gx)
dy = abs(sy - gy)
if dx % k or dy % k:
print("NO")
continue
top = sx + k * ((N - sx) // k)
if sy > gy: sy, gy = gy, sy
if query(sy, gy + 1) < top: print("YES")
else: print("NO")
solve()
|
from sys import setrecursionlimit, stdin
readline = stdin.readline
# def I(): return int(readline())
# def ST(): return readline()[:-1]
# def LI(): return list(map(int, readline().split()))
def solve():
N, m = map(int, readline().split())
A = [0] + list(map(int, readline().split()))
n = len(A)
num = 1 << (n - 1).bit_length()
tree = [0] * 2 * num
for i in range(n):
tree[num + i] = A[i]
for i in range(num - 1, 0, -1):
tree[i] = max(tree[2 * i], tree[2 * i + 1])
def query(l, r):
ret = 0
l += num
r += num
while l < r:
if l & 1:
ret = max(ret, tree[l])
l += 1
if r & 1:
{{completion}}
l >>= 1
r >>= 1
return ret
for _ in range(int(input())):
sx, sy, gx, gy, k = map(int, readline().split())
dx = abs(sx - gx)
dy = abs(sy - gy)
if dx % k or dy % k:
print("NO")
continue
top = sx + k * ((N - sx) // k)
if sy > gy: sy, gy = gy, sy
if query(sy, gy + 1) < top: print("YES")
else: print("NO")
solve()
|
ret = max(ret, tree[r - 1])
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
block_completion_002996
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
input = __import__('sys').stdin.readline
log2s = [0, 0]
for i in range(2, 200005):
log2s.append(log2s[i // 2] + 1)
n, m = map(int, input().split())
sparse = [[0] + list(map(int, input().split()))]
for j in range(20):
sparse.append([max(sparse[j][min(i + (1 << j), m)], sparse[j][i]) for i in range(m+1)])
def getmax(L, R):
j = log2s[R - L + 1]
return max(sparse[j][R - (1 << j) + 1], sparse[j][L])
for _ in range(int(input())):
x1, y1, x2, y2, k = map(int, input().split())
if (x1 - x2) % k != 0 or (y1 - y2) % k != 0:
# TODO: Your code here
# i * k + x1 <= n
i = (n - x1) // k
h = i * k + x1
while h > n:
h -= k
print('YES' if getmax(min(y1, y2), max(y1, y2)) < h else 'NO')
|
input = __import__('sys').stdin.readline
log2s = [0, 0]
for i in range(2, 200005):
log2s.append(log2s[i // 2] + 1)
n, m = map(int, input().split())
sparse = [[0] + list(map(int, input().split()))]
for j in range(20):
sparse.append([max(sparse[j][min(i + (1 << j), m)], sparse[j][i]) for i in range(m+1)])
def getmax(L, R):
j = log2s[R - L + 1]
return max(sparse[j][R - (1 << j) + 1], sparse[j][L])
for _ in range(int(input())):
x1, y1, x2, y2, k = map(int, input().split())
if (x1 - x2) % k != 0 or (y1 - y2) % k != 0:
{{completion}}
# i * k + x1 <= n
i = (n - x1) // k
h = i * k + x1
while h > n:
h -= k
print('YES' if getmax(min(y1, y2), max(y1, y2)) < h else 'NO')
|
print('NO')
continue
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
block_completion_002997
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
input = __import__('sys').stdin.readline
log2s = [0, 0]
for i in range(2, 200005):
log2s.append(log2s[i // 2] + 1)
n, m = map(int, input().split())
sparse = [[0] + list(map(int, input().split()))]
for j in range(20):
sparse.append([max(sparse[j][min(i + (1 << j), m)], sparse[j][i]) for i in range(m+1)])
def getmax(L, R):
j = log2s[R - L + 1]
return max(sparse[j][R - (1 << j) + 1], sparse[j][L])
for _ in range(int(input())):
x1, y1, x2, y2, k = map(int, input().split())
if (x1 - x2) % k != 0 or (y1 - y2) % k != 0:
print('NO')
continue
# i * k + x1 <= n
i = (n - x1) // k
h = i * k + x1
while h > n:
# TODO: Your code here
print('YES' if getmax(min(y1, y2), max(y1, y2)) < h else 'NO')
|
input = __import__('sys').stdin.readline
log2s = [0, 0]
for i in range(2, 200005):
log2s.append(log2s[i // 2] + 1)
n, m = map(int, input().split())
sparse = [[0] + list(map(int, input().split()))]
for j in range(20):
sparse.append([max(sparse[j][min(i + (1 << j), m)], sparse[j][i]) for i in range(m+1)])
def getmax(L, R):
j = log2s[R - L + 1]
return max(sparse[j][R - (1 << j) + 1], sparse[j][L])
for _ in range(int(input())):
x1, y1, x2, y2, k = map(int, input().split())
if (x1 - x2) % k != 0 or (y1 - y2) % k != 0:
print('NO')
continue
# i * k + x1 <= n
i = (n - x1) // k
h = i * k + x1
while h > n:
{{completion}}
print('YES' if getmax(min(y1, y2), max(y1, y2)) < h else 'NO')
|
h -= k
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
block_completion_002998
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
import math,sys;input=sys.stdin.readline;S=lambda:input().rstrip();I=lambda:int(S());M=lambda:map(int,S().split());L=lambda:list(M());mod1=1000000007;mod2=998244353
n,m=M();a=L();q=I()
t=[0]*(2*m)
# a is the array of initial values
def build(t,n,a):
for i in range(n):t[i+n]=a[i]
for i in range(n-1,0,-1):t[i]=max(t[i<<1],t[(i<<1)|1])
# change value at position p to v
def modify(t,n,p,v):
p+=n
t[p]=v
while p>1:
t[p>>1]=max(t[p],t[p^1])
p>>=1
# find the combined value of range [l,r)
def query(t,n,l,r):
resl=resr=0
l+=n;r+=n
while l<r:
if (l&1):# TODO: Your code here
if (r&1):r-=1;resr=max(t[r],resr)
l>>=1;r>>=1
return max(resl,resr)
build(t,m,a)
for i in range(q):
xs,ys,xf,yf,k=M()
if abs(xs-xf)%k or abs(ys-yf)%k:print("NO");continue
p=query(t,m,min(ys,yf)-1,max(ys,yf))
z=min(xs,xf)+((n-min(xs,xf))//k)*k
if z<=p:print("NO")
else:print("YES")
|
import math,sys;input=sys.stdin.readline;S=lambda:input().rstrip();I=lambda:int(S());M=lambda:map(int,S().split());L=lambda:list(M());mod1=1000000007;mod2=998244353
n,m=M();a=L();q=I()
t=[0]*(2*m)
# a is the array of initial values
def build(t,n,a):
for i in range(n):t[i+n]=a[i]
for i in range(n-1,0,-1):t[i]=max(t[i<<1],t[(i<<1)|1])
# change value at position p to v
def modify(t,n,p,v):
p+=n
t[p]=v
while p>1:
t[p>>1]=max(t[p],t[p^1])
p>>=1
# find the combined value of range [l,r)
def query(t,n,l,r):
resl=resr=0
l+=n;r+=n
while l<r:
if (l&1):{{completion}}
if (r&1):r-=1;resr=max(t[r],resr)
l>>=1;r>>=1
return max(resl,resr)
build(t,m,a)
for i in range(q):
xs,ys,xf,yf,k=M()
if abs(xs-xf)%k or abs(ys-yf)%k:print("NO");continue
p=query(t,m,min(ys,yf)-1,max(ys,yf))
z=min(xs,xf)+((n-min(xs,xf))//k)*k
if z<=p:print("NO")
else:print("YES")
|
resl=max(resl,t[l]);l+=1
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
block_completion_002999
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
import math,sys;input=sys.stdin.readline;S=lambda:input().rstrip();I=lambda:int(S());M=lambda:map(int,S().split());L=lambda:list(M());mod1=1000000007;mod2=998244353
n,m=M();a=L();q=I()
t=[0]*(2*m)
# a is the array of initial values
def build(t,n,a):
for i in range(n):t[i+n]=a[i]
for i in range(n-1,0,-1):t[i]=max(t[i<<1],t[(i<<1)|1])
# change value at position p to v
def modify(t,n,p,v):
p+=n
t[p]=v
while p>1:
t[p>>1]=max(t[p],t[p^1])
p>>=1
# find the combined value of range [l,r)
def query(t,n,l,r):
resl=resr=0
l+=n;r+=n
while l<r:
if (l&1):resl=max(resl,t[l]);l+=1
if (r&1):# TODO: Your code here
l>>=1;r>>=1
return max(resl,resr)
build(t,m,a)
for i in range(q):
xs,ys,xf,yf,k=M()
if abs(xs-xf)%k or abs(ys-yf)%k:print("NO");continue
p=query(t,m,min(ys,yf)-1,max(ys,yf))
z=min(xs,xf)+((n-min(xs,xf))//k)*k
if z<=p:print("NO")
else:print("YES")
|
import math,sys;input=sys.stdin.readline;S=lambda:input().rstrip();I=lambda:int(S());M=lambda:map(int,S().split());L=lambda:list(M());mod1=1000000007;mod2=998244353
n,m=M();a=L();q=I()
t=[0]*(2*m)
# a is the array of initial values
def build(t,n,a):
for i in range(n):t[i+n]=a[i]
for i in range(n-1,0,-1):t[i]=max(t[i<<1],t[(i<<1)|1])
# change value at position p to v
def modify(t,n,p,v):
p+=n
t[p]=v
while p>1:
t[p>>1]=max(t[p],t[p^1])
p>>=1
# find the combined value of range [l,r)
def query(t,n,l,r):
resl=resr=0
l+=n;r+=n
while l<r:
if (l&1):resl=max(resl,t[l]);l+=1
if (r&1):{{completion}}
l>>=1;r>>=1
return max(resl,resr)
build(t,m,a)
for i in range(q):
xs,ys,xf,yf,k=M()
if abs(xs-xf)%k or abs(ys-yf)%k:print("NO");continue
p=query(t,m,min(ys,yf)-1,max(ys,yf))
z=min(xs,xf)+((n-min(xs,xf))//k)*k
if z<=p:print("NO")
else:print("YES")
|
r-=1;resr=max(t[r],resr)
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
block_completion_003000
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a tree consisting of $$$n$$$ vertices. A number is written on each vertex; the number on vertex $$$i$$$ is equal to $$$a_i$$$.Recall that a simple path is a path that visits each vertex at most once. Let the weight of the path be the bitwise XOR of the values written on vertices it consists of. Let's say that a tree is good if no simple path has weight $$$0$$$.You can apply the following operation any number of times (possibly, zero): select a vertex of the tree and replace the value written on it with an arbitrary positive integer. What is the minimum number of times you have to apply this operation in order to make the tree good?
Input Specification: The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of vertices. The second line contains $$$n$$$ integers $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \le a_i < 2^{30}$$$) — the numbers written on vertices. Then $$$n - 1$$$ lines follow, each containing two integers $$$x$$$ and $$$y$$$ ($$$1 \le x, y \le n; x \ne y$$$) denoting an edge connecting vertex $$$x$$$ with vertex $$$y$$$. It is guaranteed that these edges form a tree.
Output Specification: Print a single integer — the minimum number of times you have to apply the operation in order to make the tree good.
Notes: NoteIn the first example, it is enough to replace the value on the vertex $$$1$$$ with $$$13$$$, and the value on the vertex $$$4$$$ with $$$42$$$.
Code:
import sys
input = sys.stdin.buffer.readline
N = int(input())
A = list(map(int, input().split()))
adj = [[] for _ in range(N)]
for _ in range(N - 1):
a, b = map(lambda s: int(s) - 1, input().split())
adj[a].append(b)
adj[b].append(a)
O = [0]
for i in O:
for j in adj[i]:
adj[j].remove(i)
O.append(j)
class XORSet:
def __init__(self, el=None):
self.s = set()
self.xor = 0
if el is not None:
self.s.add(el)
def add(self, el: int):
self.s.add(el ^ self.xor)
def update(self, xor: int):
self.xor ^= xor
def __len__(self) -> int:
return len(self.s)
def __iter__(self):
return (x ^ self.xor for x in self.s)
def __contains__(self, el: int) -> bool:
return (el ^ self.xor) in self.s
r = 0
D = [XORSet(a) for a in A]
for i in reversed(O):
for j in adj[i]:
if len(D[j]) > len(D[i]):
D[i], D[j] = D[j], D[i]
D[i].update(A[i])
D[j].update(A[i])
#print('merging', i, j, list(D[i]), list(D[j]))
if any(x in D[i] for x in D[j]):
r += 1
D[i].s.clear()
break
else:
for x in D[j]:
# TODO: Your code here
#assert 0 not in D[i]
#print(i, A[i], adj[i], list(D[i]))
print(r)
|
import sys
input = sys.stdin.buffer.readline
N = int(input())
A = list(map(int, input().split()))
adj = [[] for _ in range(N)]
for _ in range(N - 1):
a, b = map(lambda s: int(s) - 1, input().split())
adj[a].append(b)
adj[b].append(a)
O = [0]
for i in O:
for j in adj[i]:
adj[j].remove(i)
O.append(j)
class XORSet:
def __init__(self, el=None):
self.s = set()
self.xor = 0
if el is not None:
self.s.add(el)
def add(self, el: int):
self.s.add(el ^ self.xor)
def update(self, xor: int):
self.xor ^= xor
def __len__(self) -> int:
return len(self.s)
def __iter__(self):
return (x ^ self.xor for x in self.s)
def __contains__(self, el: int) -> bool:
return (el ^ self.xor) in self.s
r = 0
D = [XORSet(a) for a in A]
for i in reversed(O):
for j in adj[i]:
if len(D[j]) > len(D[i]):
D[i], D[j] = D[j], D[i]
D[i].update(A[i])
D[j].update(A[i])
#print('merging', i, j, list(D[i]), list(D[j]))
if any(x in D[i] for x in D[j]):
r += 1
D[i].s.clear()
break
else:
for x in D[j]:
{{completion}}
#assert 0 not in D[i]
#print(i, A[i], adj[i], list(D[i]))
print(r)
|
D[i].add(x ^ A[i])
|
[{"input": "6\n3 2 1 3 2 1\n4 5\n3 4\n1 4\n2 1\n6 1", "output": ["2"]}, {"input": "4\n2 1 1 1\n1 2\n1 3\n1 4", "output": ["0"]}, {"input": "5\n2 2 2 2 2\n1 2\n2 3\n3 4\n4 5", "output": ["2"]}]
|
block_completion_003038
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a tree consisting of $$$n$$$ vertices. A number is written on each vertex; the number on vertex $$$i$$$ is equal to $$$a_i$$$.Recall that a simple path is a path that visits each vertex at most once. Let the weight of the path be the bitwise XOR of the values written on vertices it consists of. Let's say that a tree is good if no simple path has weight $$$0$$$.You can apply the following operation any number of times (possibly, zero): select a vertex of the tree and replace the value written on it with an arbitrary positive integer. What is the minimum number of times you have to apply this operation in order to make the tree good?
Input Specification: The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of vertices. The second line contains $$$n$$$ integers $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \le a_i < 2^{30}$$$) — the numbers written on vertices. Then $$$n - 1$$$ lines follow, each containing two integers $$$x$$$ and $$$y$$$ ($$$1 \le x, y \le n; x \ne y$$$) denoting an edge connecting vertex $$$x$$$ with vertex $$$y$$$. It is guaranteed that these edges form a tree.
Output Specification: Print a single integer — the minimum number of times you have to apply the operation in order to make the tree good.
Notes: NoteIn the first example, it is enough to replace the value on the vertex $$$1$$$ with $$$13$$$, and the value on the vertex $$$4$$$ with $$$42$$$.
Code:
import sys
input = sys.stdin.buffer.readline
N = int(input())
A = list(map(int, input().split()))
adj = [[] for _ in range(N)]
for _ in range(N - 1):
a, b = map(lambda s: int(s) - 1, input().split())
adj[a].append(b)
adj[b].append(a)
O = [0]
for i in O:
for j in adj[i]:
adj[j].remove(i)
O.append(j)
class XORSet:
def __init__(self, el=None):
self.s = set()
self.xor = 0
if el is not None:
# TODO: Your code here
def add(self, el: int):
self.s.add(el ^ self.xor)
def update(self, xor: int):
self.xor ^= xor
def __len__(self) -> int:
return len(self.s)
def __iter__(self):
return (x ^ self.xor for x in self.s)
def __contains__(self, el: int) -> bool:
return (el ^ self.xor) in self.s
r = 0
D = [XORSet(a) for a in A]
for i in reversed(O):
for j in adj[i]:
if len(D[j]) > len(D[i]):
D[i], D[j] = D[j], D[i]
D[i].update(A[i])
D[j].update(A[i])
#print('merging', i, j, list(D[i]), list(D[j]))
if any(x in D[i] for x in D[j]):
r += 1
D[i].s.clear()
break
else:
for x in D[j]:
D[i].add(x ^ A[i])
#assert 0 not in D[i]
#print(i, A[i], adj[i], list(D[i]))
print(r)
|
import sys
input = sys.stdin.buffer.readline
N = int(input())
A = list(map(int, input().split()))
adj = [[] for _ in range(N)]
for _ in range(N - 1):
a, b = map(lambda s: int(s) - 1, input().split())
adj[a].append(b)
adj[b].append(a)
O = [0]
for i in O:
for j in adj[i]:
adj[j].remove(i)
O.append(j)
class XORSet:
def __init__(self, el=None):
self.s = set()
self.xor = 0
if el is not None:
{{completion}}
def add(self, el: int):
self.s.add(el ^ self.xor)
def update(self, xor: int):
self.xor ^= xor
def __len__(self) -> int:
return len(self.s)
def __iter__(self):
return (x ^ self.xor for x in self.s)
def __contains__(self, el: int) -> bool:
return (el ^ self.xor) in self.s
r = 0
D = [XORSet(a) for a in A]
for i in reversed(O):
for j in adj[i]:
if len(D[j]) > len(D[i]):
D[i], D[j] = D[j], D[i]
D[i].update(A[i])
D[j].update(A[i])
#print('merging', i, j, list(D[i]), list(D[j]))
if any(x in D[i] for x in D[j]):
r += 1
D[i].s.clear()
break
else:
for x in D[j]:
D[i].add(x ^ A[i])
#assert 0 not in D[i]
#print(i, A[i], adj[i], list(D[i]))
print(r)
|
self.s.add(el)
|
[{"input": "6\n3 2 1 3 2 1\n4 5\n3 4\n1 4\n2 1\n6 1", "output": ["2"]}, {"input": "4\n2 1 1 1\n1 2\n1 3\n1 4", "output": ["0"]}, {"input": "5\n2 2 2 2 2\n1 2\n2 3\n3 4\n4 5", "output": ["2"]}]
|
block_completion_003039
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a tree consisting of $$$n$$$ vertices. A number is written on each vertex; the number on vertex $$$i$$$ is equal to $$$a_i$$$.Recall that a simple path is a path that visits each vertex at most once. Let the weight of the path be the bitwise XOR of the values written on vertices it consists of. Let's say that a tree is good if no simple path has weight $$$0$$$.You can apply the following operation any number of times (possibly, zero): select a vertex of the tree and replace the value written on it with an arbitrary positive integer. What is the minimum number of times you have to apply this operation in order to make the tree good?
Input Specification: The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of vertices. The second line contains $$$n$$$ integers $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \le a_i < 2^{30}$$$) — the numbers written on vertices. Then $$$n - 1$$$ lines follow, each containing two integers $$$x$$$ and $$$y$$$ ($$$1 \le x, y \le n; x \ne y$$$) denoting an edge connecting vertex $$$x$$$ with vertex $$$y$$$. It is guaranteed that these edges form a tree.
Output Specification: Print a single integer — the minimum number of times you have to apply the operation in order to make the tree good.
Notes: NoteIn the first example, it is enough to replace the value on the vertex $$$1$$$ with $$$13$$$, and the value on the vertex $$$4$$$ with $$$42$$$.
Code:
import sys
from types import GeneratorType
from collections import defaultdict
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
# TODO: Your code here
to = stack[-1].send(to)
return to
return wrappedfunc
input = lambda: sys.stdin.readline().rstrip()
def solve():
N = int(input())
A = [int(x) for x in input().split()]
G = [[] for _ in range(N)]
for _ in range(N - 1):
x, y = [int(x) - 1 for x in input().split()]
G[x].append(y)
G[y].append(x)
B = [0] * N
vals = defaultdict(set)
res = [0]
@bootstrap
def fill_dfs(v, p):
B[v] = A[v]
if p != -1:
B[v] ^= B[p]
for u in G[v]:
if u != p:
yield fill_dfs(u, v)
yield
@bootstrap
def calc_dfs(v, p):
zero = False
vals[v].add(B[v])
for u in G[v]:
if u != p:
yield calc_dfs(u, v)
if len(vals[v]) < len(vals[u]):
vals[v], vals[u] = vals[u], vals[v]
for x in vals[u]:
zero |= (x ^ A[v]) in vals[v]
for x in vals[u]:
vals[v].add(x)
vals[u].clear()
if zero:
res[0] += 1
vals[v].clear()
yield
fill_dfs(0, -1)
calc_dfs(0, -1)
return res[0]
print(solve())
|
import sys
from types import GeneratorType
from collections import defaultdict
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
{{completion}}
to = stack[-1].send(to)
return to
return wrappedfunc
input = lambda: sys.stdin.readline().rstrip()
def solve():
N = int(input())
A = [int(x) for x in input().split()]
G = [[] for _ in range(N)]
for _ in range(N - 1):
x, y = [int(x) - 1 for x in input().split()]
G[x].append(y)
G[y].append(x)
B = [0] * N
vals = defaultdict(set)
res = [0]
@bootstrap
def fill_dfs(v, p):
B[v] = A[v]
if p != -1:
B[v] ^= B[p]
for u in G[v]:
if u != p:
yield fill_dfs(u, v)
yield
@bootstrap
def calc_dfs(v, p):
zero = False
vals[v].add(B[v])
for u in G[v]:
if u != p:
yield calc_dfs(u, v)
if len(vals[v]) < len(vals[u]):
vals[v], vals[u] = vals[u], vals[v]
for x in vals[u]:
zero |= (x ^ A[v]) in vals[v]
for x in vals[u]:
vals[v].add(x)
vals[u].clear()
if zero:
res[0] += 1
vals[v].clear()
yield
fill_dfs(0, -1)
calc_dfs(0, -1)
return res[0]
print(solve())
|
break
|
[{"input": "6\n3 2 1 3 2 1\n4 5\n3 4\n1 4\n2 1\n6 1", "output": ["2"]}, {"input": "4\n2 1 1 1\n1 2\n1 3\n1 4", "output": ["0"]}, {"input": "5\n2 2 2 2 2\n1 2\n2 3\n3 4\n4 5", "output": ["2"]}]
|
block_completion_003040
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a tree consisting of $$$n$$$ vertices. A number is written on each vertex; the number on vertex $$$i$$$ is equal to $$$a_i$$$.Recall that a simple path is a path that visits each vertex at most once. Let the weight of the path be the bitwise XOR of the values written on vertices it consists of. Let's say that a tree is good if no simple path has weight $$$0$$$.You can apply the following operation any number of times (possibly, zero): select a vertex of the tree and replace the value written on it with an arbitrary positive integer. What is the minimum number of times you have to apply this operation in order to make the tree good?
Input Specification: The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of vertices. The second line contains $$$n$$$ integers $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \le a_i < 2^{30}$$$) — the numbers written on vertices. Then $$$n - 1$$$ lines follow, each containing two integers $$$x$$$ and $$$y$$$ ($$$1 \le x, y \le n; x \ne y$$$) denoting an edge connecting vertex $$$x$$$ with vertex $$$y$$$. It is guaranteed that these edges form a tree.
Output Specification: Print a single integer — the minimum number of times you have to apply the operation in order to make the tree good.
Notes: NoteIn the first example, it is enough to replace the value on the vertex $$$1$$$ with $$$13$$$, and the value on the vertex $$$4$$$ with $$$42$$$.
Code:
import sys
input = sys.stdin.buffer.readline
N = int(input())
A = list(map(int, input().split()))
adj = [[] for _ in range(N)]
for _ in range(N - 1):
a, b = map(lambda s: int(s) - 1, input().split())
adj[a].append(b)
adj[b].append(a)
O = [0]
for i in O:
for j in adj[i]:
adj[j].remove(i)
O.append(j)
class XORSet:
def __init__(self, el=None):
self.s = set()
self.xor = 0
if el is not None:
self.s.add(el)
def add(self, el: int):
self.s.add(el ^ self.xor)
def update(self, xor: int):
self.xor ^= xor
def __len__(self) -> int:
return len(self.s)
def __iter__(self):
return (x ^ self.xor for x in self.s)
def __contains__(self, el: int) -> bool:
return (el ^ self.xor) in self.s
r = 0
D = [XORSet(a) for a in A]
for i in reversed(O):
for j in adj[i]:
if len(D[j]) > len(D[i]):
D[i], D[j] = D[j], D[i]
D[i].update(A[i])
D[j].update(A[i])
l = list(D[j])
for x in l:
if x in D[i]:
r += 1
D[i].s.clear()
break
else:
for x in l:
# TODO: Your code here
continue
break
#print(i, A[i], adj[i], list(D[i]))
print(r)
|
import sys
input = sys.stdin.buffer.readline
N = int(input())
A = list(map(int, input().split()))
adj = [[] for _ in range(N)]
for _ in range(N - 1):
a, b = map(lambda s: int(s) - 1, input().split())
adj[a].append(b)
adj[b].append(a)
O = [0]
for i in O:
for j in adj[i]:
adj[j].remove(i)
O.append(j)
class XORSet:
def __init__(self, el=None):
self.s = set()
self.xor = 0
if el is not None:
self.s.add(el)
def add(self, el: int):
self.s.add(el ^ self.xor)
def update(self, xor: int):
self.xor ^= xor
def __len__(self) -> int:
return len(self.s)
def __iter__(self):
return (x ^ self.xor for x in self.s)
def __contains__(self, el: int) -> bool:
return (el ^ self.xor) in self.s
r = 0
D = [XORSet(a) for a in A]
for i in reversed(O):
for j in adj[i]:
if len(D[j]) > len(D[i]):
D[i], D[j] = D[j], D[i]
D[i].update(A[i])
D[j].update(A[i])
l = list(D[j])
for x in l:
if x in D[i]:
r += 1
D[i].s.clear()
break
else:
for x in l:
{{completion}}
continue
break
#print(i, A[i], adj[i], list(D[i]))
print(r)
|
D[i].add(x ^ A[i])
|
[{"input": "6\n3 2 1 3 2 1\n4 5\n3 4\n1 4\n2 1\n6 1", "output": ["2"]}, {"input": "4\n2 1 1 1\n1 2\n1 3\n1 4", "output": ["0"]}, {"input": "5\n2 2 2 2 2\n1 2\n2 3\n3 4\n4 5", "output": ["2"]}]
|
block_completion_003041
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a tree consisting of $$$n$$$ vertices. A number is written on each vertex; the number on vertex $$$i$$$ is equal to $$$a_i$$$.Recall that a simple path is a path that visits each vertex at most once. Let the weight of the path be the bitwise XOR of the values written on vertices it consists of. Let's say that a tree is good if no simple path has weight $$$0$$$.You can apply the following operation any number of times (possibly, zero): select a vertex of the tree and replace the value written on it with an arbitrary positive integer. What is the minimum number of times you have to apply this operation in order to make the tree good?
Input Specification: The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of vertices. The second line contains $$$n$$$ integers $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \le a_i < 2^{30}$$$) — the numbers written on vertices. Then $$$n - 1$$$ lines follow, each containing two integers $$$x$$$ and $$$y$$$ ($$$1 \le x, y \le n; x \ne y$$$) denoting an edge connecting vertex $$$x$$$ with vertex $$$y$$$. It is guaranteed that these edges form a tree.
Output Specification: Print a single integer — the minimum number of times you have to apply the operation in order to make the tree good.
Notes: NoteIn the first example, it is enough to replace the value on the vertex $$$1$$$ with $$$13$$$, and the value on the vertex $$$4$$$ with $$$42$$$.
Code:
import sys
input = sys.stdin.buffer.readline
N = int(input())
A = list(map(int, input().split()))
adj = [[] for _ in range(N)]
for _ in range(N - 1):
a, b = map(lambda s: int(s) - 1, input().split())
adj[a].append(b)
adj[b].append(a)
O = [0]
for i in O:
for j in adj[i]:
adj[j].remove(i)
O.append(j)
class XORSet:
def __init__(self, el=None):
self.s = set()
self.xor = 0
if el is not None:
self.s.add(el)
def add(self, el: int):
self.s.add(el ^ self.xor)
def update(self, xor: int):
self.xor ^= xor
def __len__(self) -> int:
return len(self.s)
def __iter__(self):
return (x ^ self.xor for x in self.s)
def __contains__(self, el: int) -> bool:
return (el ^ self.xor) in self.s
r = 0
D = [XORSet(a) for a in A]
for i in reversed(O):
for j in adj[i]:
if len(D[j]) > len(D[i]):
D[i], D[j] = D[j], D[i]
D[i].update(A[i])
D[j].update(A[i])
l = list(D[j])
for x in l:
if x in D[i]:
# TODO: Your code here
else:
for x in l:
D[i].add(x ^ A[i])
continue
break
#print(i, A[i], adj[i], list(D[i]))
print(r)
|
import sys
input = sys.stdin.buffer.readline
N = int(input())
A = list(map(int, input().split()))
adj = [[] for _ in range(N)]
for _ in range(N - 1):
a, b = map(lambda s: int(s) - 1, input().split())
adj[a].append(b)
adj[b].append(a)
O = [0]
for i in O:
for j in adj[i]:
adj[j].remove(i)
O.append(j)
class XORSet:
def __init__(self, el=None):
self.s = set()
self.xor = 0
if el is not None:
self.s.add(el)
def add(self, el: int):
self.s.add(el ^ self.xor)
def update(self, xor: int):
self.xor ^= xor
def __len__(self) -> int:
return len(self.s)
def __iter__(self):
return (x ^ self.xor for x in self.s)
def __contains__(self, el: int) -> bool:
return (el ^ self.xor) in self.s
r = 0
D = [XORSet(a) for a in A]
for i in reversed(O):
for j in adj[i]:
if len(D[j]) > len(D[i]):
D[i], D[j] = D[j], D[i]
D[i].update(A[i])
D[j].update(A[i])
l = list(D[j])
for x in l:
if x in D[i]:
{{completion}}
else:
for x in l:
D[i].add(x ^ A[i])
continue
break
#print(i, A[i], adj[i], list(D[i]))
print(r)
|
r += 1
D[i].s.clear()
break
|
[{"input": "6\n3 2 1 3 2 1\n4 5\n3 4\n1 4\n2 1\n6 1", "output": ["2"]}, {"input": "4\n2 1 1 1\n1 2\n1 3\n1 4", "output": ["0"]}, {"input": "5\n2 2 2 2 2\n1 2\n2 3\n3 4\n4 5", "output": ["2"]}]
|
block_completion_003042
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Leslie and Leon entered a labyrinth. The labyrinth consists of $$$n$$$ halls and $$$m$$$ one-way passages between them. The halls are numbered from $$$1$$$ to $$$n$$$.Leslie and Leon start their journey in the hall $$$s$$$. Right away, they quarrel and decide to explore the labyrinth separately. However, they want to meet again at the end of their journey.To help Leslie and Leon, your task is to find two different paths from the given hall $$$s$$$ to some other hall $$$t$$$, such that these two paths do not share halls other than the staring hall $$$s$$$ and the ending hall $$$t$$$. The hall $$$t$$$ has not been determined yet, so you can choose any of the labyrinth's halls as $$$t$$$ except $$$s$$$.Leslie's and Leon's paths do not have to be the shortest ones, but their paths must be simple, visiting any hall at most once. Also, they cannot visit any common halls except $$$s$$$ and $$$t$$$ during their journey, even at different times.
Input Specification: The first line contains three integers $$$n$$$, $$$m$$$, and $$$s$$$, where $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) is the number of vertices, $$$m$$$ ($$$0 \le m \le 2 \cdot 10^5$$$) is the number of edges in the labyrinth, and $$$s$$$ ($$$1 \le s \le n$$$) is the starting hall. Then $$$m$$$ lines with descriptions of passages follow. Each description contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$; $$$u_i \neq v_i$$$), denoting a passage from the hall $$$u_i$$$ to the hall $$$v_i$$$. The passages are one-way. Each tuple $$$(u_i, v_i)$$$ is present in the input at most once. The labyrinth can contain cycles and is not necessarily connected in any way.
Output Specification: If it is possible to find the desired two paths, output "Possible", otherwise output "Impossible". If the answer exists, output two path descriptions. Each description occupies two lines. The first line of the description contains an integer $$$h$$$ ($$$2 \le h \le n$$$) — the number of halls in a path, and the second line contains distinct integers $$$w_1, w_2, \dots, w_h$$$ ($$$w_1 = s$$$; $$$1 \le w_j \le n$$$; $$$w_h = t$$$) — the halls in the path in the order of passing. Both paths must end at the same vertex $$$t$$$. The paths must be different, and all intermediate halls in these paths must be distinct.
Code:
import collections
def caminho(parent, sala):
resp = []
while sala is not None:
resp.append(sala)
sala = parent[sala]
return list(reversed(resp))
def solve(grafo, total, inicio):
if len(grafo[inicio]) < 2:
return
globalParent = collections.defaultdict(lambda: None)
for sala1 in grafo[inicio]:
currentParent = collections.defaultdict(lambda: None)
currentParent[sala1] = inicio
fila = collections.deque()
fila.append(sala1)
while len(fila) > 0:
y = fila.popleft()
for x in grafo[y]:
if x != inicio and currentParent[x] is None:
# TODO: Your code here
for x in currentParent:
if x in globalParent:
#Deu bom
return (caminho(globalParent, x), caminho(currentParent, x))
for x,y in currentParent.items():
globalParent[x] = y
n, m, s = map(int, input().split())
g = collections.defaultdict(list)
for i in range(m):
x,y = map(int, input().split())
g[x].append(y)
paths = solve(g,n,s)
if paths is None:
print("Impossible")
else:
print("Possible")
for i in paths:
print(len(i))
print(" ".join(map(str,i)))
|
import collections
def caminho(parent, sala):
resp = []
while sala is not None:
resp.append(sala)
sala = parent[sala]
return list(reversed(resp))
def solve(grafo, total, inicio):
if len(grafo[inicio]) < 2:
return
globalParent = collections.defaultdict(lambda: None)
for sala1 in grafo[inicio]:
currentParent = collections.defaultdict(lambda: None)
currentParent[sala1] = inicio
fila = collections.deque()
fila.append(sala1)
while len(fila) > 0:
y = fila.popleft()
for x in grafo[y]:
if x != inicio and currentParent[x] is None:
{{completion}}
for x in currentParent:
if x in globalParent:
#Deu bom
return (caminho(globalParent, x), caminho(currentParent, x))
for x,y in currentParent.items():
globalParent[x] = y
n, m, s = map(int, input().split())
g = collections.defaultdict(list)
for i in range(m):
x,y = map(int, input().split())
g[x].append(y)
paths = solve(g,n,s)
if paths is None:
print("Impossible")
else:
print("Possible")
for i in paths:
print(len(i))
print(" ".join(map(str,i)))
|
currentParent[x] = y
fila.append(x)
|
[{"input": "5 5 1\n1 2\n2 3\n1 4\n4 3\n3 5", "output": ["Possible\n3\n1 2 3\n3\n1 4 3"]}, {"input": "5 5 1\n1 2\n2 3\n3 4\n2 5\n5 4", "output": ["Impossible"]}, {"input": "3 3 2\n1 2\n2 3\n3 1", "output": ["Impossible"]}]
|
block_completion_003160
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Leslie and Leon entered a labyrinth. The labyrinth consists of $$$n$$$ halls and $$$m$$$ one-way passages between them. The halls are numbered from $$$1$$$ to $$$n$$$.Leslie and Leon start their journey in the hall $$$s$$$. Right away, they quarrel and decide to explore the labyrinth separately. However, they want to meet again at the end of their journey.To help Leslie and Leon, your task is to find two different paths from the given hall $$$s$$$ to some other hall $$$t$$$, such that these two paths do not share halls other than the staring hall $$$s$$$ and the ending hall $$$t$$$. The hall $$$t$$$ has not been determined yet, so you can choose any of the labyrinth's halls as $$$t$$$ except $$$s$$$.Leslie's and Leon's paths do not have to be the shortest ones, but their paths must be simple, visiting any hall at most once. Also, they cannot visit any common halls except $$$s$$$ and $$$t$$$ during their journey, even at different times.
Input Specification: The first line contains three integers $$$n$$$, $$$m$$$, and $$$s$$$, where $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) is the number of vertices, $$$m$$$ ($$$0 \le m \le 2 \cdot 10^5$$$) is the number of edges in the labyrinth, and $$$s$$$ ($$$1 \le s \le n$$$) is the starting hall. Then $$$m$$$ lines with descriptions of passages follow. Each description contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$; $$$u_i \neq v_i$$$), denoting a passage from the hall $$$u_i$$$ to the hall $$$v_i$$$. The passages are one-way. Each tuple $$$(u_i, v_i)$$$ is present in the input at most once. The labyrinth can contain cycles and is not necessarily connected in any way.
Output Specification: If it is possible to find the desired two paths, output "Possible", otherwise output "Impossible". If the answer exists, output two path descriptions. Each description occupies two lines. The first line of the description contains an integer $$$h$$$ ($$$2 \le h \le n$$$) — the number of halls in a path, and the second line contains distinct integers $$$w_1, w_2, \dots, w_h$$$ ($$$w_1 = s$$$; $$$1 \le w_j \le n$$$; $$$w_h = t$$$) — the halls in the path in the order of passing. Both paths must end at the same vertex $$$t$$$. The paths must be different, and all intermediate halls in these paths must be distinct.
Code:
n, m, s = [int(x) for x in input().split()]
labyrinth = {x: set() for x in range(1, n+1)}
for _ in range(m):
pt1, pt2 = [int(x) for x in input().split()]
labyrinth[pt1].add(pt2)
d_father = {}
for pt in labyrinth[s]:
d_father[pt] = s
if len(d_father) < 2: print('Impossible')
else:
for pt in labyrinth[s]:
visited = {pt, s}
to_visit = {pt}
while to_visit:
new_visit = set()
for origin in to_visit:
for new_pt in labyrinth[origin]:
if new_pt not in visited:
if new_pt in d_father:
path1 = [new_pt]
path2 = [new_pt, origin]
while path1[-1] in d_father:
# TODO: Your code here
while path2[-1] in d_father:
path2.append(d_father[path2[-1]])
path1.reverse()
path2.reverse()
print('Possible')
print(len(path1))
print(' '.join(str(x) for x in path1))
print(len(path2))
print(' '.join(str(x) for x in path2))
break
else:
d_father[new_pt] = origin
new_visit.add(new_pt)
visited.add(new_pt)
else: continue
break
else:
to_visit = new_visit
continue
break
else: continue
break
else: print('Impossible')
|
n, m, s = [int(x) for x in input().split()]
labyrinth = {x: set() for x in range(1, n+1)}
for _ in range(m):
pt1, pt2 = [int(x) for x in input().split()]
labyrinth[pt1].add(pt2)
d_father = {}
for pt in labyrinth[s]:
d_father[pt] = s
if len(d_father) < 2: print('Impossible')
else:
for pt in labyrinth[s]:
visited = {pt, s}
to_visit = {pt}
while to_visit:
new_visit = set()
for origin in to_visit:
for new_pt in labyrinth[origin]:
if new_pt not in visited:
if new_pt in d_father:
path1 = [new_pt]
path2 = [new_pt, origin]
while path1[-1] in d_father:
{{completion}}
while path2[-1] in d_father:
path2.append(d_father[path2[-1]])
path1.reverse()
path2.reverse()
print('Possible')
print(len(path1))
print(' '.join(str(x) for x in path1))
print(len(path2))
print(' '.join(str(x) for x in path2))
break
else:
d_father[new_pt] = origin
new_visit.add(new_pt)
visited.add(new_pt)
else: continue
break
else:
to_visit = new_visit
continue
break
else: continue
break
else: print('Impossible')
|
path1.append(d_father[path1[-1]])
|
[{"input": "5 5 1\n1 2\n2 3\n1 4\n4 3\n3 5", "output": ["Possible\n3\n1 2 3\n3\n1 4 3"]}, {"input": "5 5 1\n1 2\n2 3\n3 4\n2 5\n5 4", "output": ["Impossible"]}, {"input": "3 3 2\n1 2\n2 3\n3 1", "output": ["Impossible"]}]
|
block_completion_003161
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Leslie and Leon entered a labyrinth. The labyrinth consists of $$$n$$$ halls and $$$m$$$ one-way passages between them. The halls are numbered from $$$1$$$ to $$$n$$$.Leslie and Leon start their journey in the hall $$$s$$$. Right away, they quarrel and decide to explore the labyrinth separately. However, they want to meet again at the end of their journey.To help Leslie and Leon, your task is to find two different paths from the given hall $$$s$$$ to some other hall $$$t$$$, such that these two paths do not share halls other than the staring hall $$$s$$$ and the ending hall $$$t$$$. The hall $$$t$$$ has not been determined yet, so you can choose any of the labyrinth's halls as $$$t$$$ except $$$s$$$.Leslie's and Leon's paths do not have to be the shortest ones, but their paths must be simple, visiting any hall at most once. Also, they cannot visit any common halls except $$$s$$$ and $$$t$$$ during their journey, even at different times.
Input Specification: The first line contains three integers $$$n$$$, $$$m$$$, and $$$s$$$, where $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) is the number of vertices, $$$m$$$ ($$$0 \le m \le 2 \cdot 10^5$$$) is the number of edges in the labyrinth, and $$$s$$$ ($$$1 \le s \le n$$$) is the starting hall. Then $$$m$$$ lines with descriptions of passages follow. Each description contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$; $$$u_i \neq v_i$$$), denoting a passage from the hall $$$u_i$$$ to the hall $$$v_i$$$. The passages are one-way. Each tuple $$$(u_i, v_i)$$$ is present in the input at most once. The labyrinth can contain cycles and is not necessarily connected in any way.
Output Specification: If it is possible to find the desired two paths, output "Possible", otherwise output "Impossible". If the answer exists, output two path descriptions. Each description occupies two lines. The first line of the description contains an integer $$$h$$$ ($$$2 \le h \le n$$$) — the number of halls in a path, and the second line contains distinct integers $$$w_1, w_2, \dots, w_h$$$ ($$$w_1 = s$$$; $$$1 \le w_j \le n$$$; $$$w_h = t$$$) — the halls in the path in the order of passing. Both paths must end at the same vertex $$$t$$$. The paths must be different, and all intermediate halls in these paths must be distinct.
Code:
n, m, s = [int(x) for x in input().split()]
labyrinth = {x: set() for x in range(1, n+1)}
for _ in range(m):
pt1, pt2 = [int(x) for x in input().split()]
labyrinth[pt1].add(pt2)
d_father = {}
for pt in labyrinth[s]:
d_father[pt] = s
if len(d_father) < 2: print('Impossible')
else:
for pt in labyrinth[s]:
visited = {pt, s}
to_visit = {pt}
while to_visit:
new_visit = set()
for origin in to_visit:
for new_pt in labyrinth[origin]:
if new_pt not in visited:
if new_pt in d_father:
path1 = [new_pt]
path2 = [new_pt, origin]
while path1[-1] in d_father:
path1.append(d_father[path1[-1]])
while path2[-1] in d_father:
# TODO: Your code here
path1.reverse()
path2.reverse()
print('Possible')
print(len(path1))
print(' '.join(str(x) for x in path1))
print(len(path2))
print(' '.join(str(x) for x in path2))
break
else:
d_father[new_pt] = origin
new_visit.add(new_pt)
visited.add(new_pt)
else: continue
break
else:
to_visit = new_visit
continue
break
else: continue
break
else: print('Impossible')
|
n, m, s = [int(x) for x in input().split()]
labyrinth = {x: set() for x in range(1, n+1)}
for _ in range(m):
pt1, pt2 = [int(x) for x in input().split()]
labyrinth[pt1].add(pt2)
d_father = {}
for pt in labyrinth[s]:
d_father[pt] = s
if len(d_father) < 2: print('Impossible')
else:
for pt in labyrinth[s]:
visited = {pt, s}
to_visit = {pt}
while to_visit:
new_visit = set()
for origin in to_visit:
for new_pt in labyrinth[origin]:
if new_pt not in visited:
if new_pt in d_father:
path1 = [new_pt]
path2 = [new_pt, origin]
while path1[-1] in d_father:
path1.append(d_father[path1[-1]])
while path2[-1] in d_father:
{{completion}}
path1.reverse()
path2.reverse()
print('Possible')
print(len(path1))
print(' '.join(str(x) for x in path1))
print(len(path2))
print(' '.join(str(x) for x in path2))
break
else:
d_father[new_pt] = origin
new_visit.add(new_pt)
visited.add(new_pt)
else: continue
break
else:
to_visit = new_visit
continue
break
else: continue
break
else: print('Impossible')
|
path2.append(d_father[path2[-1]])
|
[{"input": "5 5 1\n1 2\n2 3\n1 4\n4 3\n3 5", "output": ["Possible\n3\n1 2 3\n3\n1 4 3"]}, {"input": "5 5 1\n1 2\n2 3\n3 4\n2 5\n5 4", "output": ["Impossible"]}, {"input": "3 3 2\n1 2\n2 3\n3 1", "output": ["Impossible"]}]
|
block_completion_003162
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Leslie and Leon entered a labyrinth. The labyrinth consists of $$$n$$$ halls and $$$m$$$ one-way passages between them. The halls are numbered from $$$1$$$ to $$$n$$$.Leslie and Leon start their journey in the hall $$$s$$$. Right away, they quarrel and decide to explore the labyrinth separately. However, they want to meet again at the end of their journey.To help Leslie and Leon, your task is to find two different paths from the given hall $$$s$$$ to some other hall $$$t$$$, such that these two paths do not share halls other than the staring hall $$$s$$$ and the ending hall $$$t$$$. The hall $$$t$$$ has not been determined yet, so you can choose any of the labyrinth's halls as $$$t$$$ except $$$s$$$.Leslie's and Leon's paths do not have to be the shortest ones, but their paths must be simple, visiting any hall at most once. Also, they cannot visit any common halls except $$$s$$$ and $$$t$$$ during their journey, even at different times.
Input Specification: The first line contains three integers $$$n$$$, $$$m$$$, and $$$s$$$, where $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) is the number of vertices, $$$m$$$ ($$$0 \le m \le 2 \cdot 10^5$$$) is the number of edges in the labyrinth, and $$$s$$$ ($$$1 \le s \le n$$$) is the starting hall. Then $$$m$$$ lines with descriptions of passages follow. Each description contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$; $$$u_i \neq v_i$$$), denoting a passage from the hall $$$u_i$$$ to the hall $$$v_i$$$. The passages are one-way. Each tuple $$$(u_i, v_i)$$$ is present in the input at most once. The labyrinth can contain cycles and is not necessarily connected in any way.
Output Specification: If it is possible to find the desired two paths, output "Possible", otherwise output "Impossible". If the answer exists, output two path descriptions. Each description occupies two lines. The first line of the description contains an integer $$$h$$$ ($$$2 \le h \le n$$$) — the number of halls in a path, and the second line contains distinct integers $$$w_1, w_2, \dots, w_h$$$ ($$$w_1 = s$$$; $$$1 \le w_j \le n$$$; $$$w_h = t$$$) — the halls in the path in the order of passing. Both paths must end at the same vertex $$$t$$$. The paths must be different, and all intermediate halls in these paths must be distinct.
Code:
import re
import sys
exit=sys.exit
from bisect import *
from collections import *
ddict=defaultdict
from functools import lru_cache
cache=lru_cache(None)
from heapq import *
from itertools import *
from math import inf
from pprint import pprint as pp
enum=enumerate
rb=lambda:list(map(int,rl()))
rfs=lambda:rln().split()
ri=lambda:int(rln())
ris=lambda:list(map(int,rfs()))
rl=lambda:rln().rstrip('\n')
rln=sys.stdin.readline
cat=''.join
catn='\n'.join
mod=1000000007
d4=[(0,-1),(1,0),(0,1),(-1,0)]
d8=[(-1,-1),(0,-1),(1,-1),(-1,0),(1,0),(-1,1),(0,1),(1,1)]
YES,NO,Yes,No,yes,no='YES','NO','Yes','No','yes','no'
########################################################################
n,m,s=ris()
adj=[[] for _ in range(n+1)]
for _ in range(m):
u,v=ris()
adj[u].append(v)
vis=[{} for _ in range(n+1)]
for i in adj[s]:
stk=[i]
vis[s][i]=0
vis[i][i]=s
while stk:
u=stk.pop()
if 1<len(vis[u]):
print('Possible')
for j in vis[u]:
x,path=u,[]
while j in vis[x]:
# TODO: Your code here
print(len(path))
print(*reversed(path))
exit()
for v in adj[u]:
if i in vis[v] or v==s:
continue
stk.append(v)
vis[v][i]=u
print('Impossible')
|
import re
import sys
exit=sys.exit
from bisect import *
from collections import *
ddict=defaultdict
from functools import lru_cache
cache=lru_cache(None)
from heapq import *
from itertools import *
from math import inf
from pprint import pprint as pp
enum=enumerate
rb=lambda:list(map(int,rl()))
rfs=lambda:rln().split()
ri=lambda:int(rln())
ris=lambda:list(map(int,rfs()))
rl=lambda:rln().rstrip('\n')
rln=sys.stdin.readline
cat=''.join
catn='\n'.join
mod=1000000007
d4=[(0,-1),(1,0),(0,1),(-1,0)]
d8=[(-1,-1),(0,-1),(1,-1),(-1,0),(1,0),(-1,1),(0,1),(1,1)]
YES,NO,Yes,No,yes,no='YES','NO','Yes','No','yes','no'
########################################################################
n,m,s=ris()
adj=[[] for _ in range(n+1)]
for _ in range(m):
u,v=ris()
adj[u].append(v)
vis=[{} for _ in range(n+1)]
for i in adj[s]:
stk=[i]
vis[s][i]=0
vis[i][i]=s
while stk:
u=stk.pop()
if 1<len(vis[u]):
print('Possible')
for j in vis[u]:
x,path=u,[]
while j in vis[x]:
{{completion}}
print(len(path))
print(*reversed(path))
exit()
for v in adj[u]:
if i in vis[v] or v==s:
continue
stk.append(v)
vis[v][i]=u
print('Impossible')
|
path.append(x)
x=vis[x][j]
|
[{"input": "5 5 1\n1 2\n2 3\n1 4\n4 3\n3 5", "output": ["Possible\n3\n1 2 3\n3\n1 4 3"]}, {"input": "5 5 1\n1 2\n2 3\n3 4\n2 5\n5 4", "output": ["Impossible"]}, {"input": "3 3 2\n1 2\n2 3\n3 1", "output": ["Impossible"]}]
|
block_completion_003163
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Leslie and Leon entered a labyrinth. The labyrinth consists of $$$n$$$ halls and $$$m$$$ one-way passages between them. The halls are numbered from $$$1$$$ to $$$n$$$.Leslie and Leon start their journey in the hall $$$s$$$. Right away, they quarrel and decide to explore the labyrinth separately. However, they want to meet again at the end of their journey.To help Leslie and Leon, your task is to find two different paths from the given hall $$$s$$$ to some other hall $$$t$$$, such that these two paths do not share halls other than the staring hall $$$s$$$ and the ending hall $$$t$$$. The hall $$$t$$$ has not been determined yet, so you can choose any of the labyrinth's halls as $$$t$$$ except $$$s$$$.Leslie's and Leon's paths do not have to be the shortest ones, but their paths must be simple, visiting any hall at most once. Also, they cannot visit any common halls except $$$s$$$ and $$$t$$$ during their journey, even at different times.
Input Specification: The first line contains three integers $$$n$$$, $$$m$$$, and $$$s$$$, where $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) is the number of vertices, $$$m$$$ ($$$0 \le m \le 2 \cdot 10^5$$$) is the number of edges in the labyrinth, and $$$s$$$ ($$$1 \le s \le n$$$) is the starting hall. Then $$$m$$$ lines with descriptions of passages follow. Each description contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$; $$$u_i \neq v_i$$$), denoting a passage from the hall $$$u_i$$$ to the hall $$$v_i$$$. The passages are one-way. Each tuple $$$(u_i, v_i)$$$ is present in the input at most once. The labyrinth can contain cycles and is not necessarily connected in any way.
Output Specification: If it is possible to find the desired two paths, output "Possible", otherwise output "Impossible". If the answer exists, output two path descriptions. Each description occupies two lines. The first line of the description contains an integer $$$h$$$ ($$$2 \le h \le n$$$) — the number of halls in a path, and the second line contains distinct integers $$$w_1, w_2, \dots, w_h$$$ ($$$w_1 = s$$$; $$$1 \le w_j \le n$$$; $$$w_h = t$$$) — the halls in the path in the order of passing. Both paths must end at the same vertex $$$t$$$. The paths must be different, and all intermediate halls in these paths must be distinct.
Code:
def DFS(start):
nodes=set()
stack=[start]
while stack:
parent=stack.pop()
if(not visited[parent]):
nodes.add(parent)
visited[parent]=True
for child in graph[parent]:
if (not visited[child]):
stack.append(child)
else:
if child not in nodes and child!=s:
# TODO: Your code here
else:
if parent not in nodes and parent != s:
return parent
return -1
def DFS_get_path(start):
stack=[start]
parent_list[start]=-1
while stack:
parent=stack.pop()
if parent==end:
visited[end]=False
return True
if(not visited[parent]):
visited[parent]=True
for child in graph[parent]:
if (not visited[child]):
stack.append(child)
parent_list[child]=parent
return False
def get_path(node):
path=[]
while node!=-1:
path.append(node)
node=parent_list[node]
path.reverse()
return path
n,m,s=map(int,input().split())
s-=1
graph=[[] for _ in range(n)]
for _ in range(m):
a,b=map(int,input().split())
a-=1
b-=1
graph[a].append(b)
visited=[False]*n
visited[s]=True
for child in graph[s]:
end=DFS(child)
if end!=-1:
visited = [False] * n
parent_list=[-1]*n
visited[s]=True
ans=[]
for child in graph[s]:
if DFS_get_path(child):
ans.append([s]+get_path(end))
if len(ans)==2:
break
print("Possible")
for i in ans:
print(len(i))
print(*[j+1 for j in i])
break
else:
print("Impossible")
# 3 3 1
# 1 2
# 2 1
# 1 3
|
def DFS(start):
nodes=set()
stack=[start]
while stack:
parent=stack.pop()
if(not visited[parent]):
nodes.add(parent)
visited[parent]=True
for child in graph[parent]:
if (not visited[child]):
stack.append(child)
else:
if child not in nodes and child!=s:
{{completion}}
else:
if parent not in nodes and parent != s:
return parent
return -1
def DFS_get_path(start):
stack=[start]
parent_list[start]=-1
while stack:
parent=stack.pop()
if parent==end:
visited[end]=False
return True
if(not visited[parent]):
visited[parent]=True
for child in graph[parent]:
if (not visited[child]):
stack.append(child)
parent_list[child]=parent
return False
def get_path(node):
path=[]
while node!=-1:
path.append(node)
node=parent_list[node]
path.reverse()
return path
n,m,s=map(int,input().split())
s-=1
graph=[[] for _ in range(n)]
for _ in range(m):
a,b=map(int,input().split())
a-=1
b-=1
graph[a].append(b)
visited=[False]*n
visited[s]=True
for child in graph[s]:
end=DFS(child)
if end!=-1:
visited = [False] * n
parent_list=[-1]*n
visited[s]=True
ans=[]
for child in graph[s]:
if DFS_get_path(child):
ans.append([s]+get_path(end))
if len(ans)==2:
break
print("Possible")
for i in ans:
print(len(i))
print(*[j+1 for j in i])
break
else:
print("Impossible")
# 3 3 1
# 1 2
# 2 1
# 1 3
|
return child
|
[{"input": "5 5 1\n1 2\n2 3\n1 4\n4 3\n3 5", "output": ["Possible\n3\n1 2 3\n3\n1 4 3"]}, {"input": "5 5 1\n1 2\n2 3\n3 4\n2 5\n5 4", "output": ["Impossible"]}, {"input": "3 3 2\n1 2\n2 3\n3 1", "output": ["Impossible"]}]
|
block_completion_003164
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Leslie and Leon entered a labyrinth. The labyrinth consists of $$$n$$$ halls and $$$m$$$ one-way passages between them. The halls are numbered from $$$1$$$ to $$$n$$$.Leslie and Leon start their journey in the hall $$$s$$$. Right away, they quarrel and decide to explore the labyrinth separately. However, they want to meet again at the end of their journey.To help Leslie and Leon, your task is to find two different paths from the given hall $$$s$$$ to some other hall $$$t$$$, such that these two paths do not share halls other than the staring hall $$$s$$$ and the ending hall $$$t$$$. The hall $$$t$$$ has not been determined yet, so you can choose any of the labyrinth's halls as $$$t$$$ except $$$s$$$.Leslie's and Leon's paths do not have to be the shortest ones, but their paths must be simple, visiting any hall at most once. Also, they cannot visit any common halls except $$$s$$$ and $$$t$$$ during their journey, even at different times.
Input Specification: The first line contains three integers $$$n$$$, $$$m$$$, and $$$s$$$, where $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) is the number of vertices, $$$m$$$ ($$$0 \le m \le 2 \cdot 10^5$$$) is the number of edges in the labyrinth, and $$$s$$$ ($$$1 \le s \le n$$$) is the starting hall. Then $$$m$$$ lines with descriptions of passages follow. Each description contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$; $$$u_i \neq v_i$$$), denoting a passage from the hall $$$u_i$$$ to the hall $$$v_i$$$. The passages are one-way. Each tuple $$$(u_i, v_i)$$$ is present in the input at most once. The labyrinth can contain cycles and is not necessarily connected in any way.
Output Specification: If it is possible to find the desired two paths, output "Possible", otherwise output "Impossible". If the answer exists, output two path descriptions. Each description occupies two lines. The first line of the description contains an integer $$$h$$$ ($$$2 \le h \le n$$$) — the number of halls in a path, and the second line contains distinct integers $$$w_1, w_2, \dots, w_h$$$ ($$$w_1 = s$$$; $$$1 \le w_j \le n$$$; $$$w_h = t$$$) — the halls in the path in the order of passing. Both paths must end at the same vertex $$$t$$$. The paths must be different, and all intermediate halls in these paths must be distinct.
Code:
def DFS(start):
nodes=set()
stack=[start]
while stack:
parent=stack.pop()
if(not visited[parent]):
nodes.add(parent)
visited[parent]=True
for child in graph[parent]:
if (not visited[child]):
# TODO: Your code here
else:
if child not in nodes and child!=s:
return child
else:
if parent not in nodes and parent != s:
return parent
return -1
def DFS_get_path(start):
stack=[start]
parent_list[start]=-1
while stack:
parent=stack.pop()
if parent==end:
visited[end]=False
return True
if(not visited[parent]):
visited[parent]=True
for child in graph[parent]:
if (not visited[child]):
stack.append(child)
parent_list[child]=parent
return False
def get_path(node):
path=[]
while node!=-1:
path.append(node)
node=parent_list[node]
path.reverse()
return path
n,m,s=map(int,input().split())
s-=1
graph=[[] for _ in range(n)]
for _ in range(m):
a,b=map(int,input().split())
a-=1
b-=1
graph[a].append(b)
visited=[False]*n
visited[s]=True
for child in graph[s]:
end=DFS(child)
if end!=-1:
visited = [False] * n
parent_list=[-1]*n
visited[s]=True
ans=[]
for child in graph[s]:
if DFS_get_path(child):
ans.append([s]+get_path(end))
if len(ans)==2:
break
print("Possible")
for i in ans:
print(len(i))
print(*[j+1 for j in i])
break
else:
print("Impossible")
# 3 3 1
# 1 2
# 2 1
# 1 3
|
def DFS(start):
nodes=set()
stack=[start]
while stack:
parent=stack.pop()
if(not visited[parent]):
nodes.add(parent)
visited[parent]=True
for child in graph[parent]:
if (not visited[child]):
{{completion}}
else:
if child not in nodes and child!=s:
return child
else:
if parent not in nodes and parent != s:
return parent
return -1
def DFS_get_path(start):
stack=[start]
parent_list[start]=-1
while stack:
parent=stack.pop()
if parent==end:
visited[end]=False
return True
if(not visited[parent]):
visited[parent]=True
for child in graph[parent]:
if (not visited[child]):
stack.append(child)
parent_list[child]=parent
return False
def get_path(node):
path=[]
while node!=-1:
path.append(node)
node=parent_list[node]
path.reverse()
return path
n,m,s=map(int,input().split())
s-=1
graph=[[] for _ in range(n)]
for _ in range(m):
a,b=map(int,input().split())
a-=1
b-=1
graph[a].append(b)
visited=[False]*n
visited[s]=True
for child in graph[s]:
end=DFS(child)
if end!=-1:
visited = [False] * n
parent_list=[-1]*n
visited[s]=True
ans=[]
for child in graph[s]:
if DFS_get_path(child):
ans.append([s]+get_path(end))
if len(ans)==2:
break
print("Possible")
for i in ans:
print(len(i))
print(*[j+1 for j in i])
break
else:
print("Impossible")
# 3 3 1
# 1 2
# 2 1
# 1 3
|
stack.append(child)
|
[{"input": "5 5 1\n1 2\n2 3\n1 4\n4 3\n3 5", "output": ["Possible\n3\n1 2 3\n3\n1 4 3"]}, {"input": "5 5 1\n1 2\n2 3\n3 4\n2 5\n5 4", "output": ["Impossible"]}, {"input": "3 3 2\n1 2\n2 3\n3 1", "output": ["Impossible"]}]
|
block_completion_003165
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given three points on a plane. You should choose some segments on the plane that are parallel to coordinate axes, so that all three points become connected. The total length of the chosen segments should be the minimal possible.Two points $$$a$$$ and $$$b$$$ are considered connected if there is a sequence of points $$$p_0 = a, p_1, \ldots, p_k = b$$$ such that points $$$p_i$$$ and $$$p_{i+1}$$$ lie on the same segment.
Input Specification: The input consists of three lines describing three points. Each line contains two integers $$$x$$$ and $$$y$$$ separated by a space — the coordinates of the point ($$$-10^9 \le x, y \le 10^9$$$). The points are pairwise distinct.
Output Specification: On the first line output $$$n$$$ — the number of segments, at most 100. The next $$$n$$$ lines should contain descriptions of segments. Output four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ on a line — the coordinates of the endpoints of the corresponding segment ($$$-10^9 \le x_1, y_1, x_2, y_2 \le 10^9$$$). Each segment should be either horizontal or vertical. It is guaranteed that the solution with the given constraints exists.
Notes: NoteThe points and the segments from the example are shown below.
Code:
l=[[*map(int,input().split())] for i in range(3)]
l=sorted(l,key=lambda x:x[1])
ans=[]
ans.append([l[0][0],l[0][1],l[0][0],l[1][1]])
l[0]=[l[0][0],l[1][1]]
if (l[1][0]<l[0][0] and l[2][0]>l[0][0]) or (l[1][0]>l[0][0] and l[2][0]<l[0][0]):
ans.append([*l[0],l[1][0],l[0][1]])
ans.append([*l[0],l[2][0],l[0][1]])
ans.append([l[2][0],l[0][1],l[2][0],l[2][1]])
else:
if max(l[1][0],l[2][0])>l[0][0]:leng=max(l[1][0],l[2][0])
else:# TODO: Your code here
ans.append([*l[0],leng,l[0][1]])
ans.append([l[2][0],l[0][1],l[2][0],l[2][1]])
print(len(ans))
for i in ans:
print(*i)
|
l=[[*map(int,input().split())] for i in range(3)]
l=sorted(l,key=lambda x:x[1])
ans=[]
ans.append([l[0][0],l[0][1],l[0][0],l[1][1]])
l[0]=[l[0][0],l[1][1]]
if (l[1][0]<l[0][0] and l[2][0]>l[0][0]) or (l[1][0]>l[0][0] and l[2][0]<l[0][0]):
ans.append([*l[0],l[1][0],l[0][1]])
ans.append([*l[0],l[2][0],l[0][1]])
ans.append([l[2][0],l[0][1],l[2][0],l[2][1]])
else:
if max(l[1][0],l[2][0])>l[0][0]:leng=max(l[1][0],l[2][0])
else:{{completion}}
ans.append([*l[0],leng,l[0][1]])
ans.append([l[2][0],l[0][1],l[2][0],l[2][1]])
print(len(ans))
for i in ans:
print(*i)
|
leng=min(l[1][0],l[2][0])
|
[{"input": "1 1\n3 5\n8 6", "output": ["3\n1 1 1 5\n1 5 8 5\n8 5 8 6"]}]
|
block_completion_003169
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given three points on a plane. You should choose some segments on the plane that are parallel to coordinate axes, so that all three points become connected. The total length of the chosen segments should be the minimal possible.Two points $$$a$$$ and $$$b$$$ are considered connected if there is a sequence of points $$$p_0 = a, p_1, \ldots, p_k = b$$$ such that points $$$p_i$$$ and $$$p_{i+1}$$$ lie on the same segment.
Input Specification: The input consists of three lines describing three points. Each line contains two integers $$$x$$$ and $$$y$$$ separated by a space — the coordinates of the point ($$$-10^9 \le x, y \le 10^9$$$). The points are pairwise distinct.
Output Specification: On the first line output $$$n$$$ — the number of segments, at most 100. The next $$$n$$$ lines should contain descriptions of segments. Output four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ on a line — the coordinates of the endpoints of the corresponding segment ($$$-10^9 \le x_1, y_1, x_2, y_2 \le 10^9$$$). Each segment should be either horizontal or vertical. It is guaranteed that the solution with the given constraints exists.
Notes: NoteThe points and the segments from the example are shown below.
Code:
result = []
t = [[0, 0], [0, 0], [0, 0]]
for i in range(3):
t[i] = list(map(int, input().split()))
t.sort()
a = t[0]
b = t[1]
c = t[2]
d1 = b[1] - a[1]
d2 = c[1] - b[1]
if d1*d2 >= 0: # 2nd between 1st and 3rd
if a[1] != b[1]:
result.append([a[0], a[1], a[0], b[1]])
if a[0] != c[0]:
result.append([a[0], b[1], c[0], b[1]])
if b[1] != c[1]:
result.append([c[0], b[1], c[0], c[1]])
else: # additional point X needed
d1 = b[1] - c[1]
d2 = c[1] - a[1]
if d1*d2 >= 0: # 3rd between 1st and 2nd
x = [b[0], c[1]]
if a[1] != x[1]:
result.append([a[0], a[1], a[0], x[1]])
if a[0] != x[0]:
result.append([a[0], x[1], x[0], x[1]])
if x[0] != c[0]:
result.append([x[0], x[1], c[0], c[1]])
if x[1] != b[1]:
result.append([x[0], x[1], b[0], b[1]])
else: # 1st between 2nd and 2rd
x = [b[0], a[1]]
if c[0] != x[0]:
# TODO: Your code here
if c[1] != x[1]:
result.append([x[0], c[1], x[0], x[1]])
if x[0] != a[0]:
result.append([x[0], x[1], a[0], a[1]])
if x[1] != b[1]:
result.append([x[0], x[1], b[0], b[1]])
print(len(result))
for i in range(len(result)):
tt = result[i]
print(tt[0], tt[1], tt[2], tt[3])
|
result = []
t = [[0, 0], [0, 0], [0, 0]]
for i in range(3):
t[i] = list(map(int, input().split()))
t.sort()
a = t[0]
b = t[1]
c = t[2]
d1 = b[1] - a[1]
d2 = c[1] - b[1]
if d1*d2 >= 0: # 2nd between 1st and 3rd
if a[1] != b[1]:
result.append([a[0], a[1], a[0], b[1]])
if a[0] != c[0]:
result.append([a[0], b[1], c[0], b[1]])
if b[1] != c[1]:
result.append([c[0], b[1], c[0], c[1]])
else: # additional point X needed
d1 = b[1] - c[1]
d2 = c[1] - a[1]
if d1*d2 >= 0: # 3rd between 1st and 2nd
x = [b[0], c[1]]
if a[1] != x[1]:
result.append([a[0], a[1], a[0], x[1]])
if a[0] != x[0]:
result.append([a[0], x[1], x[0], x[1]])
if x[0] != c[0]:
result.append([x[0], x[1], c[0], c[1]])
if x[1] != b[1]:
result.append([x[0], x[1], b[0], b[1]])
else: # 1st between 2nd and 2rd
x = [b[0], a[1]]
if c[0] != x[0]:
{{completion}}
if c[1] != x[1]:
result.append([x[0], c[1], x[0], x[1]])
if x[0] != a[0]:
result.append([x[0], x[1], a[0], a[1]])
if x[1] != b[1]:
result.append([x[0], x[1], b[0], b[1]])
print(len(result))
for i in range(len(result)):
tt = result[i]
print(tt[0], tt[1], tt[2], tt[3])
|
result.append([c[0], c[1], x[0], c[1]])
|
[{"input": "1 1\n3 5\n8 6", "output": ["3\n1 1 1 5\n1 5 8 5\n8 5 8 6"]}]
|
block_completion_003170
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given three points on a plane. You should choose some segments on the plane that are parallel to coordinate axes, so that all three points become connected. The total length of the chosen segments should be the minimal possible.Two points $$$a$$$ and $$$b$$$ are considered connected if there is a sequence of points $$$p_0 = a, p_1, \ldots, p_k = b$$$ such that points $$$p_i$$$ and $$$p_{i+1}$$$ lie on the same segment.
Input Specification: The input consists of three lines describing three points. Each line contains two integers $$$x$$$ and $$$y$$$ separated by a space — the coordinates of the point ($$$-10^9 \le x, y \le 10^9$$$). The points are pairwise distinct.
Output Specification: On the first line output $$$n$$$ — the number of segments, at most 100. The next $$$n$$$ lines should contain descriptions of segments. Output four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ on a line — the coordinates of the endpoints of the corresponding segment ($$$-10^9 \le x_1, y_1, x_2, y_2 \le 10^9$$$). Each segment should be either horizontal or vertical. It is guaranteed that the solution with the given constraints exists.
Notes: NoteThe points and the segments from the example are shown below.
Code:
result = []
t = [[0, 0], [0, 0], [0, 0]]
for i in range(3):
t[i] = list(map(int, input().split()))
t.sort()
a = t[0]
b = t[1]
c = t[2]
d1 = b[1] - a[1]
d2 = c[1] - b[1]
if d1*d2 >= 0: # 2nd between 1st and 3rd
if a[1] != b[1]:
result.append([a[0], a[1], a[0], b[1]])
if a[0] != c[0]:
result.append([a[0], b[1], c[0], b[1]])
if b[1] != c[1]:
result.append([c[0], b[1], c[0], c[1]])
else: # additional point X needed
d1 = b[1] - c[1]
d2 = c[1] - a[1]
if d1*d2 >= 0: # 3rd between 1st and 2nd
x = [b[0], c[1]]
if a[1] != x[1]:
result.append([a[0], a[1], a[0], x[1]])
if a[0] != x[0]:
result.append([a[0], x[1], x[0], x[1]])
if x[0] != c[0]:
result.append([x[0], x[1], c[0], c[1]])
if x[1] != b[1]:
result.append([x[0], x[1], b[0], b[1]])
else: # 1st between 2nd and 2rd
x = [b[0], a[1]]
if c[0] != x[0]:
result.append([c[0], c[1], x[0], c[1]])
if c[1] != x[1]:
# TODO: Your code here
if x[0] != a[0]:
result.append([x[0], x[1], a[0], a[1]])
if x[1] != b[1]:
result.append([x[0], x[1], b[0], b[1]])
print(len(result))
for i in range(len(result)):
tt = result[i]
print(tt[0], tt[1], tt[2], tt[3])
|
result = []
t = [[0, 0], [0, 0], [0, 0]]
for i in range(3):
t[i] = list(map(int, input().split()))
t.sort()
a = t[0]
b = t[1]
c = t[2]
d1 = b[1] - a[1]
d2 = c[1] - b[1]
if d1*d2 >= 0: # 2nd between 1st and 3rd
if a[1] != b[1]:
result.append([a[0], a[1], a[0], b[1]])
if a[0] != c[0]:
result.append([a[0], b[1], c[0], b[1]])
if b[1] != c[1]:
result.append([c[0], b[1], c[0], c[1]])
else: # additional point X needed
d1 = b[1] - c[1]
d2 = c[1] - a[1]
if d1*d2 >= 0: # 3rd between 1st and 2nd
x = [b[0], c[1]]
if a[1] != x[1]:
result.append([a[0], a[1], a[0], x[1]])
if a[0] != x[0]:
result.append([a[0], x[1], x[0], x[1]])
if x[0] != c[0]:
result.append([x[0], x[1], c[0], c[1]])
if x[1] != b[1]:
result.append([x[0], x[1], b[0], b[1]])
else: # 1st between 2nd and 2rd
x = [b[0], a[1]]
if c[0] != x[0]:
result.append([c[0], c[1], x[0], c[1]])
if c[1] != x[1]:
{{completion}}
if x[0] != a[0]:
result.append([x[0], x[1], a[0], a[1]])
if x[1] != b[1]:
result.append([x[0], x[1], b[0], b[1]])
print(len(result))
for i in range(len(result)):
tt = result[i]
print(tt[0], tt[1], tt[2], tt[3])
|
result.append([x[0], c[1], x[0], x[1]])
|
[{"input": "1 1\n3 5\n8 6", "output": ["3\n1 1 1 5\n1 5 8 5\n8 5 8 6"]}]
|
block_completion_003171
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Julia's $$$n$$$ friends want to organize a startup in a new country they moved to. They assigned each other numbers from 1 to $$$n$$$ according to the jobs they have, from the most front-end tasks to the most back-end ones. They also estimated a matrix $$$c$$$, where $$$c_{ij} = c_{ji}$$$ is the average number of messages per month between people doing jobs $$$i$$$ and $$$j$$$.Now they want to make a hierarchy tree. It will be a binary tree with each node containing one member of the team. Some member will be selected as a leader of the team and will be contained in the root node. In order for the leader to be able to easily reach any subordinate, for each node $$$v$$$ of the tree, the following should apply: all members in its left subtree must have smaller numbers than $$$v$$$, and all members in its right subtree must have larger numbers than $$$v$$$.After the hierarchy tree is settled, people doing jobs $$$i$$$ and $$$j$$$ will be communicating via the shortest path in the tree between their nodes. Let's denote the length of this path as $$$d_{ij}$$$. Thus, the cost of their communication is $$$c_{ij} \cdot d_{ij}$$$.Your task is to find a hierarchy tree that minimizes the total cost of communication over all pairs: $$$\sum_{1 \le i < j \le n} c_{ij} \cdot d_{ij}$$$.
Input Specification: The first line contains an integer $$$n$$$ ($$$1 \le n \le 200$$$) – the number of team members organizing a startup. The next $$$n$$$ lines contain $$$n$$$ integers each, $$$j$$$-th number in $$$i$$$-th line is $$$c_{ij}$$$ — the estimated number of messages per month between team members $$$i$$$ and $$$j$$$ ($$$0 \le c_{ij} \le 10^9; c_{ij} = c_{ji}; c_{ii} = 0$$$).
Output Specification: Output a description of a hierarchy tree that minimizes the total cost of communication. To do so, for each team member from 1 to $$$n$$$ output the number of the member in its parent node, or 0 for the leader. If there are many optimal trees, output a description of any one of them.
Notes: NoteThe minimal possible total cost is $$$566 \cdot 1+239 \cdot 1+30 \cdot 1+1 \cdot 2+1 \cdot 2=839$$$:
Code:
'''
题意：将n个人安排到二叉树中，每两人之间通过最短路径d[i,j]通信，消息数量c[i,j]通过矩阵给出，
要求构造的二叉树满足：任意节点u的左子树中节点序号<u、右子树…>，且所有的节点对之间的c[i,j]*d[i,j]的总和最小。
区间dp，对于f[l,r]，枚举根节点k，得到最大值时记住区间[l,r]的根节点k，
最后先序遍历为每个节点指定根节点。
'''
R,G=lambda:map(int,input().split()),range;n,=R();Fa=[0]*n;N=n+1
A=lambda:[[0]*N for _ in G(N)];S,Rt,f=A(),A(),A()
for i in G(n):
r=[*R()]
for j in G(n):S[i+1][j+1]=r[j]+S[i][j+1]+S[i+1][j]-S[i][j]
rec=lambda i,x:S[x][x]-S[x][i]-S[i][x]+S[i][i]
for c in G(1,N):
for l in G(N-c):
r=l+c;f[l][l+c]=float('inf')
for k in G(l,r):
C=f[l][k]+S[k][n]-S[l][n]-rec(l,k)+f[k+1][r]+S[r][n]-S[k+1][n]-rec(k+1,r)
if C<f[l][r]:# TODO: Your code here
def F(l,r,fa):
if l==r:return
mid=Rt[l][r];Fa[mid]=fa+1;F(l,mid,mid);F(mid+1,r,mid)
F(0,n,-1);print(*Fa)
|
'''
题意：将n个人安排到二叉树中，每两人之间通过最短路径d[i,j]通信，消息数量c[i,j]通过矩阵给出，
要求构造的二叉树满足：任意节点u的左子树中节点序号<u、右子树…>，且所有的节点对之间的c[i,j]*d[i,j]的总和最小。
区间dp，对于f[l,r]，枚举根节点k，得到最大值时记住区间[l,r]的根节点k，
最后先序遍历为每个节点指定根节点。
'''
R,G=lambda:map(int,input().split()),range;n,=R();Fa=[0]*n;N=n+1
A=lambda:[[0]*N for _ in G(N)];S,Rt,f=A(),A(),A()
for i in G(n):
r=[*R()]
for j in G(n):S[i+1][j+1]=r[j]+S[i][j+1]+S[i+1][j]-S[i][j]
rec=lambda i,x:S[x][x]-S[x][i]-S[i][x]+S[i][i]
for c in G(1,N):
for l in G(N-c):
r=l+c;f[l][l+c]=float('inf')
for k in G(l,r):
C=f[l][k]+S[k][n]-S[l][n]-rec(l,k)+f[k+1][r]+S[r][n]-S[k+1][n]-rec(k+1,r)
if C<f[l][r]:{{completion}}
def F(l,r,fa):
if l==r:return
mid=Rt[l][r];Fa[mid]=fa+1;F(l,mid,mid);F(mid+1,r,mid)
F(0,n,-1);print(*Fa)
|
f[l][r]=C;Rt[l][r]=k
|
[{"input": "4\n0 566 1 0\n566 0 239 30\n1 239 0 1\n0 30 1 0", "output": ["2 4 2 0"]}]
|
block_completion_003207
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Julia's $$$n$$$ friends want to organize a startup in a new country they moved to. They assigned each other numbers from 1 to $$$n$$$ according to the jobs they have, from the most front-end tasks to the most back-end ones. They also estimated a matrix $$$c$$$, where $$$c_{ij} = c_{ji}$$$ is the average number of messages per month between people doing jobs $$$i$$$ and $$$j$$$.Now they want to make a hierarchy tree. It will be a binary tree with each node containing one member of the team. Some member will be selected as a leader of the team and will be contained in the root node. In order for the leader to be able to easily reach any subordinate, for each node $$$v$$$ of the tree, the following should apply: all members in its left subtree must have smaller numbers than $$$v$$$, and all members in its right subtree must have larger numbers than $$$v$$$.After the hierarchy tree is settled, people doing jobs $$$i$$$ and $$$j$$$ will be communicating via the shortest path in the tree between their nodes. Let's denote the length of this path as $$$d_{ij}$$$. Thus, the cost of their communication is $$$c_{ij} \cdot d_{ij}$$$.Your task is to find a hierarchy tree that minimizes the total cost of communication over all pairs: $$$\sum_{1 \le i < j \le n} c_{ij} \cdot d_{ij}$$$.
Input Specification: The first line contains an integer $$$n$$$ ($$$1 \le n \le 200$$$) – the number of team members organizing a startup. The next $$$n$$$ lines contain $$$n$$$ integers each, $$$j$$$-th number in $$$i$$$-th line is $$$c_{ij}$$$ — the estimated number of messages per month between team members $$$i$$$ and $$$j$$$ ($$$0 \le c_{ij} \le 10^9; c_{ij} = c_{ji}; c_{ii} = 0$$$).
Output Specification: Output a description of a hierarchy tree that minimizes the total cost of communication. To do so, for each team member from 1 to $$$n$$$ output the number of the member in its parent node, or 0 for the leader. If there are many optimal trees, output a description of any one of them.
Notes: NoteThe minimal possible total cost is $$$566 \cdot 1+239 \cdot 1+30 \cdot 1+1 \cdot 2+1 \cdot 2=839$$$:
Code:
n = int(input().strip())
S = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
for j, c in enumerate(map(int, input().strip().split())):
S[i][j] = c
for i in range(n):
for j in range(n):
if i > 0 and j > 0:
S[i][j] += S[i][j-1] + S[i-1][j] - S[i-1][j-1]
elif i > 0:
# TODO: Your code here
elif j > 0:
S[i][j] += S[i][j-1]
def acc(i1, i2, j1, j2):
if i1 >= i2 or j1 >= j2:
return 0
a = S[i2-1][j2-1]
b = S[i2-1][j1-1] if j1 > 0 else 0
c = S[i1-1][j2-1] if i1 > 0 else 0
d = S[i1-1][j1-1] if i1 > 0 and j1 > 0 else 0
return a - b - c + d
M = [[-1 for i in range(n)] for j in range(n)]
P = [[-1 for i in range(n)] for j in range(n)]
def solve(b, e):
if e - b == 1:
M[b][e-1] = 0
return 0
if e - b == 0:
return 0
if M[b][e-1] != -1:
return M[b][e-1]
M[b][e-1] = 1e18
for i in range(b, e):
s = solve(b, i) + solve(i+1, e)
s += acc(0, b, b, i) + acc(b, i, i, n) + acc(0, i+1, i+1, e) + acc(i+1, e, e, n)
if s < M[b][e-1]:
M[b][e-1] = s
P[b][e-1] = i
return M[b][e-1]
solve(0, n)
sol = ["" for _ in range(n)]
def label(b, e, p):
if e - b == 1:
sol[b] = str(p)
return
elif e - b == 0:
return
i = P[b][e-1]
sol[i] = str(p)
label(b, i, i+1)
label(i+1, e, i+1)
label(0, n, 0)
print(" ".join(sol))
|
n = int(input().strip())
S = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
for j, c in enumerate(map(int, input().strip().split())):
S[i][j] = c
for i in range(n):
for j in range(n):
if i > 0 and j > 0:
S[i][j] += S[i][j-1] + S[i-1][j] - S[i-1][j-1]
elif i > 0:
{{completion}}
elif j > 0:
S[i][j] += S[i][j-1]
def acc(i1, i2, j1, j2):
if i1 >= i2 or j1 >= j2:
return 0
a = S[i2-1][j2-1]
b = S[i2-1][j1-1] if j1 > 0 else 0
c = S[i1-1][j2-1] if i1 > 0 else 0
d = S[i1-1][j1-1] if i1 > 0 and j1 > 0 else 0
return a - b - c + d
M = [[-1 for i in range(n)] for j in range(n)]
P = [[-1 for i in range(n)] for j in range(n)]
def solve(b, e):
if e - b == 1:
M[b][e-1] = 0
return 0
if e - b == 0:
return 0
if M[b][e-1] != -1:
return M[b][e-1]
M[b][e-1] = 1e18
for i in range(b, e):
s = solve(b, i) + solve(i+1, e)
s += acc(0, b, b, i) + acc(b, i, i, n) + acc(0, i+1, i+1, e) + acc(i+1, e, e, n)
if s < M[b][e-1]:
M[b][e-1] = s
P[b][e-1] = i
return M[b][e-1]
solve(0, n)
sol = ["" for _ in range(n)]
def label(b, e, p):
if e - b == 1:
sol[b] = str(p)
return
elif e - b == 0:
return
i = P[b][e-1]
sol[i] = str(p)
label(b, i, i+1)
label(i+1, e, i+1)
label(0, n, 0)
print(" ".join(sol))
|
S[i][j] += S[i-1][j]
|
[{"input": "4\n0 566 1 0\n566 0 239 30\n1 239 0 1\n0 30 1 0", "output": ["2 4 2 0"]}]
|
block_completion_003208
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Julia's $$$n$$$ friends want to organize a startup in a new country they moved to. They assigned each other numbers from 1 to $$$n$$$ according to the jobs they have, from the most front-end tasks to the most back-end ones. They also estimated a matrix $$$c$$$, where $$$c_{ij} = c_{ji}$$$ is the average number of messages per month between people doing jobs $$$i$$$ and $$$j$$$.Now they want to make a hierarchy tree. It will be a binary tree with each node containing one member of the team. Some member will be selected as a leader of the team and will be contained in the root node. In order for the leader to be able to easily reach any subordinate, for each node $$$v$$$ of the tree, the following should apply: all members in its left subtree must have smaller numbers than $$$v$$$, and all members in its right subtree must have larger numbers than $$$v$$$.After the hierarchy tree is settled, people doing jobs $$$i$$$ and $$$j$$$ will be communicating via the shortest path in the tree between their nodes. Let's denote the length of this path as $$$d_{ij}$$$. Thus, the cost of their communication is $$$c_{ij} \cdot d_{ij}$$$.Your task is to find a hierarchy tree that minimizes the total cost of communication over all pairs: $$$\sum_{1 \le i < j \le n} c_{ij} \cdot d_{ij}$$$.
Input Specification: The first line contains an integer $$$n$$$ ($$$1 \le n \le 200$$$) – the number of team members organizing a startup. The next $$$n$$$ lines contain $$$n$$$ integers each, $$$j$$$-th number in $$$i$$$-th line is $$$c_{ij}$$$ — the estimated number of messages per month between team members $$$i$$$ and $$$j$$$ ($$$0 \le c_{ij} \le 10^9; c_{ij} = c_{ji}; c_{ii} = 0$$$).
Output Specification: Output a description of a hierarchy tree that minimizes the total cost of communication. To do so, for each team member from 1 to $$$n$$$ output the number of the member in its parent node, or 0 for the leader. If there are many optimal trees, output a description of any one of them.
Notes: NoteThe minimal possible total cost is $$$566 \cdot 1+239 \cdot 1+30 \cdot 1+1 \cdot 2+1 \cdot 2=839$$$:
Code:
n = int(input().strip())
S = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
for j, c in enumerate(map(int, input().strip().split())):
S[i][j] = c
for i in range(n):
for j in range(n):
if i > 0 and j > 0:
S[i][j] += S[i][j-1] + S[i-1][j] - S[i-1][j-1]
elif i > 0:
S[i][j] += S[i-1][j]
elif j > 0:
# TODO: Your code here
def acc(i1, i2, j1, j2):
if i1 >= i2 or j1 >= j2:
return 0
a = S[i2-1][j2-1]
b = S[i2-1][j1-1] if j1 > 0 else 0
c = S[i1-1][j2-1] if i1 > 0 else 0
d = S[i1-1][j1-1] if i1 > 0 and j1 > 0 else 0
return a - b - c + d
M = [[-1 for i in range(n)] for j in range(n)]
P = [[-1 for i in range(n)] for j in range(n)]
def solve(b, e):
if e - b == 1:
M[b][e-1] = 0
return 0
if e - b == 0:
return 0
if M[b][e-1] != -1:
return M[b][e-1]
M[b][e-1] = 1e18
for i in range(b, e):
s = solve(b, i) + solve(i+1, e)
s += acc(0, b, b, i) + acc(b, i, i, n) + acc(0, i+1, i+1, e) + acc(i+1, e, e, n)
if s < M[b][e-1]:
M[b][e-1] = s
P[b][e-1] = i
return M[b][e-1]
solve(0, n)
sol = ["" for _ in range(n)]
def label(b, e, p):
if e - b == 1:
sol[b] = str(p)
return
elif e - b == 0:
return
i = P[b][e-1]
sol[i] = str(p)
label(b, i, i+1)
label(i+1, e, i+1)
label(0, n, 0)
print(" ".join(sol))
|
n = int(input().strip())
S = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
for j, c in enumerate(map(int, input().strip().split())):
S[i][j] = c
for i in range(n):
for j in range(n):
if i > 0 and j > 0:
S[i][j] += S[i][j-1] + S[i-1][j] - S[i-1][j-1]
elif i > 0:
S[i][j] += S[i-1][j]
elif j > 0:
{{completion}}
def acc(i1, i2, j1, j2):
if i1 >= i2 or j1 >= j2:
return 0
a = S[i2-1][j2-1]
b = S[i2-1][j1-1] if j1 > 0 else 0
c = S[i1-1][j2-1] if i1 > 0 else 0
d = S[i1-1][j1-1] if i1 > 0 and j1 > 0 else 0
return a - b - c + d
M = [[-1 for i in range(n)] for j in range(n)]
P = [[-1 for i in range(n)] for j in range(n)]
def solve(b, e):
if e - b == 1:
M[b][e-1] = 0
return 0
if e - b == 0:
return 0
if M[b][e-1] != -1:
return M[b][e-1]
M[b][e-1] = 1e18
for i in range(b, e):
s = solve(b, i) + solve(i+1, e)
s += acc(0, b, b, i) + acc(b, i, i, n) + acc(0, i+1, i+1, e) + acc(i+1, e, e, n)
if s < M[b][e-1]:
M[b][e-1] = s
P[b][e-1] = i
return M[b][e-1]
solve(0, n)
sol = ["" for _ in range(n)]
def label(b, e, p):
if e - b == 1:
sol[b] = str(p)
return
elif e - b == 0:
return
i = P[b][e-1]
sol[i] = str(p)
label(b, i, i+1)
label(i+1, e, i+1)
label(0, n, 0)
print(" ".join(sol))
|
S[i][j] += S[i][j-1]
|
[{"input": "4\n0 566 1 0\n566 0 239 30\n1 239 0 1\n0 30 1 0", "output": ["2 4 2 0"]}]
|
block_completion_003209
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Julia's $$$n$$$ friends want to organize a startup in a new country they moved to. They assigned each other numbers from 1 to $$$n$$$ according to the jobs they have, from the most front-end tasks to the most back-end ones. They also estimated a matrix $$$c$$$, where $$$c_{ij} = c_{ji}$$$ is the average number of messages per month between people doing jobs $$$i$$$ and $$$j$$$.Now they want to make a hierarchy tree. It will be a binary tree with each node containing one member of the team. Some member will be selected as a leader of the team and will be contained in the root node. In order for the leader to be able to easily reach any subordinate, for each node $$$v$$$ of the tree, the following should apply: all members in its left subtree must have smaller numbers than $$$v$$$, and all members in its right subtree must have larger numbers than $$$v$$$.After the hierarchy tree is settled, people doing jobs $$$i$$$ and $$$j$$$ will be communicating via the shortest path in the tree between their nodes. Let's denote the length of this path as $$$d_{ij}$$$. Thus, the cost of their communication is $$$c_{ij} \cdot d_{ij}$$$.Your task is to find a hierarchy tree that minimizes the total cost of communication over all pairs: $$$\sum_{1 \le i < j \le n} c_{ij} \cdot d_{ij}$$$.
Input Specification: The first line contains an integer $$$n$$$ ($$$1 \le n \le 200$$$) – the number of team members organizing a startup. The next $$$n$$$ lines contain $$$n$$$ integers each, $$$j$$$-th number in $$$i$$$-th line is $$$c_{ij}$$$ — the estimated number of messages per month between team members $$$i$$$ and $$$j$$$ ($$$0 \le c_{ij} \le 10^9; c_{ij} = c_{ji}; c_{ii} = 0$$$).
Output Specification: Output a description of a hierarchy tree that minimizes the total cost of communication. To do so, for each team member from 1 to $$$n$$$ output the number of the member in its parent node, or 0 for the leader. If there are many optimal trees, output a description of any one of them.
Notes: NoteThe minimal possible total cost is $$$566 \cdot 1+239 \cdot 1+30 \cdot 1+1 \cdot 2+1 \cdot 2=839$$$:
Code:
n=int(input())
c=[]
for _ in range(n):
c.append(tuple(map(int,input().split())))
prefix_sum=[[0]*(n+1) for _ in range(n+1)]
for i in range(1,n+1):
temp=0
for j in range(1,n+1):
temp+=c[i-1][j-1]
prefix_sum[i][j]+=prefix_sum[i-1][j]+temp
def get_rectangel_sum(x1,y1,x2,y2):
return prefix_sum[x2+1][y2+1]-prefix_sum[x1][y2+1]-prefix_sum[x2+1][y1]+prefix_sum[x1][y1]
def cost(x,y):
if x>y:
return 0
a=get_rectangel_sum(x,0,y,x-1) if x!=0 else 0
b=get_rectangel_sum(x,y+1,y,n-1) if y!=n-1 else 0
return a+b
dp=[[float("INF")]*n for _ in range(n)]
best_root_for_range=[[-1]*n for _ in range(n)]
for i in range(n):
dp[i][i]=0
best_root_for_range[i][i]=i
def get_dp_cost(x,y):
return dp[x][y] if x<=y else 0
for length in range(1,n):
# actual length is length+1
for i in range(n-length):
j=i+length
for root in range(i,j+1):
temp=cost(i,root-1)+cost(root+1,j)+get_dp_cost(i,root-1)+get_dp_cost(root+1,j)
if temp<dp[i][j]:
# TODO: Your code here
ans=[-1]*n
def assign_ans(ansecstor,x,y):
if x>y:
return
root=best_root_for_range[x][y]
ans[root]=ansecstor
assign_ans(root,x,root-1)
assign_ans(root,root+1,y)
assign_ans(-1,0,n-1)
print(*[i+1 for i in ans])
# 3
# 0 1 2
# 1 0 3
# 2 3 0
# 4
# 0 1 2 3
# 1 0 5 7
# 2 5 0 4
# 3 7 4 0
# 6
# 0 100 20 37 14 73
# 100 0 17 13 20 2
# 20 17 0 1093 900 1
# 37 13 1093 0 2 4
# 14 20 900 2 0 1
# 73 2 1 4 1 0
|
n=int(input())
c=[]
for _ in range(n):
c.append(tuple(map(int,input().split())))
prefix_sum=[[0]*(n+1) for _ in range(n+1)]
for i in range(1,n+1):
temp=0
for j in range(1,n+1):
temp+=c[i-1][j-1]
prefix_sum[i][j]+=prefix_sum[i-1][j]+temp
def get_rectangel_sum(x1,y1,x2,y2):
return prefix_sum[x2+1][y2+1]-prefix_sum[x1][y2+1]-prefix_sum[x2+1][y1]+prefix_sum[x1][y1]
def cost(x,y):
if x>y:
return 0
a=get_rectangel_sum(x,0,y,x-1) if x!=0 else 0
b=get_rectangel_sum(x,y+1,y,n-1) if y!=n-1 else 0
return a+b
dp=[[float("INF")]*n for _ in range(n)]
best_root_for_range=[[-1]*n for _ in range(n)]
for i in range(n):
dp[i][i]=0
best_root_for_range[i][i]=i
def get_dp_cost(x,y):
return dp[x][y] if x<=y else 0
for length in range(1,n):
# actual length is length+1
for i in range(n-length):
j=i+length
for root in range(i,j+1):
temp=cost(i,root-1)+cost(root+1,j)+get_dp_cost(i,root-1)+get_dp_cost(root+1,j)
if temp<dp[i][j]:
{{completion}}
ans=[-1]*n
def assign_ans(ansecstor,x,y):
if x>y:
return
root=best_root_for_range[x][y]
ans[root]=ansecstor
assign_ans(root,x,root-1)
assign_ans(root,root+1,y)
assign_ans(-1,0,n-1)
print(*[i+1 for i in ans])
# 3
# 0 1 2
# 1 0 3
# 2 3 0
# 4
# 0 1 2 3
# 1 0 5 7
# 2 5 0 4
# 3 7 4 0
# 6
# 0 100 20 37 14 73
# 100 0 17 13 20 2
# 20 17 0 1093 900 1
# 37 13 1093 0 2 4
# 14 20 900 2 0 1
# 73 2 1 4 1 0
|
dp[i][j]=temp
best_root_for_range[i][j]=root
|
[{"input": "4\n0 566 1 0\n566 0 239 30\n1 239 0 1\n0 30 1 0", "output": ["2 4 2 0"]}]
|
block_completion_003210
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Suppose you have an integer $$$v$$$. In one operation, you can: either set $$$v = (v + 1) \bmod 32768$$$ or set $$$v = (2 \cdot v) \bmod 32768$$$. You are given $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. What is the minimum number of operations you need to make each $$$a_i$$$ equal to $$$0$$$?
Input Specification: The first line contains the single integer $$$n$$$ ($$$1 \le n \le 32768$$$) — the number of integers. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i < 32768$$$).
Output Specification: Print $$$n$$$ integers. The $$$i$$$-th integer should be equal to the minimum number of operations required to make $$$a_i$$$ equal to $$$0$$$.
Notes: NoteLet's consider each $$$a_i$$$: $$$a_1 = 19$$$. You can, firstly, increase it by one to get $$$20$$$ and then multiply it by two $$$13$$$ times. You'll get $$$0$$$ in $$$1 + 13 = 14$$$ steps. $$$a_2 = 32764$$$. You can increase it by one $$$4$$$ times: $$$32764 \rightarrow 32765 \rightarrow 32766 \rightarrow 32767 \rightarrow 0$$$. $$$a_3 = 10240$$$. You can multiply it by two $$$4$$$ times: $$$10240 \rightarrow 20480 \rightarrow 8192 \rightarrow 16384 \rightarrow 0$$$. $$$a_4 = 49$$$. You can multiply it by two $$$15$$$ times.
Code:
n,s=open(0)
for x in s.split():# TODO: Your code here
|
n,s=open(0)
for x in s.split():{{completion}}
|
print(min(-int(x)%2**i-i+15for i in range(16)))
|
[{"input": "4\n19 32764 10240 49", "output": ["14 4 4 15"]}]
|
block_completion_003352
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Suppose you have an integer $$$v$$$. In one operation, you can: either set $$$v = (v + 1) \bmod 32768$$$ or set $$$v = (2 \cdot v) \bmod 32768$$$. You are given $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. What is the minimum number of operations you need to make each $$$a_i$$$ equal to $$$0$$$?
Input Specification: The first line contains the single integer $$$n$$$ ($$$1 \le n \le 32768$$$) — the number of integers. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i < 32768$$$).
Output Specification: Print $$$n$$$ integers. The $$$i$$$-th integer should be equal to the minimum number of operations required to make $$$a_i$$$ equal to $$$0$$$.
Notes: NoteLet's consider each $$$a_i$$$: $$$a_1 = 19$$$. You can, firstly, increase it by one to get $$$20$$$ and then multiply it by two $$$13$$$ times. You'll get $$$0$$$ in $$$1 + 13 = 14$$$ steps. $$$a_2 = 32764$$$. You can increase it by one $$$4$$$ times: $$$32764 \rightarrow 32765 \rightarrow 32766 \rightarrow 32767 \rightarrow 0$$$. $$$a_3 = 10240$$$. You can multiply it by two $$$4$$$ times: $$$10240 \rightarrow 20480 \rightarrow 8192 \rightarrow 16384 \rightarrow 0$$$. $$$a_4 = 49$$$. You can multiply it by two $$$15$$$ times.
Code:
n = int(input())
mod = 1 << 15
for x in map(int, input().split()):
res = 16
for a in range(15):
for b in range(15):
if (x + a) * (1 << b) % mod == 0:
# TODO: Your code here
print(res)
|
n = int(input())
mod = 1 << 15
for x in map(int, input().split()):
res = 16
for a in range(15):
for b in range(15):
if (x + a) * (1 << b) % mod == 0:
{{completion}}
print(res)
|
res = min(res, a + b)
|
[{"input": "4\n19 32764 10240 49", "output": ["14 4 4 15"]}]
|
block_completion_003353
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Suppose you have an integer $$$v$$$. In one operation, you can: either set $$$v = (v + 1) \bmod 32768$$$ or set $$$v = (2 \cdot v) \bmod 32768$$$. You are given $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. What is the minimum number of operations you need to make each $$$a_i$$$ equal to $$$0$$$?
Input Specification: The first line contains the single integer $$$n$$$ ($$$1 \le n \le 32768$$$) — the number of integers. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i < 32768$$$).
Output Specification: Print $$$n$$$ integers. The $$$i$$$-th integer should be equal to the minimum number of operations required to make $$$a_i$$$ equal to $$$0$$$.
Notes: NoteLet's consider each $$$a_i$$$: $$$a_1 = 19$$$. You can, firstly, increase it by one to get $$$20$$$ and then multiply it by two $$$13$$$ times. You'll get $$$0$$$ in $$$1 + 13 = 14$$$ steps. $$$a_2 = 32764$$$. You can increase it by one $$$4$$$ times: $$$32764 \rightarrow 32765 \rightarrow 32766 \rightarrow 32767 \rightarrow 0$$$. $$$a_3 = 10240$$$. You can multiply it by two $$$4$$$ times: $$$10240 \rightarrow 20480 \rightarrow 8192 \rightarrow 16384 \rightarrow 0$$$. $$$a_4 = 49$$$. You can multiply it by two $$$15$$$ times.
Code:
n, s = open(0)
for x in map(int, s.split()):
# TODO: Your code here
|
n, s = open(0)
for x in map(int, s.split()):
{{completion}}
|
print(min(15-i+-x % 2**i for i in range(16)))
|
[{"input": "4\n19 32764 10240 49", "output": ["14 4 4 15"]}]
|
block_completion_003354
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Suppose you have an integer $$$v$$$. In one operation, you can: either set $$$v = (v + 1) \bmod 32768$$$ or set $$$v = (2 \cdot v) \bmod 32768$$$. You are given $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. What is the minimum number of operations you need to make each $$$a_i$$$ equal to $$$0$$$?
Input Specification: The first line contains the single integer $$$n$$$ ($$$1 \le n \le 32768$$$) — the number of integers. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i < 32768$$$).
Output Specification: Print $$$n$$$ integers. The $$$i$$$-th integer should be equal to the minimum number of operations required to make $$$a_i$$$ equal to $$$0$$$.
Notes: NoteLet's consider each $$$a_i$$$: $$$a_1 = 19$$$. You can, firstly, increase it by one to get $$$20$$$ and then multiply it by two $$$13$$$ times. You'll get $$$0$$$ in $$$1 + 13 = 14$$$ steps. $$$a_2 = 32764$$$. You can increase it by one $$$4$$$ times: $$$32764 \rightarrow 32765 \rightarrow 32766 \rightarrow 32767 \rightarrow 0$$$. $$$a_3 = 10240$$$. You can multiply it by two $$$4$$$ times: $$$10240 \rightarrow 20480 \rightarrow 8192 \rightarrow 16384 \rightarrow 0$$$. $$$a_4 = 49$$$. You can multiply it by two $$$15$$$ times.
Code:
n,s=open(0)
for x in map(int,s.split()):# TODO: Your code here
|
n,s=open(0)
for x in map(int,s.split()):{{completion}}
|
print(min(-x%2**i-i+15for i in range(16)))
|
[{"input": "4\n19 32764 10240 49", "output": ["14 4 4 15"]}]
|
block_completion_003355
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Suppose you have an integer $$$v$$$. In one operation, you can: either set $$$v = (v + 1) \bmod 32768$$$ or set $$$v = (2 \cdot v) \bmod 32768$$$. You are given $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. What is the minimum number of operations you need to make each $$$a_i$$$ equal to $$$0$$$?
Input Specification: The first line contains the single integer $$$n$$$ ($$$1 \le n \le 32768$$$) — the number of integers. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i < 32768$$$).
Output Specification: Print $$$n$$$ integers. The $$$i$$$-th integer should be equal to the minimum number of operations required to make $$$a_i$$$ equal to $$$0$$$.
Notes: NoteLet's consider each $$$a_i$$$: $$$a_1 = 19$$$. You can, firstly, increase it by one to get $$$20$$$ and then multiply it by two $$$13$$$ times. You'll get $$$0$$$ in $$$1 + 13 = 14$$$ steps. $$$a_2 = 32764$$$. You can increase it by one $$$4$$$ times: $$$32764 \rightarrow 32765 \rightarrow 32766 \rightarrow 32767 \rightarrow 0$$$. $$$a_3 = 10240$$$. You can multiply it by two $$$4$$$ times: $$$10240 \rightarrow 20480 \rightarrow 8192 \rightarrow 16384 \rightarrow 0$$$. $$$a_4 = 49$$$. You can multiply it by two $$$15$$$ times.
Code:
n,s=open(0)
for x in s.split():# TODO: Your code here
|
n,s=open(0)
for x in s.split():{{completion}}
|
print(min(-int(x)%2**i-i+15for i in range(16)))
|
[{"input": "4\n19 32764 10240 49", "output": ["14 4 4 15"]}]
|
block_completion_003356
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a matrix $$$a$$$, consisting of $$$3$$$ rows and $$$n$$$ columns. Each cell of the matrix is either free or taken.A free cell $$$y$$$ is reachable from a free cell $$$x$$$ if at least one of these conditions hold: $$$x$$$ and $$$y$$$ share a side; there exists a free cell $$$z$$$ such that $$$z$$$ is reachable from $$$x$$$ and $$$y$$$ is reachable from $$$z$$$. A connected component is a set of free cells of the matrix such that all cells in it are reachable from one another, but adding any other free cell to the set violates this rule.You are asked $$$q$$$ queries about the matrix. Each query is the following: $$$l$$$ $$$r$$$ — count the number of connected components of the matrix, consisting of columns from $$$l$$$ to $$$r$$$ of the matrix $$$a$$$, inclusive. Print the answers to all queries.
Input Specification: The first line contains an integer $$$n$$$ ($$$1 \le n \le 5 \cdot 10^5$$$) — the number of columns of matrix $$$a$$$. The $$$i$$$-th of the next three lines contains a description of the $$$i$$$-th row of the matrix $$$a$$$ — a string, consisting of $$$n$$$ characters. Each character is either $$$1$$$ (denoting a free cell) or $$$0$$$ (denoting a taken cell). The next line contains an integer $$$q$$$ ($$$1 \le q \le 3 \cdot 10^5$$$) — the number of queries. The $$$j$$$-th of the next $$$q$$$ lines contains two integers $$$l_j$$$ and $$$r_j$$$ ($$$1 \le l_j \le r_j \le n$$$) — the description of the $$$j$$$-th query.
Output Specification: Print $$$q$$$ integers — the $$$j$$$-th value should be equal to the number of the connected components of the matrix, consisting of columns from $$$l_j$$$ to $$$r_j$$$ of the matrix $$$a$$$, inclusive.
Code:
import sys
def Column2Num( m, idx ):
return int(m[0][idx] != 0) | (int(m[1][idx]) << 1) | (int(m[2][idx]) << 2)
def QColumn( m, bits, idx ):
if bits[idx] == 5:
if m[0][idx] == m[2][idx]:
return True
return False
def GetIntegratedCount( m ):
ret, curr = [ 0 ], set()
for c in range( len( m[0] ) ):
if m[0][c] != 0: curr.add( m[0][c] )
if m[1][c] != 0: curr.add( m[1][c] )
if m[2][c] != 0: curr.add( m[2][c] )
ret.append( len( curr ) )
ret.append( len( curr ) )
return ret
def Print( tm ):
print( '\n', tm[0], '\n', tm[1], '\n', tm[2] )
def PrintIndexed( tm ):
for i in range( len( tm[0] ) ):
print( i+1, ':', tm[0][i], tm[1][i], tm[2][i] )
def next( b, next ):
b &= next
if b == 0:
return b
if b & 1 or b & 4:
if next & 2:
b |= 2
if b & 2:
if next & 1:
b |= 1
if next & 4:
b |= 4
return b
def setCompNumber( b, m, i, compNumber ):
if b & 1:
m[0][i] = compNumber
if b & 2:
m[1][i] = compNumber
if b & 4:
m[2][i] = compNumber
def goLeft( start, compNumber, size, m, bits, fullColumn ):
b = bits[start]
fc = start
for i in range( start - 1, -1, -1 ):
b = next( b, bits[i] )
if b == 7:
fc = i
break
if b == 0:
break
setCompNumber( b, m, i, compNumber )
if b == 5:
fullColumn[ i ] = fc
def goRight( start, compNumber, size, m, bits, fullColumn ):
b = bits[start]
for i in range( start, size ):
b = next( b, bits[i] )
if b == 7:
fc = i
if b == 0:
break
setCompNumber( b, m, i, compNumber )
if b == 5:
fullColumn[ i ] = fc
def goRight12( b, start, compNumber, size, m, bits ):
for i in range( start, size ):
b = next( b, bits[i] )
if b == 0:
break
setCompNumber( b, m, i, compNumber )
def get3Components( compNumber, size, m, bits, leftFullColumn, rightFullColumn ):
for i in range( size ):
if bits[i] == 7:
if m[ 0 ][ i ] == 1:
compNumber += 1
goRight( i, compNumber, size, m, bits, leftFullColumn )
goLeft ( i, compNumber, size, m, bits, rightFullColumn )
return compNumber
def get12Components( compNumber, size, m, bits ):
for i in range( size ):
if m[ 0 ][ i ] == 1:
compNumber += 1
goRight12( 1, i, compNumber, size, m, bits )
if m[ 1 ][ i ] == 1:
compNumber += 1
goRight12( 2, i, compNumber, size, m, bits )
if m[ 2 ][ i ] == 1:
compNumber += 1
goRight12( 4, i, compNumber, size, m, bits )
def SolveBB( m, bits, integratedCount, s, e, leftFullColumn, rightFullColumn ):
#debug = 0
#print( 'start,end =',s, e )
#if debug: Print( [ m[0][s-1:e], m[1][s-1:e], m[2][s-1:e] ] )
sol1 = 0
if s-1 == 0:
sol1 = integratedCount[e]
else:
startCnt = 1
if bits[s-1] == 0:
startCnt = 0
elif bits[s-1] == 5:
if m[0][s-1] != m[2][s-1]:
# TODO: Your code here
sol1 = startCnt + integratedCount[e] - integratedCount[s]
sQ = QColumn( m, bits, s - 1)
eQ = QColumn( m, bits, e - 1)
if sQ and eQ:
if m[0][s-1] == m[2][s-1]:
if rightFullColumn[s-1] == rightFullColumn[e-1]:
sol1 += 1
return sol1
elif leftFullColumn[s-1] == leftFullColumn[e-1]:
sol1 += 1
return sol1
if sQ:
if rightFullColumn[s-1] != -1:
if rightFullColumn[s-1] > e-1:
sol1 += 1
else:
sol1 += 1
if eQ:
if leftFullColumn[e-1] != -1:
if leftFullColumn[e-1] < s-1:
sol1 += 1
else:
sol1 += 1
return sol1
def mainBB():
debug = 0
input = sys.stdin
if len( sys.argv ) >= 2:
input = open( sys.argv[1], 'r' )
size = int( input.readline() )
m = []
for i in range( 3 ):
m.append( [ int( t ) for t in list( input.readline().strip() ) ] )
bits = [ Column2Num( m, i ) for i in range( size ) ]
leftFullColumn = [ -1 for i in range( size ) ]
rightFullColumn = list( leftFullColumn )
compNumber = get3Components( 1, size, m, bits, leftFullColumn, rightFullColumn )
get12Components( compNumber, size, m, bits )
integratedCount = GetIntegratedCount( m )
if debug: PrintIndexed( m )
if debug: Print( m )
if debug: print( integratedCount )
if debug: print( leftFullColumn )
if debug: print( rightFullColumn )
n = int( input.readline() )
for i in range( n ):
ln = input.readline().strip().split()
s = int( ln[0] )
e = int( ln[1] )
if debug: print( s, e, m )
print( SolveBB( m, bits, integratedCount, s, e, leftFullColumn, rightFullColumn ) )
if __name__ == "__main__":
mainBB()
|
import sys
def Column2Num( m, idx ):
return int(m[0][idx] != 0) | (int(m[1][idx]) << 1) | (int(m[2][idx]) << 2)
def QColumn( m, bits, idx ):
if bits[idx] == 5:
if m[0][idx] == m[2][idx]:
return True
return False
def GetIntegratedCount( m ):
ret, curr = [ 0 ], set()
for c in range( len( m[0] ) ):
if m[0][c] != 0: curr.add( m[0][c] )
if m[1][c] != 0: curr.add( m[1][c] )
if m[2][c] != 0: curr.add( m[2][c] )
ret.append( len( curr ) )
ret.append( len( curr ) )
return ret
def Print( tm ):
print( '\n', tm[0], '\n', tm[1], '\n', tm[2] )
def PrintIndexed( tm ):
for i in range( len( tm[0] ) ):
print( i+1, ':', tm[0][i], tm[1][i], tm[2][i] )
def next( b, next ):
b &= next
if b == 0:
return b
if b & 1 or b & 4:
if next & 2:
b |= 2
if b & 2:
if next & 1:
b |= 1
if next & 4:
b |= 4
return b
def setCompNumber( b, m, i, compNumber ):
if b & 1:
m[0][i] = compNumber
if b & 2:
m[1][i] = compNumber
if b & 4:
m[2][i] = compNumber
def goLeft( start, compNumber, size, m, bits, fullColumn ):
b = bits[start]
fc = start
for i in range( start - 1, -1, -1 ):
b = next( b, bits[i] )
if b == 7:
fc = i
break
if b == 0:
break
setCompNumber( b, m, i, compNumber )
if b == 5:
fullColumn[ i ] = fc
def goRight( start, compNumber, size, m, bits, fullColumn ):
b = bits[start]
for i in range( start, size ):
b = next( b, bits[i] )
if b == 7:
fc = i
if b == 0:
break
setCompNumber( b, m, i, compNumber )
if b == 5:
fullColumn[ i ] = fc
def goRight12( b, start, compNumber, size, m, bits ):
for i in range( start, size ):
b = next( b, bits[i] )
if b == 0:
break
setCompNumber( b, m, i, compNumber )
def get3Components( compNumber, size, m, bits, leftFullColumn, rightFullColumn ):
for i in range( size ):
if bits[i] == 7:
if m[ 0 ][ i ] == 1:
compNumber += 1
goRight( i, compNumber, size, m, bits, leftFullColumn )
goLeft ( i, compNumber, size, m, bits, rightFullColumn )
return compNumber
def get12Components( compNumber, size, m, bits ):
for i in range( size ):
if m[ 0 ][ i ] == 1:
compNumber += 1
goRight12( 1, i, compNumber, size, m, bits )
if m[ 1 ][ i ] == 1:
compNumber += 1
goRight12( 2, i, compNumber, size, m, bits )
if m[ 2 ][ i ] == 1:
compNumber += 1
goRight12( 4, i, compNumber, size, m, bits )
def SolveBB( m, bits, integratedCount, s, e, leftFullColumn, rightFullColumn ):
#debug = 0
#print( 'start,end =',s, e )
#if debug: Print( [ m[0][s-1:e], m[1][s-1:e], m[2][s-1:e] ] )
sol1 = 0
if s-1 == 0:
sol1 = integratedCount[e]
else:
startCnt = 1
if bits[s-1] == 0:
startCnt = 0
elif bits[s-1] == 5:
if m[0][s-1] != m[2][s-1]:
{{completion}}
sol1 = startCnt + integratedCount[e] - integratedCount[s]
sQ = QColumn( m, bits, s - 1)
eQ = QColumn( m, bits, e - 1)
if sQ and eQ:
if m[0][s-1] == m[2][s-1]:
if rightFullColumn[s-1] == rightFullColumn[e-1]:
sol1 += 1
return sol1
elif leftFullColumn[s-1] == leftFullColumn[e-1]:
sol1 += 1
return sol1
if sQ:
if rightFullColumn[s-1] != -1:
if rightFullColumn[s-1] > e-1:
sol1 += 1
else:
sol1 += 1
if eQ:
if leftFullColumn[e-1] != -1:
if leftFullColumn[e-1] < s-1:
sol1 += 1
else:
sol1 += 1
return sol1
def mainBB():
debug = 0
input = sys.stdin
if len( sys.argv ) >= 2:
input = open( sys.argv[1], 'r' )
size = int( input.readline() )
m = []
for i in range( 3 ):
m.append( [ int( t ) for t in list( input.readline().strip() ) ] )
bits = [ Column2Num( m, i ) for i in range( size ) ]
leftFullColumn = [ -1 for i in range( size ) ]
rightFullColumn = list( leftFullColumn )
compNumber = get3Components( 1, size, m, bits, leftFullColumn, rightFullColumn )
get12Components( compNumber, size, m, bits )
integratedCount = GetIntegratedCount( m )
if debug: PrintIndexed( m )
if debug: Print( m )
if debug: print( integratedCount )
if debug: print( leftFullColumn )
if debug: print( rightFullColumn )
n = int( input.readline() )
for i in range( n ):
ln = input.readline().strip().split()
s = int( ln[0] )
e = int( ln[1] )
if debug: print( s, e, m )
print( SolveBB( m, bits, integratedCount, s, e, leftFullColumn, rightFullColumn ) )
if __name__ == "__main__":
mainBB()
|
startCnt = 2
|
[{"input": "12\n100101011101\n110110010110\n010001011101\n8\n1 12\n1 1\n1 2\n9 9\n8 11\n9 12\n11 12\n4 6", "output": ["7\n1\n1\n2\n1\n3\n3\n3"]}]
|
block_completion_003393
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a matrix $$$a$$$, consisting of $$$3$$$ rows and $$$n$$$ columns. Each cell of the matrix is either free or taken.A free cell $$$y$$$ is reachable from a free cell $$$x$$$ if at least one of these conditions hold: $$$x$$$ and $$$y$$$ share a side; there exists a free cell $$$z$$$ such that $$$z$$$ is reachable from $$$x$$$ and $$$y$$$ is reachable from $$$z$$$. A connected component is a set of free cells of the matrix such that all cells in it are reachable from one another, but adding any other free cell to the set violates this rule.You are asked $$$q$$$ queries about the matrix. Each query is the following: $$$l$$$ $$$r$$$ — count the number of connected components of the matrix, consisting of columns from $$$l$$$ to $$$r$$$ of the matrix $$$a$$$, inclusive. Print the answers to all queries.
Input Specification: The first line contains an integer $$$n$$$ ($$$1 \le n \le 5 \cdot 10^5$$$) — the number of columns of matrix $$$a$$$. The $$$i$$$-th of the next three lines contains a description of the $$$i$$$-th row of the matrix $$$a$$$ — a string, consisting of $$$n$$$ characters. Each character is either $$$1$$$ (denoting a free cell) or $$$0$$$ (denoting a taken cell). The next line contains an integer $$$q$$$ ($$$1 \le q \le 3 \cdot 10^5$$$) — the number of queries. The $$$j$$$-th of the next $$$q$$$ lines contains two integers $$$l_j$$$ and $$$r_j$$$ ($$$1 \le l_j \le r_j \le n$$$) — the description of the $$$j$$$-th query.
Output Specification: Print $$$q$$$ integers — the $$$j$$$-th value should be equal to the number of the connected components of the matrix, consisting of columns from $$$l_j$$$ to $$$r_j$$$ of the matrix $$$a$$$, inclusive.
Code:
import sys
def Column2Num( m, idx ):
return int(m[0][idx] != 0) | (int(m[1][idx]) << 1) | (int(m[2][idx]) << 2)
def QColumn( m, bits, idx ):
if bits[idx] == 5:
if m[0][idx] == m[2][idx]:
return True
return False
def GetIntegratedCount( m ):
ret, curr = [ 0 ], set()
for c in range( len( m[0] ) ):
if m[0][c] != 0: curr.add( m[0][c] )
if m[1][c] != 0: curr.add( m[1][c] )
if m[2][c] != 0: curr.add( m[2][c] )
ret.append( len( curr ) )
ret.append( len( curr ) )
return ret
def Print( tm ):
print( '\n', tm[0], '\n', tm[1], '\n', tm[2] )
def PrintIndexed( tm ):
for i in range( len( tm[0] ) ):
print( i+1, ':', tm[0][i], tm[1][i], tm[2][i] )
def next( b, next ):
b &= next
if b == 0:
return b
if b & 1 or b & 4:
if next & 2:
b |= 2
if b & 2:
if next & 1:
b |= 1
if next & 4:
b |= 4
return b
def setCompNumber( b, m, i, compNumber ):
if b & 1:
m[0][i] = compNumber
if b & 2:
m[1][i] = compNumber
if b & 4:
m[2][i] = compNumber
def goLeft( start, compNumber, size, m, bits, fullColumn ):
b = bits[start]
fc = start
for i in range( start - 1, -1, -1 ):
b = next( b, bits[i] )
if b == 7:
fc = i
break
if b == 0:
break
setCompNumber( b, m, i, compNumber )
if b == 5:
fullColumn[ i ] = fc
def goRight( start, compNumber, size, m, bits, fullColumn ):
b = bits[start]
for i in range( start, size ):
b = next( b, bits[i] )
if b == 7:
fc = i
if b == 0:
break
setCompNumber( b, m, i, compNumber )
if b == 5:
fullColumn[ i ] = fc
def goRight12( b, start, compNumber, size, m, bits ):
for i in range( start, size ):
b = next( b, bits[i] )
if b == 0:
break
setCompNumber( b, m, i, compNumber )
def get3Components( compNumber, size, m, bits, leftFullColumn, rightFullColumn ):
for i in range( size ):
if bits[i] == 7:
if m[ 0 ][ i ] == 1:
compNumber += 1
goRight( i, compNumber, size, m, bits, leftFullColumn )
goLeft ( i, compNumber, size, m, bits, rightFullColumn )
return compNumber
def get12Components( compNumber, size, m, bits ):
for i in range( size ):
if m[ 0 ][ i ] == 1:
compNumber += 1
goRight12( 1, i, compNumber, size, m, bits )
if m[ 1 ][ i ] == 1:
compNumber += 1
goRight12( 2, i, compNumber, size, m, bits )
if m[ 2 ][ i ] == 1:
compNumber += 1
goRight12( 4, i, compNumber, size, m, bits )
def SolveBB( m, bits, integratedCount, s, e, leftFullColumn, rightFullColumn ):
#debug = 0
#print( 'start,end =',s, e )
#if debug: Print( [ m[0][s-1:e], m[1][s-1:e], m[2][s-1:e] ] )
sol1 = 0
if s-1 == 0:
sol1 = integratedCount[e]
else:
startCnt = 1
if bits[s-1] == 0:
startCnt = 0
elif bits[s-1] == 5:
if m[0][s-1] != m[2][s-1]:
startCnt = 2
sol1 = startCnt + integratedCount[e] - integratedCount[s]
sQ = QColumn( m, bits, s - 1)
eQ = QColumn( m, bits, e - 1)
if sQ and eQ:
if m[0][s-1] == m[2][s-1]:
if rightFullColumn[s-1] == rightFullColumn[e-1]:
sol1 += 1
return sol1
elif leftFullColumn[s-1] == leftFullColumn[e-1]:
# TODO: Your code here
if sQ:
if rightFullColumn[s-1] != -1:
if rightFullColumn[s-1] > e-1:
sol1 += 1
else:
sol1 += 1
if eQ:
if leftFullColumn[e-1] != -1:
if leftFullColumn[e-1] < s-1:
sol1 += 1
else:
sol1 += 1
return sol1
def mainBB():
debug = 0
input = sys.stdin
if len( sys.argv ) >= 2:
input = open( sys.argv[1], 'r' )
size = int( input.readline() )
m = []
for i in range( 3 ):
m.append( [ int( t ) for t in list( input.readline().strip() ) ] )
bits = [ Column2Num( m, i ) for i in range( size ) ]
leftFullColumn = [ -1 for i in range( size ) ]
rightFullColumn = list( leftFullColumn )
compNumber = get3Components( 1, size, m, bits, leftFullColumn, rightFullColumn )
get12Components( compNumber, size, m, bits )
integratedCount = GetIntegratedCount( m )
if debug: PrintIndexed( m )
if debug: Print( m )
if debug: print( integratedCount )
if debug: print( leftFullColumn )
if debug: print( rightFullColumn )
n = int( input.readline() )
for i in range( n ):
ln = input.readline().strip().split()
s = int( ln[0] )
e = int( ln[1] )
if debug: print( s, e, m )
print( SolveBB( m, bits, integratedCount, s, e, leftFullColumn, rightFullColumn ) )
if __name__ == "__main__":
mainBB()
|
import sys
def Column2Num( m, idx ):
return int(m[0][idx] != 0) | (int(m[1][idx]) << 1) | (int(m[2][idx]) << 2)
def QColumn( m, bits, idx ):
if bits[idx] == 5:
if m[0][idx] == m[2][idx]:
return True
return False
def GetIntegratedCount( m ):
ret, curr = [ 0 ], set()
for c in range( len( m[0] ) ):
if m[0][c] != 0: curr.add( m[0][c] )
if m[1][c] != 0: curr.add( m[1][c] )
if m[2][c] != 0: curr.add( m[2][c] )
ret.append( len( curr ) )
ret.append( len( curr ) )
return ret
def Print( tm ):
print( '\n', tm[0], '\n', tm[1], '\n', tm[2] )
def PrintIndexed( tm ):
for i in range( len( tm[0] ) ):
print( i+1, ':', tm[0][i], tm[1][i], tm[2][i] )
def next( b, next ):
b &= next
if b == 0:
return b
if b & 1 or b & 4:
if next & 2:
b |= 2
if b & 2:
if next & 1:
b |= 1
if next & 4:
b |= 4
return b
def setCompNumber( b, m, i, compNumber ):
if b & 1:
m[0][i] = compNumber
if b & 2:
m[1][i] = compNumber
if b & 4:
m[2][i] = compNumber
def goLeft( start, compNumber, size, m, bits, fullColumn ):
b = bits[start]
fc = start
for i in range( start - 1, -1, -1 ):
b = next( b, bits[i] )
if b == 7:
fc = i
break
if b == 0:
break
setCompNumber( b, m, i, compNumber )
if b == 5:
fullColumn[ i ] = fc
def goRight( start, compNumber, size, m, bits, fullColumn ):
b = bits[start]
for i in range( start, size ):
b = next( b, bits[i] )
if b == 7:
fc = i
if b == 0:
break
setCompNumber( b, m, i, compNumber )
if b == 5:
fullColumn[ i ] = fc
def goRight12( b, start, compNumber, size, m, bits ):
for i in range( start, size ):
b = next( b, bits[i] )
if b == 0:
break
setCompNumber( b, m, i, compNumber )
def get3Components( compNumber, size, m, bits, leftFullColumn, rightFullColumn ):
for i in range( size ):
if bits[i] == 7:
if m[ 0 ][ i ] == 1:
compNumber += 1
goRight( i, compNumber, size, m, bits, leftFullColumn )
goLeft ( i, compNumber, size, m, bits, rightFullColumn )
return compNumber
def get12Components( compNumber, size, m, bits ):
for i in range( size ):
if m[ 0 ][ i ] == 1:
compNumber += 1
goRight12( 1, i, compNumber, size, m, bits )
if m[ 1 ][ i ] == 1:
compNumber += 1
goRight12( 2, i, compNumber, size, m, bits )
if m[ 2 ][ i ] == 1:
compNumber += 1
goRight12( 4, i, compNumber, size, m, bits )
def SolveBB( m, bits, integratedCount, s, e, leftFullColumn, rightFullColumn ):
#debug = 0
#print( 'start,end =',s, e )
#if debug: Print( [ m[0][s-1:e], m[1][s-1:e], m[2][s-1:e] ] )
sol1 = 0
if s-1 == 0:
sol1 = integratedCount[e]
else:
startCnt = 1
if bits[s-1] == 0:
startCnt = 0
elif bits[s-1] == 5:
if m[0][s-1] != m[2][s-1]:
startCnt = 2
sol1 = startCnt + integratedCount[e] - integratedCount[s]
sQ = QColumn( m, bits, s - 1)
eQ = QColumn( m, bits, e - 1)
if sQ and eQ:
if m[0][s-1] == m[2][s-1]:
if rightFullColumn[s-1] == rightFullColumn[e-1]:
sol1 += 1
return sol1
elif leftFullColumn[s-1] == leftFullColumn[e-1]:
{{completion}}
if sQ:
if rightFullColumn[s-1] != -1:
if rightFullColumn[s-1] > e-1:
sol1 += 1
else:
sol1 += 1
if eQ:
if leftFullColumn[e-1] != -1:
if leftFullColumn[e-1] < s-1:
sol1 += 1
else:
sol1 += 1
return sol1
def mainBB():
debug = 0
input = sys.stdin
if len( sys.argv ) >= 2:
input = open( sys.argv[1], 'r' )
size = int( input.readline() )
m = []
for i in range( 3 ):
m.append( [ int( t ) for t in list( input.readline().strip() ) ] )
bits = [ Column2Num( m, i ) for i in range( size ) ]
leftFullColumn = [ -1 for i in range( size ) ]
rightFullColumn = list( leftFullColumn )
compNumber = get3Components( 1, size, m, bits, leftFullColumn, rightFullColumn )
get12Components( compNumber, size, m, bits )
integratedCount = GetIntegratedCount( m )
if debug: PrintIndexed( m )
if debug: Print( m )
if debug: print( integratedCount )
if debug: print( leftFullColumn )
if debug: print( rightFullColumn )
n = int( input.readline() )
for i in range( n ):
ln = input.readline().strip().split()
s = int( ln[0] )
e = int( ln[1] )
if debug: print( s, e, m )
print( SolveBB( m, bits, integratedCount, s, e, leftFullColumn, rightFullColumn ) )
if __name__ == "__main__":
mainBB()
|
sol1 += 1
return sol1
|
[{"input": "12\n100101011101\n110110010110\n010001011101\n8\n1 12\n1 1\n1 2\n9 9\n8 11\n9 12\n11 12\n4 6", "output": ["7\n1\n1\n2\n1\n3\n3\n3"]}]
|
block_completion_003394
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ trees in a park, numbered from $$$1$$$ to $$$n$$$. The initial height of the $$$i$$$-th tree is $$$h_i$$$.You want to water these trees, so they all grow to the same height.The watering process goes as follows. You start watering trees at day $$$1$$$. During the $$$j$$$-th day you can: Choose a tree and water it. If the day is odd (e.g. $$$1, 3, 5, 7, \dots$$$), then the height of the tree increases by $$$1$$$. If the day is even (e.g. $$$2, 4, 6, 8, \dots$$$), then the height of the tree increases by $$$2$$$. Or skip a day without watering any tree. Note that you can't water more than one tree in a day. Your task is to determine the minimum number of days required to water the trees so they grow to the same height.You have to answer $$$t$$$ independent test cases.
Input Specification: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$) — the number of trees. The second line of the test case contains $$$n$$$ integers $$$h_1, h_2, \ldots, h_n$$$ ($$$1 \le h_i \le 10^9$$$), where $$$h_i$$$ is the height of the $$$i$$$-th tree. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$ ($$$\sum n \le 3 \cdot 10^5$$$).
Output Specification: For each test case, print one integer — the minimum number of days required to water the trees, so they grow to the same height.
Notes: NoteConsider the first test case of the example. The initial state of the trees is $$$[1, 2, 4]$$$. During the first day, let's water the first tree, so the sequence of heights becomes $$$[2, 2, 4]$$$; during the second day, let's water the second tree, so the sequence of heights becomes $$$[2, 4, 4]$$$; let's skip the third day; during the fourth day, let's water the first tree, so the sequence of heights becomes $$$[4, 4, 4]$$$. Thus, the answer is $$$4$$$.
Code:
for test in range(int(input())):
n = int(input())
h = [int(i) for i in input().split()]
res = 2 << 69
for req in range(max(h), max(h)+3):
hm = req
d = 0
c = 0
for i in h:
# TODO: Your code here
res = min(res, max((d//3)*2+d % 3, c*2-1))
print(res)
|
for test in range(int(input())):
n = int(input())
h = [int(i) for i in input().split()]
res = 2 << 69
for req in range(max(h), max(h)+3):
hm = req
d = 0
c = 0
for i in h:
{{completion}}
res = min(res, max((d//3)*2+d % 3, c*2-1))
print(res)
|
d += req-i
c += (req-i) & 1
|
[{"input": "3\n3\n1 2 4\n5\n4 4 3 5 5\n7\n2 5 4 8 3 7 4", "output": ["4\n3\n16"]}]
|
block_completion_003415
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ trees in a park, numbered from $$$1$$$ to $$$n$$$. The initial height of the $$$i$$$-th tree is $$$h_i$$$.You want to water these trees, so they all grow to the same height.The watering process goes as follows. You start watering trees at day $$$1$$$. During the $$$j$$$-th day you can: Choose a tree and water it. If the day is odd (e.g. $$$1, 3, 5, 7, \dots$$$), then the height of the tree increases by $$$1$$$. If the day is even (e.g. $$$2, 4, 6, 8, \dots$$$), then the height of the tree increases by $$$2$$$. Or skip a day without watering any tree. Note that you can't water more than one tree in a day. Your task is to determine the minimum number of days required to water the trees so they grow to the same height.You have to answer $$$t$$$ independent test cases.
Input Specification: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$) — the number of trees. The second line of the test case contains $$$n$$$ integers $$$h_1, h_2, \ldots, h_n$$$ ($$$1 \le h_i \le 10^9$$$), where $$$h_i$$$ is the height of the $$$i$$$-th tree. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$ ($$$\sum n \le 3 \cdot 10^5$$$).
Output Specification: For each test case, print one integer — the minimum number of days required to water the trees, so they grow to the same height.
Notes: NoteConsider the first test case of the example. The initial state of the trees is $$$[1, 2, 4]$$$. During the first day, let's water the first tree, so the sequence of heights becomes $$$[2, 2, 4]$$$; during the second day, let's water the second tree, so the sequence of heights becomes $$$[2, 4, 4]$$$; let's skip the third day; during the fourth day, let's water the first tree, so the sequence of heights becomes $$$[4, 4, 4]$$$. Thus, the answer is $$$4$$$.
Code:
from bisect import bisect
t = int(input().strip())
def solve(o, t):
if t - o > 1:
# TODO: Your code here
return (o + (t > o)) * 2 - (o > t)
out = []
for _ in range(t):
n = int(input().strip())
h = list(map(int, input().strip().split()))
mx = max(h)
o, t, e = 0, 0, 0
for x in h:
e += (mx - x + 1) % 2
o += (mx - x) % 2
t += (mx - x) // 2
out.append(str(min(solve(o, t), solve(e, t + o))))
print("\n".join(out))
|
from bisect import bisect
t = int(input().strip())
def solve(o, t):
if t - o > 1:
{{completion}}
return (o + (t > o)) * 2 - (o > t)
out = []
for _ in range(t):
n = int(input().strip())
h = list(map(int, input().strip().split()))
mx = max(h)
o, t, e = 0, 0, 0
for x in h:
e += (mx - x + 1) % 2
o += (mx - x) % 2
t += (mx - x) // 2
out.append(str(min(solve(o, t), solve(e, t + o))))
print("\n".join(out))
|
d = (t - o) // 3 + ((t - o) % 3 == 2)
o, t = o + 2 * d, t - d
|
[{"input": "3\n3\n1 2 4\n5\n4 4 3 5 5\n7\n2 5 4 8 3 7 4", "output": ["4\n3\n16"]}]
|
block_completion_003416
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ trees in a park, numbered from $$$1$$$ to $$$n$$$. The initial height of the $$$i$$$-th tree is $$$h_i$$$.You want to water these trees, so they all grow to the same height.The watering process goes as follows. You start watering trees at day $$$1$$$. During the $$$j$$$-th day you can: Choose a tree and water it. If the day is odd (e.g. $$$1, 3, 5, 7, \dots$$$), then the height of the tree increases by $$$1$$$. If the day is even (e.g. $$$2, 4, 6, 8, \dots$$$), then the height of the tree increases by $$$2$$$. Or skip a day without watering any tree. Note that you can't water more than one tree in a day. Your task is to determine the minimum number of days required to water the trees so they grow to the same height.You have to answer $$$t$$$ independent test cases.
Input Specification: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$) — the number of trees. The second line of the test case contains $$$n$$$ integers $$$h_1, h_2, \ldots, h_n$$$ ($$$1 \le h_i \le 10^9$$$), where $$$h_i$$$ is the height of the $$$i$$$-th tree. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$ ($$$\sum n \le 3 \cdot 10^5$$$).
Output Specification: For each test case, print one integer — the minimum number of days required to water the trees, so they grow to the same height.
Notes: NoteConsider the first test case of the example. The initial state of the trees is $$$[1, 2, 4]$$$. During the first day, let's water the first tree, so the sequence of heights becomes $$$[2, 2, 4]$$$; during the second day, let's water the second tree, so the sequence of heights becomes $$$[2, 4, 4]$$$; let's skip the third day; during the fourth day, let's water the first tree, so the sequence of heights becomes $$$[4, 4, 4]$$$. Thus, the answer is $$$4$$$.
Code:
from bisect import bisect
t = int(input().strip())
def solve(o, t):
if t - o > 1:
d = (t - o) // 3 + ((t - o) % 3 == 2)
o, t = o + 2 * d, t - d
return (o + (t > o)) * 2 - (o > t)
out = []
for _ in range(t):
n = int(input().strip())
h = list(map(int, input().strip().split()))
mx = max(h)
o, t, e = 0, 0, 0
for x in h:
# TODO: Your code here
out.append(str(min(solve(o, t), solve(e, t + o))))
print("\n".join(out))
|
from bisect import bisect
t = int(input().strip())
def solve(o, t):
if t - o > 1:
d = (t - o) // 3 + ((t - o) % 3 == 2)
o, t = o + 2 * d, t - d
return (o + (t > o)) * 2 - (o > t)
out = []
for _ in range(t):
n = int(input().strip())
h = list(map(int, input().strip().split()))
mx = max(h)
o, t, e = 0, 0, 0
for x in h:
{{completion}}
out.append(str(min(solve(o, t), solve(e, t + o))))
print("\n".join(out))
|
e += (mx - x + 1) % 2
o += (mx - x) % 2
t += (mx - x) // 2
|
[{"input": "3\n3\n1 2 4\n5\n4 4 3 5 5\n7\n2 5 4 8 3 7 4", "output": ["4\n3\n16"]}]
|
block_completion_003417
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ trees in a park, numbered from $$$1$$$ to $$$n$$$. The initial height of the $$$i$$$-th tree is $$$h_i$$$.You want to water these trees, so they all grow to the same height.The watering process goes as follows. You start watering trees at day $$$1$$$. During the $$$j$$$-th day you can: Choose a tree and water it. If the day is odd (e.g. $$$1, 3, 5, 7, \dots$$$), then the height of the tree increases by $$$1$$$. If the day is even (e.g. $$$2, 4, 6, 8, \dots$$$), then the height of the tree increases by $$$2$$$. Or skip a day without watering any tree. Note that you can't water more than one tree in a day. Your task is to determine the minimum number of days required to water the trees so they grow to the same height.You have to answer $$$t$$$ independent test cases.
Input Specification: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$) — the number of trees. The second line of the test case contains $$$n$$$ integers $$$h_1, h_2, \ldots, h_n$$$ ($$$1 \le h_i \le 10^9$$$), where $$$h_i$$$ is the height of the $$$i$$$-th tree. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$ ($$$\sum n \le 3 \cdot 10^5$$$).
Output Specification: For each test case, print one integer — the minimum number of days required to water the trees, so they grow to the same height.
Notes: NoteConsider the first test case of the example. The initial state of the trees is $$$[1, 2, 4]$$$. During the first day, let's water the first tree, so the sequence of heights becomes $$$[2, 2, 4]$$$; during the second day, let's water the second tree, so the sequence of heights becomes $$$[2, 4, 4]$$$; let's skip the third day; during the fourth day, let's water the first tree, so the sequence of heights becomes $$$[4, 4, 4]$$$. Thus, the answer is $$$4$$$.
Code:
def solve(target,r):
k=len(r)
ones=twos=0
for i in range(k):
# TODO: Your code here
if ones>twos:return 2*ones-1
return (ones+twos*2)//3*2+(ones+twos*2)%3
for _ in [0]*int(input()):
input();r=[*map(int,input().split())]
print(min(solve(max(r),r),solve(max(r)+1,r)))
|
def solve(target,r):
k=len(r)
ones=twos=0
for i in range(k):
{{completion}}
if ones>twos:return 2*ones-1
return (ones+twos*2)//3*2+(ones+twos*2)%3
for _ in [0]*int(input()):
input();r=[*map(int,input().split())]
print(min(solve(max(r),r),solve(max(r)+1,r)))
|
ones+=(target-r[i])%2
twos+=(target-r[i])//2
|
[{"input": "3\n3\n1 2 4\n5\n4 4 3 5 5\n7\n2 5 4 8 3 7 4", "output": ["4\n3\n16"]}]
|
block_completion_003418
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ trees in a park, numbered from $$$1$$$ to $$$n$$$. The initial height of the $$$i$$$-th tree is $$$h_i$$$.You want to water these trees, so they all grow to the same height.The watering process goes as follows. You start watering trees at day $$$1$$$. During the $$$j$$$-th day you can: Choose a tree and water it. If the day is odd (e.g. $$$1, 3, 5, 7, \dots$$$), then the height of the tree increases by $$$1$$$. If the day is even (e.g. $$$2, 4, 6, 8, \dots$$$), then the height of the tree increases by $$$2$$$. Or skip a day without watering any tree. Note that you can't water more than one tree in a day. Your task is to determine the minimum number of days required to water the trees so they grow to the same height.You have to answer $$$t$$$ independent test cases.
Input Specification: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$) — the number of trees. The second line of the test case contains $$$n$$$ integers $$$h_1, h_2, \ldots, h_n$$$ ($$$1 \le h_i \le 10^9$$$), where $$$h_i$$$ is the height of the $$$i$$$-th tree. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$ ($$$\sum n \le 3 \cdot 10^5$$$).
Output Specification: For each test case, print one integer — the minimum number of days required to water the trees, so they grow to the same height.
Notes: NoteConsider the first test case of the example. The initial state of the trees is $$$[1, 2, 4]$$$. During the first day, let's water the first tree, so the sequence of heights becomes $$$[2, 2, 4]$$$; during the second day, let's water the second tree, so the sequence of heights becomes $$$[2, 4, 4]$$$; let's skip the third day; during the fourth day, let's water the first tree, so the sequence of heights becomes $$$[4, 4, 4]$$$. Thus, the answer is $$$4$$$.
Code:
def solve(target,r):
k=len(r)
ones=twos=0
for i in range(k):
ones+=(target-r[i])%2
twos+=(target-r[i])//2
if ones>twos:# TODO: Your code here
return (ones+twos*2)//3*2+(ones+twos*2)%3
for _ in [0]*int(input()):
input();r=[*map(int,input().split())]
print(min(solve(max(r),r),solve(max(r)+1,r)))
|
def solve(target,r):
k=len(r)
ones=twos=0
for i in range(k):
ones+=(target-r[i])%2
twos+=(target-r[i])//2
if ones>twos:{{completion}}
return (ones+twos*2)//3*2+(ones+twos*2)%3
for _ in [0]*int(input()):
input();r=[*map(int,input().split())]
print(min(solve(max(r),r),solve(max(r)+1,r)))
|
return 2*ones-1
|
[{"input": "3\n3\n1 2 4\n5\n4 4 3 5 5\n7\n2 5 4 8 3 7 4", "output": ["4\n3\n16"]}]
|
block_completion_003419
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ trees in a park, numbered from $$$1$$$ to $$$n$$$. The initial height of the $$$i$$$-th tree is $$$h_i$$$.You want to water these trees, so they all grow to the same height.The watering process goes as follows. You start watering trees at day $$$1$$$. During the $$$j$$$-th day you can: Choose a tree and water it. If the day is odd (e.g. $$$1, 3, 5, 7, \dots$$$), then the height of the tree increases by $$$1$$$. If the day is even (e.g. $$$2, 4, 6, 8, \dots$$$), then the height of the tree increases by $$$2$$$. Or skip a day without watering any tree. Note that you can't water more than one tree in a day. Your task is to determine the minimum number of days required to water the trees so they grow to the same height.You have to answer $$$t$$$ independent test cases.
Input Specification: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$) — the number of trees. The second line of the test case contains $$$n$$$ integers $$$h_1, h_2, \ldots, h_n$$$ ($$$1 \le h_i \le 10^9$$$), where $$$h_i$$$ is the height of the $$$i$$$-th tree. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$ ($$$\sum n \le 3 \cdot 10^5$$$).
Output Specification: For each test case, print one integer — the minimum number of days required to water the trees, so they grow to the same height.
Notes: NoteConsider the first test case of the example. The initial state of the trees is $$$[1, 2, 4]$$$. During the first day, let's water the first tree, so the sequence of heights becomes $$$[2, 2, 4]$$$; during the second day, let's water the second tree, so the sequence of heights becomes $$$[2, 4, 4]$$$; let's skip the third day; during the fourth day, let's water the first tree, so the sequence of heights becomes $$$[4, 4, 4]$$$. Thus, the answer is $$$4$$$.
Code:
for _ in range(int(input())):
n=int(input())
l=list(map(int,input().strip().split()))
l.sort()
a=l.count(l[-1])
odd,even=0,0
for i in l:
if i%2==0:
even+=1
else:
# TODO: Your code here
su=sum(l[:n-a])
needed=l[-1]*(n-a)-su
if l[-1]%2==0:
p1,p2=odd,even
else:
p1,p2=even,odd
ans=max(2*(needed//3)+needed%3,2*p1-1)
needed+=n
ans2=max(2*(needed//3)+needed%3,2*p2-1)
print(min(ans,ans2))
|
for _ in range(int(input())):
n=int(input())
l=list(map(int,input().strip().split()))
l.sort()
a=l.count(l[-1])
odd,even=0,0
for i in l:
if i%2==0:
even+=1
else:
{{completion}}
su=sum(l[:n-a])
needed=l[-1]*(n-a)-su
if l[-1]%2==0:
p1,p2=odd,even
else:
p1,p2=even,odd
ans=max(2*(needed//3)+needed%3,2*p1-1)
needed+=n
ans2=max(2*(needed//3)+needed%3,2*p2-1)
print(min(ans,ans2))
|
odd+=1
|
[{"input": "3\n3\n1 2 4\n5\n4 4 3 5 5\n7\n2 5 4 8 3 7 4", "output": ["4\n3\n16"]}]
|
block_completion_003420
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ trees in a park, numbered from $$$1$$$ to $$$n$$$. The initial height of the $$$i$$$-th tree is $$$h_i$$$.You want to water these trees, so they all grow to the same height.The watering process goes as follows. You start watering trees at day $$$1$$$. During the $$$j$$$-th day you can: Choose a tree and water it. If the day is odd (e.g. $$$1, 3, 5, 7, \dots$$$), then the height of the tree increases by $$$1$$$. If the day is even (e.g. $$$2, 4, 6, 8, \dots$$$), then the height of the tree increases by $$$2$$$. Or skip a day without watering any tree. Note that you can't water more than one tree in a day. Your task is to determine the minimum number of days required to water the trees so they grow to the same height.You have to answer $$$t$$$ independent test cases.
Input Specification: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$) — the number of trees. The second line of the test case contains $$$n$$$ integers $$$h_1, h_2, \ldots, h_n$$$ ($$$1 \le h_i \le 10^9$$$), where $$$h_i$$$ is the height of the $$$i$$$-th tree. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$ ($$$\sum n \le 3 \cdot 10^5$$$).
Output Specification: For each test case, print one integer — the minimum number of days required to water the trees, so they grow to the same height.
Notes: NoteConsider the first test case of the example. The initial state of the trees is $$$[1, 2, 4]$$$. During the first day, let's water the first tree, so the sequence of heights becomes $$$[2, 2, 4]$$$; during the second day, let's water the second tree, so the sequence of heights becomes $$$[2, 4, 4]$$$; let's skip the third day; during the fourth day, let's water the first tree, so the sequence of heights becomes $$$[4, 4, 4]$$$. Thus, the answer is $$$4$$$.
Code:
def solve(m,a):
ev=od=0
for i in a:
# TODO: Your code here
if(od>=ev):
return od*2-(od!=ev)
ev = (ev-od)*2
return od*2 + ev//3*2 + ev%3
I = lambda: map(int,input().split())
t,=I()
for _ in [1]*t:
n, = I()
b = [*I()]
mx = max(b)
print(min(solve(mx,b),solve(mx+1,b)))
|
def solve(m,a):
ev=od=0
for i in a:
{{completion}}
if(od>=ev):
return od*2-(od!=ev)
ev = (ev-od)*2
return od*2 + ev//3*2 + ev%3
I = lambda: map(int,input().split())
t,=I()
for _ in [1]*t:
n, = I()
b = [*I()]
mx = max(b)
print(min(solve(mx,b),solve(mx+1,b)))
|
ev += (m-i)//2; od += (m-i)%2
|
[{"input": "3\n3\n1 2 4\n5\n4 4 3 5 5\n7\n2 5 4 8 3 7 4", "output": ["4\n3\n16"]}]
|
block_completion_003421
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ trees in a park, numbered from $$$1$$$ to $$$n$$$. The initial height of the $$$i$$$-th tree is $$$h_i$$$.You want to water these trees, so they all grow to the same height.The watering process goes as follows. You start watering trees at day $$$1$$$. During the $$$j$$$-th day you can: Choose a tree and water it. If the day is odd (e.g. $$$1, 3, 5, 7, \dots$$$), then the height of the tree increases by $$$1$$$. If the day is even (e.g. $$$2, 4, 6, 8, \dots$$$), then the height of the tree increases by $$$2$$$. Or skip a day without watering any tree. Note that you can't water more than one tree in a day. Your task is to determine the minimum number of days required to water the trees so they grow to the same height.You have to answer $$$t$$$ independent test cases.
Input Specification: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$) — the number of trees. The second line of the test case contains $$$n$$$ integers $$$h_1, h_2, \ldots, h_n$$$ ($$$1 \le h_i \le 10^9$$$), where $$$h_i$$$ is the height of the $$$i$$$-th tree. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$ ($$$\sum n \le 3 \cdot 10^5$$$).
Output Specification: For each test case, print one integer — the minimum number of days required to water the trees, so they grow to the same height.
Notes: NoteConsider the first test case of the example. The initial state of the trees is $$$[1, 2, 4]$$$. During the first day, let's water the first tree, so the sequence of heights becomes $$$[2, 2, 4]$$$; during the second day, let's water the second tree, so the sequence of heights becomes $$$[2, 4, 4]$$$; let's skip the third day; during the fourth day, let's water the first tree, so the sequence of heights becomes $$$[4, 4, 4]$$$. Thus, the answer is $$$4$$$.
Code:
def solve(m,a):
ev=od=0
for i in a:
ev += (m-i)//2; od += (m-i)%2
if(od>=ev):
# TODO: Your code here
ev = (ev-od)*2
return od*2 + ev//3*2 + ev%3
I = lambda: map(int,input().split())
t,=I()
for _ in [1]*t:
n, = I()
b = [*I()]
mx = max(b)
print(min(solve(mx,b),solve(mx+1,b)))
|
def solve(m,a):
ev=od=0
for i in a:
ev += (m-i)//2; od += (m-i)%2
if(od>=ev):
{{completion}}
ev = (ev-od)*2
return od*2 + ev//3*2 + ev%3
I = lambda: map(int,input().split())
t,=I()
for _ in [1]*t:
n, = I()
b = [*I()]
mx = max(b)
print(min(solve(mx,b),solve(mx+1,b)))
|
return od*2-(od!=ev)
|
[{"input": "3\n3\n1 2 4\n5\n4 4 3 5 5\n7\n2 5 4 8 3 7 4", "output": ["4\n3\n16"]}]
|
block_completion_003422
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ trees in a park, numbered from $$$1$$$ to $$$n$$$. The initial height of the $$$i$$$-th tree is $$$h_i$$$.You want to water these trees, so they all grow to the same height.The watering process goes as follows. You start watering trees at day $$$1$$$. During the $$$j$$$-th day you can: Choose a tree and water it. If the day is odd (e.g. $$$1, 3, 5, 7, \dots$$$), then the height of the tree increases by $$$1$$$. If the day is even (e.g. $$$2, 4, 6, 8, \dots$$$), then the height of the tree increases by $$$2$$$. Or skip a day without watering any tree. Note that you can't water more than one tree in a day. Your task is to determine the minimum number of days required to water the trees so they grow to the same height.You have to answer $$$t$$$ independent test cases.
Input Specification: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$) — the number of trees. The second line of the test case contains $$$n$$$ integers $$$h_1, h_2, \ldots, h_n$$$ ($$$1 \le h_i \le 10^9$$$), where $$$h_i$$$ is the height of the $$$i$$$-th tree. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$ ($$$\sum n \le 3 \cdot 10^5$$$).
Output Specification: For each test case, print one integer — the minimum number of days required to water the trees, so they grow to the same height.
Notes: NoteConsider the first test case of the example. The initial state of the trees is $$$[1, 2, 4]$$$. During the first day, let's water the first tree, so the sequence of heights becomes $$$[2, 2, 4]$$$; during the second day, let's water the second tree, so the sequence of heights becomes $$$[2, 4, 4]$$$; let's skip the third day; during the fourth day, let's water the first tree, so the sequence of heights becomes $$$[4, 4, 4]$$$. Thus, the answer is $$$4$$$.
Code:
# by the authority of GOD author: Kritarth Sharma #
import sys
from collections import defaultdict,deque,Counter
from bisect import bisect_left
import math
input=sys.stdin.readline
def inp():
l=list(map(int,input().split()))
return l
for _ in range(int(input())):
n,=inp()
l=inp()
m=max(l)
a=float('inf')
for i in range(m,m+2):
x=0
y=0
for j in range(n):
# TODO: Your code here
a=min(a,max( 2*y -1, ((2*x+y)//3)*2 +((2*x+y)%3) ))
print(a)
|
# by the authority of GOD author: Kritarth Sharma #
import sys
from collections import defaultdict,deque,Counter
from bisect import bisect_left
import math
input=sys.stdin.readline
def inp():
l=list(map(int,input().split()))
return l
for _ in range(int(input())):
n,=inp()
l=inp()
m=max(l)
a=float('inf')
for i in range(m,m+2):
x=0
y=0
for j in range(n):
{{completion}}
a=min(a,max( 2*y -1, ((2*x+y)//3)*2 +((2*x+y)%3) ))
print(a)
|
x+=(i-l[j])//2
y+=(i-l[j])%2
|
[{"input": "3\n3\n1 2 4\n5\n4 4 3 5 5\n7\n2 5 4 8 3 7 4", "output": ["4\n3\n16"]}]
|
block_completion_003423
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ trees in a park, numbered from $$$1$$$ to $$$n$$$. The initial height of the $$$i$$$-th tree is $$$h_i$$$.You want to water these trees, so they all grow to the same height.The watering process goes as follows. You start watering trees at day $$$1$$$. During the $$$j$$$-th day you can: Choose a tree and water it. If the day is odd (e.g. $$$1, 3, 5, 7, \dots$$$), then the height of the tree increases by $$$1$$$. If the day is even (e.g. $$$2, 4, 6, 8, \dots$$$), then the height of the tree increases by $$$2$$$. Or skip a day without watering any tree. Note that you can't water more than one tree in a day. Your task is to determine the minimum number of days required to water the trees so they grow to the same height.You have to answer $$$t$$$ independent test cases.
Input Specification: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$) — the number of trees. The second line of the test case contains $$$n$$$ integers $$$h_1, h_2, \ldots, h_n$$$ ($$$1 \le h_i \le 10^9$$$), where $$$h_i$$$ is the height of the $$$i$$$-th tree. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$ ($$$\sum n \le 3 \cdot 10^5$$$).
Output Specification: For each test case, print one integer — the minimum number of days required to water the trees, so they grow to the same height.
Notes: NoteConsider the first test case of the example. The initial state of the trees is $$$[1, 2, 4]$$$. During the first day, let's water the first tree, so the sequence of heights becomes $$$[2, 2, 4]$$$; during the second day, let's water the second tree, so the sequence of heights becomes $$$[2, 4, 4]$$$; let's skip the third day; during the fourth day, let's water the first tree, so the sequence of heights becomes $$$[4, 4, 4]$$$. Thus, the answer is $$$4$$$.
Code:
def solve1(n, h, asdf):
max_h = max(h)
diff_h = list(map(lambda x:max_h+asdf-x, h))
required_h = sum(diff_h)
min_odd_days = sum(1 for diff in diff_h if diff % 2 == 1)
if required_h < min_odd_days * 3:
return min_odd_days * 2 - 1
else:
# TODO: Your code here
def solve():
n = int(input())
h = list(map(int, input().split()))
print(min(solve1(n, h, 0), solve1(n, h, 1)))
tc = int(input())
for _ in range(tc):
solve()
|
def solve1(n, h, asdf):
max_h = max(h)
diff_h = list(map(lambda x:max_h+asdf-x, h))
required_h = sum(diff_h)
min_odd_days = sum(1 for diff in diff_h if diff % 2 == 1)
if required_h < min_odd_days * 3:
return min_odd_days * 2 - 1
else:
{{completion}}
def solve():
n = int(input())
h = list(map(int, input().split()))
print(min(solve1(n, h, 0), solve1(n, h, 1)))
tc = int(input())
for _ in range(tc):
solve()
|
return (required_h//3) * 2 + (required_h % 3)
|
[{"input": "3\n3\n1 2 4\n5\n4 4 3 5 5\n7\n2 5 4 8 3 7 4", "output": ["4\n3\n16"]}]
|
block_completion_003424
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ trees in a park, numbered from $$$1$$$ to $$$n$$$. The initial height of the $$$i$$$-th tree is $$$h_i$$$.You want to water these trees, so they all grow to the same height.The watering process goes as follows. You start watering trees at day $$$1$$$. During the $$$j$$$-th day you can: Choose a tree and water it. If the day is odd (e.g. $$$1, 3, 5, 7, \dots$$$), then the height of the tree increases by $$$1$$$. If the day is even (e.g. $$$2, 4, 6, 8, \dots$$$), then the height of the tree increases by $$$2$$$. Or skip a day without watering any tree. Note that you can't water more than one tree in a day. Your task is to determine the minimum number of days required to water the trees so they grow to the same height.You have to answer $$$t$$$ independent test cases.
Input Specification: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$) — the number of trees. The second line of the test case contains $$$n$$$ integers $$$h_1, h_2, \ldots, h_n$$$ ($$$1 \le h_i \le 10^9$$$), where $$$h_i$$$ is the height of the $$$i$$$-th tree. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$ ($$$\sum n \le 3 \cdot 10^5$$$).
Output Specification: For each test case, print one integer — the minimum number of days required to water the trees, so they grow to the same height.
Notes: NoteConsider the first test case of the example. The initial state of the trees is $$$[1, 2, 4]$$$. During the first day, let's water the first tree, so the sequence of heights becomes $$$[2, 2, 4]$$$; during the second day, let's water the second tree, so the sequence of heights becomes $$$[2, 4, 4]$$$; let's skip the third day; during the fourth day, let's water the first tree, so the sequence of heights becomes $$$[4, 4, 4]$$$. Thus, the answer is $$$4$$$.
Code:
inp = lambda : list(map(int,input().split()))
for _ in range(int(input())):
n = int(input())
t = inp()
c = 0
m = max(t)
def mm(m):
n1 = n2 = 0
tot =0
for i in t:
# TODO: Your code here
return (n1*2-1) if n2<n1 else ((n2*2+n1)//3*2+(n2*2+n1)%3)
print(min(mm(m),mm(m+1)))
|
inp = lambda : list(map(int,input().split()))
for _ in range(int(input())):
n = int(input())
t = inp()
c = 0
m = max(t)
def mm(m):
n1 = n2 = 0
tot =0
for i in t:
{{completion}}
return (n1*2-1) if n2<n1 else ((n2*2+n1)//3*2+(n2*2+n1)%3)
print(min(mm(m),mm(m+1)))
|
n1+= (m-i)%2
n2+= (m-i)//2
|
[{"input": "3\n3\n1 2 4\n5\n4 4 3 5 5\n7\n2 5 4 8 3 7 4", "output": ["4\n3\n16"]}]
|
block_completion_003425
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ trees in a park, numbered from $$$1$$$ to $$$n$$$. The initial height of the $$$i$$$-th tree is $$$h_i$$$.You want to water these trees, so they all grow to the same height.The watering process goes as follows. You start watering trees at day $$$1$$$. During the $$$j$$$-th day you can: Choose a tree and water it. If the day is odd (e.g. $$$1, 3, 5, 7, \dots$$$), then the height of the tree increases by $$$1$$$. If the day is even (e.g. $$$2, 4, 6, 8, \dots$$$), then the height of the tree increases by $$$2$$$. Or skip a day without watering any tree. Note that you can't water more than one tree in a day. Your task is to determine the minimum number of days required to water the trees so they grow to the same height.You have to answer $$$t$$$ independent test cases.
Input Specification: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$) — the number of trees. The second line of the test case contains $$$n$$$ integers $$$h_1, h_2, \ldots, h_n$$$ ($$$1 \le h_i \le 10^9$$$), where $$$h_i$$$ is the height of the $$$i$$$-th tree. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$ ($$$\sum n \le 3 \cdot 10^5$$$).
Output Specification: For each test case, print one integer — the minimum number of days required to water the trees, so they grow to the same height.
Notes: NoteConsider the first test case of the example. The initial state of the trees is $$$[1, 2, 4]$$$. During the first day, let's water the first tree, so the sequence of heights becomes $$$[2, 2, 4]$$$; during the second day, let's water the second tree, so the sequence of heights becomes $$$[2, 4, 4]$$$; let's skip the third day; during the fourth day, let's water the first tree, so the sequence of heights becomes $$$[4, 4, 4]$$$. Thus, the answer is $$$4$$$.
Code:
for ii in range(int(input())):
n=int(input())
a = list(map(int, input().split()))
m=max(a)
ans=float("inf")
for jj in range(m,m+4):
x,y=0,0
for kk in a:
# TODO: Your code here
ans=min(max(((x+y*2)//3*2)+(x+y*2)%3,x*2-1),ans)
print(ans)
|
for ii in range(int(input())):
n=int(input())
a = list(map(int, input().split()))
m=max(a)
ans=float("inf")
for jj in range(m,m+4):
x,y=0,0
for kk in a:
{{completion}}
ans=min(max(((x+y*2)//3*2)+(x+y*2)%3,x*2-1),ans)
print(ans)
|
x+=(jj-kk)%2
y+=(jj-kk)//2
|
[{"input": "3\n3\n1 2 4\n5\n4 4 3 5 5\n7\n2 5 4 8 3 7 4", "output": ["4\n3\n16"]}]
|
block_completion_003426
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given two arrays: an array $$$a$$$ consisting of $$$n$$$ zeros and an array $$$b$$$ consisting of $$$n$$$ integers.You can apply the following operation to the array $$$a$$$ an arbitrary number of times: choose some subsegment of $$$a$$$ of length $$$k$$$ and add the arithmetic progression $$$1, 2, \ldots, k$$$ to this subsegment — i. e. add $$$1$$$ to the first element of the subsegment, $$$2$$$ to the second element, and so on. The chosen subsegment should be inside the borders of the array $$$a$$$ (i.e., if the left border of the chosen subsegment is $$$l$$$, then the condition $$$1 \le l \le l + k - 1 \le n$$$ should be satisfied). Note that the progression added is always $$$1, 2, \ldots, k$$$ but not the $$$k, k - 1, \ldots, 1$$$ or anything else (i.e., the leftmost element of the subsegment always increases by $$$1$$$, the second element always increases by $$$2$$$ and so on).Your task is to find the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Note that the condition $$$a_i \ge b_i$$$ should be satisfied for all elements at once.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 3 \cdot 10^5$$$) — the number of elements in both arrays and the length of the subsegment, respectively. The second line of the input contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i \le 10^{12}$$$), where $$$b_i$$$ is the $$$i$$$-th element of the array $$$b$$$.
Output Specification: Print one integer — the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$.
Notes: NoteConsider the first example. In this test, we don't really have any choice, so we need to add at least five progressions to make the first element equals $$$5$$$. The array $$$a$$$ becomes $$$[5, 10, 15]$$$.Consider the second example. In this test, let's add one progression on the segment $$$[1; 3]$$$ and two progressions on the segment $$$[4; 6]$$$. Then, the array $$$a$$$ becomes $$$[1, 2, 3, 2, 4, 6]$$$.
Code:
n,k = map(int, input().split())
bb = list(map(int, input().split()))
ans = 0
sofar = 0
sumprog = 0
timeq = []
for ib,b in enumerate(bb[::-1]):
kk = min(k, n-ib)
time = (max(0,b-sofar)+kk-1)//kk
ans += time
timeq.append(time)
sumprog += time
if ib >= k:
# TODO: Your code here
sofar += kk*time
sofar -= sumprog
# print(time, sofar, timeq, sumprog)
print(ans)
|
n,k = map(int, input().split())
bb = list(map(int, input().split()))
ans = 0
sofar = 0
sumprog = 0
timeq = []
for ib,b in enumerate(bb[::-1]):
kk = min(k, n-ib)
time = (max(0,b-sofar)+kk-1)//kk
ans += time
timeq.append(time)
sumprog += time
if ib >= k:
{{completion}}
sofar += kk*time
sofar -= sumprog
# print(time, sofar, timeq, sumprog)
print(ans)
|
sumprog -= timeq[ib-k]
|
[{"input": "3 3\n5 4 6", "output": ["5"]}, {"input": "6 3\n1 2 3 2 2 3", "output": ["3"]}, {"input": "6 3\n1 2 4 1 2 3", "output": ["3"]}, {"input": "7 3\n50 17 81 25 42 39 96", "output": ["92"]}]
|
block_completion_003443
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given two arrays: an array $$$a$$$ consisting of $$$n$$$ zeros and an array $$$b$$$ consisting of $$$n$$$ integers.You can apply the following operation to the array $$$a$$$ an arbitrary number of times: choose some subsegment of $$$a$$$ of length $$$k$$$ and add the arithmetic progression $$$1, 2, \ldots, k$$$ to this subsegment — i. e. add $$$1$$$ to the first element of the subsegment, $$$2$$$ to the second element, and so on. The chosen subsegment should be inside the borders of the array $$$a$$$ (i.e., if the left border of the chosen subsegment is $$$l$$$, then the condition $$$1 \le l \le l + k - 1 \le n$$$ should be satisfied). Note that the progression added is always $$$1, 2, \ldots, k$$$ but not the $$$k, k - 1, \ldots, 1$$$ or anything else (i.e., the leftmost element of the subsegment always increases by $$$1$$$, the second element always increases by $$$2$$$ and so on).Your task is to find the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Note that the condition $$$a_i \ge b_i$$$ should be satisfied for all elements at once.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 3 \cdot 10^5$$$) — the number of elements in both arrays and the length of the subsegment, respectively. The second line of the input contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i \le 10^{12}$$$), where $$$b_i$$$ is the $$$i$$$-th element of the array $$$b$$$.
Output Specification: Print one integer — the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$.
Notes: NoteConsider the first example. In this test, we don't really have any choice, so we need to add at least five progressions to make the first element equals $$$5$$$. The array $$$a$$$ becomes $$$[5, 10, 15]$$$.Consider the second example. In this test, let's add one progression on the segment $$$[1; 3]$$$ and two progressions on the segment $$$[4; 6]$$$. Then, the array $$$a$$$ becomes $$$[1, 2, 3, 2, 4, 6]$$$.
Code:
I = lambda: [int(x) for x in input().split()]
n, k = I()
B, d = I() + [0]*k, [0] * (n + k)
s = total = 0
for i in range(n-1, -1, -1):
B[i] -= total
if B[i] > 0:
# TODO: Your code here
s += d[i] - d[i + k]
total += d[i] * dd - s
print(sum(d))
|
I = lambda: [int(x) for x in input().split()]
n, k = I()
B, d = I() + [0]*k, [0] * (n + k)
s = total = 0
for i in range(n-1, -1, -1):
B[i] -= total
if B[i] > 0:
{{completion}}
s += d[i] - d[i + k]
total += d[i] * dd - s
print(sum(d))
|
dd = min(k, i + 1)
d[i] = (B[i] + dd - 1)//dd
|
[{"input": "3 3\n5 4 6", "output": ["5"]}, {"input": "6 3\n1 2 3 2 2 3", "output": ["3"]}, {"input": "6 3\n1 2 4 1 2 3", "output": ["3"]}, {"input": "7 3\n50 17 81 25 42 39 96", "output": ["92"]}]
|
block_completion_003444
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given two arrays: an array $$$a$$$ consisting of $$$n$$$ zeros and an array $$$b$$$ consisting of $$$n$$$ integers.You can apply the following operation to the array $$$a$$$ an arbitrary number of times: choose some subsegment of $$$a$$$ of length $$$k$$$ and add the arithmetic progression $$$1, 2, \ldots, k$$$ to this subsegment — i. e. add $$$1$$$ to the first element of the subsegment, $$$2$$$ to the second element, and so on. The chosen subsegment should be inside the borders of the array $$$a$$$ (i.e., if the left border of the chosen subsegment is $$$l$$$, then the condition $$$1 \le l \le l + k - 1 \le n$$$ should be satisfied). Note that the progression added is always $$$1, 2, \ldots, k$$$ but not the $$$k, k - 1, \ldots, 1$$$ or anything else (i.e., the leftmost element of the subsegment always increases by $$$1$$$, the second element always increases by $$$2$$$ and so on).Your task is to find the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Note that the condition $$$a_i \ge b_i$$$ should be satisfied for all elements at once.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 3 \cdot 10^5$$$) — the number of elements in both arrays and the length of the subsegment, respectively. The second line of the input contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i \le 10^{12}$$$), where $$$b_i$$$ is the $$$i$$$-th element of the array $$$b$$$.
Output Specification: Print one integer — the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$.
Notes: NoteConsider the first example. In this test, we don't really have any choice, so we need to add at least five progressions to make the first element equals $$$5$$$. The array $$$a$$$ becomes $$$[5, 10, 15]$$$.Consider the second example. In this test, let's add one progression on the segment $$$[1; 3]$$$ and two progressions on the segment $$$[4; 6]$$$. Then, the array $$$a$$$ becomes $$$[1, 2, 3, 2, 4, 6]$$$.
Code:
#!/usr/bin/env PyPy3
from collections import Counter, defaultdict, deque
import itertools
import re
import math
from functools import reduce
import operator
import bisect
import heapq
import functools
mod=10**9+7
import sys
input=sys.stdin.readline
n,k=map(int,input().split())
b=list(map(int,input().split()))
ans = 0
dec = 0
cnt = [0] * n
tmp = 0
for i in range(k-1,n)[::-1]:
tmp -= cnt[i]
dec -= tmp
#print(tmp,dec)
if b[i] > dec:
#print(b[i]-dec)
x = -(-(b[i]-dec) // k)
ans += x
if i - k - 1 >= 0:
# TODO: Your code here
dec += x * k
tmp += x
#print(ans)
#tmp -= cnt[i]
#print(cnt)
ma = 0
for i in range(k-1)[::-1]:
tmp -= cnt[i]
dec -= tmp
ma = max(ma,-(-(b[i]-dec) // (i+1)))
print(ans+ma)
|
#!/usr/bin/env PyPy3
from collections import Counter, defaultdict, deque
import itertools
import re
import math
from functools import reduce
import operator
import bisect
import heapq
import functools
mod=10**9+7
import sys
input=sys.stdin.readline
n,k=map(int,input().split())
b=list(map(int,input().split()))
ans = 0
dec = 0
cnt = [0] * n
tmp = 0
for i in range(k-1,n)[::-1]:
tmp -= cnt[i]
dec -= tmp
#print(tmp,dec)
if b[i] > dec:
#print(b[i]-dec)
x = -(-(b[i]-dec) // k)
ans += x
if i - k - 1 >= 0:
{{completion}}
dec += x * k
tmp += x
#print(ans)
#tmp -= cnt[i]
#print(cnt)
ma = 0
for i in range(k-1)[::-1]:
tmp -= cnt[i]
dec -= tmp
ma = max(ma,-(-(b[i]-dec) // (i+1)))
print(ans+ma)
|
cnt[i-k-1] = x
|
[{"input": "3 3\n5 4 6", "output": ["5"]}, {"input": "6 3\n1 2 3 2 2 3", "output": ["3"]}, {"input": "6 3\n1 2 4 1 2 3", "output": ["3"]}, {"input": "7 3\n50 17 81 25 42 39 96", "output": ["92"]}]
|
block_completion_003445
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given two arrays: an array $$$a$$$ consisting of $$$n$$$ zeros and an array $$$b$$$ consisting of $$$n$$$ integers.You can apply the following operation to the array $$$a$$$ an arbitrary number of times: choose some subsegment of $$$a$$$ of length $$$k$$$ and add the arithmetic progression $$$1, 2, \ldots, k$$$ to this subsegment — i. e. add $$$1$$$ to the first element of the subsegment, $$$2$$$ to the second element, and so on. The chosen subsegment should be inside the borders of the array $$$a$$$ (i.e., if the left border of the chosen subsegment is $$$l$$$, then the condition $$$1 \le l \le l + k - 1 \le n$$$ should be satisfied). Note that the progression added is always $$$1, 2, \ldots, k$$$ but not the $$$k, k - 1, \ldots, 1$$$ or anything else (i.e., the leftmost element of the subsegment always increases by $$$1$$$, the second element always increases by $$$2$$$ and so on).Your task is to find the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Note that the condition $$$a_i \ge b_i$$$ should be satisfied for all elements at once.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 3 \cdot 10^5$$$) — the number of elements in both arrays and the length of the subsegment, respectively. The second line of the input contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i \le 10^{12}$$$), where $$$b_i$$$ is the $$$i$$$-th element of the array $$$b$$$.
Output Specification: Print one integer — the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$.
Notes: NoteConsider the first example. In this test, we don't really have any choice, so we need to add at least five progressions to make the first element equals $$$5$$$. The array $$$a$$$ becomes $$$[5, 10, 15]$$$.Consider the second example. In this test, let's add one progression on the segment $$$[1; 3]$$$ and two progressions on the segment $$$[4; 6]$$$. Then, the array $$$a$$$ becomes $$$[1, 2, 3, 2, 4, 6]$$$.
Code:
n, k = [int(i) for i in input().split()]
a = [int(i) for i in input().split()]
dd = [0]*(len(a)+5)
add = 0
diff = 0
moves = 0
for key, i in reversed([*enumerate(a)]):
add += diff
i += add
diff += dd[-1]
dd.pop()
if i > 0:
# TODO: Your code here
print(moves)
|
n, k = [int(i) for i in input().split()]
a = [int(i) for i in input().split()]
dd = [0]*(len(a)+5)
add = 0
diff = 0
moves = 0
for key, i in reversed([*enumerate(a)]):
add += diff
i += add
diff += dd[-1]
dd.pop()
if i > 0:
{{completion}}
print(moves)
|
K = min(k, key+1)
dd[-K] -= (i+K-1)//K
diff += (i+K-1)//K
moves += (i+K-1)//K
add -= K*((i+K-1)//K)
|
[{"input": "3 3\n5 4 6", "output": ["5"]}, {"input": "6 3\n1 2 3 2 2 3", "output": ["3"]}, {"input": "6 3\n1 2 4 1 2 3", "output": ["3"]}, {"input": "7 3\n50 17 81 25 42 39 96", "output": ["92"]}]
|
block_completion_003446
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given two arrays: an array $$$a$$$ consisting of $$$n$$$ zeros and an array $$$b$$$ consisting of $$$n$$$ integers.You can apply the following operation to the array $$$a$$$ an arbitrary number of times: choose some subsegment of $$$a$$$ of length $$$k$$$ and add the arithmetic progression $$$1, 2, \ldots, k$$$ to this subsegment — i. e. add $$$1$$$ to the first element of the subsegment, $$$2$$$ to the second element, and so on. The chosen subsegment should be inside the borders of the array $$$a$$$ (i.e., if the left border of the chosen subsegment is $$$l$$$, then the condition $$$1 \le l \le l + k - 1 \le n$$$ should be satisfied). Note that the progression added is always $$$1, 2, \ldots, k$$$ but not the $$$k, k - 1, \ldots, 1$$$ or anything else (i.e., the leftmost element of the subsegment always increases by $$$1$$$, the second element always increases by $$$2$$$ and so on).Your task is to find the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Note that the condition $$$a_i \ge b_i$$$ should be satisfied for all elements at once.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 3 \cdot 10^5$$$) — the number of elements in both arrays and the length of the subsegment, respectively. The second line of the input contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i \le 10^{12}$$$), where $$$b_i$$$ is the $$$i$$$-th element of the array $$$b$$$.
Output Specification: Print one integer — the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$.
Notes: NoteConsider the first example. In this test, we don't really have any choice, so we need to add at least five progressions to make the first element equals $$$5$$$. The array $$$a$$$ becomes $$$[5, 10, 15]$$$.Consider the second example. In this test, let's add one progression on the segment $$$[1; 3]$$$ and two progressions on the segment $$$[4; 6]$$$. Then, the array $$$a$$$ becomes $$$[1, 2, 3, 2, 4, 6]$$$.
Code:
"""
take element as "the tail" will use more less operations,
use variables s,cnt and closed,
to avoid the inner iteration(update neighbor k elements every time),
complexity is O(n).
"""
row=lambda:map(int,input().split())
n,k=row()
a=list(row())
closed=[0]*n
s=cnt=res=0
for i in range(n-1,-1,-1):
# print(i,a[i],'s,cnt:',s,cnt,'closed:',closed)
s-=cnt
cnt-=closed[i]
a[i]-=s
if a[i]<=0:
# TODO: Your code here
th=min(i+1,k)
need=(a[i]+th-1)//th#equals ceil()
s+=need*th
cnt+=need
res+=need
if i>=th:
closed[i-th]+=need
print(res)
|
"""
take element as "the tail" will use more less operations,
use variables s,cnt and closed,
to avoid the inner iteration(update neighbor k elements every time),
complexity is O(n).
"""
row=lambda:map(int,input().split())
n,k=row()
a=list(row())
closed=[0]*n
s=cnt=res=0
for i in range(n-1,-1,-1):
# print(i,a[i],'s,cnt:',s,cnt,'closed:',closed)
s-=cnt
cnt-=closed[i]
a[i]-=s
if a[i]<=0:
{{completion}}
th=min(i+1,k)
need=(a[i]+th-1)//th#equals ceil()
s+=need*th
cnt+=need
res+=need
if i>=th:
closed[i-th]+=need
print(res)
|
continue
|
[{"input": "3 3\n5 4 6", "output": ["5"]}, {"input": "6 3\n1 2 3 2 2 3", "output": ["3"]}, {"input": "6 3\n1 2 4 1 2 3", "output": ["3"]}, {"input": "7 3\n50 17 81 25 42 39 96", "output": ["92"]}]
|
block_completion_003447
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given two arrays: an array $$$a$$$ consisting of $$$n$$$ zeros and an array $$$b$$$ consisting of $$$n$$$ integers.You can apply the following operation to the array $$$a$$$ an arbitrary number of times: choose some subsegment of $$$a$$$ of length $$$k$$$ and add the arithmetic progression $$$1, 2, \ldots, k$$$ to this subsegment — i. e. add $$$1$$$ to the first element of the subsegment, $$$2$$$ to the second element, and so on. The chosen subsegment should be inside the borders of the array $$$a$$$ (i.e., if the left border of the chosen subsegment is $$$l$$$, then the condition $$$1 \le l \le l + k - 1 \le n$$$ should be satisfied). Note that the progression added is always $$$1, 2, \ldots, k$$$ but not the $$$k, k - 1, \ldots, 1$$$ or anything else (i.e., the leftmost element of the subsegment always increases by $$$1$$$, the second element always increases by $$$2$$$ and so on).Your task is to find the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Note that the condition $$$a_i \ge b_i$$$ should be satisfied for all elements at once.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 3 \cdot 10^5$$$) — the number of elements in both arrays and the length of the subsegment, respectively. The second line of the input contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i \le 10^{12}$$$), where $$$b_i$$$ is the $$$i$$$-th element of the array $$$b$$$.
Output Specification: Print one integer — the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$.
Notes: NoteConsider the first example. In this test, we don't really have any choice, so we need to add at least five progressions to make the first element equals $$$5$$$. The array $$$a$$$ becomes $$$[5, 10, 15]$$$.Consider the second example. In this test, let's add one progression on the segment $$$[1; 3]$$$ and two progressions on the segment $$$[4; 6]$$$. Then, the array $$$a$$$ becomes $$$[1, 2, 3, 2, 4, 6]$$$.
Code:
"""
take element as "the tail" will use more less operations,
use variables s,cnt and closed,
to avoid the inner iteration(update neighbor k elements every time),
complexity is O(n).
"""
row=lambda:map(int,input().split())
n,k=row()
a=list(row())
closed=[0]*n
s=cnt=res=0
for i in range(n-1,-1,-1):
# print(i,a[i],'s,cnt:',s,cnt,'closed:',closed)
s-=cnt
cnt-=closed[i]
a[i]-=s
if a[i]<=0:
continue
th=min(i+1,k)
need=(a[i]+th-1)//th#equals ceil()
s+=need*th
cnt+=need
res+=need
if i>=th:
# TODO: Your code here
print(res)
|
"""
take element as "the tail" will use more less operations,
use variables s,cnt and closed,
to avoid the inner iteration(update neighbor k elements every time),
complexity is O(n).
"""
row=lambda:map(int,input().split())
n,k=row()
a=list(row())
closed=[0]*n
s=cnt=res=0
for i in range(n-1,-1,-1):
# print(i,a[i],'s,cnt:',s,cnt,'closed:',closed)
s-=cnt
cnt-=closed[i]
a[i]-=s
if a[i]<=0:
continue
th=min(i+1,k)
need=(a[i]+th-1)//th#equals ceil()
s+=need*th
cnt+=need
res+=need
if i>=th:
{{completion}}
print(res)
|
closed[i-th]+=need
|
[{"input": "3 3\n5 4 6", "output": ["5"]}, {"input": "6 3\n1 2 3 2 2 3", "output": ["3"]}, {"input": "6 3\n1 2 4 1 2 3", "output": ["3"]}, {"input": "7 3\n50 17 81 25 42 39 96", "output": ["92"]}]
|
block_completion_003448
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given two arrays: an array $$$a$$$ consisting of $$$n$$$ zeros and an array $$$b$$$ consisting of $$$n$$$ integers.You can apply the following operation to the array $$$a$$$ an arbitrary number of times: choose some subsegment of $$$a$$$ of length $$$k$$$ and add the arithmetic progression $$$1, 2, \ldots, k$$$ to this subsegment — i. e. add $$$1$$$ to the first element of the subsegment, $$$2$$$ to the second element, and so on. The chosen subsegment should be inside the borders of the array $$$a$$$ (i.e., if the left border of the chosen subsegment is $$$l$$$, then the condition $$$1 \le l \le l + k - 1 \le n$$$ should be satisfied). Note that the progression added is always $$$1, 2, \ldots, k$$$ but not the $$$k, k - 1, \ldots, 1$$$ or anything else (i.e., the leftmost element of the subsegment always increases by $$$1$$$, the second element always increases by $$$2$$$ and so on).Your task is to find the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Note that the condition $$$a_i \ge b_i$$$ should be satisfied for all elements at once.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 3 \cdot 10^5$$$) — the number of elements in both arrays and the length of the subsegment, respectively. The second line of the input contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i \le 10^{12}$$$), where $$$b_i$$$ is the $$$i$$$-th element of the array $$$b$$$.
Output Specification: Print one integer — the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$.
Notes: NoteConsider the first example. In this test, we don't really have any choice, so we need to add at least five progressions to make the first element equals $$$5$$$. The array $$$a$$$ becomes $$$[5, 10, 15]$$$.Consider the second example. In this test, let's add one progression on the segment $$$[1; 3]$$$ and two progressions on the segment $$$[4; 6]$$$. Then, the array $$$a$$$ becomes $$$[1, 2, 3, 2, 4, 6]$$$.
Code:
import math
n, k = list(map(int, input().split(' ')))
nums = list(map(int, input().split(' ')))
ans = res = tot = minus = 0
pre = []
prefix = []
for i in range(n)[::-1]:
if i < n - 1:
# TODO: Your code here
nums[i] -= minus
cur = max(0, math.ceil(nums[i] / k))
ans += (cur if i >= k else 0)
pre.append(cur if i >= k else 0)
tot += (cur if i >= k else 0)
if len(pre) > k:
tot -= pre[- k - 1]
prefix.append(tot)
for i in range(k):
res = max(res, math.ceil(nums[i] / (i + 1)))
print(ans + res)
|
import math
n, k = list(map(int, input().split(' ')))
nums = list(map(int, input().split(' ')))
ans = res = tot = minus = 0
pre = []
prefix = []
for i in range(n)[::-1]:
if i < n - 1:
{{completion}}
nums[i] -= minus
cur = max(0, math.ceil(nums[i] / k))
ans += (cur if i >= k else 0)
pre.append(cur if i >= k else 0)
tot += (cur if i >= k else 0)
if len(pre) > k:
tot -= pre[- k - 1]
prefix.append(tot)
for i in range(k):
res = max(res, math.ceil(nums[i] / (i + 1)))
print(ans + res)
|
minus += k * pre[-1] - prefix[-1]
|
[{"input": "3 3\n5 4 6", "output": ["5"]}, {"input": "6 3\n1 2 3 2 2 3", "output": ["3"]}, {"input": "6 3\n1 2 4 1 2 3", "output": ["3"]}, {"input": "7 3\n50 17 81 25 42 39 96", "output": ["92"]}]
|
block_completion_003449
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given two arrays: an array $$$a$$$ consisting of $$$n$$$ zeros and an array $$$b$$$ consisting of $$$n$$$ integers.You can apply the following operation to the array $$$a$$$ an arbitrary number of times: choose some subsegment of $$$a$$$ of length $$$k$$$ and add the arithmetic progression $$$1, 2, \ldots, k$$$ to this subsegment — i. e. add $$$1$$$ to the first element of the subsegment, $$$2$$$ to the second element, and so on. The chosen subsegment should be inside the borders of the array $$$a$$$ (i.e., if the left border of the chosen subsegment is $$$l$$$, then the condition $$$1 \le l \le l + k - 1 \le n$$$ should be satisfied). Note that the progression added is always $$$1, 2, \ldots, k$$$ but not the $$$k, k - 1, \ldots, 1$$$ or anything else (i.e., the leftmost element of the subsegment always increases by $$$1$$$, the second element always increases by $$$2$$$ and so on).Your task is to find the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Note that the condition $$$a_i \ge b_i$$$ should be satisfied for all elements at once.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 3 \cdot 10^5$$$) — the number of elements in both arrays and the length of the subsegment, respectively. The second line of the input contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i \le 10^{12}$$$), where $$$b_i$$$ is the $$$i$$$-th element of the array $$$b$$$.
Output Specification: Print one integer — the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$.
Notes: NoteConsider the first example. In this test, we don't really have any choice, so we need to add at least five progressions to make the first element equals $$$5$$$. The array $$$a$$$ becomes $$$[5, 10, 15]$$$.Consider the second example. In this test, let's add one progression on the segment $$$[1; 3]$$$ and two progressions on the segment $$$[4; 6]$$$. Then, the array $$$a$$$ becomes $$$[1, 2, 3, 2, 4, 6]$$$.
Code:
import math
n, k = list(map(int, input().split(' ')))
nums = list(map(int, input().split(' ')))
ans = res = tot = minus = 0
pre = []
prefix = []
for i in range(n)[::-1]:
if i < n - 1:
minus += k * pre[-1] - prefix[-1]
nums[i] -= minus
cur = max(0, math.ceil(nums[i] / k))
ans += (cur if i >= k else 0)
pre.append(cur if i >= k else 0)
tot += (cur if i >= k else 0)
if len(pre) > k:
# TODO: Your code here
prefix.append(tot)
for i in range(k):
res = max(res, math.ceil(nums[i] / (i + 1)))
print(ans + res)
|
import math
n, k = list(map(int, input().split(' ')))
nums = list(map(int, input().split(' ')))
ans = res = tot = minus = 0
pre = []
prefix = []
for i in range(n)[::-1]:
if i < n - 1:
minus += k * pre[-1] - prefix[-1]
nums[i] -= minus
cur = max(0, math.ceil(nums[i] / k))
ans += (cur if i >= k else 0)
pre.append(cur if i >= k else 0)
tot += (cur if i >= k else 0)
if len(pre) > k:
{{completion}}
prefix.append(tot)
for i in range(k):
res = max(res, math.ceil(nums[i] / (i + 1)))
print(ans + res)
|
tot -= pre[- k - 1]
|
[{"input": "3 3\n5 4 6", "output": ["5"]}, {"input": "6 3\n1 2 3 2 2 3", "output": ["3"]}, {"input": "6 3\n1 2 4 1 2 3", "output": ["3"]}, {"input": "7 3\n50 17 81 25 42 39 96", "output": ["92"]}]
|
block_completion_003450
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n+1$$$ teleporters on a straight line, located in points $$$0$$$, $$$a_1$$$, $$$a_2$$$, $$$a_3$$$, ..., $$$a_n$$$. It's possible to teleport from point $$$x$$$ to point $$$y$$$ if there are teleporters in both of those points, and it costs $$$(x-y)^2$$$ energy.You want to install some additional teleporters so that it is possible to get from the point $$$0$$$ to the point $$$a_n$$$ (possibly through some other teleporters) spending no more than $$$m$$$ energy in total. Each teleporter you install must be located in an integer point.What is the minimum number of teleporters you have to install?
Input Specification: The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_1 < a_2 < a_3 < \dots < a_n \le 10^9$$$). The third line contains one integer $$$m$$$ ($$$a_n \le m \le 10^{18}$$$).
Output Specification: Print one integer — the minimum number of teleporters you have to install so that it is possible to get from $$$0$$$ to $$$a_n$$$ spending at most $$$m$$$ energy. It can be shown that it's always possible under the constraints from the input format.
Code:
I=lambda:[*map(int,input().split())]
def F(g,i):s=g//i;b=g%i;return b*(s+1)**2+(i-b)*s**2
def f(g,c):
if c>g**2//2:return 0,g**2
s=0;b=g
while b-s>1:
m=(b+s)//2
if F(g,m)-F(g,m+1)>=c:s=m
else:# TODO: Your code here
return s,F(g,b)
n,=I()
a=[0]+I()
m,=I()
G=[a[i+1]-a[i]for i in range(n)]
s=2
b=1<<59
while b-s>1:
C=0;M=(b+s)//2
for g in G:a,c=f(g,M);C+=c
if C>m:b=M
else:s=M
A=C=0
for g in G:a,c=f(g,b);A+=a;C+=c
print(A+max(0,(C-m-1)//s+1))
|
I=lambda:[*map(int,input().split())]
def F(g,i):s=g//i;b=g%i;return b*(s+1)**2+(i-b)*s**2
def f(g,c):
if c>g**2//2:return 0,g**2
s=0;b=g
while b-s>1:
m=(b+s)//2
if F(g,m)-F(g,m+1)>=c:s=m
else:{{completion}}
return s,F(g,b)
n,=I()
a=[0]+I()
m,=I()
G=[a[i+1]-a[i]for i in range(n)]
s=2
b=1<<59
while b-s>1:
C=0;M=(b+s)//2
for g in G:a,c=f(g,M);C+=c
if C>m:b=M
else:s=M
A=C=0
for g in G:a,c=f(g,b);A+=a;C+=c
print(A+max(0,(C-m-1)//s+1))
|
b=m
|
[{"input": "2\n1 5\n7", "output": ["2"]}, {"input": "2\n1 5\n6", "output": ["3"]}, {"input": "1\n5\n5", "output": ["4"]}, {"input": "1\n1000000000\n1000000043", "output": ["999999978"]}]
|
block_completion_003461
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n+1$$$ teleporters on a straight line, located in points $$$0$$$, $$$a_1$$$, $$$a_2$$$, $$$a_3$$$, ..., $$$a_n$$$. It's possible to teleport from point $$$x$$$ to point $$$y$$$ if there are teleporters in both of those points, and it costs $$$(x-y)^2$$$ energy.You want to install some additional teleporters so that it is possible to get from the point $$$0$$$ to the point $$$a_n$$$ (possibly through some other teleporters) spending no more than $$$m$$$ energy in total. Each teleporter you install must be located in an integer point.What is the minimum number of teleporters you have to install?
Input Specification: The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_1 < a_2 < a_3 < \dots < a_n \le 10^9$$$). The third line contains one integer $$$m$$$ ($$$a_n \le m \le 10^{18}$$$).
Output Specification: Print one integer — the minimum number of teleporters you have to install so that it is possible to get from $$$0$$$ to $$$a_n$$$ spending at most $$$m$$$ energy. It can be shown that it's always possible under the constraints from the input format.
Code:
from collections import Counter
n, a, m = int(input()), [*map(int, input().split())], int(input())
def energy(l, t):
x,y = divmod(l, t+1)
return x*x*(t+1-y)+(x+1)*(x+1)*y
def getdiff(l, diff):
lo, hi = 0, l
while lo < hi:
mid = lo + hi >> 1
if energy(l, mid) - energy(l, mid+1) < diff: hi = mid
else: # TODO: Your code here
return lo, energy(l, lo)
def getsum(d, c=0): a1,a2 = zip(*[getdiff(i, d) for i,_ in a]); return (0 if c else sum(a1[i]*x for i, (_, x) in enumerate(a)), sum(a2[i]*x for i, (_, x) in enumerate(a)))
a = [0] + a
a = Counter([a[i+1]-a[i] for i in range(n)]).items()
lo, hi = 1, m
while lo < hi:
mid = lo + hi >> 1
if getsum(mid, 1)[1] > m: hi = mid
else: lo = mid + 1
lo-=1
a1, a2 = getsum(lo)
print(a1-(m-a2)//lo if lo else a1)
|
from collections import Counter
n, a, m = int(input()), [*map(int, input().split())], int(input())
def energy(l, t):
x,y = divmod(l, t+1)
return x*x*(t+1-y)+(x+1)*(x+1)*y
def getdiff(l, diff):
lo, hi = 0, l
while lo < hi:
mid = lo + hi >> 1
if energy(l, mid) - energy(l, mid+1) < diff: hi = mid
else: {{completion}}
return lo, energy(l, lo)
def getsum(d, c=0): a1,a2 = zip(*[getdiff(i, d) for i,_ in a]); return (0 if c else sum(a1[i]*x for i, (_, x) in enumerate(a)), sum(a2[i]*x for i, (_, x) in enumerate(a)))
a = [0] + a
a = Counter([a[i+1]-a[i] for i in range(n)]).items()
lo, hi = 1, m
while lo < hi:
mid = lo + hi >> 1
if getsum(mid, 1)[1] > m: hi = mid
else: lo = mid + 1
lo-=1
a1, a2 = getsum(lo)
print(a1-(m-a2)//lo if lo else a1)
|
lo = mid + 1
|
[{"input": "2\n1 5\n7", "output": ["2"]}, {"input": "2\n1 5\n6", "output": ["3"]}, {"input": "1\n5\n5", "output": ["4"]}, {"input": "1\n1000000000\n1000000043", "output": ["999999978"]}]
|
block_completion_003462
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n+1$$$ teleporters on a straight line, located in points $$$0$$$, $$$a_1$$$, $$$a_2$$$, $$$a_3$$$, ..., $$$a_n$$$. It's possible to teleport from point $$$x$$$ to point $$$y$$$ if there are teleporters in both of those points, and it costs $$$(x-y)^2$$$ energy.You want to install some additional teleporters so that it is possible to get from the point $$$0$$$ to the point $$$a_n$$$ (possibly through some other teleporters) spending no more than $$$m$$$ energy in total. Each teleporter you install must be located in an integer point.What is the minimum number of teleporters you have to install?
Input Specification: The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_1 < a_2 < a_3 < \dots < a_n \le 10^9$$$). The third line contains one integer $$$m$$$ ($$$a_n \le m \le 10^{18}$$$).
Output Specification: Print one integer — the minimum number of teleporters you have to install so that it is possible to get from $$$0$$$ to $$$a_n$$$ spending at most $$$m$$$ energy. It can be shown that it's always possible under the constraints from the input format.
Code:
# import io,os
# read = io.BytesIO(os.read(0, os.fstat(0).st_size))
# I = lambda: [*map(int, read.readline().split())]
import sys
I=lambda:[*map(int,sys.stdin.readline().split())]
def ff(gap, ints):
sml = gap // ints
bigcount = gap % ints
return bigcount * (sml + 1) ** 2 + (ints - bigcount) * sml ** 2
def f(gap, c):
if c > gap ** 2 // 2:
return 0, gap ** 2
sml = 0
big = gap
while big - sml > 1:
mid = (big + sml) // 2
a = ff(gap, mid)
b = ff(gap, mid + 1)
if a - b >= c:
sml = mid
else:
# TODO: Your code here
return sml, ff(gap, big)
n, = I()
a = I()
m, = I()
gaps = [a[0]]
for i in range(n - 1):
gaps.append(a[i + 1] - a[i])
sml = 2
big = 1 << 59 + 2
while big - sml > 1:
cost = 0
mid = (big + sml) // 2
for g in gaps:
a, c = f(g, mid)
cost += c
if cost > m:
big = mid
else:
sml = mid
abig = 0
cbig = 0
for g in gaps:
a, c = f(g, big)
abig += a
cbig += c
print(abig + max(0, (cbig - m - 1) // sml + 1))
|
# import io,os
# read = io.BytesIO(os.read(0, os.fstat(0).st_size))
# I = lambda: [*map(int, read.readline().split())]
import sys
I=lambda:[*map(int,sys.stdin.readline().split())]
def ff(gap, ints):
sml = gap // ints
bigcount = gap % ints
return bigcount * (sml + 1) ** 2 + (ints - bigcount) * sml ** 2
def f(gap, c):
if c > gap ** 2 // 2:
return 0, gap ** 2
sml = 0
big = gap
while big - sml > 1:
mid = (big + sml) // 2
a = ff(gap, mid)
b = ff(gap, mid + 1)
if a - b >= c:
sml = mid
else:
{{completion}}
return sml, ff(gap, big)
n, = I()
a = I()
m, = I()
gaps = [a[0]]
for i in range(n - 1):
gaps.append(a[i + 1] - a[i])
sml = 2
big = 1 << 59 + 2
while big - sml > 1:
cost = 0
mid = (big + sml) // 2
for g in gaps:
a, c = f(g, mid)
cost += c
if cost > m:
big = mid
else:
sml = mid
abig = 0
cbig = 0
for g in gaps:
a, c = f(g, big)
abig += a
cbig += c
print(abig + max(0, (cbig - m - 1) // sml + 1))
|
big = mid
|
[{"input": "2\n1 5\n7", "output": ["2"]}, {"input": "2\n1 5\n6", "output": ["3"]}, {"input": "1\n5\n5", "output": ["4"]}, {"input": "1\n1000000000\n1000000043", "output": ["999999978"]}]
|
block_completion_003463
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Nastya baked $$$m$$$ pancakes and spread them on $$$n$$$ dishes. The dishes are in a row and numbered from left to right. She put $$$a_i$$$ pancakes on the dish with the index $$$i$$$.Seeing the dishes, Vlad decided to bring order to the stacks and move some pancakes. In one move, he can shift one pancake from any dish to the closest one, that is, select the dish $$$i$$$ ($$$a_i > 0$$$) and do one of the following: if $$$i > 1$$$, put the pancake on a dish with the previous index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i - 1} = a_{i - 1} + 1$$$; if $$$i < n$$$, put the pancake on a dish with the following index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i + 1} = a_{i + 1} + 1$$$.Vlad wants to make the array $$$a$$$non-increasing, after moving as few pancakes as possible. Help him find the minimum number of moves needed for this.The array $$$a=[a_1, a_2,\dots,a_n]$$$ is called non-increasing if $$$a_i \ge a_{i+1}$$$ for all $$$i$$$ from $$$1$$$ to $$$n-1$$$.
Input Specification: The first line of the input contains two numbers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 250$$$) — the number of dishes and the number of pancakes, respectively. The second line contains $$$n$$$ numbers $$$a_i$$$ ($$$0 \le a_i \le m$$$), the sum of all $$$a_i$$$ is equal to $$$m$$$.
Output Specification: Print a single number: the minimum number of moves required to make the array $$$a$$$ non-increasing.
Notes: NoteIn the first example, you first need to move the pancake from the dish $$$1$$$, after which $$$a = [4, 4, 2, 3, 3, 3]$$$. After that, you need to move the pancake from the dish $$$2$$$ to the dish $$$3$$$ and the array will become non-growing $$$a = [4, 3, 3, 3, 3, 3]$$$.
Code:
from functools import cache
from math import inf
from itertools import accumulate
def solve(A, m):
n = len(A)
A.reverse()
@cache
def dp(i, val, balance):
if abs(balance) > m:
# TODO: Your code here
if (n - i) * val > m:
return inf
if i == n:
return 0 if balance == 0 else inf
curr = A[i] + balance
take = abs(curr - val) + dp(i + 1, val, curr - val)
skip = dp(i, val + 1, balance)
return min(take, skip)
return dp(0, 0, 0)
n, m = map(int, input().split())
A = list(map(int, input().split()))
print(solve(A, m))
|
from functools import cache
from math import inf
from itertools import accumulate
def solve(A, m):
n = len(A)
A.reverse()
@cache
def dp(i, val, balance):
if abs(balance) > m:
{{completion}}
if (n - i) * val > m:
return inf
if i == n:
return 0 if balance == 0 else inf
curr = A[i] + balance
take = abs(curr - val) + dp(i + 1, val, curr - val)
skip = dp(i, val + 1, balance)
return min(take, skip)
return dp(0, 0, 0)
n, m = map(int, input().split())
A = list(map(int, input().split()))
print(solve(A, m))
|
return inf
|
[{"input": "6 19\n5 3 2 3 3 3", "output": ["2"]}, {"input": "3 6\n3 2 1", "output": ["0"]}, {"input": "3 6\n2 1 3", "output": ["1"]}, {"input": "6 19\n3 4 3 3 5 1", "output": ["3"]}, {"input": "10 1\n0 0 0 0 0 0 0 0 0 1", "output": ["9"]}]
|
block_completion_003581
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Nastya baked $$$m$$$ pancakes and spread them on $$$n$$$ dishes. The dishes are in a row and numbered from left to right. She put $$$a_i$$$ pancakes on the dish with the index $$$i$$$.Seeing the dishes, Vlad decided to bring order to the stacks and move some pancakes. In one move, he can shift one pancake from any dish to the closest one, that is, select the dish $$$i$$$ ($$$a_i > 0$$$) and do one of the following: if $$$i > 1$$$, put the pancake on a dish with the previous index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i - 1} = a_{i - 1} + 1$$$; if $$$i < n$$$, put the pancake on a dish with the following index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i + 1} = a_{i + 1} + 1$$$.Vlad wants to make the array $$$a$$$non-increasing, after moving as few pancakes as possible. Help him find the minimum number of moves needed for this.The array $$$a=[a_1, a_2,\dots,a_n]$$$ is called non-increasing if $$$a_i \ge a_{i+1}$$$ for all $$$i$$$ from $$$1$$$ to $$$n-1$$$.
Input Specification: The first line of the input contains two numbers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 250$$$) — the number of dishes and the number of pancakes, respectively. The second line contains $$$n$$$ numbers $$$a_i$$$ ($$$0 \le a_i \le m$$$), the sum of all $$$a_i$$$ is equal to $$$m$$$.
Output Specification: Print a single number: the minimum number of moves required to make the array $$$a$$$ non-increasing.
Notes: NoteIn the first example, you first need to move the pancake from the dish $$$1$$$, after which $$$a = [4, 4, 2, 3, 3, 3]$$$. After that, you need to move the pancake from the dish $$$2$$$ to the dish $$$3$$$ and the array will become non-growing $$$a = [4, 3, 3, 3, 3, 3]$$$.
Code:
from functools import cache
from math import inf
from itertools import accumulate
def solve(A, m):
n = len(A)
A.reverse()
@cache
def dp(i, val, balance):
if abs(balance) > m:
return inf
if (n - i) * val > m:
# TODO: Your code here
if i == n:
return 0 if balance == 0 else inf
curr = A[i] + balance
take = abs(curr - val) + dp(i + 1, val, curr - val)
skip = dp(i, val + 1, balance)
return min(take, skip)
return dp(0, 0, 0)
n, m = map(int, input().split())
A = list(map(int, input().split()))
print(solve(A, m))
|
from functools import cache
from math import inf
from itertools import accumulate
def solve(A, m):
n = len(A)
A.reverse()
@cache
def dp(i, val, balance):
if abs(balance) > m:
return inf
if (n - i) * val > m:
{{completion}}
if i == n:
return 0 if balance == 0 else inf
curr = A[i] + balance
take = abs(curr - val) + dp(i + 1, val, curr - val)
skip = dp(i, val + 1, balance)
return min(take, skip)
return dp(0, 0, 0)
n, m = map(int, input().split())
A = list(map(int, input().split()))
print(solve(A, m))
|
return inf
|
[{"input": "6 19\n5 3 2 3 3 3", "output": ["2"]}, {"input": "3 6\n3 2 1", "output": ["0"]}, {"input": "3 6\n2 1 3", "output": ["1"]}, {"input": "6 19\n3 4 3 3 5 1", "output": ["3"]}, {"input": "10 1\n0 0 0 0 0 0 0 0 0 1", "output": ["9"]}]
|
block_completion_003582
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Nastya baked $$$m$$$ pancakes and spread them on $$$n$$$ dishes. The dishes are in a row and numbered from left to right. She put $$$a_i$$$ pancakes on the dish with the index $$$i$$$.Seeing the dishes, Vlad decided to bring order to the stacks and move some pancakes. In one move, he can shift one pancake from any dish to the closest one, that is, select the dish $$$i$$$ ($$$a_i > 0$$$) and do one of the following: if $$$i > 1$$$, put the pancake on a dish with the previous index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i - 1} = a_{i - 1} + 1$$$; if $$$i < n$$$, put the pancake on a dish with the following index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i + 1} = a_{i + 1} + 1$$$.Vlad wants to make the array $$$a$$$non-increasing, after moving as few pancakes as possible. Help him find the minimum number of moves needed for this.The array $$$a=[a_1, a_2,\dots,a_n]$$$ is called non-increasing if $$$a_i \ge a_{i+1}$$$ for all $$$i$$$ from $$$1$$$ to $$$n-1$$$.
Input Specification: The first line of the input contains two numbers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 250$$$) — the number of dishes and the number of pancakes, respectively. The second line contains $$$n$$$ numbers $$$a_i$$$ ($$$0 \le a_i \le m$$$), the sum of all $$$a_i$$$ is equal to $$$m$$$.
Output Specification: Print a single number: the minimum number of moves required to make the array $$$a$$$ non-increasing.
Notes: NoteIn the first example, you first need to move the pancake from the dish $$$1$$$, after which $$$a = [4, 4, 2, 3, 3, 3]$$$. After that, you need to move the pancake from the dish $$$2$$$ to the dish $$$3$$$ and the array will become non-growing $$$a = [4, 3, 3, 3, 3, 3]$$$.
Code:
from itertools import accumulate
I=lambda:map(int,input().split())
n,m=I()
a,inf=[*I()],float('inf')
pre=[0]+[*accumulate(a)]
dp=[[inf]*(m+1) for _ in range(m+1)]
dp[m][0]=0
for i in range(n):
cur=[[inf]*(m+1) for _ in range(m+1)]
for lst in reversed(range(m+1)):
for sums in range(m+1):
if lst<m:
# TODO: Your code here
if sums+lst<=m:
cur[lst][sums+lst]=min(cur[lst][sums+lst], dp[lst][sums]+abs(pre[i+1]-(sums+lst)))
dp,cur=cur,dp
print(min(dp[lst][m] for lst in range(m+1)))
|
from itertools import accumulate
I=lambda:map(int,input().split())
n,m=I()
a,inf=[*I()],float('inf')
pre=[0]+[*accumulate(a)]
dp=[[inf]*(m+1) for _ in range(m+1)]
dp[m][0]=0
for i in range(n):
cur=[[inf]*(m+1) for _ in range(m+1)]
for lst in reversed(range(m+1)):
for sums in range(m+1):
if lst<m:
{{completion}}
if sums+lst<=m:
cur[lst][sums+lst]=min(cur[lst][sums+lst], dp[lst][sums]+abs(pre[i+1]-(sums+lst)))
dp,cur=cur,dp
print(min(dp[lst][m] for lst in range(m+1)))
|
dp[lst][sums]=min(dp[lst][sums],dp[lst+1][sums])
|
[{"input": "6 19\n5 3 2 3 3 3", "output": ["2"]}, {"input": "3 6\n3 2 1", "output": ["0"]}, {"input": "3 6\n2 1 3", "output": ["1"]}, {"input": "6 19\n3 4 3 3 5 1", "output": ["3"]}, {"input": "10 1\n0 0 0 0 0 0 0 0 0 1", "output": ["9"]}]
|
block_completion_003583
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Nastya baked $$$m$$$ pancakes and spread them on $$$n$$$ dishes. The dishes are in a row and numbered from left to right. She put $$$a_i$$$ pancakes on the dish with the index $$$i$$$.Seeing the dishes, Vlad decided to bring order to the stacks and move some pancakes. In one move, he can shift one pancake from any dish to the closest one, that is, select the dish $$$i$$$ ($$$a_i > 0$$$) and do one of the following: if $$$i > 1$$$, put the pancake on a dish with the previous index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i - 1} = a_{i - 1} + 1$$$; if $$$i < n$$$, put the pancake on a dish with the following index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i + 1} = a_{i + 1} + 1$$$.Vlad wants to make the array $$$a$$$non-increasing, after moving as few pancakes as possible. Help him find the minimum number of moves needed for this.The array $$$a=[a_1, a_2,\dots,a_n]$$$ is called non-increasing if $$$a_i \ge a_{i+1}$$$ for all $$$i$$$ from $$$1$$$ to $$$n-1$$$.
Input Specification: The first line of the input contains two numbers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 250$$$) — the number of dishes and the number of pancakes, respectively. The second line contains $$$n$$$ numbers $$$a_i$$$ ($$$0 \le a_i \le m$$$), the sum of all $$$a_i$$$ is equal to $$$m$$$.
Output Specification: Print a single number: the minimum number of moves required to make the array $$$a$$$ non-increasing.
Notes: NoteIn the first example, you first need to move the pancake from the dish $$$1$$$, after which $$$a = [4, 4, 2, 3, 3, 3]$$$. After that, you need to move the pancake from the dish $$$2$$$ to the dish $$$3$$$ and the array will become non-growing $$$a = [4, 3, 3, 3, 3, 3]$$$.
Code:
from itertools import accumulate
I=lambda:map(int,input().split())
n,m=I()
a,inf=[*I()],float('inf')
pre=[0]+[*accumulate(a)]
dp=[[inf]*(m+1) for _ in range(m+1)]
dp[m][0]=0
for i in range(n):
cur=[[inf]*(m+1) for _ in range(m+1)]
for lst in reversed(range(m+1)):
for sums in range(m+1):
if lst<m:
dp[lst][sums]=min(dp[lst][sums],dp[lst+1][sums])
if sums+lst<=m:
# TODO: Your code here
dp,cur=cur,dp
print(min(dp[lst][m] for lst in range(m+1)))
|
from itertools import accumulate
I=lambda:map(int,input().split())
n,m=I()
a,inf=[*I()],float('inf')
pre=[0]+[*accumulate(a)]
dp=[[inf]*(m+1) for _ in range(m+1)]
dp[m][0]=0
for i in range(n):
cur=[[inf]*(m+1) for _ in range(m+1)]
for lst in reversed(range(m+1)):
for sums in range(m+1):
if lst<m:
dp[lst][sums]=min(dp[lst][sums],dp[lst+1][sums])
if sums+lst<=m:
{{completion}}
dp,cur=cur,dp
print(min(dp[lst][m] for lst in range(m+1)))
|
cur[lst][sums+lst]=min(cur[lst][sums+lst], dp[lst][sums]+abs(pre[i+1]-(sums+lst)))
|
[{"input": "6 19\n5 3 2 3 3 3", "output": ["2"]}, {"input": "3 6\n3 2 1", "output": ["0"]}, {"input": "3 6\n2 1 3", "output": ["1"]}, {"input": "6 19\n3 4 3 3 5 1", "output": ["3"]}, {"input": "10 1\n0 0 0 0 0 0 0 0 0 1", "output": ["9"]}]
|
block_completion_003584
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Nastya baked $$$m$$$ pancakes and spread them on $$$n$$$ dishes. The dishes are in a row and numbered from left to right. She put $$$a_i$$$ pancakes on the dish with the index $$$i$$$.Seeing the dishes, Vlad decided to bring order to the stacks and move some pancakes. In one move, he can shift one pancake from any dish to the closest one, that is, select the dish $$$i$$$ ($$$a_i > 0$$$) and do one of the following: if $$$i > 1$$$, put the pancake on a dish with the previous index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i - 1} = a_{i - 1} + 1$$$; if $$$i < n$$$, put the pancake on a dish with the following index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i + 1} = a_{i + 1} + 1$$$.Vlad wants to make the array $$$a$$$non-increasing, after moving as few pancakes as possible. Help him find the minimum number of moves needed for this.The array $$$a=[a_1, a_2,\dots,a_n]$$$ is called non-increasing if $$$a_i \ge a_{i+1}$$$ for all $$$i$$$ from $$$1$$$ to $$$n-1$$$.
Input Specification: The first line of the input contains two numbers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 250$$$) — the number of dishes and the number of pancakes, respectively. The second line contains $$$n$$$ numbers $$$a_i$$$ ($$$0 \le a_i \le m$$$), the sum of all $$$a_i$$$ is equal to $$$m$$$.
Output Specification: Print a single number: the minimum number of moves required to make the array $$$a$$$ non-increasing.
Notes: NoteIn the first example, you first need to move the pancake from the dish $$$1$$$, after which $$$a = [4, 4, 2, 3, 3, 3]$$$. After that, you need to move the pancake from the dish $$$2$$$ to the dish $$$3$$$ and the array will become non-growing $$$a = [4, 3, 3, 3, 3, 3]$$$.
Code:
n,m=map(int,input().split())
a=list(map(int,input().split()))[::-1]
id=[]
for i in range(n):
for _ in range(a[i]):
id.append(i)
inf=10**5
dp=[[inf]*(m+1) for i in range(m+1)]
dp[0][0]=0
for i in range(n):
cost=[]
for j in id:
cost.append(abs(i-j))
cum=[0]
tmp=0
for j in cost:
tmp+=j
cum.append(tmp)
dp_new=[[inf]*(m+1) for i in range(m+1)]
for j in range(m+1):
mx=(m-j)//(n-i)
for k in range(mx+1):
if dp[j][k]==inf:# TODO: Your code here
#print(i,j,k,mx)
for l in range(k,mx+1):
#print(l)
c=cum[l+j]-cum[j]
dp_new[j+l][l]=min(dp_new[j+l][l],dp[j][k]+c)
dp=dp_new
print(min(dp[-1]))
|
n,m=map(int,input().split())
a=list(map(int,input().split()))[::-1]
id=[]
for i in range(n):
for _ in range(a[i]):
id.append(i)
inf=10**5
dp=[[inf]*(m+1) for i in range(m+1)]
dp[0][0]=0
for i in range(n):
cost=[]
for j in id:
cost.append(abs(i-j))
cum=[0]
tmp=0
for j in cost:
tmp+=j
cum.append(tmp)
dp_new=[[inf]*(m+1) for i in range(m+1)]
for j in range(m+1):
mx=(m-j)//(n-i)
for k in range(mx+1):
if dp[j][k]==inf:{{completion}}
#print(i,j,k,mx)
for l in range(k,mx+1):
#print(l)
c=cum[l+j]-cum[j]
dp_new[j+l][l]=min(dp_new[j+l][l],dp[j][k]+c)
dp=dp_new
print(min(dp[-1]))
|
continue
|
[{"input": "6 19\n5 3 2 3 3 3", "output": ["2"]}, {"input": "3 6\n3 2 1", "output": ["0"]}, {"input": "3 6\n2 1 3", "output": ["1"]}, {"input": "6 19\n3 4 3 3 5 1", "output": ["3"]}, {"input": "10 1\n0 0 0 0 0 0 0 0 0 1", "output": ["9"]}]
|
block_completion_003585
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: Nastya baked $$$m$$$ pancakes and spread them on $$$n$$$ dishes. The dishes are in a row and numbered from left to right. She put $$$a_i$$$ pancakes on the dish with the index $$$i$$$.Seeing the dishes, Vlad decided to bring order to the stacks and move some pancakes. In one move, he can shift one pancake from any dish to the closest one, that is, select the dish $$$i$$$ ($$$a_i > 0$$$) and do one of the following: if $$$i > 1$$$, put the pancake on a dish with the previous index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i - 1} = a_{i - 1} + 1$$$; if $$$i < n$$$, put the pancake on a dish with the following index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i + 1} = a_{i + 1} + 1$$$.Vlad wants to make the array $$$a$$$non-increasing, after moving as few pancakes as possible. Help him find the minimum number of moves needed for this.The array $$$a=[a_1, a_2,\dots,a_n]$$$ is called non-increasing if $$$a_i \ge a_{i+1}$$$ for all $$$i$$$ from $$$1$$$ to $$$n-1$$$.
Input Specification: The first line of the input contains two numbers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 250$$$) — the number of dishes and the number of pancakes, respectively. The second line contains $$$n$$$ numbers $$$a_i$$$ ($$$0 \le a_i \le m$$$), the sum of all $$$a_i$$$ is equal to $$$m$$$.
Output Specification: Print a single number: the minimum number of moves required to make the array $$$a$$$ non-increasing.
Notes: NoteIn the first example, you first need to move the pancake from the dish $$$1$$$, after which $$$a = [4, 4, 2, 3, 3, 3]$$$. After that, you need to move the pancake from the dish $$$2$$$ to the dish $$$3$$$ and the array will become non-growing $$$a = [4, 3, 3, 3, 3, 3]$$$.
Code:
n,m=map(int,input().split())
a=list(map(int,input().split()))[::-1]
id=[]
for i in range(n):
for _ in range(a[i]):
id.append(i)
inf=10**5
dp=[[inf]*(m+1) for i in range(m+1)]
dp[0][0]=0
for i in range(n):
cost=[]
for j in id:
cost.append(abs(i-j))
cum=[0]
tmp=0
for j in cost:
tmp+=j
cum.append(tmp)
dp_new=[[inf]*(m+1) for i in range(m+1)]
for j in range(m+1):
mx=(m-j)//(n-i)
for k in range(mx+1):
if dp[j][k]==inf:continue
#print(i,j,k,mx)
for l in range(k,mx+1):
#print(l)
# TODO: Your code here
dp=dp_new
print(min(dp[-1]))
|
n,m=map(int,input().split())
a=list(map(int,input().split()))[::-1]
id=[]
for i in range(n):
for _ in range(a[i]):
id.append(i)
inf=10**5
dp=[[inf]*(m+1) for i in range(m+1)]
dp[0][0]=0
for i in range(n):
cost=[]
for j in id:
cost.append(abs(i-j))
cum=[0]
tmp=0
for j in cost:
tmp+=j
cum.append(tmp)
dp_new=[[inf]*(m+1) for i in range(m+1)]
for j in range(m+1):
mx=(m-j)//(n-i)
for k in range(mx+1):
if dp[j][k]==inf:continue
#print(i,j,k,mx)
for l in range(k,mx+1):
#print(l)
{{completion}}
dp=dp_new
print(min(dp[-1]))
|
c=cum[l+j]-cum[j]
dp_new[j+l][l]=min(dp_new[j+l][l],dp[j][k]+c)
|
[{"input": "6 19\n5 3 2 3 3 3", "output": ["2"]}, {"input": "3 6\n3 2 1", "output": ["0"]}, {"input": "3 6\n2 1 3", "output": ["1"]}, {"input": "6 19\n3 4 3 3 5 1", "output": ["3"]}, {"input": "10 1\n0 0 0 0 0 0 0 0 0 1", "output": ["9"]}]
|
block_completion_003586
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: This is the easy version of the problem. The only difference between the versions is the constraints on $$$n$$$, $$$k$$$, $$$a_i$$$, and the sum of $$$n$$$ over all test cases. You can make hacks only if both versions of the problem are solved.Note the unusual memory limit.You are given an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$, and an integer $$$k$$$.The cost of an array of integers $$$p_1, p_2, \ldots, p_n$$$ of length $$$n$$$ is $$$$$$\max\limits_{1 \le i \le n}\left(\left \lfloor \frac{a_i}{p_i} \right \rfloor \right) - \min\limits_{1 \le i \le n}\left(\left \lfloor \frac{a_i}{p_i} \right \rfloor \right).$$$$$$Here, $$$\lfloor \frac{x}{y} \rfloor$$$ denotes the integer part of the division of $$$x$$$ by $$$y$$$. Find the minimum cost of an array $$$p$$$ such that $$$1 \le p_i \le k$$$ for all $$$1 \le i \le n$$$.
Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n, k \le 3000$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_1 \le a_2 \le \ldots \le a_n \le 3000$$$). It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3000$$$.
Output Specification: For each test case, print a single integer — the minimum possible cost of an array $$$p$$$ satisfying the condition above.
Notes: NoteIn the first test case, the optimal array is $$$p = [1, 1, 1, 2, 2]$$$. The resulting array of values of $$$\lfloor \frac{a_i}{p_i} \rfloor$$$ is $$$[4, 5, 6, 4, 5]$$$. The cost of $$$p$$$ is $$$\max\limits_{1 \le i \le n}(\lfloor \frac{a_i}{p_i} \rfloor) - \min\limits_{1 \le i \le n}(\lfloor \frac{a_i}{p_i} \rfloor) = 6 - 4 = 2$$$. We can show that there is no array (satisfying the condition from the statement) with a smaller cost.In the second test case, one of the optimal arrays is $$$p = [12, 12, 12, 12, 12]$$$, which results in all $$$\lfloor \frac{a_i}{p_i} \rfloor$$$ being $$$0$$$.In the third test case, the only possible array is $$$p = [1, 1, 1]$$$.
Code:
import sys
tokens = ''.join(sys.stdin.readlines()).split()[::-1]
def next(): return tokens.pop()
def nextInt(): return int(next())
def nextFloat(): return float(next())
def getIntArray(n): return [nextInt() for _ in range(n)]
def getFloatArray(n): return [nextFloat() for _ in range(n)]
def getStringArray(n): return [next() for _ in range(n)]
testcase = True
def solve(testcase = 1):
N, K = map(int, getIntArray(2))
A = getIntArray(N)
s = [set() for i in range(3005)]
for i in range(N):
for k in range(1, K + 1):
s[A[i] // k].add(i)
ans = 999999999999999999999999999999999999999999999999999999999999999999999
r = 0
freq = {}
for l in range(len(s)):
while len(freq) < N and r < len(s):
for v in s[r]:
if v not in freq: # TODO: Your code here
freq[v] += 1
r += 1
if len(freq) < N: break
ans = min(ans, r - l - 1)
for v in s[l]:
if freq[v] == 1: del freq[v]
else: freq[v] -= 1
print(ans)
if testcase is None:
testcaseCount = 1
while tokens:
solve(testcaseCount)
testcaseCount += 1
else:
testcaseCount = nextInt() if testcase else 1
for tc in range(testcaseCount):
solve(tc + 1)
assert not tokens
|
import sys
tokens = ''.join(sys.stdin.readlines()).split()[::-1]
def next(): return tokens.pop()
def nextInt(): return int(next())
def nextFloat(): return float(next())
def getIntArray(n): return [nextInt() for _ in range(n)]
def getFloatArray(n): return [nextFloat() for _ in range(n)]
def getStringArray(n): return [next() for _ in range(n)]
testcase = True
def solve(testcase = 1):
N, K = map(int, getIntArray(2))
A = getIntArray(N)
s = [set() for i in range(3005)]
for i in range(N):
for k in range(1, K + 1):
s[A[i] // k].add(i)
ans = 999999999999999999999999999999999999999999999999999999999999999999999
r = 0
freq = {}
for l in range(len(s)):
while len(freq) < N and r < len(s):
for v in s[r]:
if v not in freq: {{completion}}
freq[v] += 1
r += 1
if len(freq) < N: break
ans = min(ans, r - l - 1)
for v in s[l]:
if freq[v] == 1: del freq[v]
else: freq[v] -= 1
print(ans)
if testcase is None:
testcaseCount = 1
while tokens:
solve(testcaseCount)
testcaseCount += 1
else:
testcaseCount = nextInt() if testcase else 1
for tc in range(testcaseCount):
solve(tc + 1)
assert not tokens
|
freq[v] = 0
|
[{"input": "7\n\n5 2\n\n4 5 6 8 11\n\n5 12\n\n4 5 6 8 11\n\n3 1\n\n2 9 15\n\n7 3\n\n2 3 5 5 6 9 10\n\n6 56\n\n54 286 527 1436 2450 2681\n\n3 95\n\n16 340 2241\n\n2 2\n\n1 3", "output": ["2\n0\n13\n1\n4\n7\n0"]}]
|
block_completion_003651
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: This is the easy version of the problem. The only difference between the versions is the constraints on $$$n$$$, $$$k$$$, $$$a_i$$$, and the sum of $$$n$$$ over all test cases. You can make hacks only if both versions of the problem are solved.Note the unusual memory limit.You are given an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$, and an integer $$$k$$$.The cost of an array of integers $$$p_1, p_2, \ldots, p_n$$$ of length $$$n$$$ is $$$$$$\max\limits_{1 \le i \le n}\left(\left \lfloor \frac{a_i}{p_i} \right \rfloor \right) - \min\limits_{1 \le i \le n}\left(\left \lfloor \frac{a_i}{p_i} \right \rfloor \right).$$$$$$Here, $$$\lfloor \frac{x}{y} \rfloor$$$ denotes the integer part of the division of $$$x$$$ by $$$y$$$. Find the minimum cost of an array $$$p$$$ such that $$$1 \le p_i \le k$$$ for all $$$1 \le i \le n$$$.
Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n, k \le 3000$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_1 \le a_2 \le \ldots \le a_n \le 3000$$$). It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3000$$$.
Output Specification: For each test case, print a single integer — the minimum possible cost of an array $$$p$$$ satisfying the condition above.
Notes: NoteIn the first test case, the optimal array is $$$p = [1, 1, 1, 2, 2]$$$. The resulting array of values of $$$\lfloor \frac{a_i}{p_i} \rfloor$$$ is $$$[4, 5, 6, 4, 5]$$$. The cost of $$$p$$$ is $$$\max\limits_{1 \le i \le n}(\lfloor \frac{a_i}{p_i} \rfloor) - \min\limits_{1 \le i \le n}(\lfloor \frac{a_i}{p_i} \rfloor) = 6 - 4 = 2$$$. We can show that there is no array (satisfying the condition from the statement) with a smaller cost.In the second test case, one of the optimal arrays is $$$p = [12, 12, 12, 12, 12]$$$, which results in all $$$\lfloor \frac{a_i}{p_i} \rfloor$$$ being $$$0$$$.In the third test case, the only possible array is $$$p = [1, 1, 1]$$$.
Code:
import sys
tokens = ''.join(sys.stdin.readlines()).split()[::-1]
def next(): return tokens.pop()
def nextInt(): return int(next())
def nextFloat(): return float(next())
def getIntArray(n): return [nextInt() for _ in range(n)]
def getFloatArray(n): return [nextFloat() for _ in range(n)]
def getStringArray(n): return [next() for _ in range(n)]
testcase = True
def solve(testcase = 1):
N, K = map(int, getIntArray(2))
A = getIntArray(N)
s = [set() for i in range(3005)]
for i in range(N):
for k in range(1, K + 1):
s[A[i] // k].add(i)
ans = 999999999999999999999999999999999999999999999999999999999999999999999
r = 0
freq = {}
for l in range(len(s)):
while len(freq) < N and r < len(s):
for v in s[r]:
if v not in freq: freq[v] = 0
freq[v] += 1
r += 1
if len(freq) < N: break
ans = min(ans, r - l - 1)
for v in s[l]:
if freq[v] == 1: del freq[v]
else: # TODO: Your code here
print(ans)
if testcase is None:
testcaseCount = 1
while tokens:
solve(testcaseCount)
testcaseCount += 1
else:
testcaseCount = nextInt() if testcase else 1
for tc in range(testcaseCount):
solve(tc + 1)
assert not tokens
|
import sys
tokens = ''.join(sys.stdin.readlines()).split()[::-1]
def next(): return tokens.pop()
def nextInt(): return int(next())
def nextFloat(): return float(next())
def getIntArray(n): return [nextInt() for _ in range(n)]
def getFloatArray(n): return [nextFloat() for _ in range(n)]
def getStringArray(n): return [next() for _ in range(n)]
testcase = True
def solve(testcase = 1):
N, K = map(int, getIntArray(2))
A = getIntArray(N)
s = [set() for i in range(3005)]
for i in range(N):
for k in range(1, K + 1):
s[A[i] // k].add(i)
ans = 999999999999999999999999999999999999999999999999999999999999999999999
r = 0
freq = {}
for l in range(len(s)):
while len(freq) < N and r < len(s):
for v in s[r]:
if v not in freq: freq[v] = 0
freq[v] += 1
r += 1
if len(freq) < N: break
ans = min(ans, r - l - 1)
for v in s[l]:
if freq[v] == 1: del freq[v]
else: {{completion}}
print(ans)
if testcase is None:
testcaseCount = 1
while tokens:
solve(testcaseCount)
testcaseCount += 1
else:
testcaseCount = nextInt() if testcase else 1
for tc in range(testcaseCount):
solve(tc + 1)
assert not tokens
|
freq[v] -= 1
|
[{"input": "7\n\n5 2\n\n4 5 6 8 11\n\n5 12\n\n4 5 6 8 11\n\n3 1\n\n2 9 15\n\n7 3\n\n2 3 5 5 6 9 10\n\n6 56\n\n54 286 527 1436 2450 2681\n\n3 95\n\n16 340 2241\n\n2 2\n\n1 3", "output": ["2\n0\n13\n1\n4\n7\n0"]}]
|
block_completion_003652
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You have a sequence of $$$n$$$ colored blocks. The color of the $$$i$$$-th block is $$$c_i$$$, an integer between $$$1$$$ and $$$n$$$.You will place the blocks down in sequence on an infinite coordinate grid in the following way. Initially, you place block $$$1$$$ at $$$(0, 0)$$$. For $$$2 \le i \le n$$$, if the $$$(i - 1)$$$-th block is placed at position $$$(x, y)$$$, then the $$$i$$$-th block can be placed at one of positions $$$(x + 1, y)$$$, $$$(x - 1, y)$$$, $$$(x, y + 1)$$$ (but not at position $$$(x, y - 1)$$$), as long no previous block was placed at that position. A tower is formed by $$$s$$$ blocks such that they are placed at positions $$$(x, y), (x, y + 1), \ldots, (x, y + s - 1)$$$ for some position $$$(x, y)$$$ and integer $$$s$$$. The size of the tower is $$$s$$$, the number of blocks in it. A tower of color $$$r$$$ is a tower such that all blocks in it have the color $$$r$$$.For each color $$$r$$$ from $$$1$$$ to $$$n$$$, solve the following problem independently: Find the maximum size of a tower of color $$$r$$$ that you can form by placing down the blocks according to the rules.
Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$1 \le c_i \le n$$$). It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.
Output Specification: For each test case, output $$$n$$$ integers. The $$$r$$$-th of them should be the maximum size of an tower of color $$$r$$$ you can form by following the given rules. If you cannot form any tower of color $$$r$$$, the $$$r$$$-th integer should be $$$0$$$.
Notes: NoteIn the first test case, one of the possible ways to form a tower of color $$$1$$$ and size $$$3$$$ is: place block $$$1$$$ at position $$$(0, 0)$$$; place block $$$2$$$ to the right of block $$$1$$$, at position $$$(1, 0)$$$; place block $$$3$$$ above block $$$2$$$, at position $$$(1, 1)$$$; place block $$$4$$$ to the left of block $$$3$$$, at position $$$(0, 1)$$$; place block $$$5$$$ to the left of block $$$4$$$, at position $$$(-1, 1)$$$; place block $$$6$$$ above block $$$5$$$, at position $$$(-1, 2)$$$; place block $$$7$$$ to the right of block $$$6$$$, at position $$$(0, 2)$$$. The blocks at positions $$$(0, 0)$$$, $$$(0, 1)$$$, and $$$(0, 2)$$$ all have color $$$1$$$, forming an tower of size $$$3$$$.In the second test case, note that the following placement is not valid, since you are not allowed to place block $$$6$$$ under block $$$5$$$: It can be shown that it is impossible to form a tower of color $$$4$$$ and size $$$3$$$.
Code:
import sys
tokens = ''.join(sys.stdin.readlines()).split()[::-1]
def next(): return tokens.pop()
def nextInt(): return int(next())
def nextFloat(): return float(next())
def getIntArray(n): return [nextInt() for _ in range(n)]
def getFloatArray(n): return [nextFloat() for _ in range(n)]
def getStringArray(n): return [next() for _ in range(n)]
testcase = True
def solve(testcase = 1):
N = nextInt()
A = getIntArray(N)
map = {}
for i in range(N):
if A[i] not in map: map[A[i]] = []
map[A[i]].append(i)
for color in range(1, N + 1):
if color not in map:
print(0, end=' ')
continue
ar = map[color]
oddCount = evenCount = 0
for i in ar:
if i % 2 == 0:
evenCount = max(evenCount, oddCount + 1)
else:
# TODO: Your code here
print(max(oddCount, evenCount), end=' ')
print()
if testcase is None:
testcaseCount = 1
while tokens:
solve(testcaseCount)
testcaseCount += 1
else:
testcaseCount = nextInt() if testcase else 1
for tc in range(testcaseCount):
solve(tc + 1)
assert not tokens
|
import sys
tokens = ''.join(sys.stdin.readlines()).split()[::-1]
def next(): return tokens.pop()
def nextInt(): return int(next())
def nextFloat(): return float(next())
def getIntArray(n): return [nextInt() for _ in range(n)]
def getFloatArray(n): return [nextFloat() for _ in range(n)]
def getStringArray(n): return [next() for _ in range(n)]
testcase = True
def solve(testcase = 1):
N = nextInt()
A = getIntArray(N)
map = {}
for i in range(N):
if A[i] not in map: map[A[i]] = []
map[A[i]].append(i)
for color in range(1, N + 1):
if color not in map:
print(0, end=' ')
continue
ar = map[color]
oddCount = evenCount = 0
for i in ar:
if i % 2 == 0:
evenCount = max(evenCount, oddCount + 1)
else:
{{completion}}
print(max(oddCount, evenCount), end=' ')
print()
if testcase is None:
testcaseCount = 1
while tokens:
solve(testcaseCount)
testcaseCount += 1
else:
testcaseCount = nextInt() if testcase else 1
for tc in range(testcaseCount):
solve(tc + 1)
assert not tokens
|
oddCount = max(oddCount, evenCount + 1)
|
[{"input": "6\n\n7\n\n1 2 3 1 2 3 1\n\n6\n\n4 2 2 2 4 4\n\n1\n\n1\n\n5\n\n5 4 5 3 5\n\n6\n\n3 3 3 1 3 3\n\n8\n\n1 2 3 4 4 3 2 1", "output": ["3 2 2 0 0 0 0 \n0 3 0 2 0 0 \n1 \n0 0 1 1 1 \n1 0 4 0 0 0 \n2 2 2 2 0 0 0 0"]}]
|
block_completion_003673
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: You have a sequence of $$$n$$$ colored blocks. The color of the $$$i$$$-th block is $$$c_i$$$, an integer between $$$1$$$ and $$$n$$$.You will place the blocks down in sequence on an infinite coordinate grid in the following way. Initially, you place block $$$1$$$ at $$$(0, 0)$$$. For $$$2 \le i \le n$$$, if the $$$(i - 1)$$$-th block is placed at position $$$(x, y)$$$, then the $$$i$$$-th block can be placed at one of positions $$$(x + 1, y)$$$, $$$(x - 1, y)$$$, $$$(x, y + 1)$$$ (but not at position $$$(x, y - 1)$$$), as long no previous block was placed at that position. A tower is formed by $$$s$$$ blocks such that they are placed at positions $$$(x, y), (x, y + 1), \ldots, (x, y + s - 1)$$$ for some position $$$(x, y)$$$ and integer $$$s$$$. The size of the tower is $$$s$$$, the number of blocks in it. A tower of color $$$r$$$ is a tower such that all blocks in it have the color $$$r$$$.For each color $$$r$$$ from $$$1$$$ to $$$n$$$, solve the following problem independently: Find the maximum size of a tower of color $$$r$$$ that you can form by placing down the blocks according to the rules.
Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$1 \le c_i \le n$$$). It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.
Output Specification: For each test case, output $$$n$$$ integers. The $$$r$$$-th of them should be the maximum size of an tower of color $$$r$$$ you can form by following the given rules. If you cannot form any tower of color $$$r$$$, the $$$r$$$-th integer should be $$$0$$$.
Notes: NoteIn the first test case, one of the possible ways to form a tower of color $$$1$$$ and size $$$3$$$ is: place block $$$1$$$ at position $$$(0, 0)$$$; place block $$$2$$$ to the right of block $$$1$$$, at position $$$(1, 0)$$$; place block $$$3$$$ above block $$$2$$$, at position $$$(1, 1)$$$; place block $$$4$$$ to the left of block $$$3$$$, at position $$$(0, 1)$$$; place block $$$5$$$ to the left of block $$$4$$$, at position $$$(-1, 1)$$$; place block $$$6$$$ above block $$$5$$$, at position $$$(-1, 2)$$$; place block $$$7$$$ to the right of block $$$6$$$, at position $$$(0, 2)$$$. The blocks at positions $$$(0, 0)$$$, $$$(0, 1)$$$, and $$$(0, 2)$$$ all have color $$$1$$$, forming an tower of size $$$3$$$.In the second test case, note that the following placement is not valid, since you are not allowed to place block $$$6$$$ under block $$$5$$$: It can be shown that it is impossible to form a tower of color $$$4$$$ and size $$$3$$$.
Code:
import sys
tokens = ''.join(sys.stdin.readlines()).split()[::-1]
def next(): return tokens.pop()
def nextInt(): return int(next())
def nextFloat(): return float(next())
def getIntArray(n): return [nextInt() for _ in range(n)]
def getFloatArray(n): return [nextFloat() for _ in range(n)]
def getStringArray(n): return [next() for _ in range(n)]
testcase = True
def solve(testcase = 1):
N = nextInt()
A = getIntArray(N)
map = {}
for i in range(N):
if A[i] not in map: map[A[i]] = []
map[A[i]].append(i)
for color in range(1, N + 1):
if color not in map:
print(0, end=' ')
continue
ar = map[color]
oddCount = evenCount = 0
for i in ar:
if i % 2 == 0:
# TODO: Your code here
else:
oddCount = max(oddCount, evenCount + 1)
print(max(oddCount, evenCount), end=' ')
print()
if testcase is None:
testcaseCount = 1
while tokens:
solve(testcaseCount)
testcaseCount += 1
else:
testcaseCount = nextInt() if testcase else 1
for tc in range(testcaseCount):
solve(tc + 1)
assert not tokens
|
import sys
tokens = ''.join(sys.stdin.readlines()).split()[::-1]
def next(): return tokens.pop()
def nextInt(): return int(next())
def nextFloat(): return float(next())
def getIntArray(n): return [nextInt() for _ in range(n)]
def getFloatArray(n): return [nextFloat() for _ in range(n)]
def getStringArray(n): return [next() for _ in range(n)]
testcase = True
def solve(testcase = 1):
N = nextInt()
A = getIntArray(N)
map = {}
for i in range(N):
if A[i] not in map: map[A[i]] = []
map[A[i]].append(i)
for color in range(1, N + 1):
if color not in map:
print(0, end=' ')
continue
ar = map[color]
oddCount = evenCount = 0
for i in ar:
if i % 2 == 0:
{{completion}}
else:
oddCount = max(oddCount, evenCount + 1)
print(max(oddCount, evenCount), end=' ')
print()
if testcase is None:
testcaseCount = 1
while tokens:
solve(testcaseCount)
testcaseCount += 1
else:
testcaseCount = nextInt() if testcase else 1
for tc in range(testcaseCount):
solve(tc + 1)
assert not tokens
|
evenCount = max(evenCount, oddCount + 1)
|
[{"input": "6\n\n7\n\n1 2 3 1 2 3 1\n\n6\n\n4 2 2 2 4 4\n\n1\n\n1\n\n5\n\n5 4 5 3 5\n\n6\n\n3 3 3 1 3 3\n\n8\n\n1 2 3 4 4 3 2 1", "output": ["3 2 2 0 0 0 0 \n0 3 0 2 0 0 \n1 \n0 0 1 1 1 \n1 0 4 0 0 0 \n2 2 2 2 0 0 0 0"]}]
|
block_completion_003674
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: A basketball competition is held where the number of players in a team does not have a maximum or minimum limit (not necessarily $$$5$$$ players in one team for each match). There are $$$N$$$ candidate players in the competition that will be trained by Pak Chanek, the best basketball coach on earth. The $$$i$$$-th candidate player has a power of $$$P_i$$$.Pak Chanek will form zero or more teams from the $$$N$$$ candidate players on the condition that each candidate player may only join in at most one team. Each of Pak Chanek's teams will be sent to compete once with an enemy team that has a power of $$$D$$$. In each match, the team sent is said to defeat the enemy team if the sum of powers from the formed players is strictly greater than $$$D$$$.One of Pak Chanek's skills is that when a team that has been formed plays in a match, he can change the power of each player in the team to be equal to the biggest player power from the team.Determine the maximum number of wins that can be achieved by Pak Chanek.
Input Specification: The first line contains two integers $$$N$$$ and $$$D$$$ ($$$1 \le N \le 10^5$$$, $$$1 \le D \le 10^9$$$) — the number of candidate players and the power of the enemy team. The second line contains $$$N$$$ integers $$$P_1, P_2, \ldots, P_N$$$ ($$$1 \le P_i \le 10^9$$$) — the powers of all candidate players.
Output Specification: A line containing an integer representing the maximum number of wins that can be achieved by Pak Chanek.
Notes: NoteThe $$$1$$$-st team formed is a team containing players $$$4$$$ and $$$6$$$. The power of each player in the team becomes $$$100$$$. So the total power of the team is $$$100 + 100 = 200 > 180$$$.The $$$2$$$-nd team formed is a team containing players $$$1$$$, $$$2$$$, and $$$5$$$. The power of each player in the team becomes $$$90$$$. So the total power of the team is $$$90 + 90 + 90 = 270 > 180$$$.
Code:
n, d = map(int, input().split())
p = sorted(map(int, input().split()), reverse=True)
ans = 0
for num in p:
if n >= d // num + 1:
n -= d // num + 1
ans += 1
else:
# TODO: Your code here
print(ans)
|
n, d = map(int, input().split())
p = sorted(map(int, input().split()), reverse=True)
ans = 0
for num in p:
if n >= d // num + 1:
n -= d // num + 1
ans += 1
else:
{{completion}}
print(ans)
|
break
|
[{"input": "6 180\n90 80 70 60 50 100", "output": ["2"]}]
|
block_completion_003722
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: A basketball competition is held where the number of players in a team does not have a maximum or minimum limit (not necessarily $$$5$$$ players in one team for each match). There are $$$N$$$ candidate players in the competition that will be trained by Pak Chanek, the best basketball coach on earth. The $$$i$$$-th candidate player has a power of $$$P_i$$$.Pak Chanek will form zero or more teams from the $$$N$$$ candidate players on the condition that each candidate player may only join in at most one team. Each of Pak Chanek's teams will be sent to compete once with an enemy team that has a power of $$$D$$$. In each match, the team sent is said to defeat the enemy team if the sum of powers from the formed players is strictly greater than $$$D$$$.One of Pak Chanek's skills is that when a team that has been formed plays in a match, he can change the power of each player in the team to be equal to the biggest player power from the team.Determine the maximum number of wins that can be achieved by Pak Chanek.
Input Specification: The first line contains two integers $$$N$$$ and $$$D$$$ ($$$1 \le N \le 10^5$$$, $$$1 \le D \le 10^9$$$) — the number of candidate players and the power of the enemy team. The second line contains $$$N$$$ integers $$$P_1, P_2, \ldots, P_N$$$ ($$$1 \le P_i \le 10^9$$$) — the powers of all candidate players.
Output Specification: A line containing an integer representing the maximum number of wins that can be achieved by Pak Chanek.
Notes: NoteThe $$$1$$$-st team formed is a team containing players $$$4$$$ and $$$6$$$. The power of each player in the team becomes $$$100$$$. So the total power of the team is $$$100 + 100 = 200 > 180$$$.The $$$2$$$-nd team formed is a team containing players $$$1$$$, $$$2$$$, and $$$5$$$. The power of each player in the team becomes $$$90$$$. So the total power of the team is $$$90 + 90 + 90 = 270 > 180$$$.
Code:
def solve():
n, d = [int(i) for i in input().split(' ')]
power = [int(i) for i in input().split(' ')]
power.sort()
used = 0
w = 0
for i in range(len(power)-1, -1, -1):
min_players = -(d // -power[i])
p = power[i] * min_players
if(p > d):
used += min_players
elif(p == d):
# TODO: Your code here
if(used > n):
break
w += 1
print(w)
solve()
|
def solve():
n, d = [int(i) for i in input().split(' ')]
power = [int(i) for i in input().split(' ')]
power.sort()
used = 0
w = 0
for i in range(len(power)-1, -1, -1):
min_players = -(d // -power[i])
p = power[i] * min_players
if(p > d):
used += min_players
elif(p == d):
{{completion}}
if(used > n):
break
w += 1
print(w)
solve()
|
used += min_players + 1
|
[{"input": "6 180\n90 80 70 60 50 100", "output": ["2"]}]
|
block_completion_003723
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: A basketball competition is held where the number of players in a team does not have a maximum or minimum limit (not necessarily $$$5$$$ players in one team for each match). There are $$$N$$$ candidate players in the competition that will be trained by Pak Chanek, the best basketball coach on earth. The $$$i$$$-th candidate player has a power of $$$P_i$$$.Pak Chanek will form zero or more teams from the $$$N$$$ candidate players on the condition that each candidate player may only join in at most one team. Each of Pak Chanek's teams will be sent to compete once with an enemy team that has a power of $$$D$$$. In each match, the team sent is said to defeat the enemy team if the sum of powers from the formed players is strictly greater than $$$D$$$.One of Pak Chanek's skills is that when a team that has been formed plays in a match, he can change the power of each player in the team to be equal to the biggest player power from the team.Determine the maximum number of wins that can be achieved by Pak Chanek.
Input Specification: The first line contains two integers $$$N$$$ and $$$D$$$ ($$$1 \le N \le 10^5$$$, $$$1 \le D \le 10^9$$$) — the number of candidate players and the power of the enemy team. The second line contains $$$N$$$ integers $$$P_1, P_2, \ldots, P_N$$$ ($$$1 \le P_i \le 10^9$$$) — the powers of all candidate players.
Output Specification: A line containing an integer representing the maximum number of wins that can be achieved by Pak Chanek.
Notes: NoteThe $$$1$$$-st team formed is a team containing players $$$4$$$ and $$$6$$$. The power of each player in the team becomes $$$100$$$. So the total power of the team is $$$100 + 100 = 200 > 180$$$.The $$$2$$$-nd team formed is a team containing players $$$1$$$, $$$2$$$, and $$$5$$$. The power of each player in the team becomes $$$90$$$. So the total power of the team is $$$90 + 90 + 90 = 270 > 180$$$.
Code:
import math
enemy_power=int(input().split()[1])
team=[int(i) for i in input().split()]
team.sort()
days=0
while len(team)>0:
num=enemy_power//team[-1]+1
if len(team)<num:
break;
else:
# TODO: Your code here
print(days)
|
import math
enemy_power=int(input().split()[1])
team=[int(i) for i in input().split()]
team.sort()
days=0
while len(team)>0:
num=enemy_power//team[-1]+1
if len(team)<num:
break;
else:
{{completion}}
print(days)
|
del team[-1]
del team[0:num-1]
days+=1
|
[{"input": "6 180\n90 80 70 60 50 100", "output": ["2"]}]
|
block_completion_003724
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: A basketball competition is held where the number of players in a team does not have a maximum or minimum limit (not necessarily $$$5$$$ players in one team for each match). There are $$$N$$$ candidate players in the competition that will be trained by Pak Chanek, the best basketball coach on earth. The $$$i$$$-th candidate player has a power of $$$P_i$$$.Pak Chanek will form zero or more teams from the $$$N$$$ candidate players on the condition that each candidate player may only join in at most one team. Each of Pak Chanek's teams will be sent to compete once with an enemy team that has a power of $$$D$$$. In each match, the team sent is said to defeat the enemy team if the sum of powers from the formed players is strictly greater than $$$D$$$.One of Pak Chanek's skills is that when a team that has been formed plays in a match, he can change the power of each player in the team to be equal to the biggest player power from the team.Determine the maximum number of wins that can be achieved by Pak Chanek.
Input Specification: The first line contains two integers $$$N$$$ and $$$D$$$ ($$$1 \le N \le 10^5$$$, $$$1 \le D \le 10^9$$$) — the number of candidate players and the power of the enemy team. The second line contains $$$N$$$ integers $$$P_1, P_2, \ldots, P_N$$$ ($$$1 \le P_i \le 10^9$$$) — the powers of all candidate players.
Output Specification: A line containing an integer representing the maximum number of wins that can be achieved by Pak Chanek.
Notes: NoteThe $$$1$$$-st team formed is a team containing players $$$4$$$ and $$$6$$$. The power of each player in the team becomes $$$100$$$. So the total power of the team is $$$100 + 100 = 200 > 180$$$.The $$$2$$$-nd team formed is a team containing players $$$1$$$, $$$2$$$, and $$$5$$$. The power of each player in the team becomes $$$90$$$. So the total power of the team is $$$90 + 90 + 90 = 270 > 180$$$.
Code:
n,d=map(int,input().split())
s=list(map(int,input().split()))
k=n
r=-1
s.sort()
while k-(d//s[r])-1>=0:
k-=((d//s[r])+1)
r-=1
if r<-n:
# TODO: Your code here
print(-1-r)
|
n,d=map(int,input().split())
s=list(map(int,input().split()))
k=n
r=-1
s.sort()
while k-(d//s[r])-1>=0:
k-=((d//s[r])+1)
r-=1
if r<-n:
{{completion}}
print(-1-r)
|
break
|
[{"input": "6 180\n90 80 70 60 50 100", "output": ["2"]}]
|
block_completion_003725
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: A basketball competition is held where the number of players in a team does not have a maximum or minimum limit (not necessarily $$$5$$$ players in one team for each match). There are $$$N$$$ candidate players in the competition that will be trained by Pak Chanek, the best basketball coach on earth. The $$$i$$$-th candidate player has a power of $$$P_i$$$.Pak Chanek will form zero or more teams from the $$$N$$$ candidate players on the condition that each candidate player may only join in at most one team. Each of Pak Chanek's teams will be sent to compete once with an enemy team that has a power of $$$D$$$. In each match, the team sent is said to defeat the enemy team if the sum of powers from the formed players is strictly greater than $$$D$$$.One of Pak Chanek's skills is that when a team that has been formed plays in a match, he can change the power of each player in the team to be equal to the biggest player power from the team.Determine the maximum number of wins that can be achieved by Pak Chanek.
Input Specification: The first line contains two integers $$$N$$$ and $$$D$$$ ($$$1 \le N \le 10^5$$$, $$$1 \le D \le 10^9$$$) — the number of candidate players and the power of the enemy team. The second line contains $$$N$$$ integers $$$P_1, P_2, \ldots, P_N$$$ ($$$1 \le P_i \le 10^9$$$) — the powers of all candidate players.
Output Specification: A line containing an integer representing the maximum number of wins that can be achieved by Pak Chanek.
Notes: NoteThe $$$1$$$-st team formed is a team containing players $$$4$$$ and $$$6$$$. The power of each player in the team becomes $$$100$$$. So the total power of the team is $$$100 + 100 = 200 > 180$$$.The $$$2$$$-nd team formed is a team containing players $$$1$$$, $$$2$$$, and $$$5$$$. The power of each player in the team becomes $$$90$$$. So the total power of the team is $$$90 + 90 + 90 = 270 > 180$$$.
Code:
import sys,math
n,team=map(int,sys.stdin.readline().split())
arr=sorted(map(int,sys.stdin.readline().split()),reverse=True)
# print(arr)
all=n+1
count=0
for i in range(n):
sub=int(math.floor(team/arr[i])+1)
all-=sub
if all>0:
count+=1
else:# TODO: Your code here
print(count)
|
import sys,math
n,team=map(int,sys.stdin.readline().split())
arr=sorted(map(int,sys.stdin.readline().split()),reverse=True)
# print(arr)
all=n+1
count=0
for i in range(n):
sub=int(math.floor(team/arr[i])+1)
all-=sub
if all>0:
count+=1
else:{{completion}}
print(count)
|
break
|
[{"input": "6 180\n90 80 70 60 50 100", "output": ["2"]}]
|
block_completion_003726
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: A basketball competition is held where the number of players in a team does not have a maximum or minimum limit (not necessarily $$$5$$$ players in one team for each match). There are $$$N$$$ candidate players in the competition that will be trained by Pak Chanek, the best basketball coach on earth. The $$$i$$$-th candidate player has a power of $$$P_i$$$.Pak Chanek will form zero or more teams from the $$$N$$$ candidate players on the condition that each candidate player may only join in at most one team. Each of Pak Chanek's teams will be sent to compete once with an enemy team that has a power of $$$D$$$. In each match, the team sent is said to defeat the enemy team if the sum of powers from the formed players is strictly greater than $$$D$$$.One of Pak Chanek's skills is that when a team that has been formed plays in a match, he can change the power of each player in the team to be equal to the biggest player power from the team.Determine the maximum number of wins that can be achieved by Pak Chanek.
Input Specification: The first line contains two integers $$$N$$$ and $$$D$$$ ($$$1 \le N \le 10^5$$$, $$$1 \le D \le 10^9$$$) — the number of candidate players and the power of the enemy team. The second line contains $$$N$$$ integers $$$P_1, P_2, \ldots, P_N$$$ ($$$1 \le P_i \le 10^9$$$) — the powers of all candidate players.
Output Specification: A line containing an integer representing the maximum number of wins that can be achieved by Pak Chanek.
Notes: NoteThe $$$1$$$-st team formed is a team containing players $$$4$$$ and $$$6$$$. The power of each player in the team becomes $$$100$$$. So the total power of the team is $$$100 + 100 = 200 > 180$$$.The $$$2$$$-nd team formed is a team containing players $$$1$$$, $$$2$$$, and $$$5$$$. The power of each player in the team becomes $$$90$$$. So the total power of the team is $$$90 + 90 + 90 = 270 > 180$$$.
Code:
def solve():
n,d=map(int,input().split())
a=sorted([*map(int,input().split())])[::-1]
i,j,r=0,len(a),0
while i<j:
x=a[i]
while x<=d:
j-=1
if i<j:
x+=a[i]
else:
# TODO: Your code here
else:
r+=1
i+=1
return r
print(solve())
|
def solve():
n,d=map(int,input().split())
a=sorted([*map(int,input().split())])[::-1]
i,j,r=0,len(a),0
while i<j:
x=a[i]
while x<=d:
j-=1
if i<j:
x+=a[i]
else:
{{completion}}
else:
r+=1
i+=1
return r
print(solve())
|
return r
|
[{"input": "6 180\n90 80 70 60 50 100", "output": ["2"]}]
|
block_completion_003727
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: A basketball competition is held where the number of players in a team does not have a maximum or minimum limit (not necessarily $$$5$$$ players in one team for each match). There are $$$N$$$ candidate players in the competition that will be trained by Pak Chanek, the best basketball coach on earth. The $$$i$$$-th candidate player has a power of $$$P_i$$$.Pak Chanek will form zero or more teams from the $$$N$$$ candidate players on the condition that each candidate player may only join in at most one team. Each of Pak Chanek's teams will be sent to compete once with an enemy team that has a power of $$$D$$$. In each match, the team sent is said to defeat the enemy team if the sum of powers from the formed players is strictly greater than $$$D$$$.One of Pak Chanek's skills is that when a team that has been formed plays in a match, he can change the power of each player in the team to be equal to the biggest player power from the team.Determine the maximum number of wins that can be achieved by Pak Chanek.
Input Specification: The first line contains two integers $$$N$$$ and $$$D$$$ ($$$1 \le N \le 10^5$$$, $$$1 \le D \le 10^9$$$) — the number of candidate players and the power of the enemy team. The second line contains $$$N$$$ integers $$$P_1, P_2, \ldots, P_N$$$ ($$$1 \le P_i \le 10^9$$$) — the powers of all candidate players.
Output Specification: A line containing an integer representing the maximum number of wins that can be achieved by Pak Chanek.
Notes: NoteThe $$$1$$$-st team formed is a team containing players $$$4$$$ and $$$6$$$. The power of each player in the team becomes $$$100$$$. So the total power of the team is $$$100 + 100 = 200 > 180$$$.The $$$2$$$-nd team formed is a team containing players $$$1$$$, $$$2$$$, and $$$5$$$. The power of each player in the team becomes $$$90$$$. So the total power of the team is $$$90 + 90 + 90 = 270 > 180$$$.
Code:
from sys import stdin,stdout
def ans():
n,d=map(int,stdin.readline().strip().split())
p=list(map(int,stdin.readline().strip().split()))
temp=int(n)
ans=0
for x in sorted(p,reverse=True):
if temp>=((d//x)+1):
# TODO: Your code here
print(ans)
if __name__=='__main__':
ans()
|
from sys import stdin,stdout
def ans():
n,d=map(int,stdin.readline().strip().split())
p=list(map(int,stdin.readline().strip().split()))
temp=int(n)
ans=0
for x in sorted(p,reverse=True):
if temp>=((d//x)+1):
{{completion}}
print(ans)
if __name__=='__main__':
ans()
|
temp-=((d//x)+1)
ans+=1
|
[{"input": "6 180\n90 80 70 60 50 100", "output": ["2"]}]
|
block_completion_003728
|
block
|
python
|
Complete the code in python to solve this programming problem:
Description: A basketball competition is held where the number of players in a team does not have a maximum or minimum limit (not necessarily $$$5$$$ players in one team for each match). There are $$$N$$$ candidate players in the competition that will be trained by Pak Chanek, the best basketball coach on earth. The $$$i$$$-th candidate player has a power of $$$P_i$$$.Pak Chanek will form zero or more teams from the $$$N$$$ candidate players on the condition that each candidate player may only join in at most one team. Each of Pak Chanek's teams will be sent to compete once with an enemy team that has a power of $$$D$$$. In each match, the team sent is said to defeat the enemy team if the sum of powers from the formed players is strictly greater than $$$D$$$.One of Pak Chanek's skills is that when a team that has been formed plays in a match, he can change the power of each player in the team to be equal to the biggest player power from the team.Determine the maximum number of wins that can be achieved by Pak Chanek.
Input Specification: The first line contains two integers $$$N$$$ and $$$D$$$ ($$$1 \le N \le 10^5$$$, $$$1 \le D \le 10^9$$$) — the number of candidate players and the power of the enemy team. The second line contains $$$N$$$ integers $$$P_1, P_2, \ldots, P_N$$$ ($$$1 \le P_i \le 10^9$$$) — the powers of all candidate players.
Output Specification: A line containing an integer representing the maximum number of wins that can be achieved by Pak Chanek.
Notes: NoteThe $$$1$$$-st team formed is a team containing players $$$4$$$ and $$$6$$$. The power of each player in the team becomes $$$100$$$. So the total power of the team is $$$100 + 100 = 200 > 180$$$.The $$$2$$$-nd team formed is a team containing players $$$1$$$, $$$2$$$, and $$$5$$$. The power of each player in the team becomes $$$90$$$. So the total power of the team is $$$90 + 90 + 90 = 270 > 180$$$.
Code:
d = int(input().split(" ")[1])
p = sorted(map(int, input().split(" ")))
c = 0
l = 0
r = len(p) - 1
s = p[r]
while r > l:
while s <= d:
# TODO: Your code here
if l > r:
break
r -= 1
s = p[r]
c += 1
if p[0] > d:
c += 1
print(c)
|
d = int(input().split(" ")[1])
p = sorted(map(int, input().split(" ")))
c = 0
l = 0
r = len(p) - 1
s = p[r]
while r > l:
while s <= d:
{{completion}}
if l > r:
break
r -= 1
s = p[r]
c += 1
if p[0] > d:
c += 1
print(c)
|
s += p[r]
l += 1
|
[{"input": "6 180\n90 80 70 60 50 100", "output": ["2"]}]
|
block_completion_003729
|
block
|
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