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
values |
|---|---|---|---|---|---|---|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ candies put from left to right on a table. The candies are numbered from left to right. The $$$i$$$-th candy has weight $$$w_i$$$. Alice and Bob eat candies. Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row). Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row). Of course, if Alice ate a candy, Bob can't eat it (and vice versa).They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the number of candies on the table. The second line of each test case contains $$$n$$$ integers $$$w_1, w_2, \dots, w_n$$$ ($$$1 \leq w_i \leq 10^4$$$) — the weights of candies from left to right. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case, print a single integer — the maximum number of candies Alice and Bob can eat in total while satisfying the condition.
Notes: NoteFor the first test case, Alice will eat one candy from the left and Bob will eat one candy from the right. There is no better way for them to eat the same total amount of weight. The answer is $$$2$$$ because they eat two candies in total.For the second test case, Alice will eat the first three candies from the left (with total weight $$$7$$$) and Bob will eat the first three candies from the right (with total weight $$$7$$$). They cannot eat more candies since all the candies have been eaten, so the answer is $$$6$$$ (because they eat six candies in total).For the third test case, there is no way Alice and Bob will eat the same non-zero weight so the answer is $$$0$$$.For the fourth test case, Alice will eat candies with weights $$$[7, 3, 20]$$$ and Bob will eat candies with weights $$$[10, 8, 11, 1]$$$, they each eat $$$30$$$ weight. There is no better partition so the answer is $$$7$$$.
Code:
for _ in range(int(input())):
n = int(input())
a = [*map(int, input().split())]
x = sum(a) // 2
s, d = 0, {}
for idx, i in enumerate(a):
s += i
if s > x: break
d[s] = idx + 1
s, r = 0, 0
for idx, i in enumerate(a[::-1]):
s += i
if # TODO: Your code here: r = idx + 1 + d[s]
print(r)
|
for _ in range(int(input())):
n = int(input())
a = [*map(int, input().split())]
x = sum(a) // 2
s, d = 0, {}
for idx, i in enumerate(a):
s += i
if s > x: break
d[s] = idx + 1
s, r = 0, 0
for idx, i in enumerate(a[::-1]):
s += i
if {{completion}}: r = idx + 1 + d[s]
print(r)
|
s in d
|
[{"input": "4\n3\n10 20 10\n6\n2 1 4 2 4 1\n5\n1 2 4 8 16\n9\n7 3 20 5 15 1 11 8 10", "output": ["2\n6\n0\n7"]}]
|
control_completion_000791
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ candies put from left to right on a table. The candies are numbered from left to right. The $$$i$$$-th candy has weight $$$w_i$$$. Alice and Bob eat candies. Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row). Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row). Of course, if Alice ate a candy, Bob can't eat it (and vice versa).They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the number of candies on the table. The second line of each test case contains $$$n$$$ integers $$$w_1, w_2, \dots, w_n$$$ ($$$1 \leq w_i \leq 10^4$$$) — the weights of candies from left to right. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case, print a single integer — the maximum number of candies Alice and Bob can eat in total while satisfying the condition.
Notes: NoteFor the first test case, Alice will eat one candy from the left and Bob will eat one candy from the right. There is no better way for them to eat the same total amount of weight. The answer is $$$2$$$ because they eat two candies in total.For the second test case, Alice will eat the first three candies from the left (with total weight $$$7$$$) and Bob will eat the first three candies from the right (with total weight $$$7$$$). They cannot eat more candies since all the candies have been eaten, so the answer is $$$6$$$ (because they eat six candies in total).For the third test case, there is no way Alice and Bob will eat the same non-zero weight so the answer is $$$0$$$.For the fourth test case, Alice will eat candies with weights $$$[7, 3, 20]$$$ and Bob will eat candies with weights $$$[10, 8, 11, 1]$$$, they each eat $$$30$$$ weight. There is no better partition so the answer is $$$7$$$.
Code:
n = int(input())
for i in range(n):
h = 1
g = 1
counter = 0
m = int(input())
lst = list(map(int, input().split()))
index =0
e= 0
r =m-1
highest = lst[m-1]
lowest = lst[0]
while h+g <= m:
if lowest<highest:
lowest+=lst[e+1]
e+=1
g +=1
elif # TODO: Your code here:
highest+=lst[r-1]
r-=1
h+=1
elif highest == lowest:
lowest+=lst[e+1]
e+=1
g+=1
index = e + (m - r)
print(index)
|
n = int(input())
for i in range(n):
h = 1
g = 1
counter = 0
m = int(input())
lst = list(map(int, input().split()))
index =0
e= 0
r =m-1
highest = lst[m-1]
lowest = lst[0]
while h+g <= m:
if lowest<highest:
lowest+=lst[e+1]
e+=1
g +=1
elif {{completion}}:
highest+=lst[r-1]
r-=1
h+=1
elif highest == lowest:
lowest+=lst[e+1]
e+=1
g+=1
index = e + (m - r)
print(index)
|
highest<lowest
|
[{"input": "4\n3\n10 20 10\n6\n2 1 4 2 4 1\n5\n1 2 4 8 16\n9\n7 3 20 5 15 1 11 8 10", "output": ["2\n6\n0\n7"]}]
|
control_completion_000792
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ candies put from left to right on a table. The candies are numbered from left to right. The $$$i$$$-th candy has weight $$$w_i$$$. Alice and Bob eat candies. Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row). Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row). Of course, if Alice ate a candy, Bob can't eat it (and vice versa).They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the number of candies on the table. The second line of each test case contains $$$n$$$ integers $$$w_1, w_2, \dots, w_n$$$ ($$$1 \leq w_i \leq 10^4$$$) — the weights of candies from left to right. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case, print a single integer — the maximum number of candies Alice and Bob can eat in total while satisfying the condition.
Notes: NoteFor the first test case, Alice will eat one candy from the left and Bob will eat one candy from the right. There is no better way for them to eat the same total amount of weight. The answer is $$$2$$$ because they eat two candies in total.For the second test case, Alice will eat the first three candies from the left (with total weight $$$7$$$) and Bob will eat the first three candies from the right (with total weight $$$7$$$). They cannot eat more candies since all the candies have been eaten, so the answer is $$$6$$$ (because they eat six candies in total).For the third test case, there is no way Alice and Bob will eat the same non-zero weight so the answer is $$$0$$$.For the fourth test case, Alice will eat candies with weights $$$[7, 3, 20]$$$ and Bob will eat candies with weights $$$[10, 8, 11, 1]$$$, they each eat $$$30$$$ weight. There is no better partition so the answer is $$$7$$$.
Code:
n = int(input())
for i in range(n):
h = 1
g = 1
counter = 0
m = int(input())
lst = list(map(int, input().split()))
index =0
e= 0
r =m-1
highest = lst[m-1]
lowest = lst[0]
while h+g <= m:
if lowest<highest:
lowest+=lst[e+1]
e+=1
g +=1
elif highest<lowest:
highest+=lst[r-1]
r-=1
h+=1
elif # TODO: Your code here:
lowest+=lst[e+1]
e+=1
g+=1
index = e + (m - r)
print(index)
|
n = int(input())
for i in range(n):
h = 1
g = 1
counter = 0
m = int(input())
lst = list(map(int, input().split()))
index =0
e= 0
r =m-1
highest = lst[m-1]
lowest = lst[0]
while h+g <= m:
if lowest<highest:
lowest+=lst[e+1]
e+=1
g +=1
elif highest<lowest:
highest+=lst[r-1]
r-=1
h+=1
elif {{completion}}:
lowest+=lst[e+1]
e+=1
g+=1
index = e + (m - r)
print(index)
|
highest == lowest
|
[{"input": "4\n3\n10 20 10\n6\n2 1 4 2 4 1\n5\n1 2 4 8 16\n9\n7 3 20 5 15 1 11 8 10", "output": ["2\n6\n0\n7"]}]
|
control_completion_000793
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ candies put from left to right on a table. The candies are numbered from left to right. The $$$i$$$-th candy has weight $$$w_i$$$. Alice and Bob eat candies. Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row). Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row). Of course, if Alice ate a candy, Bob can't eat it (and vice versa).They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the number of candies on the table. The second line of each test case contains $$$n$$$ integers $$$w_1, w_2, \dots, w_n$$$ ($$$1 \leq w_i \leq 10^4$$$) — the weights of candies from left to right. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case, print a single integer — the maximum number of candies Alice and Bob can eat in total while satisfying the condition.
Notes: NoteFor the first test case, Alice will eat one candy from the left and Bob will eat one candy from the right. There is no better way for them to eat the same total amount of weight. The answer is $$$2$$$ because they eat two candies in total.For the second test case, Alice will eat the first three candies from the left (with total weight $$$7$$$) and Bob will eat the first three candies from the right (with total weight $$$7$$$). They cannot eat more candies since all the candies have been eaten, so the answer is $$$6$$$ (because they eat six candies in total).For the third test case, there is no way Alice and Bob will eat the same non-zero weight so the answer is $$$0$$$.For the fourth test case, Alice will eat candies with weights $$$[7, 3, 20]$$$ and Bob will eat candies with weights $$$[10, 8, 11, 1]$$$, they each eat $$$30$$$ weight. There is no better partition so the answer is $$$7$$$.
Code:
def read():
return int(input())
def readline():
return list(map(int,input().split()))
def solve():
n=read()
arr=readline()
ans,cur=0,0
a,suma=-1,0
b,sumb=n,0
while True:
if a>=b: break
elif # TODO: Your code here:
b-=1
sumb+=arr[b]
cur+=1
elif suma<sumb:
a+=1
suma+=arr[a]
cur+=1
else :
ans=cur
a+=1
b-=1
suma+=arr[a]
sumb+=arr[b]
cur+=2
print(ans)
if __name__ == "__main__":
T=read()
for i in range(T):
solve()
|
def read():
return int(input())
def readline():
return list(map(int,input().split()))
def solve():
n=read()
arr=readline()
ans,cur=0,0
a,suma=-1,0
b,sumb=n,0
while True:
if a>=b: break
elif {{completion}}:
b-=1
sumb+=arr[b]
cur+=1
elif suma<sumb:
a+=1
suma+=arr[a]
cur+=1
else :
ans=cur
a+=1
b-=1
suma+=arr[a]
sumb+=arr[b]
cur+=2
print(ans)
if __name__ == "__main__":
T=read()
for i in range(T):
solve()
|
suma>sumb
|
[{"input": "4\n3\n10 20 10\n6\n2 1 4 2 4 1\n5\n1 2 4 8 16\n9\n7 3 20 5 15 1 11 8 10", "output": ["2\n6\n0\n7"]}]
|
control_completion_000794
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ candies put from left to right on a table. The candies are numbered from left to right. The $$$i$$$-th candy has weight $$$w_i$$$. Alice and Bob eat candies. Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row). Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row). Of course, if Alice ate a candy, Bob can't eat it (and vice versa).They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the number of candies on the table. The second line of each test case contains $$$n$$$ integers $$$w_1, w_2, \dots, w_n$$$ ($$$1 \leq w_i \leq 10^4$$$) — the weights of candies from left to right. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case, print a single integer — the maximum number of candies Alice and Bob can eat in total while satisfying the condition.
Notes: NoteFor the first test case, Alice will eat one candy from the left and Bob will eat one candy from the right. There is no better way for them to eat the same total amount of weight. The answer is $$$2$$$ because they eat two candies in total.For the second test case, Alice will eat the first three candies from the left (with total weight $$$7$$$) and Bob will eat the first three candies from the right (with total weight $$$7$$$). They cannot eat more candies since all the candies have been eaten, so the answer is $$$6$$$ (because they eat six candies in total).For the third test case, there is no way Alice and Bob will eat the same non-zero weight so the answer is $$$0$$$.For the fourth test case, Alice will eat candies with weights $$$[7, 3, 20]$$$ and Bob will eat candies with weights $$$[10, 8, 11, 1]$$$, they each eat $$$30$$$ weight. There is no better partition so the answer is $$$7$$$.
Code:
def read():
return int(input())
def readline():
return list(map(int,input().split()))
def solve():
n=read()
arr=readline()
ans,cur=0,0
a,suma=-1,0
b,sumb=n,0
while True:
if a>=b: break
elif suma>sumb:
b-=1
sumb+=arr[b]
cur+=1
elif # TODO: Your code here:
a+=1
suma+=arr[a]
cur+=1
else :
ans=cur
a+=1
b-=1
suma+=arr[a]
sumb+=arr[b]
cur+=2
print(ans)
if __name__ == "__main__":
T=read()
for i in range(T):
solve()
|
def read():
return int(input())
def readline():
return list(map(int,input().split()))
def solve():
n=read()
arr=readline()
ans,cur=0,0
a,suma=-1,0
b,sumb=n,0
while True:
if a>=b: break
elif suma>sumb:
b-=1
sumb+=arr[b]
cur+=1
elif {{completion}}:
a+=1
suma+=arr[a]
cur+=1
else :
ans=cur
a+=1
b-=1
suma+=arr[a]
sumb+=arr[b]
cur+=2
print(ans)
if __name__ == "__main__":
T=read()
for i in range(T):
solve()
|
suma<sumb
|
[{"input": "4\n3\n10 20 10\n6\n2 1 4 2 4 1\n5\n1 2 4 8 16\n9\n7 3 20 5 15 1 11 8 10", "output": ["2\n6\n0\n7"]}]
|
control_completion_000795
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid with $$$n$$$ rows and $$$m$$$ columns, and three types of cells: An empty cell, denoted with '.'. A stone, denoted with '*'. An obstacle, denoted with the lowercase Latin letter 'o'. All stones fall down until they meet the floor (the bottom row), an obstacle, or other stone which is already immovable. (In other words, all the stones just fall down as long as they can fall.)Simulate the process. What does the resulting grid look like?
Input Specification: The input consists of multiple test cases. The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 50$$$) — the number of rows and the number of columns in the grid, respectively. Then $$$n$$$ lines follow, each containing $$$m$$$ characters. Each of these characters is either '.', '*', or 'o' — an empty cell, a stone, or an obstacle, respectively.
Output Specification: For each test case, output a grid with $$$n$$$ rows and $$$m$$$ columns, showing the result of the process. You don't need to output a new line after each test, it is in the samples just for clarity.
Code:
for _ in range(int(input())):
n, _ = map(int, input().split())
a = map("".join, zip(*(input() for _ in range(n))))
a = ("o".join("".join(sorted(y, reverse=True)) for y in x.split("o")) for x in a)
for # TODO: Your code here:
print("".join(x))
|
for _ in range(int(input())):
n, _ = map(int, input().split())
a = map("".join, zip(*(input() for _ in range(n))))
a = ("o".join("".join(sorted(y, reverse=True)) for y in x.split("o")) for x in a)
for {{completion}}:
print("".join(x))
|
x in zip(*a)
|
[{"input": "3\n6 10\n.*.*....*.\n.*.......*\n...o....o.\n.*.*....*.\n..........\n.o......o*\n2 9\n...***ooo\n.*o.*o.*o\n5 5\n*****\n*....\n*****\n....*\n*****", "output": ["..........\n...*....*.\n.*.o....o.\n.*........\n.*......**\n.o.*....o*\n\n....**ooo\n.*o**o.*o\n\n.....\n*...*\n*****\n*****\n*****"]}]
|
control_completion_000829
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid with $$$n$$$ rows and $$$m$$$ columns, and three types of cells: An empty cell, denoted with '.'. A stone, denoted with '*'. An obstacle, denoted with the lowercase Latin letter 'o'. All stones fall down until they meet the floor (the bottom row), an obstacle, or other stone which is already immovable. (In other words, all the stones just fall down as long as they can fall.)Simulate the process. What does the resulting grid look like?
Input Specification: The input consists of multiple test cases. The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 50$$$) — the number of rows and the number of columns in the grid, respectively. Then $$$n$$$ lines follow, each containing $$$m$$$ characters. Each of these characters is either '.', '*', or 'o' — an empty cell, a stone, or an obstacle, respectively.
Output Specification: For each test case, output a grid with $$$n$$$ rows and $$$m$$$ columns, showing the result of the process. You don't need to output a new line after each test, it is in the samples just for clarity.
Code:
# Write your code here :-)
# Fall Down
def solution():
n, m = [int(i) for i in input().split()]
grid = [list(input()) for _ in range(n)]
for i in range(m):
for j in range(n - 1, -1, -1):
if grid[j][i] == "*":
grid[j][i] = "."
pos = j
while # TODO: Your code here:
pos += 1
grid[pos][i] = "*"
for row in grid:
print(*row, sep="")
t = int(input())
for _ in range(t):
solution()
|
# Write your code here :-)
# Fall Down
def solution():
n, m = [int(i) for i in input().split()]
grid = [list(input()) for _ in range(n)]
for i in range(m):
for j in range(n - 1, -1, -1):
if grid[j][i] == "*":
grid[j][i] = "."
pos = j
while {{completion}}:
pos += 1
grid[pos][i] = "*"
for row in grid:
print(*row, sep="")
t = int(input())
for _ in range(t):
solution()
|
pos < n - 1 and grid[pos + 1][i] == "."
|
[{"input": "3\n6 10\n.*.*....*.\n.*.......*\n...o....o.\n.*.*....*.\n..........\n.o......o*\n2 9\n...***ooo\n.*o.*o.*o\n5 5\n*****\n*....\n*****\n....*\n*****", "output": ["..........\n...*....*.\n.*.o....o.\n.*........\n.*......**\n.o.*....o*\n\n....**ooo\n.*o**o.*o\n\n.....\n*...*\n*****\n*****\n*****"]}]
|
control_completion_000830
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid with $$$n$$$ rows and $$$m$$$ columns, and three types of cells: An empty cell, denoted with '.'. A stone, denoted with '*'. An obstacle, denoted with the lowercase Latin letter 'o'. All stones fall down until they meet the floor (the bottom row), an obstacle, or other stone which is already immovable. (In other words, all the stones just fall down as long as they can fall.)Simulate the process. What does the resulting grid look like?
Input Specification: The input consists of multiple test cases. The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 50$$$) — the number of rows and the number of columns in the grid, respectively. Then $$$n$$$ lines follow, each containing $$$m$$$ characters. Each of these characters is either '.', '*', or 'o' — an empty cell, a stone, or an obstacle, respectively.
Output Specification: For each test case, output a grid with $$$n$$$ rows and $$$m$$$ columns, showing the result of the process. You don't need to output a new line after each test, it is in the samples just for clarity.
Code:
def res(s):
a=s.split('o');t=''
for i in a:t+=i.count('*')*'*'+i.count('.')*'.'+'o'
return t[:-1]
for _ in[0]*int(input()):
n,m=map(int,input().split())
a=[[*input()] for x in[0]*n]
b=[]
for i in range(m):b+=res(''.join([a[~j][i] for j in range(n)])),
for i in range(n):
for # TODO: Your code here:print(b[j][~i],end='')
print()
print()
|
def res(s):
a=s.split('o');t=''
for i in a:t+=i.count('*')*'*'+i.count('.')*'.'+'o'
return t[:-1]
for _ in[0]*int(input()):
n,m=map(int,input().split())
a=[[*input()] for x in[0]*n]
b=[]
for i in range(m):b+=res(''.join([a[~j][i] for j in range(n)])),
for i in range(n):
for {{completion}}:print(b[j][~i],end='')
print()
print()
|
j in range(m)
|
[{"input": "3\n6 10\n.*.*....*.\n.*.......*\n...o....o.\n.*.*....*.\n..........\n.o......o*\n2 9\n...***ooo\n.*o.*o.*o\n5 5\n*****\n*....\n*****\n....*\n*****", "output": ["..........\n...*....*.\n.*.o....o.\n.*........\n.*......**\n.o.*....o*\n\n....**ooo\n.*o**o.*o\n\n.....\n*...*\n*****\n*****\n*****"]}]
|
control_completion_000831
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid with $$$n$$$ rows and $$$m$$$ columns, and three types of cells: An empty cell, denoted with '.'. A stone, denoted with '*'. An obstacle, denoted with the lowercase Latin letter 'o'. All stones fall down until they meet the floor (the bottom row), an obstacle, or other stone which is already immovable. (In other words, all the stones just fall down as long as they can fall.)Simulate the process. What does the resulting grid look like?
Input Specification: The input consists of multiple test cases. The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 50$$$) — the number of rows and the number of columns in the grid, respectively. Then $$$n$$$ lines follow, each containing $$$m$$$ characters. Each of these characters is either '.', '*', or 'o' — an empty cell, a stone, or an obstacle, respectively.
Output Specification: For each test case, output a grid with $$$n$$$ rows and $$$m$$$ columns, showing the result of the process. You don't need to output a new line after each test, it is in the samples just for clarity.
Code:
c=input()
for _ in range(int(c)):
b=input().split()
a=[]
for i in range(int(b[0])):
a.append(list(input()))
for i in range(int(b[1])):
count=0
row=int(b[0])-1
for j in range(int(b[0])):
if a[row][i]=='.':
count+=1
elif # TODO: Your code here:
count=0
else:
a[row][i],a[row+count][i]='.',a[row][i]
row-=1
for i in range(int(b[0])):
print("".join(a[i]))
|
c=input()
for _ in range(int(c)):
b=input().split()
a=[]
for i in range(int(b[0])):
a.append(list(input()))
for i in range(int(b[1])):
count=0
row=int(b[0])-1
for j in range(int(b[0])):
if a[row][i]=='.':
count+=1
elif {{completion}}:
count=0
else:
a[row][i],a[row+count][i]='.',a[row][i]
row-=1
for i in range(int(b[0])):
print("".join(a[i]))
|
a[row][i]=='o'
|
[{"input": "3\n6 10\n.*.*....*.\n.*.......*\n...o....o.\n.*.*....*.\n..........\n.o......o*\n2 9\n...***ooo\n.*o.*o.*o\n5 5\n*****\n*....\n*****\n....*\n*****", "output": ["..........\n...*....*.\n.*.o....o.\n.*........\n.*......**\n.o.*....o*\n\n....**ooo\n.*o**o.*o\n\n.....\n*...*\n*****\n*****\n*****"]}]
|
control_completion_000832
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid with $$$n$$$ rows and $$$m$$$ columns, and three types of cells: An empty cell, denoted with '.'. A stone, denoted with '*'. An obstacle, denoted with the lowercase Latin letter 'o'. All stones fall down until they meet the floor (the bottom row), an obstacle, or other stone which is already immovable. (In other words, all the stones just fall down as long as they can fall.)Simulate the process. What does the resulting grid look like?
Input Specification: The input consists of multiple test cases. The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 50$$$) — the number of rows and the number of columns in the grid, respectively. Then $$$n$$$ lines follow, each containing $$$m$$$ characters. Each of these characters is either '.', '*', or 'o' — an empty cell, a stone, or an obstacle, respectively.
Output Specification: For each test case, output a grid with $$$n$$$ rows and $$$m$$$ columns, showing the result of the process. You don't need to output a new line after each test, it is in the samples just for clarity.
Code:
import sys
input = sys.stdin.readline
for _ in [0]*int(input()):
n,m=map(int,input().split())
s=['']*m
for _ in [0]*n:
s=[s[i]+j for i,j in zip(range(m),input())]
col=['']*n
k=0
for i in s:
bl=0
x=''
for j in i[::-1]:
if j=='o':
x=j+'.'*bl+x
bl=0
elif # TODO: Your code here:
x=j+x
else:
bl=bl+1
x='.'*bl+x
col=[col[ind]+val for val,ind in zip(x,range(n))]
k=k+1
for j in col:
print(j)
|
import sys
input = sys.stdin.readline
for _ in [0]*int(input()):
n,m=map(int,input().split())
s=['']*m
for _ in [0]*n:
s=[s[i]+j for i,j in zip(range(m),input())]
col=['']*n
k=0
for i in s:
bl=0
x=''
for j in i[::-1]:
if j=='o':
x=j+'.'*bl+x
bl=0
elif {{completion}}:
x=j+x
else:
bl=bl+1
x='.'*bl+x
col=[col[ind]+val for val,ind in zip(x,range(n))]
k=k+1
for j in col:
print(j)
|
j=='*'
|
[{"input": "3\n6 10\n.*.*....*.\n.*.......*\n...o....o.\n.*.*....*.\n..........\n.o......o*\n2 9\n...***ooo\n.*o.*o.*o\n5 5\n*****\n*....\n*****\n....*\n*****", "output": ["..........\n...*....*.\n.*.o....o.\n.*........\n.*......**\n.o.*....o*\n\n....**ooo\n.*o**o.*o\n\n.....\n*...*\n*****\n*****\n*****"]}]
|
control_completion_000833
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid with $$$n$$$ rows and $$$m$$$ columns, and three types of cells: An empty cell, denoted with '.'. A stone, denoted with '*'. An obstacle, denoted with the lowercase Latin letter 'o'. All stones fall down until they meet the floor (the bottom row), an obstacle, or other stone which is already immovable. (In other words, all the stones just fall down as long as they can fall.)Simulate the process. What does the resulting grid look like?
Input Specification: The input consists of multiple test cases. The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 50$$$) — the number of rows and the number of columns in the grid, respectively. Then $$$n$$$ lines follow, each containing $$$m$$$ characters. Each of these characters is either '.', '*', or 'o' — an empty cell, a stone, or an obstacle, respectively.
Output Specification: For each test case, output a grid with $$$n$$$ rows and $$$m$$$ columns, showing the result of the process. You don't need to output a new line after each test, it is in the samples just for clarity.
Code:
for i in range(int(input())):
n,m=map(int,input().split())
s=[list(input()) for j in range(n)]
for _ in range(n):
for a in reversed(range(n-1)):
for b in range(m):
if s[a][b]=='*':
if # TODO: Your code here:
continue
else:
s[a][b]='.'
s[a+1][b]='*'
for a in range(n):
print(*s[a],sep='')
|
for i in range(int(input())):
n,m=map(int,input().split())
s=[list(input()) for j in range(n)]
for _ in range(n):
for a in reversed(range(n-1)):
for b in range(m):
if s[a][b]=='*':
if {{completion}}:
continue
else:
s[a][b]='.'
s[a+1][b]='*'
for a in range(n):
print(*s[a],sep='')
|
s[a+1][b]=='o' or s[a+1][b]=='*'
|
[{"input": "3\n6 10\n.*.*....*.\n.*.......*\n...o....o.\n.*.*....*.\n..........\n.o......o*\n2 9\n...***ooo\n.*o.*o.*o\n5 5\n*****\n*....\n*****\n....*\n*****", "output": ["..........\n...*....*.\n.*.o....o.\n.*........\n.*......**\n.o.*....o*\n\n....**ooo\n.*o**o.*o\n\n.....\n*...*\n*****\n*****\n*****"]}]
|
control_completion_000834
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid with $$$n$$$ rows and $$$m$$$ columns, and three types of cells: An empty cell, denoted with '.'. A stone, denoted with '*'. An obstacle, denoted with the lowercase Latin letter 'o'. All stones fall down until they meet the floor (the bottom row), an obstacle, or other stone which is already immovable. (In other words, all the stones just fall down as long as they can fall.)Simulate the process. What does the resulting grid look like?
Input Specification: The input consists of multiple test cases. The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 50$$$) — the number of rows and the number of columns in the grid, respectively. Then $$$n$$$ lines follow, each containing $$$m$$$ characters. Each of these characters is either '.', '*', or 'o' — an empty cell, a stone, or an obstacle, respectively.
Output Specification: For each test case, output a grid with $$$n$$$ rows and $$$m$$$ columns, showing the result of the process. You don't need to output a new line after each test, it is in the samples just for clarity.
Code:
I=lambda:map(int,input().split())
for _ in range(int(input())):
n,m=I()
a=[input() for _ in range(n)]
at=[''.join(col).split('o') for col in zip(*a)]
f=lambda s:''.join(sorted(s,reverse=True))
at=['o'.join(map(f, col)) for col in at]
for # TODO: Your code here:
print(''.join(row))
|
I=lambda:map(int,input().split())
for _ in range(int(input())):
n,m=I()
a=[input() for _ in range(n)]
at=[''.join(col).split('o') for col in zip(*a)]
f=lambda s:''.join(sorted(s,reverse=True))
at=['o'.join(map(f, col)) for col in at]
for {{completion}}:
print(''.join(row))
|
row in zip(*at)
|
[{"input": "3\n6 10\n.*.*....*.\n.*.......*\n...o....o.\n.*.*....*.\n..........\n.o......o*\n2 9\n...***ooo\n.*o.*o.*o\n5 5\n*****\n*....\n*****\n....*\n*****", "output": ["..........\n...*....*.\n.*.o....o.\n.*........\n.*......**\n.o.*....o*\n\n....**ooo\n.*o**o.*o\n\n.....\n*...*\n*****\n*****\n*****"]}]
|
control_completion_000835
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid with $$$n$$$ rows and $$$m$$$ columns, and three types of cells: An empty cell, denoted with '.'. A stone, denoted with '*'. An obstacle, denoted with the lowercase Latin letter 'o'. All stones fall down until they meet the floor (the bottom row), an obstacle, or other stone which is already immovable. (In other words, all the stones just fall down as long as they can fall.)Simulate the process. What does the resulting grid look like?
Input Specification: The input consists of multiple test cases. The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 50$$$) — the number of rows and the number of columns in the grid, respectively. Then $$$n$$$ lines follow, each containing $$$m$$$ characters. Each of these characters is either '.', '*', or 'o' — an empty cell, a stone, or an obstacle, respectively.
Output Specification: For each test case, output a grid with $$$n$$$ rows and $$$m$$$ columns, showing the result of the process. You don't need to output a new line after each test, it is in the samples just for clarity.
Code:
I = input
for _ in range(int(I())):
n,m = map(int,I().split())
grid = [I().strip() for __ in range(n)]
res = []
for col in range(m):
newcol = ''
for # TODO: Your code here:
newcol += '.'*seg.count('.')+'*'*seg.count('*')+'o'
res.append(newcol[0:-1])
for row in range(n):
print(''.join(res[col][row] for col in range(m)))
|
I = input
for _ in range(int(I())):
n,m = map(int,I().split())
grid = [I().strip() for __ in range(n)]
res = []
for col in range(m):
newcol = ''
for {{completion}}:
newcol += '.'*seg.count('.')+'*'*seg.count('*')+'o'
res.append(newcol[0:-1])
for row in range(n):
print(''.join(res[col][row] for col in range(m)))
|
seg in (''.join(grid[row][col] for row in range(n))).split('o')
|
[{"input": "3\n6 10\n.*.*....*.\n.*.......*\n...o....o.\n.*.*....*.\n..........\n.o......o*\n2 9\n...***ooo\n.*o.*o.*o\n5 5\n*****\n*....\n*****\n....*\n*****", "output": ["..........\n...*....*.\n.*.o....o.\n.*........\n.*......**\n.o.*....o*\n\n....**ooo\n.*o**o.*o\n\n.....\n*...*\n*****\n*****\n*****"]}]
|
control_completion_000836
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid with $$$n$$$ rows and $$$m$$$ columns, and three types of cells: An empty cell, denoted with '.'. A stone, denoted with '*'. An obstacle, denoted with the lowercase Latin letter 'o'. All stones fall down until they meet the floor (the bottom row), an obstacle, or other stone which is already immovable. (In other words, all the stones just fall down as long as they can fall.)Simulate the process. What does the resulting grid look like?
Input Specification: The input consists of multiple test cases. The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 50$$$) — the number of rows and the number of columns in the grid, respectively. Then $$$n$$$ lines follow, each containing $$$m$$$ characters. Each of these characters is either '.', '*', or 'o' — an empty cell, a stone, or an obstacle, respectively.
Output Specification: For each test case, output a grid with $$$n$$$ rows and $$$m$$$ columns, showing the result of the process. You don't need to output a new line after each test, it is in the samples just for clarity.
Code:
for ii in range(int(input())):
n,m = map(int, input().split())
mat=[]
r=[0]*m
for jj in range(n):
a=list(input())
for kk in range(m):
if a[kk]=="*":
r[kk]+=1
a[kk]="."
elif a[kk]=="o":
while # TODO: Your code here:
mat[jj-r[kk]][kk]="*"
r[kk]-=1
mat.append(a)
for jj in range(m):
while r[jj]:
mat[n-r[jj]][jj]="*"
r[jj]-=1
for jj in range(n):
print("".join(mat[jj]))
|
for ii in range(int(input())):
n,m = map(int, input().split())
mat=[]
r=[0]*m
for jj in range(n):
a=list(input())
for kk in range(m):
if a[kk]=="*":
r[kk]+=1
a[kk]="."
elif a[kk]=="o":
while {{completion}}:
mat[jj-r[kk]][kk]="*"
r[kk]-=1
mat.append(a)
for jj in range(m):
while r[jj]:
mat[n-r[jj]][jj]="*"
r[jj]-=1
for jj in range(n):
print("".join(mat[jj]))
|
r[kk]
|
[{"input": "3\n6 10\n.*.*....*.\n.*.......*\n...o....o.\n.*.*....*.\n..........\n.o......o*\n2 9\n...***ooo\n.*o.*o.*o\n5 5\n*****\n*....\n*****\n....*\n*****", "output": ["..........\n...*....*.\n.*.o....o.\n.*........\n.*......**\n.o.*....o*\n\n....**ooo\n.*o**o.*o\n\n.....\n*...*\n*****\n*****\n*****"]}]
|
control_completion_000837
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid with $$$n$$$ rows and $$$m$$$ columns, and three types of cells: An empty cell, denoted with '.'. A stone, denoted with '*'. An obstacle, denoted with the lowercase Latin letter 'o'. All stones fall down until they meet the floor (the bottom row), an obstacle, or other stone which is already immovable. (In other words, all the stones just fall down as long as they can fall.)Simulate the process. What does the resulting grid look like?
Input Specification: The input consists of multiple test cases. The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 50$$$) — the number of rows and the number of columns in the grid, respectively. Then $$$n$$$ lines follow, each containing $$$m$$$ characters. Each of these characters is either '.', '*', or 'o' — an empty cell, a stone, or an obstacle, respectively.
Output Specification: For each test case, output a grid with $$$n$$$ rows and $$$m$$$ columns, showing the result of the process. You don't need to output a new line after each test, it is in the samples just for clarity.
Code:
t = int(input())
for i in range (t):
n, m = map(int,input().split())
arr = [[0]*m]*n
for j in range(n):
arr[j] = list(input())
# for h in range(m):
# print(arr[j][h])
for k in range(m):
for l in range(n-1, -1, -1):
if arr[l][k]=='.':
# print("yes")
for f in range(l-1,-1,-1):
if arr[f][k]=='o':
break
elif # TODO: Your code here:
# print("yes")
arr[f][k]='.'
arr[l][k]='*'
break
for g in range(n):
for h in range(m-1):
print(arr[g][h],end="")
print(arr[g][m-1],end="\n")
|
t = int(input())
for i in range (t):
n, m = map(int,input().split())
arr = [[0]*m]*n
for j in range(n):
arr[j] = list(input())
# for h in range(m):
# print(arr[j][h])
for k in range(m):
for l in range(n-1, -1, -1):
if arr[l][k]=='.':
# print("yes")
for f in range(l-1,-1,-1):
if arr[f][k]=='o':
break
elif {{completion}}:
# print("yes")
arr[f][k]='.'
arr[l][k]='*'
break
for g in range(n):
for h in range(m-1):
print(arr[g][h],end="")
print(arr[g][m-1],end="\n")
|
arr[f][k]=='*'
|
[{"input": "3\n6 10\n.*.*....*.\n.*.......*\n...o....o.\n.*.*....*.\n..........\n.o......o*\n2 9\n...***ooo\n.*o.*o.*o\n5 5\n*****\n*....\n*****\n....*\n*****", "output": ["..........\n...*....*.\n.*.o....o.\n.*........\n.*......**\n.o.*....o*\n\n....**ooo\n.*o**o.*o\n\n.....\n*...*\n*****\n*****\n*****"]}]
|
control_completion_000838
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Given $$$n$$$ strings, each of length $$$2$$$, consisting of lowercase Latin alphabet letters from 'a' to 'k', output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and the $$$i$$$-th string and the $$$j$$$-th string differ in exactly one position.In other words, count the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$.The answer may not fit into 32-bit integer type, so you should use 64-bit integers like long long in C++ to avoid integer overflow.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of strings. Then follows $$$n$$$ lines, the $$$i$$$-th of which containing a single string $$$s_i$$$ of length $$$2$$$, consisting of lowercase Latin letters from 'a' to 'k'. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: For each test case, print a single integer — the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$. Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++).
Notes: NoteFor the first test case the pairs that differ in exactly one position are: ("ab", "cb"), ("ab", "db"), ("ab", "aa"), ("cb", "db") and ("cb", "cc").For the second test case the pairs that differ in exactly one position are: ("aa", "ac"), ("aa", "ca"), ("cc", "ac"), ("cc", "ca"), ("ac", "aa") and ("ca", "aa").For the third test case, the are no pairs satisfying the conditions.
Code:
from collections import Counter
for _ in range(int(input())):
n = int(input())
num = Counter(input() for x in [1]*n)
cnt = 0
for x in num:
for y in num:
if # TODO: Your code here:
cnt+=num[x]*num[y]
print(cnt//2)
|
from collections import Counter
for _ in range(int(input())):
n = int(input())
num = Counter(input() for x in [1]*n)
cnt = 0
for x in num:
for y in num:
if {{completion}}:
cnt+=num[x]*num[y]
print(cnt//2)
|
x!=y and (x[0] == y[0] or x[1] == y[1])
|
[{"input": "4\n6\nab\ncb\ndb\naa\ncc\nef\n7\naa\nbb\ncc\nac\nca\nbb\naa\n4\nkk\nkk\nab\nab\n5\njf\njf\njk\njk\njk", "output": ["5\n6\n0\n6"]}]
|
control_completion_000864
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Given $$$n$$$ strings, each of length $$$2$$$, consisting of lowercase Latin alphabet letters from 'a' to 'k', output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and the $$$i$$$-th string and the $$$j$$$-th string differ in exactly one position.In other words, count the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$.The answer may not fit into 32-bit integer type, so you should use 64-bit integers like long long in C++ to avoid integer overflow.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of strings. Then follows $$$n$$$ lines, the $$$i$$$-th of which containing a single string $$$s_i$$$ of length $$$2$$$, consisting of lowercase Latin letters from 'a' to 'k'. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: For each test case, print a single integer — the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$. Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++).
Notes: NoteFor the first test case the pairs that differ in exactly one position are: ("ab", "cb"), ("ab", "db"), ("ab", "aa"), ("cb", "db") and ("cb", "cc").For the second test case the pairs that differ in exactly one position are: ("aa", "ac"), ("aa", "ca"), ("cc", "ac"), ("cc", "ca"), ("ac", "aa") and ("ca", "aa").For the third test case, the are no pairs satisfying the conditions.
Code:
from collections import Counter
from itertools import islice
from sys import stdin
LETTERS = 'abcdefghijk'
data = (line.strip() for line in stdin.readlines()[1:])
res = []
for line in data:
n = int(line)
s = 0
ctr = Counter()
for ab in islice(data, n):
a, b = ab
ctr[ab] += 1
for l in LETTERS:
if # TODO: Your code here:
s += ctr[f'{l}{b}']
if l != b:
s += ctr[f'{a}{l}']
res.append(s)
print('\n'.join(str(x) for x in res))
|
from collections import Counter
from itertools import islice
from sys import stdin
LETTERS = 'abcdefghijk'
data = (line.strip() for line in stdin.readlines()[1:])
res = []
for line in data:
n = int(line)
s = 0
ctr = Counter()
for ab in islice(data, n):
a, b = ab
ctr[ab] += 1
for l in LETTERS:
if {{completion}}:
s += ctr[f'{l}{b}']
if l != b:
s += ctr[f'{a}{l}']
res.append(s)
print('\n'.join(str(x) for x in res))
|
l != a
|
[{"input": "4\n6\nab\ncb\ndb\naa\ncc\nef\n7\naa\nbb\ncc\nac\nca\nbb\naa\n4\nkk\nkk\nab\nab\n5\njf\njf\njk\njk\njk", "output": ["5\n6\n0\n6"]}]
|
control_completion_000865
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Given $$$n$$$ strings, each of length $$$2$$$, consisting of lowercase Latin alphabet letters from 'a' to 'k', output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and the $$$i$$$-th string and the $$$j$$$-th string differ in exactly one position.In other words, count the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$.The answer may not fit into 32-bit integer type, so you should use 64-bit integers like long long in C++ to avoid integer overflow.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of strings. Then follows $$$n$$$ lines, the $$$i$$$-th of which containing a single string $$$s_i$$$ of length $$$2$$$, consisting of lowercase Latin letters from 'a' to 'k'. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: For each test case, print a single integer — the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$. Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++).
Notes: NoteFor the first test case the pairs that differ in exactly one position are: ("ab", "cb"), ("ab", "db"), ("ab", "aa"), ("cb", "db") and ("cb", "cc").For the second test case the pairs that differ in exactly one position are: ("aa", "ac"), ("aa", "ca"), ("cc", "ac"), ("cc", "ca"), ("ac", "aa") and ("ca", "aa").For the third test case, the are no pairs satisfying the conditions.
Code:
from collections import Counter
from itertools import islice
from sys import stdin
LETTERS = 'abcdefghijk'
data = (line.strip() for line in stdin.readlines()[1:])
res = []
for line in data:
n = int(line)
s = 0
ctr = Counter()
for ab in islice(data, n):
a, b = ab
ctr[ab] += 1
for l in LETTERS:
if l != a:
s += ctr[f'{l}{b}']
if # TODO: Your code here:
s += ctr[f'{a}{l}']
res.append(s)
print('\n'.join(str(x) for x in res))
|
from collections import Counter
from itertools import islice
from sys import stdin
LETTERS = 'abcdefghijk'
data = (line.strip() for line in stdin.readlines()[1:])
res = []
for line in data:
n = int(line)
s = 0
ctr = Counter()
for ab in islice(data, n):
a, b = ab
ctr[ab] += 1
for l in LETTERS:
if l != a:
s += ctr[f'{l}{b}']
if {{completion}}:
s += ctr[f'{a}{l}']
res.append(s)
print('\n'.join(str(x) for x in res))
|
l != b
|
[{"input": "4\n6\nab\ncb\ndb\naa\ncc\nef\n7\naa\nbb\ncc\nac\nca\nbb\naa\n4\nkk\nkk\nab\nab\n5\njf\njf\njk\njk\njk", "output": ["5\n6\n0\n6"]}]
|
control_completion_000866
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Given $$$n$$$ strings, each of length $$$2$$$, consisting of lowercase Latin alphabet letters from 'a' to 'k', output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and the $$$i$$$-th string and the $$$j$$$-th string differ in exactly one position.In other words, count the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$.The answer may not fit into 32-bit integer type, so you should use 64-bit integers like long long in C++ to avoid integer overflow.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of strings. Then follows $$$n$$$ lines, the $$$i$$$-th of which containing a single string $$$s_i$$$ of length $$$2$$$, consisting of lowercase Latin letters from 'a' to 'k'. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: For each test case, print a single integer — the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$. Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++).
Notes: NoteFor the first test case the pairs that differ in exactly one position are: ("ab", "cb"), ("ab", "db"), ("ab", "aa"), ("cb", "db") and ("cb", "cc").For the second test case the pairs that differ in exactly one position are: ("aa", "ac"), ("aa", "ca"), ("cc", "ac"), ("cc", "ca"), ("ac", "aa") and ("ca", "aa").For the third test case, the are no pairs satisfying the conditions.
Code:
for i in range(int(input())):
data = [[0 for l in range(11)] for k in range(11)]
for j in range(int(input())):
first, second = input()
data[ord(first)-ord('a')][ord(second)-ord('a')] += 1
answer = 0
for j in range(11):
for k in range(11):
for l in range(11):
if # TODO: Your code here:
answer += data[j][k]*data[l][k]
if k != l:
answer += data[j][k]*data[j][l]
print(answer//2)
|
for i in range(int(input())):
data = [[0 for l in range(11)] for k in range(11)]
for j in range(int(input())):
first, second = input()
data[ord(first)-ord('a')][ord(second)-ord('a')] += 1
answer = 0
for j in range(11):
for k in range(11):
for l in range(11):
if {{completion}}:
answer += data[j][k]*data[l][k]
if k != l:
answer += data[j][k]*data[j][l]
print(answer//2)
|
j != l
|
[{"input": "4\n6\nab\ncb\ndb\naa\ncc\nef\n7\naa\nbb\ncc\nac\nca\nbb\naa\n4\nkk\nkk\nab\nab\n5\njf\njf\njk\njk\njk", "output": ["5\n6\n0\n6"]}]
|
control_completion_000867
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Given $$$n$$$ strings, each of length $$$2$$$, consisting of lowercase Latin alphabet letters from 'a' to 'k', output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and the $$$i$$$-th string and the $$$j$$$-th string differ in exactly one position.In other words, count the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$.The answer may not fit into 32-bit integer type, so you should use 64-bit integers like long long in C++ to avoid integer overflow.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of strings. Then follows $$$n$$$ lines, the $$$i$$$-th of which containing a single string $$$s_i$$$ of length $$$2$$$, consisting of lowercase Latin letters from 'a' to 'k'. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: For each test case, print a single integer — the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$. Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++).
Notes: NoteFor the first test case the pairs that differ in exactly one position are: ("ab", "cb"), ("ab", "db"), ("ab", "aa"), ("cb", "db") and ("cb", "cc").For the second test case the pairs that differ in exactly one position are: ("aa", "ac"), ("aa", "ca"), ("cc", "ac"), ("cc", "ca"), ("ac", "aa") and ("ca", "aa").For the third test case, the are no pairs satisfying the conditions.
Code:
for i in range(int(input())):
data = [[0 for l in range(11)] for k in range(11)]
for j in range(int(input())):
first, second = input()
data[ord(first)-ord('a')][ord(second)-ord('a')] += 1
answer = 0
for j in range(11):
for k in range(11):
for l in range(11):
if j != l:
answer += data[j][k]*data[l][k]
if # TODO: Your code here:
answer += data[j][k]*data[j][l]
print(answer//2)
|
for i in range(int(input())):
data = [[0 for l in range(11)] for k in range(11)]
for j in range(int(input())):
first, second = input()
data[ord(first)-ord('a')][ord(second)-ord('a')] += 1
answer = 0
for j in range(11):
for k in range(11):
for l in range(11):
if j != l:
answer += data[j][k]*data[l][k]
if {{completion}}:
answer += data[j][k]*data[j][l]
print(answer//2)
|
k != l
|
[{"input": "4\n6\nab\ncb\ndb\naa\ncc\nef\n7\naa\nbb\ncc\nac\nca\nbb\naa\n4\nkk\nkk\nab\nab\n5\njf\njf\njk\njk\njk", "output": ["5\n6\n0\n6"]}]
|
control_completion_000868
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Given $$$n$$$ strings, each of length $$$2$$$, consisting of lowercase Latin alphabet letters from 'a' to 'k', output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and the $$$i$$$-th string and the $$$j$$$-th string differ in exactly one position.In other words, count the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$.The answer may not fit into 32-bit integer type, so you should use 64-bit integers like long long in C++ to avoid integer overflow.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of strings. Then follows $$$n$$$ lines, the $$$i$$$-th of which containing a single string $$$s_i$$$ of length $$$2$$$, consisting of lowercase Latin letters from 'a' to 'k'. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: For each test case, print a single integer — the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$. Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++).
Notes: NoteFor the first test case the pairs that differ in exactly one position are: ("ab", "cb"), ("ab", "db"), ("ab", "aa"), ("cb", "db") and ("cb", "cc").For the second test case the pairs that differ in exactly one position are: ("aa", "ac"), ("aa", "ca"), ("cc", "ac"), ("cc", "ca"), ("ac", "aa") and ("ca", "aa").For the third test case, the are no pairs satisfying the conditions.
Code:
from collections import defaultdict
ak = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k"]
t = int(input())
for _ in range(t):
count = 0
d = defaultdict(int)
n = int(input())
for i in range(n):
s = input()
for c in ak:
if c != s[0]:
if # TODO: Your code here:
count += d[c + s[1]]
if c != s[1]:
if d[s[0] + c] > 0:
count += d[s[0] + c]
d[s] += 1
print(count)
|
from collections import defaultdict
ak = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k"]
t = int(input())
for _ in range(t):
count = 0
d = defaultdict(int)
n = int(input())
for i in range(n):
s = input()
for c in ak:
if c != s[0]:
if {{completion}}:
count += d[c + s[1]]
if c != s[1]:
if d[s[0] + c] > 0:
count += d[s[0] + c]
d[s] += 1
print(count)
|
d[c + s[1]] > 0
|
[{"input": "4\n6\nab\ncb\ndb\naa\ncc\nef\n7\naa\nbb\ncc\nac\nca\nbb\naa\n4\nkk\nkk\nab\nab\n5\njf\njf\njk\njk\njk", "output": ["5\n6\n0\n6"]}]
|
control_completion_000869
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Given $$$n$$$ strings, each of length $$$2$$$, consisting of lowercase Latin alphabet letters from 'a' to 'k', output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and the $$$i$$$-th string and the $$$j$$$-th string differ in exactly one position.In other words, count the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$.The answer may not fit into 32-bit integer type, so you should use 64-bit integers like long long in C++ to avoid integer overflow.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of strings. Then follows $$$n$$$ lines, the $$$i$$$-th of which containing a single string $$$s_i$$$ of length $$$2$$$, consisting of lowercase Latin letters from 'a' to 'k'. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: For each test case, print a single integer — the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$. Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++).
Notes: NoteFor the first test case the pairs that differ in exactly one position are: ("ab", "cb"), ("ab", "db"), ("ab", "aa"), ("cb", "db") and ("cb", "cc").For the second test case the pairs that differ in exactly one position are: ("aa", "ac"), ("aa", "ca"), ("cc", "ac"), ("cc", "ca"), ("ac", "aa") and ("ca", "aa").For the third test case, the are no pairs satisfying the conditions.
Code:
from collections import defaultdict
ak = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k"]
t = int(input())
for _ in range(t):
count = 0
d = defaultdict(int)
n = int(input())
for i in range(n):
s = input()
for c in ak:
if c != s[0]:
if d[c + s[1]] > 0:
count += d[c + s[1]]
if c != s[1]:
if # TODO: Your code here:
count += d[s[0] + c]
d[s] += 1
print(count)
|
from collections import defaultdict
ak = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k"]
t = int(input())
for _ in range(t):
count = 0
d = defaultdict(int)
n = int(input())
for i in range(n):
s = input()
for c in ak:
if c != s[0]:
if d[c + s[1]] > 0:
count += d[c + s[1]]
if c != s[1]:
if {{completion}}:
count += d[s[0] + c]
d[s] += 1
print(count)
|
d[s[0] + c] > 0
|
[{"input": "4\n6\nab\ncb\ndb\naa\ncc\nef\n7\naa\nbb\ncc\nac\nca\nbb\naa\n4\nkk\nkk\nab\nab\n5\njf\njf\njk\njk\njk", "output": ["5\n6\n0\n6"]}]
|
control_completion_000870
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Given $$$n$$$ strings, each of length $$$2$$$, consisting of lowercase Latin alphabet letters from 'a' to 'k', output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and the $$$i$$$-th string and the $$$j$$$-th string differ in exactly one position.In other words, count the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$.The answer may not fit into 32-bit integer type, so you should use 64-bit integers like long long in C++ to avoid integer overflow.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of strings. Then follows $$$n$$$ lines, the $$$i$$$-th of which containing a single string $$$s_i$$$ of length $$$2$$$, consisting of lowercase Latin letters from 'a' to 'k'. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: For each test case, print a single integer — the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$. Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++).
Notes: NoteFor the first test case the pairs that differ in exactly one position are: ("ab", "cb"), ("ab", "db"), ("ab", "aa"), ("cb", "db") and ("cb", "cc").For the second test case the pairs that differ in exactly one position are: ("aa", "ac"), ("aa", "ca"), ("cc", "ac"), ("cc", "ca"), ("ac", "aa") and ("ca", "aa").For the third test case, the are no pairs satisfying the conditions.
Code:
for ii in range(int(input())):
n=int(input())
a=[]
co=0
x=set()
for jj in range(n):
a.append(input())
for jj in range(n):
mul=1
if jj not in x:
for kk in range(jj+1,n):
if a[jj][0]!=a[kk][0] and a[jj][1]==a[kk][1]:
co+=mul
elif # TODO: Your code here:
co+=mul
elif a[jj][0]==a[kk][0] and a[jj][1]==a[kk][1]:
mul+=1
x.add(kk)
print(co)
|
for ii in range(int(input())):
n=int(input())
a=[]
co=0
x=set()
for jj in range(n):
a.append(input())
for jj in range(n):
mul=1
if jj not in x:
for kk in range(jj+1,n):
if a[jj][0]!=a[kk][0] and a[jj][1]==a[kk][1]:
co+=mul
elif {{completion}}:
co+=mul
elif a[jj][0]==a[kk][0] and a[jj][1]==a[kk][1]:
mul+=1
x.add(kk)
print(co)
|
a[jj][0]==a[kk][0] and a[jj][1]!=a[kk][1]
|
[{"input": "4\n6\nab\ncb\ndb\naa\ncc\nef\n7\naa\nbb\ncc\nac\nca\nbb\naa\n4\nkk\nkk\nab\nab\n5\njf\njf\njk\njk\njk", "output": ["5\n6\n0\n6"]}]
|
control_completion_000871
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Given $$$n$$$ strings, each of length $$$2$$$, consisting of lowercase Latin alphabet letters from 'a' to 'k', output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and the $$$i$$$-th string and the $$$j$$$-th string differ in exactly one position.In other words, count the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$.The answer may not fit into 32-bit integer type, so you should use 64-bit integers like long long in C++ to avoid integer overflow.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of strings. Then follows $$$n$$$ lines, the $$$i$$$-th of which containing a single string $$$s_i$$$ of length $$$2$$$, consisting of lowercase Latin letters from 'a' to 'k'. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: For each test case, print a single integer — the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$. Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++).
Notes: NoteFor the first test case the pairs that differ in exactly one position are: ("ab", "cb"), ("ab", "db"), ("ab", "aa"), ("cb", "db") and ("cb", "cc").For the second test case the pairs that differ in exactly one position are: ("aa", "ac"), ("aa", "ca"), ("cc", "ac"), ("cc", "ca"), ("ac", "aa") and ("ca", "aa").For the third test case, the are no pairs satisfying the conditions.
Code:
t=int(input())
for i in range(t):
n=int(input())
result=0
dic1={}
dic2={}
dic3={}
for i in range(n):
S=input()
if # TODO: Your code here:
result+=dic1[S[0]]
dic1[S[0]]+=1
else:
dic1[S[0]]=1
if S[1] in dic2:
result+=dic2[S[1]]
dic2[S[1]]+=1
else:
dic2[S[1]]=1
if S in dic3:
result-=dic3[S]*2
dic3[S]+=1
else:
dic3[S]=1
print(result)
|
t=int(input())
for i in range(t):
n=int(input())
result=0
dic1={}
dic2={}
dic3={}
for i in range(n):
S=input()
if {{completion}}:
result+=dic1[S[0]]
dic1[S[0]]+=1
else:
dic1[S[0]]=1
if S[1] in dic2:
result+=dic2[S[1]]
dic2[S[1]]+=1
else:
dic2[S[1]]=1
if S in dic3:
result-=dic3[S]*2
dic3[S]+=1
else:
dic3[S]=1
print(result)
|
S[0] in dic1
|
[{"input": "4\n6\nab\ncb\ndb\naa\ncc\nef\n7\naa\nbb\ncc\nac\nca\nbb\naa\n4\nkk\nkk\nab\nab\n5\njf\njf\njk\njk\njk", "output": ["5\n6\n0\n6"]}]
|
control_completion_000872
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Given $$$n$$$ strings, each of length $$$2$$$, consisting of lowercase Latin alphabet letters from 'a' to 'k', output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and the $$$i$$$-th string and the $$$j$$$-th string differ in exactly one position.In other words, count the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$.The answer may not fit into 32-bit integer type, so you should use 64-bit integers like long long in C++ to avoid integer overflow.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of strings. Then follows $$$n$$$ lines, the $$$i$$$-th of which containing a single string $$$s_i$$$ of length $$$2$$$, consisting of lowercase Latin letters from 'a' to 'k'. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: For each test case, print a single integer — the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$. Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++).
Notes: NoteFor the first test case the pairs that differ in exactly one position are: ("ab", "cb"), ("ab", "db"), ("ab", "aa"), ("cb", "db") and ("cb", "cc").For the second test case the pairs that differ in exactly one position are: ("aa", "ac"), ("aa", "ca"), ("cc", "ac"), ("cc", "ca"), ("ac", "aa") and ("ca", "aa").For the third test case, the are no pairs satisfying the conditions.
Code:
t=int(input())
for i in range(t):
n=int(input())
result=0
dic1={}
dic2={}
dic3={}
for i in range(n):
S=input()
if S[0] in dic1:
result+=dic1[S[0]]
dic1[S[0]]+=1
else:
dic1[S[0]]=1
if # TODO: Your code here:
result+=dic2[S[1]]
dic2[S[1]]+=1
else:
dic2[S[1]]=1
if S in dic3:
result-=dic3[S]*2
dic3[S]+=1
else:
dic3[S]=1
print(result)
|
t=int(input())
for i in range(t):
n=int(input())
result=0
dic1={}
dic2={}
dic3={}
for i in range(n):
S=input()
if S[0] in dic1:
result+=dic1[S[0]]
dic1[S[0]]+=1
else:
dic1[S[0]]=1
if {{completion}}:
result+=dic2[S[1]]
dic2[S[1]]+=1
else:
dic2[S[1]]=1
if S in dic3:
result-=dic3[S]*2
dic3[S]+=1
else:
dic3[S]=1
print(result)
|
S[1] in dic2
|
[{"input": "4\n6\nab\ncb\ndb\naa\ncc\nef\n7\naa\nbb\ncc\nac\nca\nbb\naa\n4\nkk\nkk\nab\nab\n5\njf\njf\njk\njk\njk", "output": ["5\n6\n0\n6"]}]
|
control_completion_000873
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Given $$$n$$$ strings, each of length $$$2$$$, consisting of lowercase Latin alphabet letters from 'a' to 'k', output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and the $$$i$$$-th string and the $$$j$$$-th string differ in exactly one position.In other words, count the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$.The answer may not fit into 32-bit integer type, so you should use 64-bit integers like long long in C++ to avoid integer overflow.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of strings. Then follows $$$n$$$ lines, the $$$i$$$-th of which containing a single string $$$s_i$$$ of length $$$2$$$, consisting of lowercase Latin letters from 'a' to 'k'. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: For each test case, print a single integer — the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$. Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++).
Notes: NoteFor the first test case the pairs that differ in exactly one position are: ("ab", "cb"), ("ab", "db"), ("ab", "aa"), ("cb", "db") and ("cb", "cc").For the second test case the pairs that differ in exactly one position are: ("aa", "ac"), ("aa", "ca"), ("cc", "ac"), ("cc", "ca"), ("ac", "aa") and ("ca", "aa").For the third test case, the are no pairs satisfying the conditions.
Code:
for i in range(int(input())):
n= int(input())
a = dict()
b = dict()
c = dict()
ans = 0
for j in range(n):
d,e = str(input())
try:
ans += a[d]
a[d] += 1
except KeyError:
a[d] = 1
try:
ans += b[e]
b[e] += 1
except KeyError:
b[e] = 1
if # TODO: Your code here:
c[d+e] = 0
else:
ans -= c[d+e]
c[d+e] += 2
print(ans)
|
for i in range(int(input())):
n= int(input())
a = dict()
b = dict()
c = dict()
ans = 0
for j in range(n):
d,e = str(input())
try:
ans += a[d]
a[d] += 1
except KeyError:
a[d] = 1
try:
ans += b[e]
b[e] += 1
except KeyError:
b[e] = 1
if {{completion}}:
c[d+e] = 0
else:
ans -= c[d+e]
c[d+e] += 2
print(ans)
|
d+e not in c
|
[{"input": "4\n6\nab\ncb\ndb\naa\ncc\nef\n7\naa\nbb\ncc\nac\nca\nbb\naa\n4\nkk\nkk\nab\nab\n5\njf\njf\njk\njk\njk", "output": ["5\n6\n0\n6"]}]
|
control_completion_000874
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Given $$$n$$$ strings, each of length $$$2$$$, consisting of lowercase Latin alphabet letters from 'a' to 'k', output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and the $$$i$$$-th string and the $$$j$$$-th string differ in exactly one position.In other words, count the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$.The answer may not fit into 32-bit integer type, so you should use 64-bit integers like long long in C++ to avoid integer overflow.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of strings. Then follows $$$n$$$ lines, the $$$i$$$-th of which containing a single string $$$s_i$$$ of length $$$2$$$, consisting of lowercase Latin letters from 'a' to 'k'. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: For each test case, print a single integer — the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$. Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++).
Notes: NoteFor the first test case the pairs that differ in exactly one position are: ("ab", "cb"), ("ab", "db"), ("ab", "aa"), ("cb", "db") and ("cb", "cc").For the second test case the pairs that differ in exactly one position are: ("aa", "ac"), ("aa", "ca"), ("cc", "ac"), ("cc", "ca"), ("ac", "aa") and ("ca", "aa").For the third test case, the are no pairs satisfying the conditions.
Code:
from collections import Counter
t=int(input())
while(t!=0):
n=int(input())
s = Counter(input() for x in [1]*n)
cnt = 0
for x in s:
for y in s:
if# TODO: Your code here: cnt += s[x]*s[y]
print(cnt//2)
t-=1
|
from collections import Counter
t=int(input())
while(t!=0):
n=int(input())
s = Counter(input() for x in [1]*n)
cnt = 0
for x in s:
for y in s:
if{{completion}}: cnt += s[x]*s[y]
print(cnt//2)
t-=1
|
(x!=y and (x[1]==y[1] or x[0]==y[0]))
|
[{"input": "4\n6\nab\ncb\ndb\naa\ncc\nef\n7\naa\nbb\ncc\nac\nca\nbb\naa\n4\nkk\nkk\nab\nab\n5\njf\njf\njk\njk\njk", "output": ["5\n6\n0\n6"]}]
|
control_completion_000875
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Given $$$n$$$ strings, each of length $$$2$$$, consisting of lowercase Latin alphabet letters from 'a' to 'k', output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and the $$$i$$$-th string and the $$$j$$$-th string differ in exactly one position.In other words, count the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$.The answer may not fit into 32-bit integer type, so you should use 64-bit integers like long long in C++ to avoid integer overflow.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of strings. Then follows $$$n$$$ lines, the $$$i$$$-th of which containing a single string $$$s_i$$$ of length $$$2$$$, consisting of lowercase Latin letters from 'a' to 'k'. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: For each test case, print a single integer — the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$. Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++).
Notes: NoteFor the first test case the pairs that differ in exactly one position are: ("ab", "cb"), ("ab", "db"), ("ab", "aa"), ("cb", "db") and ("cb", "cc").For the second test case the pairs that differ in exactly one position are: ("aa", "ac"), ("aa", "ca"), ("cc", "ac"), ("cc", "ca"), ("ac", "aa") and ("ca", "aa").For the third test case, the are no pairs satisfying the conditions.
Code:
t = int(input())
for x in range(t):
n = int(input())
d1 = {}
for i in range(97,109):
for j in range(97,109):
d1[chr(i)+chr(j)] = 0
ans1 = 0
for y in range(n):
s = input()
for l in range(2):
for m in range(97,109):
a = list(s)
a[l] = chr(m)
a = ''.join(a)
if # TODO: Your code here:
continue
ans1+=d1[a]
d1[s]+=1
print(ans1)
|
t = int(input())
for x in range(t):
n = int(input())
d1 = {}
for i in range(97,109):
for j in range(97,109):
d1[chr(i)+chr(j)] = 0
ans1 = 0
for y in range(n):
s = input()
for l in range(2):
for m in range(97,109):
a = list(s)
a[l] = chr(m)
a = ''.join(a)
if {{completion}}:
continue
ans1+=d1[a]
d1[s]+=1
print(ans1)
|
a == s
|
[{"input": "4\n6\nab\ncb\ndb\naa\ncc\nef\n7\naa\nbb\ncc\nac\nca\nbb\naa\n4\nkk\nkk\nab\nab\n5\njf\njf\njk\njk\njk", "output": ["5\n6\n0\n6"]}]
|
control_completion_000876
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Given $$$n$$$ strings, each of length $$$2$$$, consisting of lowercase Latin alphabet letters from 'a' to 'k', output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and the $$$i$$$-th string and the $$$j$$$-th string differ in exactly one position.In other words, count the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$.The answer may not fit into 32-bit integer type, so you should use 64-bit integers like long long in C++ to avoid integer overflow.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of strings. Then follows $$$n$$$ lines, the $$$i$$$-th of which containing a single string $$$s_i$$$ of length $$$2$$$, consisting of lowercase Latin letters from 'a' to 'k'. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: For each test case, print a single integer — the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$. Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++).
Notes: NoteFor the first test case the pairs that differ in exactly one position are: ("ab", "cb"), ("ab", "db"), ("ab", "aa"), ("cb", "db") and ("cb", "cc").For the second test case the pairs that differ in exactly one position are: ("aa", "ac"), ("aa", "ca"), ("cc", "ac"), ("cc", "ca"), ("ac", "aa") and ("ca", "aa").For the third test case, the are no pairs satisfying the conditions.
Code:
for n in range(int(input())):
a = {}
for j in range(int(input())):
c = input()
if c not in a:
a[c] = 1
elif c in a:
a[c] += 1
count = 0
for i in a.keys():
for j in a.keys():
if # TODO: Your code here:
count += a[i] * a[j]
print(count // 2)
|
for n in range(int(input())):
a = {}
for j in range(int(input())):
c = input()
if c not in a:
a[c] = 1
elif c in a:
a[c] += 1
count = 0
for i in a.keys():
for j in a.keys():
if {{completion}}:
count += a[i] * a[j]
print(count // 2)
|
i != j and (i[0] == j[0] or i[1] == j[1])
|
[{"input": "4\n6\nab\ncb\ndb\naa\ncc\nef\n7\naa\nbb\ncc\nac\nca\nbb\naa\n4\nkk\nkk\nab\nab\n5\njf\njf\njk\njk\njk", "output": ["5\n6\n0\n6"]}]
|
control_completion_000877
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: A row of $$$n$$$ cells is given, all initially white. Using a stamp, you can stamp any two neighboring cells such that one becomes red and the other becomes blue. A stamp can be rotated, i.e. it can be used in both ways: as $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}$$$ and as $$$\color{red}{\texttt{R}}\color{blue}{\texttt{B}}$$$.During use, the stamp must completely fit on the given $$$n$$$ cells (it cannot be partially outside the cells). The stamp can be applied multiple times to the same cell. Each usage of the stamp recolors both cells that are under the stamp.For example, one possible sequence of stamps to make the picture $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\color{blue}{\texttt{B}}\texttt{W}$$$ could be $$$\texttt{WWWWW} \to \texttt{WW}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\texttt{W} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\texttt{W} \to \color{blue}{\texttt{B}}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}\texttt{W}$$$. Here $$$\texttt{W}$$$, $$$\color{red}{\texttt{R}}$$$, and $$$\color{blue}{\texttt{B}}$$$ represent a white, red, or blue cell, respectively, and the cells that the stamp is used on are marked with an underline.Given a final picture, is it possible to make it using the stamp zero or more times?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 10^5$$$) — the length of the picture. The second line of each test case contains a string $$$s$$$ — the picture you need to make. It is guaranteed that the length of $$$s$$$ is $$$n$$$ and that $$$s$$$ only consists of the characters $$$\texttt{W}$$$, $$$\texttt{R}$$$, and $$$\texttt{B}$$$, representing a white, red, or blue cell, respectively. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: Output $$$t$$$ lines, each of which contains the answer to the corresponding test case. As an answer, output "YES" if it possible to make the picture using the stamp zero or more times, and "NO" otherwise. You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).
Notes: NoteThe first test case is explained in the statement.For the second, third, and fourth test cases, it is not possible to stamp a single cell, so the answer is "NO".For the fifth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{blue}{\texttt{B}}$$$.For the sixth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}$$$.For the seventh test case, you don't need to use the stamp at all.
Code:
for # TODO: Your code here:print('YNEOS'[1in map(len,map(set,s[:-1].split('W')))::2])
|
for {{completion}}:print('YNEOS'[1in map(len,map(set,s[:-1].split('W')))::2])
|
s in[*open(0)][2::2]
|
[{"input": "12\n5\nBRBBW\n1\nB\n2\nWB\n2\nRW\n3\nBRB\n3\nRBB\n7\nWWWWWWW\n9\nRBWBWRRBW\n10\nBRBRBRBRRB\n12\nBBBRWWRRRWBR\n10\nBRBRBRBRBW\n5\nRBWBW", "output": ["YES\nNO\nNO\nNO\nYES\nYES\nYES\nNO\nYES\nNO\nYES\nNO"]}]
|
control_completion_000905
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: A row of $$$n$$$ cells is given, all initially white. Using a stamp, you can stamp any two neighboring cells such that one becomes red and the other becomes blue. A stamp can be rotated, i.e. it can be used in both ways: as $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}$$$ and as $$$\color{red}{\texttt{R}}\color{blue}{\texttt{B}}$$$.During use, the stamp must completely fit on the given $$$n$$$ cells (it cannot be partially outside the cells). The stamp can be applied multiple times to the same cell. Each usage of the stamp recolors both cells that are under the stamp.For example, one possible sequence of stamps to make the picture $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\color{blue}{\texttt{B}}\texttt{W}$$$ could be $$$\texttt{WWWWW} \to \texttt{WW}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\texttt{W} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\texttt{W} \to \color{blue}{\texttt{B}}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}\texttt{W}$$$. Here $$$\texttt{W}$$$, $$$\color{red}{\texttt{R}}$$$, and $$$\color{blue}{\texttt{B}}$$$ represent a white, red, or blue cell, respectively, and the cells that the stamp is used on are marked with an underline.Given a final picture, is it possible to make it using the stamp zero or more times?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 10^5$$$) — the length of the picture. The second line of each test case contains a string $$$s$$$ — the picture you need to make. It is guaranteed that the length of $$$s$$$ is $$$n$$$ and that $$$s$$$ only consists of the characters $$$\texttt{W}$$$, $$$\texttt{R}$$$, and $$$\texttt{B}$$$, representing a white, red, or blue cell, respectively. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: Output $$$t$$$ lines, each of which contains the answer to the corresponding test case. As an answer, output "YES" if it possible to make the picture using the stamp zero or more times, and "NO" otherwise. You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).
Notes: NoteThe first test case is explained in the statement.For the second, third, and fourth test cases, it is not possible to stamp a single cell, so the answer is "NO".For the fifth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{blue}{\texttt{B}}$$$.For the sixth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}$$$.For the seventh test case, you don't need to use the stamp at all.
Code:
for # TODO: Your code here:
l = int(input())
print("NO" if any (len(set(x)) == 1 for x in input().split('W') ) else "YES")
|
for {{completion}}:
l = int(input())
print("NO" if any (len(set(x)) == 1 for x in input().split('W') ) else "YES")
|
_ in range(int(input()))
|
[{"input": "12\n5\nBRBBW\n1\nB\n2\nWB\n2\nRW\n3\nBRB\n3\nRBB\n7\nWWWWWWW\n9\nRBWBWRRBW\n10\nBRBRBRBRRB\n12\nBBBRWWRRRWBR\n10\nBRBRBRBRBW\n5\nRBWBW", "output": ["YES\nNO\nNO\nNO\nYES\nYES\nYES\nNO\nYES\nNO\nYES\nNO"]}]
|
control_completion_000906
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: A row of $$$n$$$ cells is given, all initially white. Using a stamp, you can stamp any two neighboring cells such that one becomes red and the other becomes blue. A stamp can be rotated, i.e. it can be used in both ways: as $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}$$$ and as $$$\color{red}{\texttt{R}}\color{blue}{\texttt{B}}$$$.During use, the stamp must completely fit on the given $$$n$$$ cells (it cannot be partially outside the cells). The stamp can be applied multiple times to the same cell. Each usage of the stamp recolors both cells that are under the stamp.For example, one possible sequence of stamps to make the picture $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\color{blue}{\texttt{B}}\texttt{W}$$$ could be $$$\texttt{WWWWW} \to \texttt{WW}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\texttt{W} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\texttt{W} \to \color{blue}{\texttt{B}}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}\texttt{W}$$$. Here $$$\texttt{W}$$$, $$$\color{red}{\texttt{R}}$$$, and $$$\color{blue}{\texttt{B}}$$$ represent a white, red, or blue cell, respectively, and the cells that the stamp is used on are marked with an underline.Given a final picture, is it possible to make it using the stamp zero or more times?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 10^5$$$) — the length of the picture. The second line of each test case contains a string $$$s$$$ — the picture you need to make. It is guaranteed that the length of $$$s$$$ is $$$n$$$ and that $$$s$$$ only consists of the characters $$$\texttt{W}$$$, $$$\texttt{R}$$$, and $$$\texttt{B}$$$, representing a white, red, or blue cell, respectively. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: Output $$$t$$$ lines, each of which contains the answer to the corresponding test case. As an answer, output "YES" if it possible to make the picture using the stamp zero or more times, and "NO" otherwise. You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).
Notes: NoteThe first test case is explained in the statement.For the second, third, and fourth test cases, it is not possible to stamp a single cell, so the answer is "NO".For the fifth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{blue}{\texttt{B}}$$$.For the sixth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}$$$.For the seventh test case, you don't need to use the stamp at all.
Code:
def solve():
n = int(input())
s = input().split('W')
for i in s:
bs = 'B' in i
rs = 'R' in i
if # TODO: Your code here:
print('NO')
return
print('YES')
for t in range(int(input())):
solve()
|
def solve():
n = int(input())
s = input().split('W')
for i in s:
bs = 'B' in i
rs = 'R' in i
if {{completion}}:
print('NO')
return
print('YES')
for t in range(int(input())):
solve()
|
bs ^ rs
|
[{"input": "12\n5\nBRBBW\n1\nB\n2\nWB\n2\nRW\n3\nBRB\n3\nRBB\n7\nWWWWWWW\n9\nRBWBWRRBW\n10\nBRBRBRBRRB\n12\nBBBRWWRRRWBR\n10\nBRBRBRBRBW\n5\nRBWBW", "output": ["YES\nNO\nNO\nNO\nYES\nYES\nYES\nNO\nYES\nNO\nYES\nNO"]}]
|
control_completion_000907
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: A row of $$$n$$$ cells is given, all initially white. Using a stamp, you can stamp any two neighboring cells such that one becomes red and the other becomes blue. A stamp can be rotated, i.e. it can be used in both ways: as $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}$$$ and as $$$\color{red}{\texttt{R}}\color{blue}{\texttt{B}}$$$.During use, the stamp must completely fit on the given $$$n$$$ cells (it cannot be partially outside the cells). The stamp can be applied multiple times to the same cell. Each usage of the stamp recolors both cells that are under the stamp.For example, one possible sequence of stamps to make the picture $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\color{blue}{\texttt{B}}\texttt{W}$$$ could be $$$\texttt{WWWWW} \to \texttt{WW}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\texttt{W} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\texttt{W} \to \color{blue}{\texttt{B}}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}\texttt{W}$$$. Here $$$\texttt{W}$$$, $$$\color{red}{\texttt{R}}$$$, and $$$\color{blue}{\texttt{B}}$$$ represent a white, red, or blue cell, respectively, and the cells that the stamp is used on are marked with an underline.Given a final picture, is it possible to make it using the stamp zero or more times?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 10^5$$$) — the length of the picture. The second line of each test case contains a string $$$s$$$ — the picture you need to make. It is guaranteed that the length of $$$s$$$ is $$$n$$$ and that $$$s$$$ only consists of the characters $$$\texttt{W}$$$, $$$\texttt{R}$$$, and $$$\texttt{B}$$$, representing a white, red, or blue cell, respectively. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: Output $$$t$$$ lines, each of which contains the answer to the corresponding test case. As an answer, output "YES" if it possible to make the picture using the stamp zero or more times, and "NO" otherwise. You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).
Notes: NoteThe first test case is explained in the statement.For the second, third, and fourth test cases, it is not possible to stamp a single cell, so the answer is "NO".For the fifth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{blue}{\texttt{B}}$$$.For the sixth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}$$$.For the seventh test case, you don't need to use the stamp at all.
Code:
for # TODO: Your code here:print('YNEOS'[1in[len({*x})for x in s[:-1].split('W')]::2])
|
for {{completion}}:print('YNEOS'[1in[len({*x})for x in s[:-1].split('W')]::2])
|
s in[*open(0)][2::2]
|
[{"input": "12\n5\nBRBBW\n1\nB\n2\nWB\n2\nRW\n3\nBRB\n3\nRBB\n7\nWWWWWWW\n9\nRBWBWRRBW\n10\nBRBRBRBRRB\n12\nBBBRWWRRRWBR\n10\nBRBRBRBRBW\n5\nRBWBW", "output": ["YES\nNO\nNO\nNO\nYES\nYES\nYES\nNO\nYES\nNO\nYES\nNO"]}]
|
control_completion_000908
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: A row of $$$n$$$ cells is given, all initially white. Using a stamp, you can stamp any two neighboring cells such that one becomes red and the other becomes blue. A stamp can be rotated, i.e. it can be used in both ways: as $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}$$$ and as $$$\color{red}{\texttt{R}}\color{blue}{\texttt{B}}$$$.During use, the stamp must completely fit on the given $$$n$$$ cells (it cannot be partially outside the cells). The stamp can be applied multiple times to the same cell. Each usage of the stamp recolors both cells that are under the stamp.For example, one possible sequence of stamps to make the picture $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\color{blue}{\texttt{B}}\texttt{W}$$$ could be $$$\texttt{WWWWW} \to \texttt{WW}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\texttt{W} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\texttt{W} \to \color{blue}{\texttt{B}}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}\texttt{W}$$$. Here $$$\texttt{W}$$$, $$$\color{red}{\texttt{R}}$$$, and $$$\color{blue}{\texttt{B}}$$$ represent a white, red, or blue cell, respectively, and the cells that the stamp is used on are marked with an underline.Given a final picture, is it possible to make it using the stamp zero or more times?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 10^5$$$) — the length of the picture. The second line of each test case contains a string $$$s$$$ — the picture you need to make. It is guaranteed that the length of $$$s$$$ is $$$n$$$ and that $$$s$$$ only consists of the characters $$$\texttt{W}$$$, $$$\texttt{R}$$$, and $$$\texttt{B}$$$, representing a white, red, or blue cell, respectively. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: Output $$$t$$$ lines, each of which contains the answer to the corresponding test case. As an answer, output "YES" if it possible to make the picture using the stamp zero or more times, and "NO" otherwise. You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).
Notes: NoteThe first test case is explained in the statement.For the second, third, and fourth test cases, it is not possible to stamp a single cell, so the answer is "NO".For the fifth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{blue}{\texttt{B}}$$$.For the sixth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}$$$.For the seventh test case, you don't need to use the stamp at all.
Code:
g = input()
for i in range(int(g)):
input()
numb = input().split('W')
ans = 'yes'
for z in numb:
if z == '':
pass
else:
if # TODO: Your code here:
pass
else:
ans = 'no'
print(ans)
|
g = input()
for i in range(int(g)):
input()
numb = input().split('W')
ans = 'yes'
for z in numb:
if z == '':
pass
else:
if {{completion}}:
pass
else:
ans = 'no'
print(ans)
|
('R' in z) and ('B' in z)
|
[{"input": "12\n5\nBRBBW\n1\nB\n2\nWB\n2\nRW\n3\nBRB\n3\nRBB\n7\nWWWWWWW\n9\nRBWBWRRBW\n10\nBRBRBRBRRB\n12\nBBBRWWRRRWBR\n10\nBRBRBRBRBW\n5\nRBWBW", "output": ["YES\nNO\nNO\nNO\nYES\nYES\nYES\nNO\nYES\nNO\nYES\nNO"]}]
|
control_completion_000909
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: A row of $$$n$$$ cells is given, all initially white. Using a stamp, you can stamp any two neighboring cells such that one becomes red and the other becomes blue. A stamp can be rotated, i.e. it can be used in both ways: as $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}$$$ and as $$$\color{red}{\texttt{R}}\color{blue}{\texttt{B}}$$$.During use, the stamp must completely fit on the given $$$n$$$ cells (it cannot be partially outside the cells). The stamp can be applied multiple times to the same cell. Each usage of the stamp recolors both cells that are under the stamp.For example, one possible sequence of stamps to make the picture $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\color{blue}{\texttt{B}}\texttt{W}$$$ could be $$$\texttt{WWWWW} \to \texttt{WW}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\texttt{W} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\texttt{W} \to \color{blue}{\texttt{B}}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}\texttt{W}$$$. Here $$$\texttt{W}$$$, $$$\color{red}{\texttt{R}}$$$, and $$$\color{blue}{\texttt{B}}$$$ represent a white, red, or blue cell, respectively, and the cells that the stamp is used on are marked with an underline.Given a final picture, is it possible to make it using the stamp zero or more times?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 10^5$$$) — the length of the picture. The second line of each test case contains a string $$$s$$$ — the picture you need to make. It is guaranteed that the length of $$$s$$$ is $$$n$$$ and that $$$s$$$ only consists of the characters $$$\texttt{W}$$$, $$$\texttt{R}$$$, and $$$\texttt{B}$$$, representing a white, red, or blue cell, respectively. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: Output $$$t$$$ lines, each of which contains the answer to the corresponding test case. As an answer, output "YES" if it possible to make the picture using the stamp zero or more times, and "NO" otherwise. You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).
Notes: NoteThe first test case is explained in the statement.For the second, third, and fourth test cases, it is not possible to stamp a single cell, so the answer is "NO".For the fifth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{blue}{\texttt{B}}$$$.For the sixth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}$$$.For the seventh test case, you don't need to use the stamp at all.
Code:
for s in[*open(0)][2::2]:
b=0
for # TODO: Your code here:b|=len({*i})%2
print('YNEOS'[b::2])
|
for s in[*open(0)][2::2]:
b=0
for {{completion}}:b|=len({*i})%2
print('YNEOS'[b::2])
|
i in s[:-1].split('W')
|
[{"input": "12\n5\nBRBBW\n1\nB\n2\nWB\n2\nRW\n3\nBRB\n3\nRBB\n7\nWWWWWWW\n9\nRBWBWRRBW\n10\nBRBRBRBRRB\n12\nBBBRWWRRRWBR\n10\nBRBRBRBRBW\n5\nRBWBW", "output": ["YES\nNO\nNO\nNO\nYES\nYES\nYES\nNO\nYES\nNO\nYES\nNO"]}]
|
control_completion_000910
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: A row of $$$n$$$ cells is given, all initially white. Using a stamp, you can stamp any two neighboring cells such that one becomes red and the other becomes blue. A stamp can be rotated, i.e. it can be used in both ways: as $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}$$$ and as $$$\color{red}{\texttt{R}}\color{blue}{\texttt{B}}$$$.During use, the stamp must completely fit on the given $$$n$$$ cells (it cannot be partially outside the cells). The stamp can be applied multiple times to the same cell. Each usage of the stamp recolors both cells that are under the stamp.For example, one possible sequence of stamps to make the picture $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\color{blue}{\texttt{B}}\texttt{W}$$$ could be $$$\texttt{WWWWW} \to \texttt{WW}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\texttt{W} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\texttt{W} \to \color{blue}{\texttt{B}}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}\texttt{W}$$$. Here $$$\texttt{W}$$$, $$$\color{red}{\texttt{R}}$$$, and $$$\color{blue}{\texttt{B}}$$$ represent a white, red, or blue cell, respectively, and the cells that the stamp is used on are marked with an underline.Given a final picture, is it possible to make it using the stamp zero or more times?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 10^5$$$) — the length of the picture. The second line of each test case contains a string $$$s$$$ — the picture you need to make. It is guaranteed that the length of $$$s$$$ is $$$n$$$ and that $$$s$$$ only consists of the characters $$$\texttt{W}$$$, $$$\texttt{R}$$$, and $$$\texttt{B}$$$, representing a white, red, or blue cell, respectively. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: Output $$$t$$$ lines, each of which contains the answer to the corresponding test case. As an answer, output "YES" if it possible to make the picture using the stamp zero or more times, and "NO" otherwise. You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).
Notes: NoteThe first test case is explained in the statement.For the second, third, and fourth test cases, it is not possible to stamp a single cell, so the answer is "NO".For the fifth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{blue}{\texttt{B}}$$$.For the sixth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}$$$.For the seventh test case, you don't need to use the stamp at all.
Code:
t=int(input())
for i in range(t):
n=int(input())
s=input()
s=s.strip("W")
temp=list(s.split('W'))
for i in temp:
if i:
if # TODO: Your code here:
print("NO")
break
else:
print("YES")
|
t=int(input())
for i in range(t):
n=int(input())
s=input()
s=s.strip("W")
temp=list(s.split('W'))
for i in temp:
if i:
if {{completion}}:
print("NO")
break
else:
print("YES")
|
'B' not in i or 'R' not in i
|
[{"input": "12\n5\nBRBBW\n1\nB\n2\nWB\n2\nRW\n3\nBRB\n3\nRBB\n7\nWWWWWWW\n9\nRBWBWRRBW\n10\nBRBRBRBRRB\n12\nBBBRWWRRRWBR\n10\nBRBRBRBRBW\n5\nRBWBW", "output": ["YES\nNO\nNO\nNO\nYES\nYES\nYES\nNO\nYES\nNO\nYES\nNO"]}]
|
control_completion_000911
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: A row of $$$n$$$ cells is given, all initially white. Using a stamp, you can stamp any two neighboring cells such that one becomes red and the other becomes blue. A stamp can be rotated, i.e. it can be used in both ways: as $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}$$$ and as $$$\color{red}{\texttt{R}}\color{blue}{\texttt{B}}$$$.During use, the stamp must completely fit on the given $$$n$$$ cells (it cannot be partially outside the cells). The stamp can be applied multiple times to the same cell. Each usage of the stamp recolors both cells that are under the stamp.For example, one possible sequence of stamps to make the picture $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\color{blue}{\texttt{B}}\texttt{W}$$$ could be $$$\texttt{WWWWW} \to \texttt{WW}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\texttt{W} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\texttt{W} \to \color{blue}{\texttt{B}}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}\texttt{W}$$$. Here $$$\texttt{W}$$$, $$$\color{red}{\texttt{R}}$$$, and $$$\color{blue}{\texttt{B}}$$$ represent a white, red, or blue cell, respectively, and the cells that the stamp is used on are marked with an underline.Given a final picture, is it possible to make it using the stamp zero or more times?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 10^5$$$) — the length of the picture. The second line of each test case contains a string $$$s$$$ — the picture you need to make. It is guaranteed that the length of $$$s$$$ is $$$n$$$ and that $$$s$$$ only consists of the characters $$$\texttt{W}$$$, $$$\texttt{R}$$$, and $$$\texttt{B}$$$, representing a white, red, or blue cell, respectively. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: Output $$$t$$$ lines, each of which contains the answer to the corresponding test case. As an answer, output "YES" if it possible to make the picture using the stamp zero or more times, and "NO" otherwise. You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).
Notes: NoteThe first test case is explained in the statement.For the second, third, and fourth test cases, it is not possible to stamp a single cell, so the answer is "NO".For the fifth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{blue}{\texttt{B}}$$$.For the sixth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}$$$.For the seventh test case, you don't need to use the stamp at all.
Code:
for # TODO: Your code here:
num = int(input())
line = [elem for elem in input().split("W") if elem != ""]
print("YES" if all(["B" in elem and "R" in elem for elem in line]) else "NO")
|
for {{completion}}:
num = int(input())
line = [elem for elem in input().split("W") if elem != ""]
print("YES" if all(["B" in elem and "R" in elem for elem in line]) else "NO")
|
i in range(int(input()))
|
[{"input": "12\n5\nBRBBW\n1\nB\n2\nWB\n2\nRW\n3\nBRB\n3\nRBB\n7\nWWWWWWW\n9\nRBWBWRRBW\n10\nBRBRBRBRRB\n12\nBBBRWWRRRWBR\n10\nBRBRBRBRBW\n5\nRBWBW", "output": ["YES\nNO\nNO\nNO\nYES\nYES\nYES\nNO\nYES\nNO\nYES\nNO"]}]
|
control_completion_000912
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: A row of $$$n$$$ cells is given, all initially white. Using a stamp, you can stamp any two neighboring cells such that one becomes red and the other becomes blue. A stamp can be rotated, i.e. it can be used in both ways: as $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}$$$ and as $$$\color{red}{\texttt{R}}\color{blue}{\texttt{B}}$$$.During use, the stamp must completely fit on the given $$$n$$$ cells (it cannot be partially outside the cells). The stamp can be applied multiple times to the same cell. Each usage of the stamp recolors both cells that are under the stamp.For example, one possible sequence of stamps to make the picture $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\color{blue}{\texttt{B}}\texttt{W}$$$ could be $$$\texttt{WWWWW} \to \texttt{WW}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\texttt{W} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\texttt{W} \to \color{blue}{\texttt{B}}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}\texttt{W}$$$. Here $$$\texttt{W}$$$, $$$\color{red}{\texttt{R}}$$$, and $$$\color{blue}{\texttt{B}}$$$ represent a white, red, or blue cell, respectively, and the cells that the stamp is used on are marked with an underline.Given a final picture, is it possible to make it using the stamp zero or more times?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 10^5$$$) — the length of the picture. The second line of each test case contains a string $$$s$$$ — the picture you need to make. It is guaranteed that the length of $$$s$$$ is $$$n$$$ and that $$$s$$$ only consists of the characters $$$\texttt{W}$$$, $$$\texttt{R}$$$, and $$$\texttt{B}$$$, representing a white, red, or blue cell, respectively. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: Output $$$t$$$ lines, each of which contains the answer to the corresponding test case. As an answer, output "YES" if it possible to make the picture using the stamp zero or more times, and "NO" otherwise. You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).
Notes: NoteThe first test case is explained in the statement.For the second, third, and fourth test cases, it is not possible to stamp a single cell, so the answer is "NO".For the fifth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{blue}{\texttt{B}}$$$.For the sixth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}$$$.For the seventh test case, you don't need to use the stamp at all.
Code:
for s in[*open(0)][2::2]:
b = 0
for # TODO: Your code here:
b|=(len(set(i))==1)
print('YNEOS '[b::2])
|
for s in[*open(0)][2::2]:
b = 0
for {{completion}}:
b|=(len(set(i))==1)
print('YNEOS '[b::2])
|
i in s[:-1].split("W")
|
[{"input": "12\n5\nBRBBW\n1\nB\n2\nWB\n2\nRW\n3\nBRB\n3\nRBB\n7\nWWWWWWW\n9\nRBWBWRRBW\n10\nBRBRBRBRRB\n12\nBBBRWWRRRWBR\n10\nBRBRBRBRBW\n5\nRBWBW", "output": ["YES\nNO\nNO\nNO\nYES\nYES\nYES\nNO\nYES\nNO\nYES\nNO"]}]
|
control_completion_000913
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: A row of $$$n$$$ cells is given, all initially white. Using a stamp, you can stamp any two neighboring cells such that one becomes red and the other becomes blue. A stamp can be rotated, i.e. it can be used in both ways: as $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}$$$ and as $$$\color{red}{\texttt{R}}\color{blue}{\texttt{B}}$$$.During use, the stamp must completely fit on the given $$$n$$$ cells (it cannot be partially outside the cells). The stamp can be applied multiple times to the same cell. Each usage of the stamp recolors both cells that are under the stamp.For example, one possible sequence of stamps to make the picture $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\color{blue}{\texttt{B}}\texttt{W}$$$ could be $$$\texttt{WWWWW} \to \texttt{WW}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\texttt{W} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\texttt{W} \to \color{blue}{\texttt{B}}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}\texttt{W}$$$. Here $$$\texttt{W}$$$, $$$\color{red}{\texttt{R}}$$$, and $$$\color{blue}{\texttt{B}}$$$ represent a white, red, or blue cell, respectively, and the cells that the stamp is used on are marked with an underline.Given a final picture, is it possible to make it using the stamp zero or more times?
Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 10^5$$$) — the length of the picture. The second line of each test case contains a string $$$s$$$ — the picture you need to make. It is guaranteed that the length of $$$s$$$ is $$$n$$$ and that $$$s$$$ only consists of the characters $$$\texttt{W}$$$, $$$\texttt{R}$$$, and $$$\texttt{B}$$$, representing a white, red, or blue cell, respectively. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: Output $$$t$$$ lines, each of which contains the answer to the corresponding test case. As an answer, output "YES" if it possible to make the picture using the stamp zero or more times, and "NO" otherwise. You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).
Notes: NoteThe first test case is explained in the statement.For the second, third, and fourth test cases, it is not possible to stamp a single cell, so the answer is "NO".For the fifth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{blue}{\texttt{B}}$$$.For the sixth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}$$$.For the seventh test case, you don't need to use the stamp at all.
Code:
t = int(input())
Ans = [-1]*t
for z in range(t):
n = int(input())
l = input().split('W')
bad = False
for s in l:
b1 = 'R' in s
b2 = 'B' in s
if # TODO: Your code here:
bad = True
print("NO" if bad else "YES")
|
t = int(input())
Ans = [-1]*t
for z in range(t):
n = int(input())
l = input().split('W')
bad = False
for s in l:
b1 = 'R' in s
b2 = 'B' in s
if {{completion}}:
bad = True
print("NO" if bad else "YES")
|
(b1 ^ b2)
|
[{"input": "12\n5\nBRBBW\n1\nB\n2\nWB\n2\nRW\n3\nBRB\n3\nRBB\n7\nWWWWWWW\n9\nRBWBWRRBW\n10\nBRBRBRBRRB\n12\nBBBRWWRRRWBR\n10\nBRBRBRBRBW\n5\nRBWBW", "output": ["YES\nNO\nNO\nNO\nYES\nYES\nYES\nNO\nYES\nNO\nYES\nNO"]}]
|
control_completion_000914
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given an array $$$a$$$ consisting of $$$n$$$ positive integers, and an array $$$b$$$, with length $$$n$$$. Initially $$$b_i=0$$$ for each $$$1 \leq i \leq n$$$.In one move you can choose an integer $$$i$$$ ($$$1 \leq i \leq n$$$), and add $$$a_i$$$ to $$$b_i$$$ or subtract $$$a_i$$$ from $$$b_i$$$. What is the minimum number of moves needed to make $$$b$$$ increasing (that is, every element is strictly greater than every element before it)?
Input Specification: The first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 5000$$$). The second line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the elements of the array $$$a$$$.
Output Specification: Print a single integer, the minimum number of moves to make $$$b$$$ increasing.
Notes: NoteExample $$$1$$$: you can subtract $$$a_1$$$ from $$$b_1$$$, and add $$$a_3$$$, $$$a_4$$$, and $$$a_5$$$ to $$$b_3$$$, $$$b_4$$$, and $$$b_5$$$ respectively. The final array will be [$$$-1$$$, $$$0$$$, $$$3$$$, $$$4$$$, $$$5$$$] after $$$4$$$ moves.Example $$$2$$$: you can reach [$$$-3$$$, $$$-2$$$, $$$-1$$$, $$$0$$$, $$$1$$$, $$$2$$$, $$$3$$$] in $$$10$$$ moves.
Code:
n=int(input())
l=[int(i) for i in input().split()]
def f(l):
cur = 0
n = 0
for # TODO: Your code here:
n += cur // i + 1
cur = i * (cur // i + 1)
return n
print(min(f(l[i+1:])+f(l[:i][::-1]) for i in range(n)))
|
n=int(input())
l=[int(i) for i in input().split()]
def f(l):
cur = 0
n = 0
for {{completion}}:
n += cur // i + 1
cur = i * (cur // i + 1)
return n
print(min(f(l[i+1:])+f(l[:i][::-1]) for i in range(n)))
|
i in l
|
[{"input": "5\n1 2 3 4 5", "output": ["4"]}, {"input": "7\n1 2 1 2 1 2 1", "output": ["10"]}, {"input": "8\n1 8 2 7 3 6 4 5", "output": ["16"]}]
|
control_completion_000959
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given an array $$$a$$$ consisting of $$$n$$$ positive integers, and an array $$$b$$$, with length $$$n$$$. Initially $$$b_i=0$$$ for each $$$1 \leq i \leq n$$$.In one move you can choose an integer $$$i$$$ ($$$1 \leq i \leq n$$$), and add $$$a_i$$$ to $$$b_i$$$ or subtract $$$a_i$$$ from $$$b_i$$$. What is the minimum number of moves needed to make $$$b$$$ increasing (that is, every element is strictly greater than every element before it)?
Input Specification: The first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 5000$$$). The second line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the elements of the array $$$a$$$.
Output Specification: Print a single integer, the minimum number of moves to make $$$b$$$ increasing.
Notes: NoteExample $$$1$$$: you can subtract $$$a_1$$$ from $$$b_1$$$, and add $$$a_3$$$, $$$a_4$$$, and $$$a_5$$$ to $$$b_3$$$, $$$b_4$$$, and $$$b_5$$$ respectively. The final array will be [$$$-1$$$, $$$0$$$, $$$3$$$, $$$4$$$, $$$5$$$] after $$$4$$$ moves.Example $$$2$$$: you can reach [$$$-3$$$, $$$-2$$$, $$$-1$$$, $$$0$$$, $$$1$$$, $$$2$$$, $$$3$$$] in $$$10$$$ moves.
Code:
n = int(input().strip())
a = list(map(int, input().strip().split()))
ans = None
for i in range(n):
acc, p = 0, 0
for # TODO: Your code here:
x = (p - 1) // a[j]
acc += -x
p = x * a[j]
p = 0
for j in range(i+1, n):
x = (p + a[j]) // a[j]
acc += x
p = x * a[j]
ans = min(ans, acc) if ans is not None else acc
print(ans)
|
n = int(input().strip())
a = list(map(int, input().strip().split()))
ans = None
for i in range(n):
acc, p = 0, 0
for {{completion}}:
x = (p - 1) // a[j]
acc += -x
p = x * a[j]
p = 0
for j in range(i+1, n):
x = (p + a[j]) // a[j]
acc += x
p = x * a[j]
ans = min(ans, acc) if ans is not None else acc
print(ans)
|
j in range(i-1, -1, -1)
|
[{"input": "5\n1 2 3 4 5", "output": ["4"]}, {"input": "7\n1 2 1 2 1 2 1", "output": ["10"]}, {"input": "8\n1 8 2 7 3 6 4 5", "output": ["16"]}]
|
control_completion_000960
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given an array $$$a$$$ consisting of $$$n$$$ positive integers, and an array $$$b$$$, with length $$$n$$$. Initially $$$b_i=0$$$ for each $$$1 \leq i \leq n$$$.In one move you can choose an integer $$$i$$$ ($$$1 \leq i \leq n$$$), and add $$$a_i$$$ to $$$b_i$$$ or subtract $$$a_i$$$ from $$$b_i$$$. What is the minimum number of moves needed to make $$$b$$$ increasing (that is, every element is strictly greater than every element before it)?
Input Specification: The first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 5000$$$). The second line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the elements of the array $$$a$$$.
Output Specification: Print a single integer, the minimum number of moves to make $$$b$$$ increasing.
Notes: NoteExample $$$1$$$: you can subtract $$$a_1$$$ from $$$b_1$$$, and add $$$a_3$$$, $$$a_4$$$, and $$$a_5$$$ to $$$b_3$$$, $$$b_4$$$, and $$$b_5$$$ respectively. The final array will be [$$$-1$$$, $$$0$$$, $$$3$$$, $$$4$$$, $$$5$$$] after $$$4$$$ moves.Example $$$2$$$: you can reach [$$$-3$$$, $$$-2$$$, $$$-1$$$, $$$0$$$, $$$1$$$, $$$2$$$, $$$3$$$] in $$$10$$$ moves.
Code:
n = int(input().strip())
a = list(map(int, input().strip().split()))
ans = None
for i in range(n):
acc, p = 0, 0
for j in range(i-1, -1, -1):
x = (p - 1) // a[j]
acc += -x
p = x * a[j]
p = 0
for # TODO: Your code here:
x = (p + a[j]) // a[j]
acc += x
p = x * a[j]
ans = min(ans, acc) if ans is not None else acc
print(ans)
|
n = int(input().strip())
a = list(map(int, input().strip().split()))
ans = None
for i in range(n):
acc, p = 0, 0
for j in range(i-1, -1, -1):
x = (p - 1) // a[j]
acc += -x
p = x * a[j]
p = 0
for {{completion}}:
x = (p + a[j]) // a[j]
acc += x
p = x * a[j]
ans = min(ans, acc) if ans is not None else acc
print(ans)
|
j in range(i+1, n)
|
[{"input": "5\n1 2 3 4 5", "output": ["4"]}, {"input": "7\n1 2 1 2 1 2 1", "output": ["10"]}, {"input": "8\n1 8 2 7 3 6 4 5", "output": ["16"]}]
|
control_completion_000961
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given an array $$$a$$$ consisting of $$$n$$$ positive integers, and an array $$$b$$$, with length $$$n$$$. Initially $$$b_i=0$$$ for each $$$1 \leq i \leq n$$$.In one move you can choose an integer $$$i$$$ ($$$1 \leq i \leq n$$$), and add $$$a_i$$$ to $$$b_i$$$ or subtract $$$a_i$$$ from $$$b_i$$$. What is the minimum number of moves needed to make $$$b$$$ increasing (that is, every element is strictly greater than every element before it)?
Input Specification: The first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 5000$$$). The second line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the elements of the array $$$a$$$.
Output Specification: Print a single integer, the minimum number of moves to make $$$b$$$ increasing.
Notes: NoteExample $$$1$$$: you can subtract $$$a_1$$$ from $$$b_1$$$, and add $$$a_3$$$, $$$a_4$$$, and $$$a_5$$$ to $$$b_3$$$, $$$b_4$$$, and $$$b_5$$$ respectively. The final array will be [$$$-1$$$, $$$0$$$, $$$3$$$, $$$4$$$, $$$5$$$] after $$$4$$$ moves.Example $$$2$$$: you can reach [$$$-3$$$, $$$-2$$$, $$$-1$$$, $$$0$$$, $$$1$$$, $$$2$$$, $$$3$$$] in $$$10$$$ moves.
Code:
from math import ceil
n=int(input())
a=list(map(int,input().split()))
ans=float("inf")
for i in range(len(a)):
t=[0]*n
temp=0
j=i-1
prev =0
while # TODO: Your code here:
x=(ceil((prev+1)/a[j]))
temp+=x
prev=(a[j]*x)
j-=1
k=i+1
prev=0
while k<len(a):
x=(ceil((prev+1)/a[k]))
temp+=x
prev=(a[k]*x)
k+=1
ans=min(ans,temp)
print(int(ans))
|
from math import ceil
n=int(input())
a=list(map(int,input().split()))
ans=float("inf")
for i in range(len(a)):
t=[0]*n
temp=0
j=i-1
prev =0
while {{completion}}:
x=(ceil((prev+1)/a[j]))
temp+=x
prev=(a[j]*x)
j-=1
k=i+1
prev=0
while k<len(a):
x=(ceil((prev+1)/a[k]))
temp+=x
prev=(a[k]*x)
k+=1
ans=min(ans,temp)
print(int(ans))
|
j>=0
|
[{"input": "5\n1 2 3 4 5", "output": ["4"]}, {"input": "7\n1 2 1 2 1 2 1", "output": ["10"]}, {"input": "8\n1 8 2 7 3 6 4 5", "output": ["16"]}]
|
control_completion_000962
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given an array $$$a$$$ consisting of $$$n$$$ positive integers, and an array $$$b$$$, with length $$$n$$$. Initially $$$b_i=0$$$ for each $$$1 \leq i \leq n$$$.In one move you can choose an integer $$$i$$$ ($$$1 \leq i \leq n$$$), and add $$$a_i$$$ to $$$b_i$$$ or subtract $$$a_i$$$ from $$$b_i$$$. What is the minimum number of moves needed to make $$$b$$$ increasing (that is, every element is strictly greater than every element before it)?
Input Specification: The first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 5000$$$). The second line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the elements of the array $$$a$$$.
Output Specification: Print a single integer, the minimum number of moves to make $$$b$$$ increasing.
Notes: NoteExample $$$1$$$: you can subtract $$$a_1$$$ from $$$b_1$$$, and add $$$a_3$$$, $$$a_4$$$, and $$$a_5$$$ to $$$b_3$$$, $$$b_4$$$, and $$$b_5$$$ respectively. The final array will be [$$$-1$$$, $$$0$$$, $$$3$$$, $$$4$$$, $$$5$$$] after $$$4$$$ moves.Example $$$2$$$: you can reach [$$$-3$$$, $$$-2$$$, $$$-1$$$, $$$0$$$, $$$1$$$, $$$2$$$, $$$3$$$] in $$$10$$$ moves.
Code:
from math import ceil
n=int(input())
a=list(map(int,input().split()))
ans=float("inf")
for i in range(len(a)):
t=[0]*n
temp=0
j=i-1
prev =0
while j>=0:
x=(ceil((prev+1)/a[j]))
temp+=x
prev=(a[j]*x)
j-=1
k=i+1
prev=0
while # TODO: Your code here:
x=(ceil((prev+1)/a[k]))
temp+=x
prev=(a[k]*x)
k+=1
ans=min(ans,temp)
print(int(ans))
|
from math import ceil
n=int(input())
a=list(map(int,input().split()))
ans=float("inf")
for i in range(len(a)):
t=[0]*n
temp=0
j=i-1
prev =0
while j>=0:
x=(ceil((prev+1)/a[j]))
temp+=x
prev=(a[j]*x)
j-=1
k=i+1
prev=0
while {{completion}}:
x=(ceil((prev+1)/a[k]))
temp+=x
prev=(a[k]*x)
k+=1
ans=min(ans,temp)
print(int(ans))
|
k<len(a)
|
[{"input": "5\n1 2 3 4 5", "output": ["4"]}, {"input": "7\n1 2 1 2 1 2 1", "output": ["10"]}, {"input": "8\n1 8 2 7 3 6 4 5", "output": ["16"]}]
|
control_completion_000963
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given an array $$$a$$$ consisting of $$$n$$$ positive integers, and an array $$$b$$$, with length $$$n$$$. Initially $$$b_i=0$$$ for each $$$1 \leq i \leq n$$$.In one move you can choose an integer $$$i$$$ ($$$1 \leq i \leq n$$$), and add $$$a_i$$$ to $$$b_i$$$ or subtract $$$a_i$$$ from $$$b_i$$$. What is the minimum number of moves needed to make $$$b$$$ increasing (that is, every element is strictly greater than every element before it)?
Input Specification: The first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 5000$$$). The second line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the elements of the array $$$a$$$.
Output Specification: Print a single integer, the minimum number of moves to make $$$b$$$ increasing.
Notes: NoteExample $$$1$$$: you can subtract $$$a_1$$$ from $$$b_1$$$, and add $$$a_3$$$, $$$a_4$$$, and $$$a_5$$$ to $$$b_3$$$, $$$b_4$$$, and $$$b_5$$$ respectively. The final array will be [$$$-1$$$, $$$0$$$, $$$3$$$, $$$4$$$, $$$5$$$] after $$$4$$$ moves.Example $$$2$$$: you can reach [$$$-3$$$, $$$-2$$$, $$$-1$$$, $$$0$$$, $$$1$$$, $$$2$$$, $$$3$$$] in $$$10$$$ moves.
Code:
for _ in range(1):
n = int(input())
a = list(map(int, input().split()))
Min = 1e18
for l in range(n):
m = a[l]
answer = 1
for i in range(l-1, -1, -1):
answer += (m + a[i]) // a[i]
m = a[i] * ((m + a[i]) // a[i])
if l + 1 < n:
m = 0
for # TODO: Your code here:
answer += (m + a[i]) // a[i]
m = a[i] * ((m + a[i]) // a[i])
Min = min(answer, Min)
print(Min)
|
for _ in range(1):
n = int(input())
a = list(map(int, input().split()))
Min = 1e18
for l in range(n):
m = a[l]
answer = 1
for i in range(l-1, -1, -1):
answer += (m + a[i]) // a[i]
m = a[i] * ((m + a[i]) // a[i])
if l + 1 < n:
m = 0
for {{completion}}:
answer += (m + a[i]) // a[i]
m = a[i] * ((m + a[i]) // a[i])
Min = min(answer, Min)
print(Min)
|
i in range(l + 2, n)
|
[{"input": "5\n1 2 3 4 5", "output": ["4"]}, {"input": "7\n1 2 1 2 1 2 1", "output": ["10"]}, {"input": "8\n1 8 2 7 3 6 4 5", "output": ["16"]}]
|
control_completion_000964
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given an array $$$a$$$ consisting of $$$n$$$ positive integers, and an array $$$b$$$, with length $$$n$$$. Initially $$$b_i=0$$$ for each $$$1 \leq i \leq n$$$.In one move you can choose an integer $$$i$$$ ($$$1 \leq i \leq n$$$), and add $$$a_i$$$ to $$$b_i$$$ or subtract $$$a_i$$$ from $$$b_i$$$. What is the minimum number of moves needed to make $$$b$$$ increasing (that is, every element is strictly greater than every element before it)?
Input Specification: The first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 5000$$$). The second line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the elements of the array $$$a$$$.
Output Specification: Print a single integer, the minimum number of moves to make $$$b$$$ increasing.
Notes: NoteExample $$$1$$$: you can subtract $$$a_1$$$ from $$$b_1$$$, and add $$$a_3$$$, $$$a_4$$$, and $$$a_5$$$ to $$$b_3$$$, $$$b_4$$$, and $$$b_5$$$ respectively. The final array will be [$$$-1$$$, $$$0$$$, $$$3$$$, $$$4$$$, $$$5$$$] after $$$4$$$ moves.Example $$$2$$$: you can reach [$$$-3$$$, $$$-2$$$, $$$-1$$$, $$$0$$$, $$$1$$$, $$$2$$$, $$$3$$$] in $$$10$$$ moves.
Code:
m=int(input())
a=[int(i)for i in input().split()]
t1,min=0,10**20
while(t1<m):
t2=t1
k,t=0,0
while# TODO: Your code here:
t+=(k//a[t2+1]+1)
k=a[t2+1]*(k//a[t2+1]+1)
t2+=1
t2=t1
k=0
while(t2>0):
t+=(k//a[t2-1]+1)
k=a[t2-1]*(k//a[t2-1]+1)
t2-=1
if(min>t):
min=t
t1+=1
print(min)
|
m=int(input())
a=[int(i)for i in input().split()]
t1,min=0,10**20
while(t1<m):
t2=t1
k,t=0,0
while{{completion}}:
t+=(k//a[t2+1]+1)
k=a[t2+1]*(k//a[t2+1]+1)
t2+=1
t2=t1
k=0
while(t2>0):
t+=(k//a[t2-1]+1)
k=a[t2-1]*(k//a[t2-1]+1)
t2-=1
if(min>t):
min=t
t1+=1
print(min)
|
(t2<m-1)
|
[{"input": "5\n1 2 3 4 5", "output": ["4"]}, {"input": "7\n1 2 1 2 1 2 1", "output": ["10"]}, {"input": "8\n1 8 2 7 3 6 4 5", "output": ["16"]}]
|
control_completion_000965
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given an array $$$a$$$ consisting of $$$n$$$ positive integers, and an array $$$b$$$, with length $$$n$$$. Initially $$$b_i=0$$$ for each $$$1 \leq i \leq n$$$.In one move you can choose an integer $$$i$$$ ($$$1 \leq i \leq n$$$), and add $$$a_i$$$ to $$$b_i$$$ or subtract $$$a_i$$$ from $$$b_i$$$. What is the minimum number of moves needed to make $$$b$$$ increasing (that is, every element is strictly greater than every element before it)?
Input Specification: The first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 5000$$$). The second line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the elements of the array $$$a$$$.
Output Specification: Print a single integer, the minimum number of moves to make $$$b$$$ increasing.
Notes: NoteExample $$$1$$$: you can subtract $$$a_1$$$ from $$$b_1$$$, and add $$$a_3$$$, $$$a_4$$$, and $$$a_5$$$ to $$$b_3$$$, $$$b_4$$$, and $$$b_5$$$ respectively. The final array will be [$$$-1$$$, $$$0$$$, $$$3$$$, $$$4$$$, $$$5$$$] after $$$4$$$ moves.Example $$$2$$$: you can reach [$$$-3$$$, $$$-2$$$, $$$-1$$$, $$$0$$$, $$$1$$$, $$$2$$$, $$$3$$$] in $$$10$$$ moves.
Code:
m=int(input())
a=[int(i)for i in input().split()]
t1,min=0,10**20
while(t1<m):
t2=t1
k,t=0,0
while(t2<m-1):
t+=(k//a[t2+1]+1)
k=a[t2+1]*(k//a[t2+1]+1)
t2+=1
t2=t1
k=0
while# TODO: Your code here:
t+=(k//a[t2-1]+1)
k=a[t2-1]*(k//a[t2-1]+1)
t2-=1
if(min>t):
min=t
t1+=1
print(min)
|
m=int(input())
a=[int(i)for i in input().split()]
t1,min=0,10**20
while(t1<m):
t2=t1
k,t=0,0
while(t2<m-1):
t+=(k//a[t2+1]+1)
k=a[t2+1]*(k//a[t2+1]+1)
t2+=1
t2=t1
k=0
while{{completion}}:
t+=(k//a[t2-1]+1)
k=a[t2-1]*(k//a[t2-1]+1)
t2-=1
if(min>t):
min=t
t1+=1
print(min)
|
(t2>0)
|
[{"input": "5\n1 2 3 4 5", "output": ["4"]}, {"input": "7\n1 2 1 2 1 2 1", "output": ["10"]}, {"input": "8\n1 8 2 7 3 6 4 5", "output": ["16"]}]
|
control_completion_000966
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given an array $$$a$$$ consisting of $$$n$$$ positive integers, and an array $$$b$$$, with length $$$n$$$. Initially $$$b_i=0$$$ for each $$$1 \leq i \leq n$$$.In one move you can choose an integer $$$i$$$ ($$$1 \leq i \leq n$$$), and add $$$a_i$$$ to $$$b_i$$$ or subtract $$$a_i$$$ from $$$b_i$$$. What is the minimum number of moves needed to make $$$b$$$ increasing (that is, every element is strictly greater than every element before it)?
Input Specification: The first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 5000$$$). The second line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the elements of the array $$$a$$$.
Output Specification: Print a single integer, the minimum number of moves to make $$$b$$$ increasing.
Notes: NoteExample $$$1$$$: you can subtract $$$a_1$$$ from $$$b_1$$$, and add $$$a_3$$$, $$$a_4$$$, and $$$a_5$$$ to $$$b_3$$$, $$$b_4$$$, and $$$b_5$$$ respectively. The final array will be [$$$-1$$$, $$$0$$$, $$$3$$$, $$$4$$$, $$$5$$$] after $$$4$$$ moves.Example $$$2$$$: you can reach [$$$-3$$$, $$$-2$$$, $$$-1$$$, $$$0$$$, $$$1$$$, $$$2$$$, $$$3$$$] in $$$10$$$ moves.
Code:
import math
n = int(input())
a = list(map(int, input().split(' '))) # numbers w/ ws
c = None
d = 0
for i in range(len(a)):
p = 0
t = 0
for # TODO: Your code here:
d = math.ceil((t+1)/k)
t = k*d
p += d
t = 0
for k in reversed(a[:i]):
d = math.ceil((t+1)/k)
t = k*d
p += d
if c == None or p < c:
c = p
print(c)
|
import math
n = int(input())
a = list(map(int, input().split(' '))) # numbers w/ ws
c = None
d = 0
for i in range(len(a)):
p = 0
t = 0
for {{completion}}:
d = math.ceil((t+1)/k)
t = k*d
p += d
t = 0
for k in reversed(a[:i]):
d = math.ceil((t+1)/k)
t = k*d
p += d
if c == None or p < c:
c = p
print(c)
|
k in a[i+1:]
|
[{"input": "5\n1 2 3 4 5", "output": ["4"]}, {"input": "7\n1 2 1 2 1 2 1", "output": ["10"]}, {"input": "8\n1 8 2 7 3 6 4 5", "output": ["16"]}]
|
control_completion_000967
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given an array $$$a$$$ consisting of $$$n$$$ positive integers, and an array $$$b$$$, with length $$$n$$$. Initially $$$b_i=0$$$ for each $$$1 \leq i \leq n$$$.In one move you can choose an integer $$$i$$$ ($$$1 \leq i \leq n$$$), and add $$$a_i$$$ to $$$b_i$$$ or subtract $$$a_i$$$ from $$$b_i$$$. What is the minimum number of moves needed to make $$$b$$$ increasing (that is, every element is strictly greater than every element before it)?
Input Specification: The first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 5000$$$). The second line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the elements of the array $$$a$$$.
Output Specification: Print a single integer, the minimum number of moves to make $$$b$$$ increasing.
Notes: NoteExample $$$1$$$: you can subtract $$$a_1$$$ from $$$b_1$$$, and add $$$a_3$$$, $$$a_4$$$, and $$$a_5$$$ to $$$b_3$$$, $$$b_4$$$, and $$$b_5$$$ respectively. The final array will be [$$$-1$$$, $$$0$$$, $$$3$$$, $$$4$$$, $$$5$$$] after $$$4$$$ moves.Example $$$2$$$: you can reach [$$$-3$$$, $$$-2$$$, $$$-1$$$, $$$0$$$, $$$1$$$, $$$2$$$, $$$3$$$] in $$$10$$$ moves.
Code:
import math
n = int(input())
a = list(map(int, input().split(' '))) # numbers w/ ws
c = None
d = 0
for i in range(len(a)):
p = 0
t = 0
for k in a[i+1:]:
d = math.ceil((t+1)/k)
t = k*d
p += d
t = 0
for # TODO: Your code here:
d = math.ceil((t+1)/k)
t = k*d
p += d
if c == None or p < c:
c = p
print(c)
|
import math
n = int(input())
a = list(map(int, input().split(' '))) # numbers w/ ws
c = None
d = 0
for i in range(len(a)):
p = 0
t = 0
for k in a[i+1:]:
d = math.ceil((t+1)/k)
t = k*d
p += d
t = 0
for {{completion}}:
d = math.ceil((t+1)/k)
t = k*d
p += d
if c == None or p < c:
c = p
print(c)
|
k in reversed(a[:i])
|
[{"input": "5\n1 2 3 4 5", "output": ["4"]}, {"input": "7\n1 2 1 2 1 2 1", "output": ["10"]}, {"input": "8\n1 8 2 7 3 6 4 5", "output": ["16"]}]
|
control_completion_000968
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given an array $$$a$$$ consisting of $$$n$$$ positive integers, and an array $$$b$$$, with length $$$n$$$. Initially $$$b_i=0$$$ for each $$$1 \leq i \leq n$$$.In one move you can choose an integer $$$i$$$ ($$$1 \leq i \leq n$$$), and add $$$a_i$$$ to $$$b_i$$$ or subtract $$$a_i$$$ from $$$b_i$$$. What is the minimum number of moves needed to make $$$b$$$ increasing (that is, every element is strictly greater than every element before it)?
Input Specification: The first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 5000$$$). The second line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the elements of the array $$$a$$$.
Output Specification: Print a single integer, the minimum number of moves to make $$$b$$$ increasing.
Notes: NoteExample $$$1$$$: you can subtract $$$a_1$$$ from $$$b_1$$$, and add $$$a_3$$$, $$$a_4$$$, and $$$a_5$$$ to $$$b_3$$$, $$$b_4$$$, and $$$b_5$$$ respectively. The final array will be [$$$-1$$$, $$$0$$$, $$$3$$$, $$$4$$$, $$$5$$$] after $$$4$$$ moves.Example $$$2$$$: you can reach [$$$-3$$$, $$$-2$$$, $$$-1$$$, $$$0$$$, $$$1$$$, $$$2$$$, $$$3$$$] in $$$10$$$ moves.
Code:
def f(b, i):
return e(b[::-1], i)
def e(b, i):
if b == []:
return 0
count = 0
ggg = [0] * len(b)
for # TODO: Your code here:
ggg[i] = (b[i - 1] * ggg[i - 1]) // b[i] + 1
count += ggg[i]
return count
def c(b, i):
return e(b[i + 1:], 0) + f(b[:i], 0)
a = int(input())
b = input().split()
for i in range(a):
b[i] = int(b[i])
d = c(b, 1)
for i in range(2, a - 1):
d = min(d, c(b, i))
print(d)
|
def f(b, i):
return e(b[::-1], i)
def e(b, i):
if b == []:
return 0
count = 0
ggg = [0] * len(b)
for {{completion}}:
ggg[i] = (b[i - 1] * ggg[i - 1]) // b[i] + 1
count += ggg[i]
return count
def c(b, i):
return e(b[i + 1:], 0) + f(b[:i], 0)
a = int(input())
b = input().split()
for i in range(a):
b[i] = int(b[i])
d = c(b, 1)
for i in range(2, a - 1):
d = min(d, c(b, i))
print(d)
|
i in range(len(b))
|
[{"input": "5\n1 2 3 4 5", "output": ["4"]}, {"input": "7\n1 2 1 2 1 2 1", "output": ["10"]}, {"input": "8\n1 8 2 7 3 6 4 5", "output": ["16"]}]
|
control_completion_000969
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given an array $$$a$$$ consisting of $$$n$$$ positive integers, and an array $$$b$$$, with length $$$n$$$. Initially $$$b_i=0$$$ for each $$$1 \leq i \leq n$$$.In one move you can choose an integer $$$i$$$ ($$$1 \leq i \leq n$$$), and add $$$a_i$$$ to $$$b_i$$$ or subtract $$$a_i$$$ from $$$b_i$$$. What is the minimum number of moves needed to make $$$b$$$ increasing (that is, every element is strictly greater than every element before it)?
Input Specification: The first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 5000$$$). The second line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the elements of the array $$$a$$$.
Output Specification: Print a single integer, the minimum number of moves to make $$$b$$$ increasing.
Notes: NoteExample $$$1$$$: you can subtract $$$a_1$$$ from $$$b_1$$$, and add $$$a_3$$$, $$$a_4$$$, and $$$a_5$$$ to $$$b_3$$$, $$$b_4$$$, and $$$b_5$$$ respectively. The final array will be [$$$-1$$$, $$$0$$$, $$$3$$$, $$$4$$$, $$$5$$$] after $$$4$$$ moves.Example $$$2$$$: you can reach [$$$-3$$$, $$$-2$$$, $$$-1$$$, $$$0$$$, $$$1$$$, $$$2$$$, $$$3$$$] in $$$10$$$ moves.
Code:
n=int(input())
a=list(map(int,input().split()))
b=[int(0) for _ in range(n)]
m=1e18
for i in range(n):
c=0
p=0
for # TODO: Your code here:
p+=a[j]-p%a[j]
c+=p//a[j]
p=0
for j in range(i-1,-1,-1):
p+=a[j]-p%a[j]
c+=p//a[j]
m=min(m,c)
print(m)
|
n=int(input())
a=list(map(int,input().split()))
b=[int(0) for _ in range(n)]
m=1e18
for i in range(n):
c=0
p=0
for {{completion}}:
p+=a[j]-p%a[j]
c+=p//a[j]
p=0
for j in range(i-1,-1,-1):
p+=a[j]-p%a[j]
c+=p//a[j]
m=min(m,c)
print(m)
|
j in range(i+1,len(b))
|
[{"input": "5\n1 2 3 4 5", "output": ["4"]}, {"input": "7\n1 2 1 2 1 2 1", "output": ["10"]}, {"input": "8\n1 8 2 7 3 6 4 5", "output": ["16"]}]
|
control_completion_000970
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given an array $$$a$$$ consisting of $$$n$$$ positive integers, and an array $$$b$$$, with length $$$n$$$. Initially $$$b_i=0$$$ for each $$$1 \leq i \leq n$$$.In one move you can choose an integer $$$i$$$ ($$$1 \leq i \leq n$$$), and add $$$a_i$$$ to $$$b_i$$$ or subtract $$$a_i$$$ from $$$b_i$$$. What is the minimum number of moves needed to make $$$b$$$ increasing (that is, every element is strictly greater than every element before it)?
Input Specification: The first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 5000$$$). The second line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the elements of the array $$$a$$$.
Output Specification: Print a single integer, the minimum number of moves to make $$$b$$$ increasing.
Notes: NoteExample $$$1$$$: you can subtract $$$a_1$$$ from $$$b_1$$$, and add $$$a_3$$$, $$$a_4$$$, and $$$a_5$$$ to $$$b_3$$$, $$$b_4$$$, and $$$b_5$$$ respectively. The final array will be [$$$-1$$$, $$$0$$$, $$$3$$$, $$$4$$$, $$$5$$$] after $$$4$$$ moves.Example $$$2$$$: you can reach [$$$-3$$$, $$$-2$$$, $$$-1$$$, $$$0$$$, $$$1$$$, $$$2$$$, $$$3$$$] in $$$10$$$ moves.
Code:
n=int(input())
a=list(map(int,input().split()))
b=[int(0) for _ in range(n)]
m=1e18
for i in range(n):
c=0
p=0
for j in range(i+1,len(b)):
p+=a[j]-p%a[j]
c+=p//a[j]
p=0
for # TODO: Your code here:
p+=a[j]-p%a[j]
c+=p//a[j]
m=min(m,c)
print(m)
|
n=int(input())
a=list(map(int,input().split()))
b=[int(0) for _ in range(n)]
m=1e18
for i in range(n):
c=0
p=0
for j in range(i+1,len(b)):
p+=a[j]-p%a[j]
c+=p//a[j]
p=0
for {{completion}}:
p+=a[j]-p%a[j]
c+=p//a[j]
m=min(m,c)
print(m)
|
j in range(i-1,-1,-1)
|
[{"input": "5\n1 2 3 4 5", "output": ["4"]}, {"input": "7\n1 2 1 2 1 2 1", "output": ["10"]}, {"input": "8\n1 8 2 7 3 6 4 5", "output": ["16"]}]
|
control_completion_000971
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: $$$m$$$ chairs are arranged in a circle sequentially. The chairs are numbered from $$$0$$$ to $$$m-1$$$. $$$n$$$ people want to sit in these chairs. The $$$i$$$-th of them wants at least $$$a[i]$$$ empty chairs both on his right and left side. More formally, if the $$$i$$$-th person sits in the $$$j$$$-th chair, then no one else should sit in the following chairs: $$$(j-a[i]) \bmod m$$$, $$$(j-a[i]+1) \bmod m$$$, ... $$$(j+a[i]-1) \bmod m$$$, $$$(j+a[i]) \bmod m$$$.Decide if it is possible to sit down for all of them, under the given limitations.
Input Specification: The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 5 \cdot 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq m \leq 10^9$$$) — the number of people and the number of chairs. The next line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ... $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the minimum number of empty chairs, on both sides of the $$$i$$$-th person. It is guaranteed that the sum of $$$n$$$ over all test cases will not exceed $$$10^5$$$.
Output Specification: For each test case print "YES" (without quotes) if it is possible for everyone to sit down and fulfil the restrictions, and "NO" (without quotes) otherwise. You may print every letter in any case you want (so, for example, the strings "yEs", "yes", "Yes" and "YES" will all be recognized as positive answers).
Notes: NoteTest case $$$1$$$: $$$n>m$$$, so they can not sit down.Test case $$$2$$$: the first person can sit $$$2$$$-nd and the second person can sit in the $$$0$$$-th chair. Both of them want at least $$$1$$$ empty chair on both sides, chairs $$$1$$$ and $$$3$$$ are free, so this is a good solution.Test case $$$3$$$: if the second person sits down somewhere, he needs $$$2$$$ empty chairs, both on his right and on his left side, so it is impossible to find a place for the first person, because there are only $$$5$$$ chairs.Test case $$$4$$$: they can sit in the $$$1$$$-st, $$$4$$$-th, $$$7$$$-th chairs respectively.
Code:
t=lambda:map(int,input().split())
for # TODO: Your code here:n,m=t();a=[*t()];print("YNEOS"[sum(a)+max(a)-min(a)+n>m::2])
|
t=lambda:map(int,input().split())
for {{completion}}:n,m=t();a=[*t()];print("YNEOS"[sum(a)+max(a)-min(a)+n>m::2])
|
_ in range(int(input()))
|
[{"input": "6\n3 2\n1 1 1\n2 4\n1 1\n2 5\n2 1\n3 8\n1 2 1\n4 12\n1 2 1 3\n4 19\n1 2 1 3", "output": ["NO\nYES\nNO\nYES\nNO\nYES"]}]
|
control_completion_000997
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: $$$m$$$ chairs are arranged in a circle sequentially. The chairs are numbered from $$$0$$$ to $$$m-1$$$. $$$n$$$ people want to sit in these chairs. The $$$i$$$-th of them wants at least $$$a[i]$$$ empty chairs both on his right and left side. More formally, if the $$$i$$$-th person sits in the $$$j$$$-th chair, then no one else should sit in the following chairs: $$$(j-a[i]) \bmod m$$$, $$$(j-a[i]+1) \bmod m$$$, ... $$$(j+a[i]-1) \bmod m$$$, $$$(j+a[i]) \bmod m$$$.Decide if it is possible to sit down for all of them, under the given limitations.
Input Specification: The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 5 \cdot 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq m \leq 10^9$$$) — the number of people and the number of chairs. The next line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ... $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the minimum number of empty chairs, on both sides of the $$$i$$$-th person. It is guaranteed that the sum of $$$n$$$ over all test cases will not exceed $$$10^5$$$.
Output Specification: For each test case print "YES" (without quotes) if it is possible for everyone to sit down and fulfil the restrictions, and "NO" (without quotes) otherwise. You may print every letter in any case you want (so, for example, the strings "yEs", "yes", "Yes" and "YES" will all be recognized as positive answers).
Notes: NoteTest case $$$1$$$: $$$n>m$$$, so they can not sit down.Test case $$$2$$$: the first person can sit $$$2$$$-nd and the second person can sit in the $$$0$$$-th chair. Both of them want at least $$$1$$$ empty chair on both sides, chairs $$$1$$$ and $$$3$$$ are free, so this is a good solution.Test case $$$3$$$: if the second person sits down somewhere, he needs $$$2$$$ empty chairs, both on his right and on his left side, so it is impossible to find a place for the first person, because there are only $$$5$$$ chairs.Test case $$$4$$$: they can sit in the $$$1$$$-st, $$$4$$$-th, $$$7$$$-th chairs respectively.
Code:
for _t in range(int(input())):
n,m = map(int, input().split(' ')) # numbers w/ ws
a = sorted(map(int, input().split(' ')))
tot = 0
dis = 0
p_i = a[-1]
for i in a:
tot += 2*i+1
if # TODO: Your code here:
dis += p_i
else:
dis += i
p_i = i
if tot-dis <= m:
print("YES")
else:
print("NO")
|
for _t in range(int(input())):
n,m = map(int, input().split(' ')) # numbers w/ ws
a = sorted(map(int, input().split(' ')))
tot = 0
dis = 0
p_i = a[-1]
for i in a:
tot += 2*i+1
if {{completion}}:
dis += p_i
else:
dis += i
p_i = i
if tot-dis <= m:
print("YES")
else:
print("NO")
|
p_i < i
|
[{"input": "6\n3 2\n1 1 1\n2 4\n1 1\n2 5\n2 1\n3 8\n1 2 1\n4 12\n1 2 1 3\n4 19\n1 2 1 3", "output": ["NO\nYES\nNO\nYES\nNO\nYES"]}]
|
control_completion_000998
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: $$$m$$$ chairs are arranged in a circle sequentially. The chairs are numbered from $$$0$$$ to $$$m-1$$$. $$$n$$$ people want to sit in these chairs. The $$$i$$$-th of them wants at least $$$a[i]$$$ empty chairs both on his right and left side. More formally, if the $$$i$$$-th person sits in the $$$j$$$-th chair, then no one else should sit in the following chairs: $$$(j-a[i]) \bmod m$$$, $$$(j-a[i]+1) \bmod m$$$, ... $$$(j+a[i]-1) \bmod m$$$, $$$(j+a[i]) \bmod m$$$.Decide if it is possible to sit down for all of them, under the given limitations.
Input Specification: The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 5 \cdot 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq m \leq 10^9$$$) — the number of people and the number of chairs. The next line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ... $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the minimum number of empty chairs, on both sides of the $$$i$$$-th person. It is guaranteed that the sum of $$$n$$$ over all test cases will not exceed $$$10^5$$$.
Output Specification: For each test case print "YES" (without quotes) if it is possible for everyone to sit down and fulfil the restrictions, and "NO" (without quotes) otherwise. You may print every letter in any case you want (so, for example, the strings "yEs", "yes", "Yes" and "YES" will all be recognized as positive answers).
Notes: NoteTest case $$$1$$$: $$$n>m$$$, so they can not sit down.Test case $$$2$$$: the first person can sit $$$2$$$-nd and the second person can sit in the $$$0$$$-th chair. Both of them want at least $$$1$$$ empty chair on both sides, chairs $$$1$$$ and $$$3$$$ are free, so this is a good solution.Test case $$$3$$$: if the second person sits down somewhere, he needs $$$2$$$ empty chairs, both on his right and on his left side, so it is impossible to find a place for the first person, because there are only $$$5$$$ chairs.Test case $$$4$$$: they can sit in the $$$1$$$-st, $$$4$$$-th, $$$7$$$-th chairs respectively.
Code:
import sys
def solve():
n, m = map(int, input().split())
num = list(map(int , input().split()))
num.sort()
s = sum(num[1:]) + num[-1] + n
print("YES" if s <= m else "NO")
for # TODO: Your code here:
solve()
|
import sys
def solve():
n, m = map(int, input().split())
num = list(map(int , input().split()))
num.sort()
s = sum(num[1:]) + num[-1] + n
print("YES" if s <= m else "NO")
for {{completion}}:
solve()
|
_ in range(int(input()))
|
[{"input": "6\n3 2\n1 1 1\n2 4\n1 1\n2 5\n2 1\n3 8\n1 2 1\n4 12\n1 2 1 3\n4 19\n1 2 1 3", "output": ["NO\nYES\nNO\nYES\nNO\nYES"]}]
|
control_completion_000999
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: $$$m$$$ chairs are arranged in a circle sequentially. The chairs are numbered from $$$0$$$ to $$$m-1$$$. $$$n$$$ people want to sit in these chairs. The $$$i$$$-th of them wants at least $$$a[i]$$$ empty chairs both on his right and left side. More formally, if the $$$i$$$-th person sits in the $$$j$$$-th chair, then no one else should sit in the following chairs: $$$(j-a[i]) \bmod m$$$, $$$(j-a[i]+1) \bmod m$$$, ... $$$(j+a[i]-1) \bmod m$$$, $$$(j+a[i]) \bmod m$$$.Decide if it is possible to sit down for all of them, under the given limitations.
Input Specification: The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 5 \cdot 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq m \leq 10^9$$$) — the number of people and the number of chairs. The next line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ... $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the minimum number of empty chairs, on both sides of the $$$i$$$-th person. It is guaranteed that the sum of $$$n$$$ over all test cases will not exceed $$$10^5$$$.
Output Specification: For each test case print "YES" (without quotes) if it is possible for everyone to sit down and fulfil the restrictions, and "NO" (without quotes) otherwise. You may print every letter in any case you want (so, for example, the strings "yEs", "yes", "Yes" and "YES" will all be recognized as positive answers).
Notes: NoteTest case $$$1$$$: $$$n>m$$$, so they can not sit down.Test case $$$2$$$: the first person can sit $$$2$$$-nd and the second person can sit in the $$$0$$$-th chair. Both of them want at least $$$1$$$ empty chair on both sides, chairs $$$1$$$ and $$$3$$$ are free, so this is a good solution.Test case $$$3$$$: if the second person sits down somewhere, he needs $$$2$$$ empty chairs, both on his right and on his left side, so it is impossible to find a place for the first person, because there are only $$$5$$$ chairs.Test case $$$4$$$: they can sit in the $$$1$$$-st, $$$4$$$-th, $$$7$$$-th chairs respectively.
Code:
import sys
for t in range(int(sys.stdin.readline())):
n,m = map(int, sys.stdin.readline().strip().split())
a = list(map(int, sys.stdin.readline().strip().split()))
if # TODO: Your code here:print('yes')
else:print('no')
|
import sys
for t in range(int(sys.stdin.readline())):
n,m = map(int, sys.stdin.readline().strip().split())
a = list(map(int, sys.stdin.readline().strip().split()))
if {{completion}}:print('yes')
else:print('no')
|
sum(a)-min(a)+max(a) + n <= m
|
[{"input": "6\n3 2\n1 1 1\n2 4\n1 1\n2 5\n2 1\n3 8\n1 2 1\n4 12\n1 2 1 3\n4 19\n1 2 1 3", "output": ["NO\nYES\nNO\nYES\nNO\nYES"]}]
|
control_completion_001000
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: $$$m$$$ chairs are arranged in a circle sequentially. The chairs are numbered from $$$0$$$ to $$$m-1$$$. $$$n$$$ people want to sit in these chairs. The $$$i$$$-th of them wants at least $$$a[i]$$$ empty chairs both on his right and left side. More formally, if the $$$i$$$-th person sits in the $$$j$$$-th chair, then no one else should sit in the following chairs: $$$(j-a[i]) \bmod m$$$, $$$(j-a[i]+1) \bmod m$$$, ... $$$(j+a[i]-1) \bmod m$$$, $$$(j+a[i]) \bmod m$$$.Decide if it is possible to sit down for all of them, under the given limitations.
Input Specification: The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 5 \cdot 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq m \leq 10^9$$$) — the number of people and the number of chairs. The next line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ... $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the minimum number of empty chairs, on both sides of the $$$i$$$-th person. It is guaranteed that the sum of $$$n$$$ over all test cases will not exceed $$$10^5$$$.
Output Specification: For each test case print "YES" (without quotes) if it is possible for everyone to sit down and fulfil the restrictions, and "NO" (without quotes) otherwise. You may print every letter in any case you want (so, for example, the strings "yEs", "yes", "Yes" and "YES" will all be recognized as positive answers).
Notes: NoteTest case $$$1$$$: $$$n>m$$$, so they can not sit down.Test case $$$2$$$: the first person can sit $$$2$$$-nd and the second person can sit in the $$$0$$$-th chair. Both of them want at least $$$1$$$ empty chair on both sides, chairs $$$1$$$ and $$$3$$$ are free, so this is a good solution.Test case $$$3$$$: if the second person sits down somewhere, he needs $$$2$$$ empty chairs, both on his right and on his left side, so it is impossible to find a place for the first person, because there are only $$$5$$$ chairs.Test case $$$4$$$: they can sit in the $$$1$$$-st, $$$4$$$-th, $$$7$$$-th chairs respectively.
Code:
x = lambda: map(int,input().split())
t,= x()
for # TODO: Your code here:
p,n = x()
a = [*x()]
s = sum(a) + (p-1) - min(a)
print("YNEOS"[n-1-s<max(a)::2])
|
x = lambda: map(int,input().split())
t,= x()
for {{completion}}:
p,n = x()
a = [*x()]
s = sum(a) + (p-1) - min(a)
print("YNEOS"[n-1-s<max(a)::2])
|
_ in [1]*t
|
[{"input": "6\n3 2\n1 1 1\n2 4\n1 1\n2 5\n2 1\n3 8\n1 2 1\n4 12\n1 2 1 3\n4 19\n1 2 1 3", "output": ["NO\nYES\nNO\nYES\nNO\nYES"]}]
|
control_completion_001001
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: $$$m$$$ chairs are arranged in a circle sequentially. The chairs are numbered from $$$0$$$ to $$$m-1$$$. $$$n$$$ people want to sit in these chairs. The $$$i$$$-th of them wants at least $$$a[i]$$$ empty chairs both on his right and left side. More formally, if the $$$i$$$-th person sits in the $$$j$$$-th chair, then no one else should sit in the following chairs: $$$(j-a[i]) \bmod m$$$, $$$(j-a[i]+1) \bmod m$$$, ... $$$(j+a[i]-1) \bmod m$$$, $$$(j+a[i]) \bmod m$$$.Decide if it is possible to sit down for all of them, under the given limitations.
Input Specification: The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 5 \cdot 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq m \leq 10^9$$$) — the number of people and the number of chairs. The next line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ... $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the minimum number of empty chairs, on both sides of the $$$i$$$-th person. It is guaranteed that the sum of $$$n$$$ over all test cases will not exceed $$$10^5$$$.
Output Specification: For each test case print "YES" (without quotes) if it is possible for everyone to sit down and fulfil the restrictions, and "NO" (without quotes) otherwise. You may print every letter in any case you want (so, for example, the strings "yEs", "yes", "Yes" and "YES" will all be recognized as positive answers).
Notes: NoteTest case $$$1$$$: $$$n>m$$$, so they can not sit down.Test case $$$2$$$: the first person can sit $$$2$$$-nd and the second person can sit in the $$$0$$$-th chair. Both of them want at least $$$1$$$ empty chair on both sides, chairs $$$1$$$ and $$$3$$$ are free, so this is a good solution.Test case $$$3$$$: if the second person sits down somewhere, he needs $$$2$$$ empty chairs, both on his right and on his left side, so it is impossible to find a place for the first person, because there are only $$$5$$$ chairs.Test case $$$4$$$: they can sit in the $$$1$$$-st, $$$4$$$-th, $$$7$$$-th chairs respectively.
Code:
def Dist():
num_nm = input().split()
m = int(num_nm[1])
n = int(num_nm[0])
a = input().split()
a = list(map(int, a))
wish = n + sum(a) - min(a) + max(a)
print("NO" if wish >m else "YES")
num_iter = int(input())
for # TODO: Your code here:
Dist()
|
def Dist():
num_nm = input().split()
m = int(num_nm[1])
n = int(num_nm[0])
a = input().split()
a = list(map(int, a))
wish = n + sum(a) - min(a) + max(a)
print("NO" if wish >m else "YES")
num_iter = int(input())
for {{completion}}:
Dist()
|
_ in range(num_iter)
|
[{"input": "6\n3 2\n1 1 1\n2 4\n1 1\n2 5\n2 1\n3 8\n1 2 1\n4 12\n1 2 1 3\n4 19\n1 2 1 3", "output": ["NO\nYES\nNO\nYES\nNO\nYES"]}]
|
control_completion_001002
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: $$$m$$$ chairs are arranged in a circle sequentially. The chairs are numbered from $$$0$$$ to $$$m-1$$$. $$$n$$$ people want to sit in these chairs. The $$$i$$$-th of them wants at least $$$a[i]$$$ empty chairs both on his right and left side. More formally, if the $$$i$$$-th person sits in the $$$j$$$-th chair, then no one else should sit in the following chairs: $$$(j-a[i]) \bmod m$$$, $$$(j-a[i]+1) \bmod m$$$, ... $$$(j+a[i]-1) \bmod m$$$, $$$(j+a[i]) \bmod m$$$.Decide if it is possible to sit down for all of them, under the given limitations.
Input Specification: The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 5 \cdot 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq m \leq 10^9$$$) — the number of people and the number of chairs. The next line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ... $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the minimum number of empty chairs, on both sides of the $$$i$$$-th person. It is guaranteed that the sum of $$$n$$$ over all test cases will not exceed $$$10^5$$$.
Output Specification: For each test case print "YES" (without quotes) if it is possible for everyone to sit down and fulfil the restrictions, and "NO" (without quotes) otherwise. You may print every letter in any case you want (so, for example, the strings "yEs", "yes", "Yes" and "YES" will all be recognized as positive answers).
Notes: NoteTest case $$$1$$$: $$$n>m$$$, so they can not sit down.Test case $$$2$$$: the first person can sit $$$2$$$-nd and the second person can sit in the $$$0$$$-th chair. Both of them want at least $$$1$$$ empty chair on both sides, chairs $$$1$$$ and $$$3$$$ are free, so this is a good solution.Test case $$$3$$$: if the second person sits down somewhere, he needs $$$2$$$ empty chairs, both on his right and on his left side, so it is impossible to find a place for the first person, because there are only $$$5$$$ chairs.Test case $$$4$$$: they can sit in the $$$1$$$-st, $$$4$$$-th, $$$7$$$-th chairs respectively.
Code:
for T in range (int(input())) :
n,m = map(int, input().strip().split())
a = sorted(list(map(int,input().strip().split())),reverse=True)
m -= 2*a[0] + 1
cont = 0
for i in range(1,n) :
if # TODO: Your code here: break
m -= a[i] + 1
cont +=1
if cont == n-1 : print('YES')
else : print ('NO')
|
for T in range (int(input())) :
n,m = map(int, input().strip().split())
a = sorted(list(map(int,input().strip().split())),reverse=True)
m -= 2*a[0] + 1
cont = 0
for i in range(1,n) :
if {{completion}}: break
m -= a[i] + 1
cont +=1
if cont == n-1 : print('YES')
else : print ('NO')
|
m <= 0
|
[{"input": "6\n3 2\n1 1 1\n2 4\n1 1\n2 5\n2 1\n3 8\n1 2 1\n4 12\n1 2 1 3\n4 19\n1 2 1 3", "output": ["NO\nYES\nNO\nYES\nNO\nYES"]}]
|
control_completion_001003
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: $$$m$$$ chairs are arranged in a circle sequentially. The chairs are numbered from $$$0$$$ to $$$m-1$$$. $$$n$$$ people want to sit in these chairs. The $$$i$$$-th of them wants at least $$$a[i]$$$ empty chairs both on his right and left side. More formally, if the $$$i$$$-th person sits in the $$$j$$$-th chair, then no one else should sit in the following chairs: $$$(j-a[i]) \bmod m$$$, $$$(j-a[i]+1) \bmod m$$$, ... $$$(j+a[i]-1) \bmod m$$$, $$$(j+a[i]) \bmod m$$$.Decide if it is possible to sit down for all of them, under the given limitations.
Input Specification: The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 5 \cdot 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq m \leq 10^9$$$) — the number of people and the number of chairs. The next line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ... $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the minimum number of empty chairs, on both sides of the $$$i$$$-th person. It is guaranteed that the sum of $$$n$$$ over all test cases will not exceed $$$10^5$$$.
Output Specification: For each test case print "YES" (without quotes) if it is possible for everyone to sit down and fulfil the restrictions, and "NO" (without quotes) otherwise. You may print every letter in any case you want (so, for example, the strings "yEs", "yes", "Yes" and "YES" will all be recognized as positive answers).
Notes: NoteTest case $$$1$$$: $$$n>m$$$, so they can not sit down.Test case $$$2$$$: the first person can sit $$$2$$$-nd and the second person can sit in the $$$0$$$-th chair. Both of them want at least $$$1$$$ empty chair on both sides, chairs $$$1$$$ and $$$3$$$ are free, so this is a good solution.Test case $$$3$$$: if the second person sits down somewhere, he needs $$$2$$$ empty chairs, both on his right and on his left side, so it is impossible to find a place for the first person, because there are only $$$5$$$ chairs.Test case $$$4$$$: they can sit in the $$$1$$$-st, $$$4$$$-th, $$$7$$$-th chairs respectively.
Code:
I=lambda:[*map(int,input().split())]
t,=I()
while # TODO: Your code here:t-=1;n,m=I();a=sorted(I());print('YNEOS'[sum(max(a[i-1],a[i])for i in range(n))+n>m::2])
|
I=lambda:[*map(int,input().split())]
t,=I()
while {{completion}}:t-=1;n,m=I();a=sorted(I());print('YNEOS'[sum(max(a[i-1],a[i])for i in range(n))+n>m::2])
|
t
|
[{"input": "6\n3 2\n1 1 1\n2 4\n1 1\n2 5\n2 1\n3 8\n1 2 1\n4 12\n1 2 1 3\n4 19\n1 2 1 3", "output": ["NO\nYES\nNO\nYES\nNO\nYES"]}]
|
control_completion_001004
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: $$$m$$$ chairs are arranged in a circle sequentially. The chairs are numbered from $$$0$$$ to $$$m-1$$$. $$$n$$$ people want to sit in these chairs. The $$$i$$$-th of them wants at least $$$a[i]$$$ empty chairs both on his right and left side. More formally, if the $$$i$$$-th person sits in the $$$j$$$-th chair, then no one else should sit in the following chairs: $$$(j-a[i]) \bmod m$$$, $$$(j-a[i]+1) \bmod m$$$, ... $$$(j+a[i]-1) \bmod m$$$, $$$(j+a[i]) \bmod m$$$.Decide if it is possible to sit down for all of them, under the given limitations.
Input Specification: The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 5 \cdot 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq m \leq 10^9$$$) — the number of people and the number of chairs. The next line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ... $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the minimum number of empty chairs, on both sides of the $$$i$$$-th person. It is guaranteed that the sum of $$$n$$$ over all test cases will not exceed $$$10^5$$$.
Output Specification: For each test case print "YES" (without quotes) if it is possible for everyone to sit down and fulfil the restrictions, and "NO" (without quotes) otherwise. You may print every letter in any case you want (so, for example, the strings "yEs", "yes", "Yes" and "YES" will all be recognized as positive answers).
Notes: NoteTest case $$$1$$$: $$$n>m$$$, so they can not sit down.Test case $$$2$$$: the first person can sit $$$2$$$-nd and the second person can sit in the $$$0$$$-th chair. Both of them want at least $$$1$$$ empty chair on both sides, chairs $$$1$$$ and $$$3$$$ are free, so this is a good solution.Test case $$$3$$$: if the second person sits down somewhere, he needs $$$2$$$ empty chairs, both on his right and on his left side, so it is impossible to find a place for the first person, because there are only $$$5$$$ chairs.Test case $$$4$$$: they can sit in the $$$1$$$-st, $$$4$$$-th, $$$7$$$-th chairs respectively.
Code:
for i in range(int(input())):
n,m=map(int,input().split())
a=list(map(int,input().split()))
if # TODO: Your code here:
print("no")
else:
print("yes")
|
for i in range(int(input())):
n,m=map(int,input().split())
a=list(map(int,input().split()))
if {{completion}}:
print("no")
else:
print("yes")
|
n+sum(a)+max(a)-min(a)>m
|
[{"input": "6\n3 2\n1 1 1\n2 4\n1 1\n2 5\n2 1\n3 8\n1 2 1\n4 12\n1 2 1 3\n4 19\n1 2 1 3", "output": ["NO\nYES\nNO\nYES\nNO\nYES"]}]
|
control_completion_001005
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given an array $$$a$$$ consisting of $$$n$$$ integers. You should divide $$$a$$$ into continuous non-empty subarrays (there are $$$2^{n-1}$$$ ways to do that).Let $$$s=a_l+a_{l+1}+\ldots+a_r$$$. The value of a subarray $$$a_l, a_{l+1}, \ldots, a_r$$$ is: $$$(r-l+1)$$$ if $$$s>0$$$, $$$0$$$ if $$$s=0$$$, $$$-(r-l+1)$$$ if $$$s<0$$$. What is the maximum sum of values you can get with a partition?
Input Specification: The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 5 \cdot 10^5$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 5 \cdot 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$). It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \cdot 10^5$$$.
Output Specification: For each test case print a single integer — the maximum sum of values you can get with an optimal parition.
Notes: NoteTest case $$$1$$$: one optimal partition is $$$[1, 2]$$$, $$$[-3]$$$. $$$1+2>0$$$ so the value of $$$[1, 2]$$$ is $$$2$$$. $$$-3<0$$$, so the value of $$$[-3]$$$ is $$$-1$$$. $$$2+(-1)=1$$$.Test case $$$2$$$: the optimal partition is $$$[0, -2, 3]$$$, $$$[-4]$$$, and the sum of values is $$$3+(-1)=2$$$.
Code:
from collections import Counter, defaultdict, deque
import bisect
from sys import stdin, stdout
from itertools import repeat
import math
MOD = 998244353
input = stdin.readline
finp = [int(x) for x in stdin.buffer.read().split()]
def inp(force_list=False):
re = list(map(int, input().split()))
if len(re) == 1 and not force_list:
return re[0]
return re
def inst():
return input().strip()
def gcd(x, y):
while(y):
x, y = y, x % y
return x
def qmod(a, b, mod=MOD):
res = 1
while b:
if b&1:
res = (res*a)%mod
b >>= 1
a = (a*a)%mod
return res
def inv(a):
return qmod(a, MOD-2)
INF = 1<<30
class Seg(object):
def __init__(self, n):
self._da = [-INF] * (n * 5)
self._op = [-INF] * (n * 5)
def update(self, p):
self._op[p] = max(self._op[p*2], self._op[p*2+1])
def modify(self, pos, x, p, l, r):
if l==r-1:
self._da[p] = self._op[p] = x
return
mid = (l+r)//2
if pos < mid:
self.modify(pos, x, p*2, l, mid)
else:
self.modify(pos, x, p*2 + 1, mid, r)
self.update(p)
def query(self, x, y, p, l, r):
if x <= l and r <= y:
return self._op[p]
if x >= r or y<=l:
return -INF
mid = (l+r)//2
return max(self.query(x, y, p*2, l, mid), self.query(x, y, p*2+1, mid, r))
class Fenwick(object):
def __init__(self, n):
self._da = [-INF] * (n+2)
self._mx = n+2
def max(self, x):
res = -INF
while # TODO: Your code here:
res = max(res, self._da[x])
x = (x&(x+1))-1
return res
def modify(self, p, x):
while p < self._mx:
self._da[p] = max(self._da[p], x)
p |= p+1
def my_main():
# print(500000)
# for i in range(500000):
# print(1)
# print(-1000000000)
ii = 0
kase = finp[ii];ii+=1
pans = []
for skase in range(kase):
# print("Case #%d: " % (skase+1), end='')
n = finp[ii];ii+=1
da = finp[ii:ii+n];ii+=n
pref = [0]
for i in da:
pref.append(pref[-1] + i)
spos, sneg = sorted([(pref[i], -i) for i, v in enumerate(pref)]), sorted([(pref[i], i) for i, v in enumerate(pref)])
ordpos, ordneg = [0] * (n+1), [0] * (n+1)
pfen, nfen = Fenwick(n), Fenwick(n)
dmx = {}
for i in range(n+1):
ordpos[-spos[i][-1]] = i
ordneg[sneg[i][-1]] = i
dp = [0] * (n+1)
dmx[0] = 0
pfen.modify(ordpos[0], 0)
nfen.modify(n+1-ordneg[0], 0)
for i in range(1, n+1):
dp[i] = max(i+pfen.max(ordpos[i]), nfen.max(n+1-ordneg[i])-i, dmx.get(pref[i], -INF))
pfen.modify(ordpos[i], dp[i]-i)
nfen.modify(n+1-ordneg[i], dp[i]+i)
if dp[i] > dmx.get(pref[i], -INF):
dmx[pref[i]] = dp[i]
pans.append(str(dp[n]))
print('\n'.join(pans))
my_main()
|
from collections import Counter, defaultdict, deque
import bisect
from sys import stdin, stdout
from itertools import repeat
import math
MOD = 998244353
input = stdin.readline
finp = [int(x) for x in stdin.buffer.read().split()]
def inp(force_list=False):
re = list(map(int, input().split()))
if len(re) == 1 and not force_list:
return re[0]
return re
def inst():
return input().strip()
def gcd(x, y):
while(y):
x, y = y, x % y
return x
def qmod(a, b, mod=MOD):
res = 1
while b:
if b&1:
res = (res*a)%mod
b >>= 1
a = (a*a)%mod
return res
def inv(a):
return qmod(a, MOD-2)
INF = 1<<30
class Seg(object):
def __init__(self, n):
self._da = [-INF] * (n * 5)
self._op = [-INF] * (n * 5)
def update(self, p):
self._op[p] = max(self._op[p*2], self._op[p*2+1])
def modify(self, pos, x, p, l, r):
if l==r-1:
self._da[p] = self._op[p] = x
return
mid = (l+r)//2
if pos < mid:
self.modify(pos, x, p*2, l, mid)
else:
self.modify(pos, x, p*2 + 1, mid, r)
self.update(p)
def query(self, x, y, p, l, r):
if x <= l and r <= y:
return self._op[p]
if x >= r or y<=l:
return -INF
mid = (l+r)//2
return max(self.query(x, y, p*2, l, mid), self.query(x, y, p*2+1, mid, r))
class Fenwick(object):
def __init__(self, n):
self._da = [-INF] * (n+2)
self._mx = n+2
def max(self, x):
res = -INF
while {{completion}}:
res = max(res, self._da[x])
x = (x&(x+1))-1
return res
def modify(self, p, x):
while p < self._mx:
self._da[p] = max(self._da[p], x)
p |= p+1
def my_main():
# print(500000)
# for i in range(500000):
# print(1)
# print(-1000000000)
ii = 0
kase = finp[ii];ii+=1
pans = []
for skase in range(kase):
# print("Case #%d: " % (skase+1), end='')
n = finp[ii];ii+=1
da = finp[ii:ii+n];ii+=n
pref = [0]
for i in da:
pref.append(pref[-1] + i)
spos, sneg = sorted([(pref[i], -i) for i, v in enumerate(pref)]), sorted([(pref[i], i) for i, v in enumerate(pref)])
ordpos, ordneg = [0] * (n+1), [0] * (n+1)
pfen, nfen = Fenwick(n), Fenwick(n)
dmx = {}
for i in range(n+1):
ordpos[-spos[i][-1]] = i
ordneg[sneg[i][-1]] = i
dp = [0] * (n+1)
dmx[0] = 0
pfen.modify(ordpos[0], 0)
nfen.modify(n+1-ordneg[0], 0)
for i in range(1, n+1):
dp[i] = max(i+pfen.max(ordpos[i]), nfen.max(n+1-ordneg[i])-i, dmx.get(pref[i], -INF))
pfen.modify(ordpos[i], dp[i]-i)
nfen.modify(n+1-ordneg[i], dp[i]+i)
if dp[i] > dmx.get(pref[i], -INF):
dmx[pref[i]] = dp[i]
pans.append(str(dp[n]))
print('\n'.join(pans))
my_main()
|
x>0
|
[{"input": "5\n\n3\n\n1 2 -3\n\n4\n\n0 -2 3 -4\n\n5\n\n-1 -2 3 -1 -1\n\n6\n\n-1 2 -3 4 -5 6\n\n7\n\n1 -1 -1 1 -1 -1 1", "output": ["1\n2\n1\n6\n-1"]}]
|
control_completion_001028
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given an array $$$a$$$ consisting of $$$n$$$ integers. You should divide $$$a$$$ into continuous non-empty subarrays (there are $$$2^{n-1}$$$ ways to do that).Let $$$s=a_l+a_{l+1}+\ldots+a_r$$$. The value of a subarray $$$a_l, a_{l+1}, \ldots, a_r$$$ is: $$$(r-l+1)$$$ if $$$s>0$$$, $$$0$$$ if $$$s=0$$$, $$$-(r-l+1)$$$ if $$$s<0$$$. What is the maximum sum of values you can get with a partition?
Input Specification: The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 5 \cdot 10^5$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 5 \cdot 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$). It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \cdot 10^5$$$.
Output Specification: For each test case print a single integer — the maximum sum of values you can get with an optimal parition.
Notes: NoteTest case $$$1$$$: one optimal partition is $$$[1, 2]$$$, $$$[-3]$$$. $$$1+2>0$$$ so the value of $$$[1, 2]$$$ is $$$2$$$. $$$-3<0$$$, so the value of $$$[-3]$$$ is $$$-1$$$. $$$2+(-1)=1$$$.Test case $$$2$$$: the optimal partition is $$$[0, -2, 3]$$$, $$$[-4]$$$, and the sum of values is $$$3+(-1)=2$$$.
Code:
from collections import Counter, defaultdict, deque
import bisect
from sys import stdin, stdout
from itertools import repeat
import math
MOD = 998244353
input = stdin.readline
finp = [int(x) for x in stdin.buffer.read().split()]
def inp(force_list=False):
re = list(map(int, input().split()))
if len(re) == 1 and not force_list:
return re[0]
return re
def inst():
return input().strip()
def gcd(x, y):
while(y):
x, y = y, x % y
return x
def qmod(a, b, mod=MOD):
res = 1
while b:
if b&1:
res = (res*a)%mod
b >>= 1
a = (a*a)%mod
return res
def inv(a):
return qmod(a, MOD-2)
INF = 1<<30
class Seg(object):
def __init__(self, n):
self._da = [-INF] * (n * 5)
self._op = [-INF] * (n * 5)
def update(self, p):
self._op[p] = max(self._op[p*2], self._op[p*2+1])
def modify(self, pos, x, p, l, r):
if l==r-1:
self._da[p] = self._op[p] = x
return
mid = (l+r)//2
if pos < mid:
self.modify(pos, x, p*2, l, mid)
else:
self.modify(pos, x, p*2 + 1, mid, r)
self.update(p)
def query(self, x, y, p, l, r):
if x <= l and r <= y:
return self._op[p]
if x >= r or y<=l:
return -INF
mid = (l+r)//2
return max(self.query(x, y, p*2, l, mid), self.query(x, y, p*2+1, mid, r))
class Fenwick(object):
def __init__(self, n):
self._da = [-INF] * (n+2)
self._mx = n+2
def max(self, x):
res = -INF
while x>0:
res = max(res, self._da[x])
x = (x&(x+1))-1
return res
def modify(self, p, x):
while # TODO: Your code here:
self._da[p] = max(self._da[p], x)
p |= p+1
def my_main():
# print(500000)
# for i in range(500000):
# print(1)
# print(-1000000000)
ii = 0
kase = finp[ii];ii+=1
pans = []
for skase in range(kase):
# print("Case #%d: " % (skase+1), end='')
n = finp[ii];ii+=1
da = finp[ii:ii+n];ii+=n
pref = [0]
for i in da:
pref.append(pref[-1] + i)
spos, sneg = sorted([(pref[i], -i) for i, v in enumerate(pref)]), sorted([(pref[i], i) for i, v in enumerate(pref)])
ordpos, ordneg = [0] * (n+1), [0] * (n+1)
pfen, nfen = Fenwick(n), Fenwick(n)
dmx = {}
for i in range(n+1):
ordpos[-spos[i][-1]] = i
ordneg[sneg[i][-1]] = i
dp = [0] * (n+1)
dmx[0] = 0
pfen.modify(ordpos[0], 0)
nfen.modify(n+1-ordneg[0], 0)
for i in range(1, n+1):
dp[i] = max(i+pfen.max(ordpos[i]), nfen.max(n+1-ordneg[i])-i, dmx.get(pref[i], -INF))
pfen.modify(ordpos[i], dp[i]-i)
nfen.modify(n+1-ordneg[i], dp[i]+i)
if dp[i] > dmx.get(pref[i], -INF):
dmx[pref[i]] = dp[i]
pans.append(str(dp[n]))
print('\n'.join(pans))
my_main()
|
from collections import Counter, defaultdict, deque
import bisect
from sys import stdin, stdout
from itertools import repeat
import math
MOD = 998244353
input = stdin.readline
finp = [int(x) for x in stdin.buffer.read().split()]
def inp(force_list=False):
re = list(map(int, input().split()))
if len(re) == 1 and not force_list:
return re[0]
return re
def inst():
return input().strip()
def gcd(x, y):
while(y):
x, y = y, x % y
return x
def qmod(a, b, mod=MOD):
res = 1
while b:
if b&1:
res = (res*a)%mod
b >>= 1
a = (a*a)%mod
return res
def inv(a):
return qmod(a, MOD-2)
INF = 1<<30
class Seg(object):
def __init__(self, n):
self._da = [-INF] * (n * 5)
self._op = [-INF] * (n * 5)
def update(self, p):
self._op[p] = max(self._op[p*2], self._op[p*2+1])
def modify(self, pos, x, p, l, r):
if l==r-1:
self._da[p] = self._op[p] = x
return
mid = (l+r)//2
if pos < mid:
self.modify(pos, x, p*2, l, mid)
else:
self.modify(pos, x, p*2 + 1, mid, r)
self.update(p)
def query(self, x, y, p, l, r):
if x <= l and r <= y:
return self._op[p]
if x >= r or y<=l:
return -INF
mid = (l+r)//2
return max(self.query(x, y, p*2, l, mid), self.query(x, y, p*2+1, mid, r))
class Fenwick(object):
def __init__(self, n):
self._da = [-INF] * (n+2)
self._mx = n+2
def max(self, x):
res = -INF
while x>0:
res = max(res, self._da[x])
x = (x&(x+1))-1
return res
def modify(self, p, x):
while {{completion}}:
self._da[p] = max(self._da[p], x)
p |= p+1
def my_main():
# print(500000)
# for i in range(500000):
# print(1)
# print(-1000000000)
ii = 0
kase = finp[ii];ii+=1
pans = []
for skase in range(kase):
# print("Case #%d: " % (skase+1), end='')
n = finp[ii];ii+=1
da = finp[ii:ii+n];ii+=n
pref = [0]
for i in da:
pref.append(pref[-1] + i)
spos, sneg = sorted([(pref[i], -i) for i, v in enumerate(pref)]), sorted([(pref[i], i) for i, v in enumerate(pref)])
ordpos, ordneg = [0] * (n+1), [0] * (n+1)
pfen, nfen = Fenwick(n), Fenwick(n)
dmx = {}
for i in range(n+1):
ordpos[-spos[i][-1]] = i
ordneg[sneg[i][-1]] = i
dp = [0] * (n+1)
dmx[0] = 0
pfen.modify(ordpos[0], 0)
nfen.modify(n+1-ordneg[0], 0)
for i in range(1, n+1):
dp[i] = max(i+pfen.max(ordpos[i]), nfen.max(n+1-ordneg[i])-i, dmx.get(pref[i], -INF))
pfen.modify(ordpos[i], dp[i]-i)
nfen.modify(n+1-ordneg[i], dp[i]+i)
if dp[i] > dmx.get(pref[i], -INF):
dmx[pref[i]] = dp[i]
pans.append(str(dp[n]))
print('\n'.join(pans))
my_main()
|
p < self._mx
|
[{"input": "5\n\n3\n\n1 2 -3\n\n4\n\n0 -2 3 -4\n\n5\n\n-1 -2 3 -1 -1\n\n6\n\n-1 2 -3 4 -5 6\n\n7\n\n1 -1 -1 1 -1 -1 1", "output": ["1\n2\n1\n6\n-1"]}]
|
control_completion_001029
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Today, like every year at SWERC, the $$$n^2$$$ contestants have gathered outside the venue to take a drone photo. Jennifer, the social media manager for the event, has arranged them into an $$$n\times n$$$ square. Being very good at her job, she knows that the contestant standing on the intersection of the $$$i$$$-th row with the $$$j$$$-th column is $$$a_{i,j}$$$ years old. Coincidentally, she notices that no two contestants have the same age, and that everyone is between $$$1$$$ and $$$n^2$$$ years old.Jennifer is planning to have some contestants hold a banner with the ICPC logo parallel to the ground, so that it is clearly visible in the aerial picture. Here are the steps that she is going to follow in order to take the perfect SWERC drone photo. First of all, Jennifer is going to select four contestants standing on the vertices of an axis-aligned rectangle. Then, she will have the two younger contestants hold one of the poles, while the two older contestants will hold the other pole. Finally, she will unfold the banner, using the poles to support its two ends. Obviously, this can only be done if the two poles are parallel and do not cross, as shown in the pictures below. Being very indecisive, Jennifer would like to try out all possible arrangements for the banner, but she is worried that this may cause the contestants to be late for the competition. How many different ways are there to choose the four contestants holding the poles in order to take a perfect photo? Two choices are considered different if at least one contestant is included in one but not the other.
Input Specification: The first line contains a single integer $$$n$$$ ($$$2\le n \le 1500$$$). The next $$$n$$$ lines describe the ages of the contestants. Specifically, the $$$i$$$-th line contains the integers $$$a_{i,1},a_{i,2},\ldots,a_{i,n}$$$ ($$$1\le a_{i,j}\le n^2$$$). It is guaranteed that $$$a_{i,j}\neq a_{k,l}$$$ if $$$i\neq k$$$ or $$$j\neq l$$$.
Output Specification: Print the number of ways for Jennifer to choose the four contestants holding the poles.
Notes: NoteIn the first sample, there are $$$4$$$ contestants, arranged as follows. There is only one way to choose four contestants, with one pole held by contestants aged $$$1$$$ and $$$2$$$ and the other one by contestants aged $$$3$$$ and $$$4$$$. But then, as we can see in the picture, the poles cross. Since there is no valid way to choose four contestants, the answer is $$$0$$$.In the second sample, the $$$4$$$ contestants are arranged as follows. Once again, there is only one way to choose four contestants, but this time the poles don't cross. Therefore, the answer is $$$1$$$.In the third sample, the $$$9$$$ contestants are arranged as follows. There are $$$6$$$ ways of choosing four contestants so that the poles don't cross, as shown in the following pictures.
Code:
import sys
input = sys.stdin.readline
n = int(input())
o1 = [0] * (n * n)
o2 = [0] * (n * n)
for i in range(n):
curr = (list(map(int, input().split())))
for # TODO: Your code here:
o1[curr[j] - 1] = i
o2[curr[j] - 1] = j
row_count = [0] * n
col_count = [0] * n
ct = 0
for u in range(n * n):
i = o1[u]
j = o2[u]
ct += row_count[i] * col_count[j]
row_count[i] += 1
col_count[j] += 1
n2 = (n * n - n)//2
ct -= n2 * n2
print(n2 * n2 - ct)
|
import sys
input = sys.stdin.readline
n = int(input())
o1 = [0] * (n * n)
o2 = [0] * (n * n)
for i in range(n):
curr = (list(map(int, input().split())))
for {{completion}}:
o1[curr[j] - 1] = i
o2[curr[j] - 1] = j
row_count = [0] * n
col_count = [0] * n
ct = 0
for u in range(n * n):
i = o1[u]
j = o2[u]
ct += row_count[i] * col_count[j]
row_count[i] += 1
col_count[j] += 1
n2 = (n * n - n)//2
ct -= n2 * n2
print(n2 * n2 - ct)
|
j in range(n)
|
[{"input": "2\n1 3\n4 2", "output": ["0"]}, {"input": "2\n3 2\n4 1", "output": ["1"]}, {"input": "3\n9 2 4\n1 5 3\n7 8 6", "output": ["6"]}]
|
control_completion_001071
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Today, like every year at SWERC, the $$$n^2$$$ contestants have gathered outside the venue to take a drone photo. Jennifer, the social media manager for the event, has arranged them into an $$$n\times n$$$ square. Being very good at her job, she knows that the contestant standing on the intersection of the $$$i$$$-th row with the $$$j$$$-th column is $$$a_{i,j}$$$ years old. Coincidentally, she notices that no two contestants have the same age, and that everyone is between $$$1$$$ and $$$n^2$$$ years old.Jennifer is planning to have some contestants hold a banner with the ICPC logo parallel to the ground, so that it is clearly visible in the aerial picture. Here are the steps that she is going to follow in order to take the perfect SWERC drone photo. First of all, Jennifer is going to select four contestants standing on the vertices of an axis-aligned rectangle. Then, she will have the two younger contestants hold one of the poles, while the two older contestants will hold the other pole. Finally, she will unfold the banner, using the poles to support its two ends. Obviously, this can only be done if the two poles are parallel and do not cross, as shown in the pictures below. Being very indecisive, Jennifer would like to try out all possible arrangements for the banner, but she is worried that this may cause the contestants to be late for the competition. How many different ways are there to choose the four contestants holding the poles in order to take a perfect photo? Two choices are considered different if at least one contestant is included in one but not the other.
Input Specification: The first line contains a single integer $$$n$$$ ($$$2\le n \le 1500$$$). The next $$$n$$$ lines describe the ages of the contestants. Specifically, the $$$i$$$-th line contains the integers $$$a_{i,1},a_{i,2},\ldots,a_{i,n}$$$ ($$$1\le a_{i,j}\le n^2$$$). It is guaranteed that $$$a_{i,j}\neq a_{k,l}$$$ if $$$i\neq k$$$ or $$$j\neq l$$$.
Output Specification: Print the number of ways for Jennifer to choose the four contestants holding the poles.
Notes: NoteIn the first sample, there are $$$4$$$ contestants, arranged as follows. There is only one way to choose four contestants, with one pole held by contestants aged $$$1$$$ and $$$2$$$ and the other one by contestants aged $$$3$$$ and $$$4$$$. But then, as we can see in the picture, the poles cross. Since there is no valid way to choose four contestants, the answer is $$$0$$$.In the second sample, the $$$4$$$ contestants are arranged as follows. Once again, there is only one way to choose four contestants, but this time the poles don't cross. Therefore, the answer is $$$1$$$.In the third sample, the $$$9$$$ contestants are arranged as follows. There are $$$6$$$ ways of choosing four contestants so that the poles don't cross, as shown in the following pictures.
Code:
import sys
import random
input = sys.stdin.buffer.readline
read = sys.stdin.buffer.read
N = int(input())
As = [list(map(int, input().split())) for _ in range(N)]
# N = 1500
# As = list(range(1, N ** 2 + 1))
# random.shuffle(As)
# As = [As[i * N:(i + 1) * N] for i in range(N)]
ijs = [0] * (N ** 2)
for i in range(N):
for # TODO: Your code here:
ijs[As[i][j] - 1] = (i, j)
answer = 0
row_sum = [0] * N
col_sum = [0] * N
for i, j in ijs:
l_row = row_sum[i]
g_row = N - 1 - row_sum[i]
l_col = col_sum[j]
g_col = N - 1 - col_sum[j]
answer += l_col * g_row + g_col * l_row
row_sum[i] += 1
col_sum[j] += 1
assert answer % 2 == 0
print(answer // 2)
|
import sys
import random
input = sys.stdin.buffer.readline
read = sys.stdin.buffer.read
N = int(input())
As = [list(map(int, input().split())) for _ in range(N)]
# N = 1500
# As = list(range(1, N ** 2 + 1))
# random.shuffle(As)
# As = [As[i * N:(i + 1) * N] for i in range(N)]
ijs = [0] * (N ** 2)
for i in range(N):
for {{completion}}:
ijs[As[i][j] - 1] = (i, j)
answer = 0
row_sum = [0] * N
col_sum = [0] * N
for i, j in ijs:
l_row = row_sum[i]
g_row = N - 1 - row_sum[i]
l_col = col_sum[j]
g_col = N - 1 - col_sum[j]
answer += l_col * g_row + g_col * l_row
row_sum[i] += 1
col_sum[j] += 1
assert answer % 2 == 0
print(answer // 2)
|
j in range(N)
|
[{"input": "2\n1 3\n4 2", "output": ["0"]}, {"input": "2\n3 2\n4 1", "output": ["1"]}, {"input": "3\n9 2 4\n1 5 3\n7 8 6", "output": ["6"]}]
|
control_completion_001072
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: The derby between Milan and Inter is happening soon, and you have been chosen as the assistant referee for the match, also known as linesman. Your task is to move along the touch-line, namely the side of the field, always looking very carefully at the match to check for offside positions and other offences.Football is an extremely serious matter in Italy, and thus it is fundamental that you keep very close track of the ball for as much time as possible. This means that you want to maximise the number of kicks which you monitor closely. You are able to monitor closely a kick if, when it happens, you are in the position along the touch-line with minimum distance from the place where the kick happens.Fortunately, expert analysts have been able to accurately predict all the kicks which will occur during the game. That is, you have been given two lists of integers, $$$t_1, \ldots, t_n$$$ and $$$a_1, \ldots, a_n$$$, indicating that $$$t_i$$$ seconds after the beginning of the match the ball will be kicked and you can monitor closely such kick if you are at the position $$$a_i$$$ along the touch-line. At the beginning of the game you start at position $$$0$$$ and the maximum speed at which you can walk along the touch-line is $$$v$$$ units per second (i.e., you can change your position by at most $$$v$$$ each second). What is the maximum number of kicks that you can monitor closely?
Input Specification: The first line contains two integers $$$n$$$ and $$$v$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le v \le 10^6$$$) — the number of kicks that will take place and your maximum speed. The second line contains $$$n$$$ integers $$$t_1, \ldots, t_n$$$ ($$$1 \le t_i \le 10^9$$$) — the times of the kicks in the match. The sequence of times is guaranteed to be strictly increasing, i.e., $$$t_1 < t_2 < \cdots < t_n$$$. The third line contains $$$n$$$ integers $$$a_1, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the positions along the touch-line where you have to be to monitor closely each kick.
Output Specification: Print the maximum number of kicks that you can monitor closely.
Notes: NoteIn the first sample, it is possible to move to the right at maximum speed for the first $$$3.5$$$ seconds and stay at position $$$7$$$ until the first kick happens, and then immediately move right also at maximum speed to watch the second kick at position $$$17$$$. There is no way to monitor closely the third kick after the second kick, so at most $$$2$$$ kicks can be seen.
Code:
import sys
import bisect
input = sys.stdin.buffer.readline
read = sys.stdin.buffer.read
N, V = map(int, input().split())
Ts = list(map(int, input().split()))
As = list(map(int, input().split()))
points = []
for T, A in zip(Ts, As):
B = T * V
x = B - A
y = B + A
if # TODO: Your code here:
continue
points.append((x, y))
points.sort()
# print(points)
lis = []
for _, w in points:
index = bisect.bisect_right(lis, w)
if index < len(lis):
lis[index] = w
else:
lis.append(w)
print(len(lis))
|
import sys
import bisect
input = sys.stdin.buffer.readline
read = sys.stdin.buffer.read
N, V = map(int, input().split())
Ts = list(map(int, input().split()))
As = list(map(int, input().split()))
points = []
for T, A in zip(Ts, As):
B = T * V
x = B - A
y = B + A
if {{completion}}:
continue
points.append((x, y))
points.sort()
# print(points)
lis = []
for _, w in points:
index = bisect.bisect_right(lis, w)
if index < len(lis):
lis[index] = w
else:
lis.append(w)
print(len(lis))
|
x < 0 or y < 0
|
[{"input": "3 2\n5 10 15\n7 17 29", "output": ["2"]}, {"input": "5 1\n5 7 8 11 13\n3 3 -2 -2 4", "output": ["3"]}, {"input": "1 2\n3\n7", "output": ["0"]}]
|
control_completion_001081
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: The derby between Milan and Inter is happening soon, and you have been chosen as the assistant referee for the match, also known as linesman. Your task is to move along the touch-line, namely the side of the field, always looking very carefully at the match to check for offside positions and other offences.Football is an extremely serious matter in Italy, and thus it is fundamental that you keep very close track of the ball for as much time as possible. This means that you want to maximise the number of kicks which you monitor closely. You are able to monitor closely a kick if, when it happens, you are in the position along the touch-line with minimum distance from the place where the kick happens.Fortunately, expert analysts have been able to accurately predict all the kicks which will occur during the game. That is, you have been given two lists of integers, $$$t_1, \ldots, t_n$$$ and $$$a_1, \ldots, a_n$$$, indicating that $$$t_i$$$ seconds after the beginning of the match the ball will be kicked and you can monitor closely such kick if you are at the position $$$a_i$$$ along the touch-line. At the beginning of the game you start at position $$$0$$$ and the maximum speed at which you can walk along the touch-line is $$$v$$$ units per second (i.e., you can change your position by at most $$$v$$$ each second). What is the maximum number of kicks that you can monitor closely?
Input Specification: The first line contains two integers $$$n$$$ and $$$v$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le v \le 10^6$$$) — the number of kicks that will take place and your maximum speed. The second line contains $$$n$$$ integers $$$t_1, \ldots, t_n$$$ ($$$1 \le t_i \le 10^9$$$) — the times of the kicks in the match. The sequence of times is guaranteed to be strictly increasing, i.e., $$$t_1 < t_2 < \cdots < t_n$$$. The third line contains $$$n$$$ integers $$$a_1, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the positions along the touch-line where you have to be to monitor closely each kick.
Output Specification: Print the maximum number of kicks that you can monitor closely.
Notes: NoteIn the first sample, it is possible to move to the right at maximum speed for the first $$$3.5$$$ seconds and stay at position $$$7$$$ until the first kick happens, and then immediately move right also at maximum speed to watch the second kick at position $$$17$$$. There is no way to monitor closely the third kick after the second kick, so at most $$$2$$$ kicks can be seen.
Code:
import sys
import bisect
input = sys.stdin.buffer.readline
read = sys.stdin.buffer.read
N, V = map(int, input().split())
Ts = list(map(int, input().split()))
As = list(map(int, input().split()))
points = []
for T, A in zip(Ts, As):
B = T * V
x = B - A
y = B + A
if x < 0 or y < 0:
continue
points.append((x, y))
points.sort()
# print(points)
lis = []
for _, w in points:
index = bisect.bisect_right(lis, w)
if # TODO: Your code here:
lis[index] = w
else:
lis.append(w)
print(len(lis))
|
import sys
import bisect
input = sys.stdin.buffer.readline
read = sys.stdin.buffer.read
N, V = map(int, input().split())
Ts = list(map(int, input().split()))
As = list(map(int, input().split()))
points = []
for T, A in zip(Ts, As):
B = T * V
x = B - A
y = B + A
if x < 0 or y < 0:
continue
points.append((x, y))
points.sort()
# print(points)
lis = []
for _, w in points:
index = bisect.bisect_right(lis, w)
if {{completion}}:
lis[index] = w
else:
lis.append(w)
print(len(lis))
|
index < len(lis)
|
[{"input": "3 2\n5 10 15\n7 17 29", "output": ["2"]}, {"input": "5 1\n5 7 8 11 13\n3 3 -2 -2 4", "output": ["3"]}, {"input": "1 2\n3\n7", "output": ["0"]}]
|
control_completion_001082
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: The derby between Milan and Inter is happening soon, and you have been chosen as the assistant referee for the match, also known as linesman. Your task is to move along the touch-line, namely the side of the field, always looking very carefully at the match to check for offside positions and other offences.Football is an extremely serious matter in Italy, and thus it is fundamental that you keep very close track of the ball for as much time as possible. This means that you want to maximise the number of kicks which you monitor closely. You are able to monitor closely a kick if, when it happens, you are in the position along the touch-line with minimum distance from the place where the kick happens.Fortunately, expert analysts have been able to accurately predict all the kicks which will occur during the game. That is, you have been given two lists of integers, $$$t_1, \ldots, t_n$$$ and $$$a_1, \ldots, a_n$$$, indicating that $$$t_i$$$ seconds after the beginning of the match the ball will be kicked and you can monitor closely such kick if you are at the position $$$a_i$$$ along the touch-line. At the beginning of the game you start at position $$$0$$$ and the maximum speed at which you can walk along the touch-line is $$$v$$$ units per second (i.e., you can change your position by at most $$$v$$$ each second). What is the maximum number of kicks that you can monitor closely?
Input Specification: The first line contains two integers $$$n$$$ and $$$v$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le v \le 10^6$$$) — the number of kicks that will take place and your maximum speed. The second line contains $$$n$$$ integers $$$t_1, \ldots, t_n$$$ ($$$1 \le t_i \le 10^9$$$) — the times of the kicks in the match. The sequence of times is guaranteed to be strictly increasing, i.e., $$$t_1 < t_2 < \cdots < t_n$$$. The third line contains $$$n$$$ integers $$$a_1, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the positions along the touch-line where you have to be to monitor closely each kick.
Output Specification: Print the maximum number of kicks that you can monitor closely.
Notes: NoteIn the first sample, it is possible to move to the right at maximum speed for the first $$$3.5$$$ seconds and stay at position $$$7$$$ until the first kick happens, and then immediately move right also at maximum speed to watch the second kick at position $$$17$$$. There is no way to monitor closely the third kick after the second kick, so at most $$$2$$$ kicks can be seen.
Code:
from bisect import bisect_right,bisect_left
n,v = map(int,input().split())
t = [*map(int,input().split())]
a = [*map(int,input().split())]
res = []
for i in range(n):
xi,yi = t[i]*v+a[i],t[i]*v-a[i]
if# TODO: Your code here:
res.append((xi,yi))
res.sort()
dp = [float("inf")]*(n+3)
dp[0] = 0
dp[n+2] = 0
for i in range(len(res)):
pos = bisect_right(dp,res[i][1],0,n+2)
dp[pos] = res[i][1]
for i in range(n,-1,-1):
if(dp[i]!=float("inf")):
print(i)
break
|
from bisect import bisect_right,bisect_left
n,v = map(int,input().split())
t = [*map(int,input().split())]
a = [*map(int,input().split())]
res = []
for i in range(n):
xi,yi = t[i]*v+a[i],t[i]*v-a[i]
if{{completion}}:
res.append((xi,yi))
res.sort()
dp = [float("inf")]*(n+3)
dp[0] = 0
dp[n+2] = 0
for i in range(len(res)):
pos = bisect_right(dp,res[i][1],0,n+2)
dp[pos] = res[i][1]
for i in range(n,-1,-1):
if(dp[i]!=float("inf")):
print(i)
break
|
(xi>=0 and yi>=0)
|
[{"input": "3 2\n5 10 15\n7 17 29", "output": ["2"]}, {"input": "5 1\n5 7 8 11 13\n3 3 -2 -2 4", "output": ["3"]}, {"input": "1 2\n3\n7", "output": ["0"]}]
|
control_completion_001083
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: The derby between Milan and Inter is happening soon, and you have been chosen as the assistant referee for the match, also known as linesman. Your task is to move along the touch-line, namely the side of the field, always looking very carefully at the match to check for offside positions and other offences.Football is an extremely serious matter in Italy, and thus it is fundamental that you keep very close track of the ball for as much time as possible. This means that you want to maximise the number of kicks which you monitor closely. You are able to monitor closely a kick if, when it happens, you are in the position along the touch-line with minimum distance from the place where the kick happens.Fortunately, expert analysts have been able to accurately predict all the kicks which will occur during the game. That is, you have been given two lists of integers, $$$t_1, \ldots, t_n$$$ and $$$a_1, \ldots, a_n$$$, indicating that $$$t_i$$$ seconds after the beginning of the match the ball will be kicked and you can monitor closely such kick if you are at the position $$$a_i$$$ along the touch-line. At the beginning of the game you start at position $$$0$$$ and the maximum speed at which you can walk along the touch-line is $$$v$$$ units per second (i.e., you can change your position by at most $$$v$$$ each second). What is the maximum number of kicks that you can monitor closely?
Input Specification: The first line contains two integers $$$n$$$ and $$$v$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le v \le 10^6$$$) — the number of kicks that will take place and your maximum speed. The second line contains $$$n$$$ integers $$$t_1, \ldots, t_n$$$ ($$$1 \le t_i \le 10^9$$$) — the times of the kicks in the match. The sequence of times is guaranteed to be strictly increasing, i.e., $$$t_1 < t_2 < \cdots < t_n$$$. The third line contains $$$n$$$ integers $$$a_1, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the positions along the touch-line where you have to be to monitor closely each kick.
Output Specification: Print the maximum number of kicks that you can monitor closely.
Notes: NoteIn the first sample, it is possible to move to the right at maximum speed for the first $$$3.5$$$ seconds and stay at position $$$7$$$ until the first kick happens, and then immediately move right also at maximum speed to watch the second kick at position $$$17$$$. There is no way to monitor closely the third kick after the second kick, so at most $$$2$$$ kicks can be seen.
Code:
from bisect import bisect_right,bisect_left
n,v = map(int,input().split())
t = [*map(int,input().split())]
a = [*map(int,input().split())]
res = []
for i in range(n):
xi,yi = t[i]*v+a[i],t[i]*v-a[i]
if(xi>=0 and yi>=0):
res.append((xi,yi))
res.sort()
dp = [float("inf")]*(n+3)
dp[0] = 0
dp[n+2] = 0
for i in range(len(res)):
pos = bisect_right(dp,res[i][1],0,n+2)
dp[pos] = res[i][1]
for i in range(n,-1,-1):
if# TODO: Your code here:
print(i)
break
|
from bisect import bisect_right,bisect_left
n,v = map(int,input().split())
t = [*map(int,input().split())]
a = [*map(int,input().split())]
res = []
for i in range(n):
xi,yi = t[i]*v+a[i],t[i]*v-a[i]
if(xi>=0 and yi>=0):
res.append((xi,yi))
res.sort()
dp = [float("inf")]*(n+3)
dp[0] = 0
dp[n+2] = 0
for i in range(len(res)):
pos = bisect_right(dp,res[i][1],0,n+2)
dp[pos] = res[i][1]
for i in range(n,-1,-1):
if{{completion}}:
print(i)
break
|
(dp[i]!=float("inf"))
|
[{"input": "3 2\n5 10 15\n7 17 29", "output": ["2"]}, {"input": "5 1\n5 7 8 11 13\n3 3 -2 -2 4", "output": ["3"]}, {"input": "1 2\n3\n7", "output": ["0"]}]
|
control_completion_001084
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a circular maze such as the ones shown in the figures. Determine if it can be solved, i.e., if there is a path which goes from the center to the outside of the maze which does not touch any wall. The maze is described by $$$n$$$ walls. Each wall can be either circular or straight. Circular walls are described by a radius $$$r$$$, the distance from the center, and two angles $$$\theta_1, \theta_2$$$ describing the beginning and the end of the wall in the clockwise direction. Notice that swapping the two angles changes the wall. Straight walls are described by an angle $$$\theta$$$, the direction of the wall, and two radii $$$r_1 < r_2$$$ describing the beginning and the end of the wall. Angles are measured in degrees; the angle $$$0$$$ corresponds to the upward pointing direction; and angles increase clockwise (hence the east direction corresponds to the angle $$$90$$$).
Input Specification: Each test contains multiple test cases. The first line contains an integer $$$t$$$ ($$$1\le t\le 20$$$) — the number of test cases. The descriptions of the $$$t$$$ test cases follow. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 5000$$$) — the number of walls. Each of the following $$$n$$$ lines each contains a character (C for circular, and S for straight) and three integers: either $$$r, \theta_1, \theta_2$$$ ($$$1 \leq r \leq 20$$$ and $$$0 \leq \theta_1,\theta_2 < 360$$$ with $$$\theta_1 \neq \theta_2$$$) if the wall is circular, or $$$r_1$$$, $$$r_2$$$ and $$$\theta$$$ ($$$1 \leq r_1 < r_2 \leq 20$$$ and $$$0 \leq \theta < 360$$$) if the wall is straight. It is guaranteed that circular walls do not overlap (but two circular walls may intersect at one or two points), and that straight walls do not overlap (but two straight walls may intersect at one point). However, circular and straight walls can intersect arbitrarily.
Output Specification: For each test case, print YES if the maze can be solved and NO otherwise.
Notes: NoteThe two sample test cases correspond to the two mazes in the picture.
Code:
t = int(input())
for _ in range(t):
field = [[0 for _ in range(2*360)] for _ in range(42)]
vis = [[False for _ in range(2*360)] for _ in range(42)]
n = int(input())
for _ in range(n):
line = input().split()
a, b, c = map(int, line[1:])
if line[0] == "C":
y = 2*a
x = 2*b
while x != 2*c:
field[y][x] = -1
x = (x + 1) % 720
field[y][x] = -1
else:
x = 2*c
for y in range(2*a, 2*b+1):
field[y][x] = -1
# for row in field: print(*row)
def check():
st = [(0, 0)]
while st:
y, x = st.pop(-1)
x = (x + 720) % 720
if y < 0 or y >= 42 or field[y][x] < 0: continue
if vis[y][x]: continue
vis[y][x] = True
if y > 40: return True
for ny in range(y-1, y+1+1):
for # TODO: Your code here:
st.append((ny, nx))
return False
print("YES" if check() else "NO")
|
t = int(input())
for _ in range(t):
field = [[0 for _ in range(2*360)] for _ in range(42)]
vis = [[False for _ in range(2*360)] for _ in range(42)]
n = int(input())
for _ in range(n):
line = input().split()
a, b, c = map(int, line[1:])
if line[0] == "C":
y = 2*a
x = 2*b
while x != 2*c:
field[y][x] = -1
x = (x + 1) % 720
field[y][x] = -1
else:
x = 2*c
for y in range(2*a, 2*b+1):
field[y][x] = -1
# for row in field: print(*row)
def check():
st = [(0, 0)]
while st:
y, x = st.pop(-1)
x = (x + 720) % 720
if y < 0 or y >= 42 or field[y][x] < 0: continue
if vis[y][x]: continue
vis[y][x] = True
if y > 40: return True
for ny in range(y-1, y+1+1):
for {{completion}}:
st.append((ny, nx))
return False
print("YES" if check() else "NO")
|
nx in range(x-1, x+1+1)
|
[{"input": "2\n5\nC 1 180 90\nC 5 250 230\nC 10 150 140\nC 20 185 180\nS 1 20 180\n6\nC 1 180 90\nC 5 250 230\nC 10 150 140\nC 20 185 180\nS 1 20 180\nS 5 10 0", "output": ["YES\nNO"]}]
|
control_completion_001093
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a circular maze such as the ones shown in the figures. Determine if it can be solved, i.e., if there is a path which goes from the center to the outside of the maze which does not touch any wall. The maze is described by $$$n$$$ walls. Each wall can be either circular or straight. Circular walls are described by a radius $$$r$$$, the distance from the center, and two angles $$$\theta_1, \theta_2$$$ describing the beginning and the end of the wall in the clockwise direction. Notice that swapping the two angles changes the wall. Straight walls are described by an angle $$$\theta$$$, the direction of the wall, and two radii $$$r_1 < r_2$$$ describing the beginning and the end of the wall. Angles are measured in degrees; the angle $$$0$$$ corresponds to the upward pointing direction; and angles increase clockwise (hence the east direction corresponds to the angle $$$90$$$).
Input Specification: Each test contains multiple test cases. The first line contains an integer $$$t$$$ ($$$1\le t\le 20$$$) — the number of test cases. The descriptions of the $$$t$$$ test cases follow. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 5000$$$) — the number of walls. Each of the following $$$n$$$ lines each contains a character (C for circular, and S for straight) and three integers: either $$$r, \theta_1, \theta_2$$$ ($$$1 \leq r \leq 20$$$ and $$$0 \leq \theta_1,\theta_2 < 360$$$ with $$$\theta_1 \neq \theta_2$$$) if the wall is circular, or $$$r_1$$$, $$$r_2$$$ and $$$\theta$$$ ($$$1 \leq r_1 < r_2 \leq 20$$$ and $$$0 \leq \theta < 360$$$) if the wall is straight. It is guaranteed that circular walls do not overlap (but two circular walls may intersect at one or two points), and that straight walls do not overlap (but two straight walls may intersect at one point). However, circular and straight walls can intersect arbitrarily.
Output Specification: For each test case, print YES if the maze can be solved and NO otherwise.
Notes: NoteThe two sample test cases correspond to the two mazes in the picture.
Code:
t = int(input())
for _ in range(t):
field = [[0 for _ in range(2*360)] for _ in range(42)]
vis = [[False for _ in range(2*360)] for _ in range(42)]
n = int(input())
for _ in range(n):
line = input().split()
a, b, c = map(int, line[1:])
if line[0] == "C":
y = 2*a
x = 2*b
while x != 2*c:
field[y][x] = -1
x = (x + 1) % 720
field[y][x] = -1
else:
x = 2*c
for # TODO: Your code here:
field[y][x] = -1
# for row in field: print(*row)
def check():
st = [(0, 0)]
while st:
y, x = st.pop(-1)
x = (x + 720) % 720
if y < 0 or y >= 42 or field[y][x] < 0: continue
if vis[y][x]: continue
vis[y][x] = True
if y > 40: return True
for ny in range(y-1, y+1+1):
for nx in range(x-1, x+1+1):
st.append((ny, nx))
return False
print("YES" if check() else "NO")
|
t = int(input())
for _ in range(t):
field = [[0 for _ in range(2*360)] for _ in range(42)]
vis = [[False for _ in range(2*360)] for _ in range(42)]
n = int(input())
for _ in range(n):
line = input().split()
a, b, c = map(int, line[1:])
if line[0] == "C":
y = 2*a
x = 2*b
while x != 2*c:
field[y][x] = -1
x = (x + 1) % 720
field[y][x] = -1
else:
x = 2*c
for {{completion}}:
field[y][x] = -1
# for row in field: print(*row)
def check():
st = [(0, 0)]
while st:
y, x = st.pop(-1)
x = (x + 720) % 720
if y < 0 or y >= 42 or field[y][x] < 0: continue
if vis[y][x]: continue
vis[y][x] = True
if y > 40: return True
for ny in range(y-1, y+1+1):
for nx in range(x-1, x+1+1):
st.append((ny, nx))
return False
print("YES" if check() else "NO")
|
y in range(2*a, 2*b+1)
|
[{"input": "2\n5\nC 1 180 90\nC 5 250 230\nC 10 150 140\nC 20 185 180\nS 1 20 180\n6\nC 1 180 90\nC 5 250 230\nC 10 150 140\nC 20 185 180\nS 1 20 180\nS 5 10 0", "output": ["YES\nNO"]}]
|
control_completion_001094
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a circular maze such as the ones shown in the figures. Determine if it can be solved, i.e., if there is a path which goes from the center to the outside of the maze which does not touch any wall. The maze is described by $$$n$$$ walls. Each wall can be either circular or straight. Circular walls are described by a radius $$$r$$$, the distance from the center, and two angles $$$\theta_1, \theta_2$$$ describing the beginning and the end of the wall in the clockwise direction. Notice that swapping the two angles changes the wall. Straight walls are described by an angle $$$\theta$$$, the direction of the wall, and two radii $$$r_1 < r_2$$$ describing the beginning and the end of the wall. Angles are measured in degrees; the angle $$$0$$$ corresponds to the upward pointing direction; and angles increase clockwise (hence the east direction corresponds to the angle $$$90$$$).
Input Specification: Each test contains multiple test cases. The first line contains an integer $$$t$$$ ($$$1\le t\le 20$$$) — the number of test cases. The descriptions of the $$$t$$$ test cases follow. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 5000$$$) — the number of walls. Each of the following $$$n$$$ lines each contains a character (C for circular, and S for straight) and three integers: either $$$r, \theta_1, \theta_2$$$ ($$$1 \leq r \leq 20$$$ and $$$0 \leq \theta_1,\theta_2 < 360$$$ with $$$\theta_1 \neq \theta_2$$$) if the wall is circular, or $$$r_1$$$, $$$r_2$$$ and $$$\theta$$$ ($$$1 \leq r_1 < r_2 \leq 20$$$ and $$$0 \leq \theta < 360$$$) if the wall is straight. It is guaranteed that circular walls do not overlap (but two circular walls may intersect at one or two points), and that straight walls do not overlap (but two straight walls may intersect at one point). However, circular and straight walls can intersect arbitrarily.
Output Specification: For each test case, print YES if the maze can be solved and NO otherwise.
Notes: NoteThe two sample test cases correspond to the two mazes in the picture.
Code:
from collections import deque
def bfs(graph, start):
visited = set()
queue = deque([start])
while queue:
vertex = queue.popleft()
if vertex not in visited:
visited.add(vertex)
queue.extend(graph[vertex] - visited)
return visited
for tc in range(int(input())):
graph = {}
for r in range(0, 22):
for angle in range(0, 360):
graph[(r,angle)] = set([
(r, (angle+1)%360),
(r, (angle-1)%360)])
if r < 21:
graph[(r,angle)].add((r+1, angle))
if r > 0:
graph[(r,angle)].add((r-1, angle))
nwalls = int(input())
for wallid in range(nwalls):
typ, a, b, c = input().split()
if typ == 'C':
rad, t1, t2 = map(int, (a,b,c))
th = t1
while th != t2:
graph[(rad, th)].remove((rad-1, th))
graph[(rad-1, th)].remove((rad, th))
th = (th + 1) % 360
#print(th)
#print((rad, th%360), (rad-1, th%360))
else:
r1, r2, th = map(int, (a,b,c))
for # TODO: Your code here:
graph[(rad, th)].remove((rad, (th-1)%360))
graph[(rad, (th-1)%360)].remove((rad, th))
if (0,0) in bfs(graph, (21, 0)):
print('YES')
else:
print('NO')
|
from collections import deque
def bfs(graph, start):
visited = set()
queue = deque([start])
while queue:
vertex = queue.popleft()
if vertex not in visited:
visited.add(vertex)
queue.extend(graph[vertex] - visited)
return visited
for tc in range(int(input())):
graph = {}
for r in range(0, 22):
for angle in range(0, 360):
graph[(r,angle)] = set([
(r, (angle+1)%360),
(r, (angle-1)%360)])
if r < 21:
graph[(r,angle)].add((r+1, angle))
if r > 0:
graph[(r,angle)].add((r-1, angle))
nwalls = int(input())
for wallid in range(nwalls):
typ, a, b, c = input().split()
if typ == 'C':
rad, t1, t2 = map(int, (a,b,c))
th = t1
while th != t2:
graph[(rad, th)].remove((rad-1, th))
graph[(rad-1, th)].remove((rad, th))
th = (th + 1) % 360
#print(th)
#print((rad, th%360), (rad-1, th%360))
else:
r1, r2, th = map(int, (a,b,c))
for {{completion}}:
graph[(rad, th)].remove((rad, (th-1)%360))
graph[(rad, (th-1)%360)].remove((rad, th))
if (0,0) in bfs(graph, (21, 0)):
print('YES')
else:
print('NO')
|
rad in range(r1, r2)
|
[{"input": "2\n5\nC 1 180 90\nC 5 250 230\nC 10 150 140\nC 20 185 180\nS 1 20 180\n6\nC 1 180 90\nC 5 250 230\nC 10 150 140\nC 20 185 180\nS 1 20 180\nS 5 10 0", "output": ["YES\nNO"]}]
|
control_completion_001095
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a circular maze such as the ones shown in the figures. Determine if it can be solved, i.e., if there is a path which goes from the center to the outside of the maze which does not touch any wall. The maze is described by $$$n$$$ walls. Each wall can be either circular or straight. Circular walls are described by a radius $$$r$$$, the distance from the center, and two angles $$$\theta_1, \theta_2$$$ describing the beginning and the end of the wall in the clockwise direction. Notice that swapping the two angles changes the wall. Straight walls are described by an angle $$$\theta$$$, the direction of the wall, and two radii $$$r_1 < r_2$$$ describing the beginning and the end of the wall. Angles are measured in degrees; the angle $$$0$$$ corresponds to the upward pointing direction; and angles increase clockwise (hence the east direction corresponds to the angle $$$90$$$).
Input Specification: Each test contains multiple test cases. The first line contains an integer $$$t$$$ ($$$1\le t\le 20$$$) — the number of test cases. The descriptions of the $$$t$$$ test cases follow. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 5000$$$) — the number of walls. Each of the following $$$n$$$ lines each contains a character (C for circular, and S for straight) and three integers: either $$$r, \theta_1, \theta_2$$$ ($$$1 \leq r \leq 20$$$ and $$$0 \leq \theta_1,\theta_2 < 360$$$ with $$$\theta_1 \neq \theta_2$$$) if the wall is circular, or $$$r_1$$$, $$$r_2$$$ and $$$\theta$$$ ($$$1 \leq r_1 < r_2 \leq 20$$$ and $$$0 \leq \theta < 360$$$) if the wall is straight. It is guaranteed that circular walls do not overlap (but two circular walls may intersect at one or two points), and that straight walls do not overlap (but two straight walls may intersect at one point). However, circular and straight walls can intersect arbitrarily.
Output Specification: For each test case, print YES if the maze can be solved and NO otherwise.
Notes: NoteThe two sample test cases correspond to the two mazes in the picture.
Code:
from collections import deque
def bfs(graph, start):
visited = set()
queue = deque([start])
while queue:
vertex = queue.popleft()
if vertex not in visited:
visited.add(vertex)
queue.extend(graph[vertex] - visited)
return visited
for tc in range(int(input())):
graph = {}
for r in range(0, 22):
for angle in range(0, 360):
graph[(r,angle)] = set([
(r, (angle+1)%360),
(r, (angle-1)%360)])
if # TODO: Your code here:
graph[(r,angle)].add((r+1, angle))
if r > 0:
graph[(r,angle)].add((r-1, angle))
nwalls = int(input())
for wallid in range(nwalls):
typ, a, b, c = input().split()
if typ == 'C':
rad, t1, t2 = map(int, (a,b,c))
th = t1
while th != t2:
graph[(rad, th)].remove((rad-1, th))
graph[(rad-1, th)].remove((rad, th))
th = (th + 1) % 360
#print(th)
#print((rad, th%360), (rad-1, th%360))
else:
r1, r2, th = map(int, (a,b,c))
for rad in range(r1, r2):
graph[(rad, th)].remove((rad, (th-1)%360))
graph[(rad, (th-1)%360)].remove((rad, th))
if (0,0) in bfs(graph, (21, 0)):
print('YES')
else:
print('NO')
|
from collections import deque
def bfs(graph, start):
visited = set()
queue = deque([start])
while queue:
vertex = queue.popleft()
if vertex not in visited:
visited.add(vertex)
queue.extend(graph[vertex] - visited)
return visited
for tc in range(int(input())):
graph = {}
for r in range(0, 22):
for angle in range(0, 360):
graph[(r,angle)] = set([
(r, (angle+1)%360),
(r, (angle-1)%360)])
if {{completion}}:
graph[(r,angle)].add((r+1, angle))
if r > 0:
graph[(r,angle)].add((r-1, angle))
nwalls = int(input())
for wallid in range(nwalls):
typ, a, b, c = input().split()
if typ == 'C':
rad, t1, t2 = map(int, (a,b,c))
th = t1
while th != t2:
graph[(rad, th)].remove((rad-1, th))
graph[(rad-1, th)].remove((rad, th))
th = (th + 1) % 360
#print(th)
#print((rad, th%360), (rad-1, th%360))
else:
r1, r2, th = map(int, (a,b,c))
for rad in range(r1, r2):
graph[(rad, th)].remove((rad, (th-1)%360))
graph[(rad, (th-1)%360)].remove((rad, th))
if (0,0) in bfs(graph, (21, 0)):
print('YES')
else:
print('NO')
|
r < 21
|
[{"input": "2\n5\nC 1 180 90\nC 5 250 230\nC 10 150 140\nC 20 185 180\nS 1 20 180\n6\nC 1 180 90\nC 5 250 230\nC 10 150 140\nC 20 185 180\nS 1 20 180\nS 5 10 0", "output": ["YES\nNO"]}]
|
control_completion_001096
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a circular maze such as the ones shown in the figures. Determine if it can be solved, i.e., if there is a path which goes from the center to the outside of the maze which does not touch any wall. The maze is described by $$$n$$$ walls. Each wall can be either circular or straight. Circular walls are described by a radius $$$r$$$, the distance from the center, and two angles $$$\theta_1, \theta_2$$$ describing the beginning and the end of the wall in the clockwise direction. Notice that swapping the two angles changes the wall. Straight walls are described by an angle $$$\theta$$$, the direction of the wall, and two radii $$$r_1 < r_2$$$ describing the beginning and the end of the wall. Angles are measured in degrees; the angle $$$0$$$ corresponds to the upward pointing direction; and angles increase clockwise (hence the east direction corresponds to the angle $$$90$$$).
Input Specification: Each test contains multiple test cases. The first line contains an integer $$$t$$$ ($$$1\le t\le 20$$$) — the number of test cases. The descriptions of the $$$t$$$ test cases follow. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 5000$$$) — the number of walls. Each of the following $$$n$$$ lines each contains a character (C for circular, and S for straight) and three integers: either $$$r, \theta_1, \theta_2$$$ ($$$1 \leq r \leq 20$$$ and $$$0 \leq \theta_1,\theta_2 < 360$$$ with $$$\theta_1 \neq \theta_2$$$) if the wall is circular, or $$$r_1$$$, $$$r_2$$$ and $$$\theta$$$ ($$$1 \leq r_1 < r_2 \leq 20$$$ and $$$0 \leq \theta < 360$$$) if the wall is straight. It is guaranteed that circular walls do not overlap (but two circular walls may intersect at one or two points), and that straight walls do not overlap (but two straight walls may intersect at one point). However, circular and straight walls can intersect arbitrarily.
Output Specification: For each test case, print YES if the maze can be solved and NO otherwise.
Notes: NoteThe two sample test cases correspond to the two mazes in the picture.
Code:
from itertools import islice, chain
from sys import stdin
MAX_RADIUS = 20
lines = iter(stdin.readlines()[1:])
for line in lines:
n = int(line)
circular_wall, straight_wall = ([[False] * 360 for _ in range(MAX_RADIUS)] for _ in range(2))
for shape, *params in map(str.split, islice(lines, n)):
params = map(int, params)
if shape == 'C':
r, theta_1, theta_2 = params
r -= 1
theta_range = range(theta_1, theta_2) if theta_1 <= theta_2 \
else chain(range(theta_1, 360), range(0, theta_2))
for theta in theta_range:
circular_wall[r][theta] = True
else:
assert shape == 'S'
r1, r2, theta = params
r1 -= 1
r2 -= 1
for # TODO: Your code here:
straight_wall[r][theta] = True
queue = [(0, i) for i, inner_wall in enumerate(circular_wall[0]) if not inner_wall]
seen = set(queue)
while queue:
row, col = queue.pop()
# print(row, col)
neighbors = []
if row >= 1 and not circular_wall[row][col]:
neighbors.append((row - 1, col))
right_col = (col + 1) % 360
if not straight_wall[row][right_col]:
neighbors.append((row, right_col))
if not straight_wall[row][col]:
neighbors.append((row, (col - 1) % 360))
next_row = row + 1
if not circular_wall[next_row][col]:
if next_row == MAX_RADIUS - 1:
print('YES')
break
neighbors.append((next_row, col))
for neighbor in neighbors:
if neighbor in seen:
continue
queue.append(neighbor)
seen.add(neighbor)
else: # no break
print('NO')
|
from itertools import islice, chain
from sys import stdin
MAX_RADIUS = 20
lines = iter(stdin.readlines()[1:])
for line in lines:
n = int(line)
circular_wall, straight_wall = ([[False] * 360 for _ in range(MAX_RADIUS)] for _ in range(2))
for shape, *params in map(str.split, islice(lines, n)):
params = map(int, params)
if shape == 'C':
r, theta_1, theta_2 = params
r -= 1
theta_range = range(theta_1, theta_2) if theta_1 <= theta_2 \
else chain(range(theta_1, 360), range(0, theta_2))
for theta in theta_range:
circular_wall[r][theta] = True
else:
assert shape == 'S'
r1, r2, theta = params
r1 -= 1
r2 -= 1
for {{completion}}:
straight_wall[r][theta] = True
queue = [(0, i) for i, inner_wall in enumerate(circular_wall[0]) if not inner_wall]
seen = set(queue)
while queue:
row, col = queue.pop()
# print(row, col)
neighbors = []
if row >= 1 and not circular_wall[row][col]:
neighbors.append((row - 1, col))
right_col = (col + 1) % 360
if not straight_wall[row][right_col]:
neighbors.append((row, right_col))
if not straight_wall[row][col]:
neighbors.append((row, (col - 1) % 360))
next_row = row + 1
if not circular_wall[next_row][col]:
if next_row == MAX_RADIUS - 1:
print('YES')
break
neighbors.append((next_row, col))
for neighbor in neighbors:
if neighbor in seen:
continue
queue.append(neighbor)
seen.add(neighbor)
else: # no break
print('NO')
|
r in range(r1, r2)
|
[{"input": "2\n5\nC 1 180 90\nC 5 250 230\nC 10 150 140\nC 20 185 180\nS 1 20 180\n6\nC 1 180 90\nC 5 250 230\nC 10 150 140\nC 20 185 180\nS 1 20 180\nS 5 10 0", "output": ["YES\nNO"]}]
|
control_completion_001097
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a circular maze such as the ones shown in the figures. Determine if it can be solved, i.e., if there is a path which goes from the center to the outside of the maze which does not touch any wall. The maze is described by $$$n$$$ walls. Each wall can be either circular or straight. Circular walls are described by a radius $$$r$$$, the distance from the center, and two angles $$$\theta_1, \theta_2$$$ describing the beginning and the end of the wall in the clockwise direction. Notice that swapping the two angles changes the wall. Straight walls are described by an angle $$$\theta$$$, the direction of the wall, and two radii $$$r_1 < r_2$$$ describing the beginning and the end of the wall. Angles are measured in degrees; the angle $$$0$$$ corresponds to the upward pointing direction; and angles increase clockwise (hence the east direction corresponds to the angle $$$90$$$).
Input Specification: Each test contains multiple test cases. The first line contains an integer $$$t$$$ ($$$1\le t\le 20$$$) — the number of test cases. The descriptions of the $$$t$$$ test cases follow. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 5000$$$) — the number of walls. Each of the following $$$n$$$ lines each contains a character (C for circular, and S for straight) and three integers: either $$$r, \theta_1, \theta_2$$$ ($$$1 \leq r \leq 20$$$ and $$$0 \leq \theta_1,\theta_2 < 360$$$ with $$$\theta_1 \neq \theta_2$$$) if the wall is circular, or $$$r_1$$$, $$$r_2$$$ and $$$\theta$$$ ($$$1 \leq r_1 < r_2 \leq 20$$$ and $$$0 \leq \theta < 360$$$) if the wall is straight. It is guaranteed that circular walls do not overlap (but two circular walls may intersect at one or two points), and that straight walls do not overlap (but two straight walls may intersect at one point). However, circular and straight walls can intersect arbitrarily.
Output Specification: For each test case, print YES if the maze can be solved and NO otherwise.
Notes: NoteThe two sample test cases correspond to the two mazes in the picture.
Code:
from itertools import islice, chain
from sys import stdin
MAX_RADIUS = 20
lines = iter(stdin.readlines()[1:])
for line in lines:
n = int(line)
circular_wall, straight_wall = ([[False] * 360 for _ in range(MAX_RADIUS)] for _ in range(2))
for shape, *params in map(str.split, islice(lines, n)):
params = map(int, params)
if shape == 'C':
r, theta_1, theta_2 = params
r -= 1
theta_range = range(theta_1, theta_2) if theta_1 <= theta_2 \
else chain(range(theta_1, 360), range(0, theta_2))
for # TODO: Your code here:
circular_wall[r][theta] = True
else:
assert shape == 'S'
r1, r2, theta = params
r1 -= 1
r2 -= 1
for r in range(r1, r2):
straight_wall[r][theta] = True
queue = [(0, i) for i, inner_wall in enumerate(circular_wall[0]) if not inner_wall]
seen = set(queue)
while queue:
row, col = queue.pop()
# print(row, col)
neighbors = []
if row >= 1 and not circular_wall[row][col]:
neighbors.append((row - 1, col))
right_col = (col + 1) % 360
if not straight_wall[row][right_col]:
neighbors.append((row, right_col))
if not straight_wall[row][col]:
neighbors.append((row, (col - 1) % 360))
next_row = row + 1
if not circular_wall[next_row][col]:
if next_row == MAX_RADIUS - 1:
print('YES')
break
neighbors.append((next_row, col))
for neighbor in neighbors:
if neighbor in seen:
continue
queue.append(neighbor)
seen.add(neighbor)
else: # no break
print('NO')
|
from itertools import islice, chain
from sys import stdin
MAX_RADIUS = 20
lines = iter(stdin.readlines()[1:])
for line in lines:
n = int(line)
circular_wall, straight_wall = ([[False] * 360 for _ in range(MAX_RADIUS)] for _ in range(2))
for shape, *params in map(str.split, islice(lines, n)):
params = map(int, params)
if shape == 'C':
r, theta_1, theta_2 = params
r -= 1
theta_range = range(theta_1, theta_2) if theta_1 <= theta_2 \
else chain(range(theta_1, 360), range(0, theta_2))
for {{completion}}:
circular_wall[r][theta] = True
else:
assert shape == 'S'
r1, r2, theta = params
r1 -= 1
r2 -= 1
for r in range(r1, r2):
straight_wall[r][theta] = True
queue = [(0, i) for i, inner_wall in enumerate(circular_wall[0]) if not inner_wall]
seen = set(queue)
while queue:
row, col = queue.pop()
# print(row, col)
neighbors = []
if row >= 1 and not circular_wall[row][col]:
neighbors.append((row - 1, col))
right_col = (col + 1) % 360
if not straight_wall[row][right_col]:
neighbors.append((row, right_col))
if not straight_wall[row][col]:
neighbors.append((row, (col - 1) % 360))
next_row = row + 1
if not circular_wall[next_row][col]:
if next_row == MAX_RADIUS - 1:
print('YES')
break
neighbors.append((next_row, col))
for neighbor in neighbors:
if neighbor in seen:
continue
queue.append(neighbor)
seen.add(neighbor)
else: # no break
print('NO')
|
theta in theta_range
|
[{"input": "2\n5\nC 1 180 90\nC 5 250 230\nC 10 150 140\nC 20 185 180\nS 1 20 180\n6\nC 1 180 90\nC 5 250 230\nC 10 150 140\nC 20 185 180\nS 1 20 180\nS 5 10 0", "output": ["YES\nNO"]}]
|
control_completion_001098
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: On a beach there are $$$n$$$ huts in a perfect line, hut $$$1$$$ being at the left and hut $$$i+1$$$ being $$$100$$$ meters to the right of hut $$$i$$$, for all $$$1 \le i \le n - 1$$$. In hut $$$i$$$ there are $$$p_i$$$ people.There are $$$m$$$ ice cream sellers, also aligned in a perfect line with all the huts. The $$$i$$$-th ice cream seller has their shop $$$x_i$$$ meters to the right of the first hut. All ice cream shops are at distinct locations, but they may be at the same location as a hut.You want to open a new ice cream shop and you wonder what the best location for your shop is. You can place your ice cream shop anywhere on the beach (not necessarily at an integer distance from the first hut) as long as it is aligned with the huts and the other ice cream shops, even if there is already another ice cream shop or a hut at that location. You know that people would come to your shop only if it is strictly closer to their hut than any other ice cream shop.If every person living in the huts wants to buy exactly one ice cream, what is the maximum number of ice creams that you can sell if you place the shop optimally?
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 200\,000$$$, $$$1 \le m \le 200\,000$$$) — the number of huts and the number of ice cream sellers. The second line contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le 10^9$$$) — the number of people in each hut. The third line contains $$$m$$$ integers $$$x_1, x_2, \ldots, x_m$$$ ($$$0 \le x_i \le 10^9$$$, $$$x_i \ne x_j$$$ for $$$i \ne j$$$) — the location of each ice cream shop.
Output Specification: Print the maximum number of ice creams that can be sold by choosing optimally the location of the new shop.
Notes: NoteIn the first sample, you can place the shop (coloured orange in the picture below) $$$150$$$ meters to the right of the first hut (for example) so that it is the closest shop to the first two huts, which have $$$2$$$ and $$$5$$$ people, for a total of $$$7$$$ sold ice creams. In the second sample, you can place the shop $$$170$$$ meters to the right of the first hut (for example) so that it is the closest shop to the last two huts, which have $$$7$$$ and $$$8$$$ people, for a total of $$$15$$$ sold ice creams.
Code:
from itertools import chain
from sys import stdin
(n, m), population, shops = [[int(x) for x in line.split()] for line in stdin.readlines()]
shops.sort()
shops = chain([float('-inf')], (v / 100 for v in shops), [float('inf')])
shop_left, shop_right = next(shops), next(shops)
hut_left_idx = max_score = score = 0
for hut_right_idx, hut_right_score in enumerate(population):
score += hut_right_score
# print(f'{score=}')
while # TODO: Your code here:
shop_left, shop_right = shop_right, next(shops)
# print(f'{hut_right_idx=} {shop_left=} {shop_right=}')
shop_delta = shop_right - shop_left
while shop_left >= hut_left_idx or 2 * (hut_right_idx - hut_left_idx) >= shop_delta:
score -= population[hut_left_idx]
hut_left_idx += 1
# print(f'{score=} {hut_left_idx=} {hut_right_idx=} {shop_delta=}')
if score > max_score:
max_score = score
print(max_score)
|
from itertools import chain
from sys import stdin
(n, m), population, shops = [[int(x) for x in line.split()] for line in stdin.readlines()]
shops.sort()
shops = chain([float('-inf')], (v / 100 for v in shops), [float('inf')])
shop_left, shop_right = next(shops), next(shops)
hut_left_idx = max_score = score = 0
for hut_right_idx, hut_right_score in enumerate(population):
score += hut_right_score
# print(f'{score=}')
while {{completion}}:
shop_left, shop_right = shop_right, next(shops)
# print(f'{hut_right_idx=} {shop_left=} {shop_right=}')
shop_delta = shop_right - shop_left
while shop_left >= hut_left_idx or 2 * (hut_right_idx - hut_left_idx) >= shop_delta:
score -= population[hut_left_idx]
hut_left_idx += 1
# print(f'{score=} {hut_left_idx=} {hut_right_idx=} {shop_delta=}')
if score > max_score:
max_score = score
print(max_score)
|
shop_right <= hut_right_idx
|
[{"input": "3 1\n2 5 6\n169", "output": ["7"]}, {"input": "4 2\n1 2 7 8\n35 157", "output": ["15"]}, {"input": "4 1\n272203905 348354708 848256926 939404176\n20", "output": ["2136015810"]}, {"input": "3 2\n1 1 1\n300 99", "output": ["2"]}]
|
control_completion_001125
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: On a beach there are $$$n$$$ huts in a perfect line, hut $$$1$$$ being at the left and hut $$$i+1$$$ being $$$100$$$ meters to the right of hut $$$i$$$, for all $$$1 \le i \le n - 1$$$. In hut $$$i$$$ there are $$$p_i$$$ people.There are $$$m$$$ ice cream sellers, also aligned in a perfect line with all the huts. The $$$i$$$-th ice cream seller has their shop $$$x_i$$$ meters to the right of the first hut. All ice cream shops are at distinct locations, but they may be at the same location as a hut.You want to open a new ice cream shop and you wonder what the best location for your shop is. You can place your ice cream shop anywhere on the beach (not necessarily at an integer distance from the first hut) as long as it is aligned with the huts and the other ice cream shops, even if there is already another ice cream shop or a hut at that location. You know that people would come to your shop only if it is strictly closer to their hut than any other ice cream shop.If every person living in the huts wants to buy exactly one ice cream, what is the maximum number of ice creams that you can sell if you place the shop optimally?
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 200\,000$$$, $$$1 \le m \le 200\,000$$$) — the number of huts and the number of ice cream sellers. The second line contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le 10^9$$$) — the number of people in each hut. The third line contains $$$m$$$ integers $$$x_1, x_2, \ldots, x_m$$$ ($$$0 \le x_i \le 10^9$$$, $$$x_i \ne x_j$$$ for $$$i \ne j$$$) — the location of each ice cream shop.
Output Specification: Print the maximum number of ice creams that can be sold by choosing optimally the location of the new shop.
Notes: NoteIn the first sample, you can place the shop (coloured orange in the picture below) $$$150$$$ meters to the right of the first hut (for example) so that it is the closest shop to the first two huts, which have $$$2$$$ and $$$5$$$ people, for a total of $$$7$$$ sold ice creams. In the second sample, you can place the shop $$$170$$$ meters to the right of the first hut (for example) so that it is the closest shop to the last two huts, which have $$$7$$$ and $$$8$$$ people, for a total of $$$15$$$ sold ice creams.
Code:
from itertools import chain
from sys import stdin
(n, m), population, shops = [[int(x) for x in line.split()] for line in stdin.readlines()]
shops.sort()
shops = chain([float('-inf')], (v / 100 for v in shops), [float('inf')])
shop_left, shop_right = next(shops), next(shops)
hut_left_idx = max_score = score = 0
for hut_right_idx, hut_right_score in enumerate(population):
score += hut_right_score
# print(f'{score=}')
while shop_right <= hut_right_idx:
shop_left, shop_right = shop_right, next(shops)
# print(f'{hut_right_idx=} {shop_left=} {shop_right=}')
shop_delta = shop_right - shop_left
while # TODO: Your code here:
score -= population[hut_left_idx]
hut_left_idx += 1
# print(f'{score=} {hut_left_idx=} {hut_right_idx=} {shop_delta=}')
if score > max_score:
max_score = score
print(max_score)
|
from itertools import chain
from sys import stdin
(n, m), population, shops = [[int(x) for x in line.split()] for line in stdin.readlines()]
shops.sort()
shops = chain([float('-inf')], (v / 100 for v in shops), [float('inf')])
shop_left, shop_right = next(shops), next(shops)
hut_left_idx = max_score = score = 0
for hut_right_idx, hut_right_score in enumerate(population):
score += hut_right_score
# print(f'{score=}')
while shop_right <= hut_right_idx:
shop_left, shop_right = shop_right, next(shops)
# print(f'{hut_right_idx=} {shop_left=} {shop_right=}')
shop_delta = shop_right - shop_left
while {{completion}}:
score -= population[hut_left_idx]
hut_left_idx += 1
# print(f'{score=} {hut_left_idx=} {hut_right_idx=} {shop_delta=}')
if score > max_score:
max_score = score
print(max_score)
|
shop_left >= hut_left_idx or 2 * (hut_right_idx - hut_left_idx) >= shop_delta
|
[{"input": "3 1\n2 5 6\n169", "output": ["7"]}, {"input": "4 2\n1 2 7 8\n35 157", "output": ["15"]}, {"input": "4 1\n272203905 348354708 848256926 939404176\n20", "output": ["2136015810"]}, {"input": "3 2\n1 1 1\n300 99", "output": ["2"]}]
|
control_completion_001126
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: On a beach there are $$$n$$$ huts in a perfect line, hut $$$1$$$ being at the left and hut $$$i+1$$$ being $$$100$$$ meters to the right of hut $$$i$$$, for all $$$1 \le i \le n - 1$$$. In hut $$$i$$$ there are $$$p_i$$$ people.There are $$$m$$$ ice cream sellers, also aligned in a perfect line with all the huts. The $$$i$$$-th ice cream seller has their shop $$$x_i$$$ meters to the right of the first hut. All ice cream shops are at distinct locations, but they may be at the same location as a hut.You want to open a new ice cream shop and you wonder what the best location for your shop is. You can place your ice cream shop anywhere on the beach (not necessarily at an integer distance from the first hut) as long as it is aligned with the huts and the other ice cream shops, even if there is already another ice cream shop or a hut at that location. You know that people would come to your shop only if it is strictly closer to their hut than any other ice cream shop.If every person living in the huts wants to buy exactly one ice cream, what is the maximum number of ice creams that you can sell if you place the shop optimally?
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 200\,000$$$, $$$1 \le m \le 200\,000$$$) — the number of huts and the number of ice cream sellers. The second line contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le 10^9$$$) — the number of people in each hut. The third line contains $$$m$$$ integers $$$x_1, x_2, \ldots, x_m$$$ ($$$0 \le x_i \le 10^9$$$, $$$x_i \ne x_j$$$ for $$$i \ne j$$$) — the location of each ice cream shop.
Output Specification: Print the maximum number of ice creams that can be sold by choosing optimally the location of the new shop.
Notes: NoteIn the first sample, you can place the shop (coloured orange in the picture below) $$$150$$$ meters to the right of the first hut (for example) so that it is the closest shop to the first two huts, which have $$$2$$$ and $$$5$$$ people, for a total of $$$7$$$ sold ice creams. In the second sample, you can place the shop $$$170$$$ meters to the right of the first hut (for example) so that it is the closest shop to the last two huts, which have $$$7$$$ and $$$8$$$ people, for a total of $$$15$$$ sold ice creams.
Code:
n,m=map(int,input().split())
p=list(map(int,input().split()))
x=sorted(list(map(int,input().split())))
s=sum(p[:-(-(x[0])//100)])
for i in range(len(x)-1):
if x[i]//100+1>=n:
break
num=int(((x[i+1]-x[i])/2)//(100)+1)
l=x[i]//100+1
r=-(-(x[i+1])//100)
r=min(r,n)
prefs=0
if # TODO: Your code here:
prefs=sum(p[l:l+num])
s=max(s,prefs)
while l+num<r:
prefs-=p[l]
prefs+=p[l+num] if l+num<n else 0
s=max(s,prefs)
l+=1
s=max(s,sum(p[x[-1]//100+1:]))
print(s)
|
n,m=map(int,input().split())
p=list(map(int,input().split()))
x=sorted(list(map(int,input().split())))
s=sum(p[:-(-(x[0])//100)])
for i in range(len(x)-1):
if x[i]//100+1>=n:
break
num=int(((x[i+1]-x[i])/2)//(100)+1)
l=x[i]//100+1
r=-(-(x[i+1])//100)
r=min(r,n)
prefs=0
if {{completion}}:
prefs=sum(p[l:l+num])
s=max(s,prefs)
while l+num<r:
prefs-=p[l]
prefs+=p[l+num] if l+num<n else 0
s=max(s,prefs)
l+=1
s=max(s,sum(p[x[-1]//100+1:]))
print(s)
|
l+num<=r
|
[{"input": "3 1\n2 5 6\n169", "output": ["7"]}, {"input": "4 2\n1 2 7 8\n35 157", "output": ["15"]}, {"input": "4 1\n272203905 348354708 848256926 939404176\n20", "output": ["2136015810"]}, {"input": "3 2\n1 1 1\n300 99", "output": ["2"]}]
|
control_completion_001127
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: On a beach there are $$$n$$$ huts in a perfect line, hut $$$1$$$ being at the left and hut $$$i+1$$$ being $$$100$$$ meters to the right of hut $$$i$$$, for all $$$1 \le i \le n - 1$$$. In hut $$$i$$$ there are $$$p_i$$$ people.There are $$$m$$$ ice cream sellers, also aligned in a perfect line with all the huts. The $$$i$$$-th ice cream seller has their shop $$$x_i$$$ meters to the right of the first hut. All ice cream shops are at distinct locations, but they may be at the same location as a hut.You want to open a new ice cream shop and you wonder what the best location for your shop is. You can place your ice cream shop anywhere on the beach (not necessarily at an integer distance from the first hut) as long as it is aligned with the huts and the other ice cream shops, even if there is already another ice cream shop or a hut at that location. You know that people would come to your shop only if it is strictly closer to their hut than any other ice cream shop.If every person living in the huts wants to buy exactly one ice cream, what is the maximum number of ice creams that you can sell if you place the shop optimally?
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 200\,000$$$, $$$1 \le m \le 200\,000$$$) — the number of huts and the number of ice cream sellers. The second line contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le 10^9$$$) — the number of people in each hut. The third line contains $$$m$$$ integers $$$x_1, x_2, \ldots, x_m$$$ ($$$0 \le x_i \le 10^9$$$, $$$x_i \ne x_j$$$ for $$$i \ne j$$$) — the location of each ice cream shop.
Output Specification: Print the maximum number of ice creams that can be sold by choosing optimally the location of the new shop.
Notes: NoteIn the first sample, you can place the shop (coloured orange in the picture below) $$$150$$$ meters to the right of the first hut (for example) so that it is the closest shop to the first two huts, which have $$$2$$$ and $$$5$$$ people, for a total of $$$7$$$ sold ice creams. In the second sample, you can place the shop $$$170$$$ meters to the right of the first hut (for example) so that it is the closest shop to the last two huts, which have $$$7$$$ and $$$8$$$ people, for a total of $$$15$$$ sold ice creams.
Code:
N, M = [int(x) for x in input().split()]
hut = [int(x) for x in input().split()]
shop = [int(x) for x in input().split()]
shop = sorted([-1e9] + shop + [1e9])
events = []
j = 0
for i in range(N):
while # TODO: Your code here:
j += 1
if shop[j] != 100 * i:
d = min(100*i - shop[j-1], shop[j] - 100*i)
events.append((100*i-d, hut[i]))
events.append((100*i+d, -hut[i]))
events.sort()
cont = 0
max = 0
for a in events:
cont += a[1]
if cont > max:
max = cont
print(max)
|
N, M = [int(x) for x in input().split()]
hut = [int(x) for x in input().split()]
shop = [int(x) for x in input().split()]
shop = sorted([-1e9] + shop + [1e9])
events = []
j = 0
for i in range(N):
while {{completion}}:
j += 1
if shop[j] != 100 * i:
d = min(100*i - shop[j-1], shop[j] - 100*i)
events.append((100*i-d, hut[i]))
events.append((100*i+d, -hut[i]))
events.sort()
cont = 0
max = 0
for a in events:
cont += a[1]
if cont > max:
max = cont
print(max)
|
shop[j] < 100*i
|
[{"input": "3 1\n2 5 6\n169", "output": ["7"]}, {"input": "4 2\n1 2 7 8\n35 157", "output": ["15"]}, {"input": "4 1\n272203905 348354708 848256926 939404176\n20", "output": ["2136015810"]}, {"input": "3 2\n1 1 1\n300 99", "output": ["2"]}]
|
control_completion_001128
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: On a beach there are $$$n$$$ huts in a perfect line, hut $$$1$$$ being at the left and hut $$$i+1$$$ being $$$100$$$ meters to the right of hut $$$i$$$, for all $$$1 \le i \le n - 1$$$. In hut $$$i$$$ there are $$$p_i$$$ people.There are $$$m$$$ ice cream sellers, also aligned in a perfect line with all the huts. The $$$i$$$-th ice cream seller has their shop $$$x_i$$$ meters to the right of the first hut. All ice cream shops are at distinct locations, but they may be at the same location as a hut.You want to open a new ice cream shop and you wonder what the best location for your shop is. You can place your ice cream shop anywhere on the beach (not necessarily at an integer distance from the first hut) as long as it is aligned with the huts and the other ice cream shops, even if there is already another ice cream shop or a hut at that location. You know that people would come to your shop only if it is strictly closer to their hut than any other ice cream shop.If every person living in the huts wants to buy exactly one ice cream, what is the maximum number of ice creams that you can sell if you place the shop optimally?
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 200\,000$$$, $$$1 \le m \le 200\,000$$$) — the number of huts and the number of ice cream sellers. The second line contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le 10^9$$$) — the number of people in each hut. The third line contains $$$m$$$ integers $$$x_1, x_2, \ldots, x_m$$$ ($$$0 \le x_i \le 10^9$$$, $$$x_i \ne x_j$$$ for $$$i \ne j$$$) — the location of each ice cream shop.
Output Specification: Print the maximum number of ice creams that can be sold by choosing optimally the location of the new shop.
Notes: NoteIn the first sample, you can place the shop (coloured orange in the picture below) $$$150$$$ meters to the right of the first hut (for example) so that it is the closest shop to the first two huts, which have $$$2$$$ and $$$5$$$ people, for a total of $$$7$$$ sold ice creams. In the second sample, you can place the shop $$$170$$$ meters to the right of the first hut (for example) so that it is the closest shop to the last two huts, which have $$$7$$$ and $$$8$$$ people, for a total of $$$15$$$ sold ice creams.
Code:
N, M = [int(x) for x in input().split()]
hut = [int(x) for x in input().split()]
shop = [int(x) for x in input().split()]
shop = sorted([-1e9] + shop + [1e9])
events = []
j = 0
for i in range(N):
while shop[j] < 100*i:
j += 1
if # TODO: Your code here:
d = min(100*i - shop[j-1], shop[j] - 100*i)
events.append((100*i-d, hut[i]))
events.append((100*i+d, -hut[i]))
events.sort()
cont = 0
max = 0
for a in events:
cont += a[1]
if cont > max:
max = cont
print(max)
|
N, M = [int(x) for x in input().split()]
hut = [int(x) for x in input().split()]
shop = [int(x) for x in input().split()]
shop = sorted([-1e9] + shop + [1e9])
events = []
j = 0
for i in range(N):
while shop[j] < 100*i:
j += 1
if {{completion}}:
d = min(100*i - shop[j-1], shop[j] - 100*i)
events.append((100*i-d, hut[i]))
events.append((100*i+d, -hut[i]))
events.sort()
cont = 0
max = 0
for a in events:
cont += a[1]
if cont > max:
max = cont
print(max)
|
shop[j] != 100 * i
|
[{"input": "3 1\n2 5 6\n169", "output": ["7"]}, {"input": "4 2\n1 2 7 8\n35 157", "output": ["15"]}, {"input": "4 1\n272203905 348354708 848256926 939404176\n20", "output": ["2136015810"]}, {"input": "3 2\n1 1 1\n300 99", "output": ["2"]}]
|
control_completion_001129
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Bethany would like to tile her bathroom. The bathroom has width $$$w$$$ centimeters and length $$$l$$$ centimeters. If Bethany simply used the basic tiles of size $$$1 \times 1$$$ centimeters, she would use $$$w \cdot l$$$ of them. However, she has something different in mind. On the interior of the floor she wants to use the $$$1 \times 1$$$ tiles. She needs exactly $$$(w-2) \cdot (l-2)$$$ of these. On the floor boundary she wants to use tiles of size $$$1 \times a$$$ for some positive integer $$$a$$$. The tiles can also be rotated by $$$90$$$ degrees. For which values of $$$a$$$ can Bethany tile the bathroom floor as described? Note that $$$a$$$ can also be $$$1$$$.
Input Specification: Each test contains multiple test cases. The first line contains an integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The descriptions of the $$$t$$$ test cases follow. Each test case consist of a single line, which contains two integers $$$w$$$, $$$l$$$ ($$$3 \leq w, l \leq 10^{9}$$$) — the dimensions of the bathroom.
Output Specification: For each test case, print an integer $$$k$$$ ($$$0\le k$$$) — the number of valid values of $$$a$$$ for the given test case — followed by $$$k$$$ integers $$$a_1, a_2,\dots, a_k$$$ ($$$1\le a_i$$$) — the valid values of $$$a$$$. The values $$$a_1, a_2, \dots, a_k$$$ have to be sorted from smallest to largest. It is guaranteed that under the problem constraints, the output contains at most $$$200\,000$$$ integers.
Notes: NoteIn the first test case, the bathroom is $$$3$$$ centimeters wide and $$$5$$$ centimeters long. There are three values of $$$a$$$ such that Bethany can tile the floor as described in the statement, namely $$$a=1$$$, $$$a=2$$$ and $$$a=3$$$. The three tilings are represented in the following pictures.
Code:
from math import sqrt, floor
from sys import stdin
data = [int(x) for x in stdin.read().split()[1:]]
res = []
for w, l in zip(data[::2], data[1::2]):
half_perimeter = w + l - 2
solutions = {1, 2}
for i in range(2, floor(sqrt(half_perimeter)) + 1):
div, mod_i = divmod(half_perimeter, i)
if mod_i != 0:
continue
for a in [i, div]:
mod_a = w % a
if # TODO: Your code here:
assert (l - 2 + mod_a) % a == 0
solutions.add(a)
res.append(f"{len(solutions)} {' '.join(map(str, sorted(solutions)))}")
print('\n'.join(res))
|
from math import sqrt, floor
from sys import stdin
data = [int(x) for x in stdin.read().split()[1:]]
res = []
for w, l in zip(data[::2], data[1::2]):
half_perimeter = w + l - 2
solutions = {1, 2}
for i in range(2, floor(sqrt(half_perimeter)) + 1):
div, mod_i = divmod(half_perimeter, i)
if mod_i != 0:
continue
for a in [i, div]:
mod_a = w % a
if {{completion}}:
assert (l - 2 + mod_a) % a == 0
solutions.add(a)
res.append(f"{len(solutions)} {' '.join(map(str, sorted(solutions)))}")
print('\n'.join(res))
|
mod_a <= 2
|
[{"input": "3\n\n3 5\n\n12 12\n\n314159265 358979323", "output": ["3 1 2 3\n3 1 2 11\n2 1 2"]}]
|
control_completion_001138
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Alina has discovered a weird language, which contains only $$$4$$$ words: $$$\texttt{A}$$$, $$$\texttt{B}$$$, $$$\texttt{AB}$$$, $$$\texttt{BA}$$$. It also turned out that there are no spaces in this language: a sentence is written by just concatenating its words into a single string.Alina has found one such sentence $$$s$$$ and she is curious: is it possible that it consists of precisely $$$a$$$ words $$$\texttt{A}$$$, $$$b$$$ words $$$\texttt{B}$$$, $$$c$$$ words $$$\texttt{AB}$$$, and $$$d$$$ words $$$\texttt{BA}$$$?In other words, determine, if it's possible to concatenate these $$$a+b+c+d$$$ words in some order so that the resulting string is $$$s$$$. Each of the $$$a+b+c+d$$$ words must be used exactly once in the concatenation, but you can choose the order in which they are concatenated.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains four integers $$$a$$$, $$$b$$$, $$$c$$$, $$$d$$$ ($$$0\le a,b,c,d\le 2\cdot 10^5$$$) — the number of times that words $$$\texttt{A}$$$, $$$\texttt{B}$$$, $$$\texttt{AB}$$$, $$$\texttt{BA}$$$ respectively must be used in the sentence. The second line contains the string $$$s$$$ ($$$s$$$ consists only of the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$, $$$1\le |s| \le 2\cdot 10^5$$$, $$$|s|=a+b+2c+2d$$$) — the sentence. Notice that the condition $$$|s|=a+b+2c+2d$$$ (here $$$|s|$$$ denotes the length of the string $$$s$$$) is equivalent to the fact that $$$s$$$ is as long as the concatenation of the $$$a+b+c+d$$$ words. The sum of the lengths of $$$s$$$ over all test cases doesn't exceed $$$2\cdot 10^5$$$.
Output Specification: For each test case output $$$\texttt{YES}$$$ if it is possible that the sentence $$$s$$$ consists of precisely $$$a$$$ words $$$\texttt{A}$$$, $$$b$$$ words $$$\texttt{B}$$$, $$$c$$$ words $$$\texttt{AB}$$$, and $$$d$$$ words $$$\texttt{BA}$$$, and $$$\texttt{NO}$$$ otherwise. You can output each letter in any case.
Notes: NoteIn the first test case, the sentence $$$s$$$ is $$$\texttt{B}$$$. Clearly, it can't consist of a single word $$$\texttt{A}$$$, so the answer is $$$\texttt{NO}$$$.In the second test case, the sentence $$$s$$$ is $$$\texttt{AB}$$$, and it's possible that it consists of a single word $$$\texttt{AB}$$$, so the answer is $$$\texttt{YES}$$$.In the third test case, the sentence $$$s$$$ is $$$\texttt{ABAB}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{B}$$$, and one word $$$\texttt{BA}$$$, as $$$\texttt{A} + \texttt{BA} + \texttt{B} = \texttt{ABAB}$$$.In the fourth test case, the sentence $$$s$$$ is $$$\texttt{ABAAB}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{AB}$$$, and one word $$$\texttt{BA}$$$, as $$$\texttt{A} + \texttt{BA} + \texttt{AB} = \texttt{ABAAB}$$$. In the fifth test case, the sentence $$$s$$$ is $$$\texttt{BAABBABBAA}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{B}$$$, two words $$$\texttt{AB}$$$, and two words $$$\texttt{BA}$$$, as $$$\texttt{BA} + \texttt{AB} + \texttt{B} + \texttt{AB} + \texttt{BA} + \texttt{A}= \texttt{BAABBABBAA}$$$.
Code:
t = int(input())
for _ in range(t):
a, b, c, d = [int(i) for i in input().split()]
s = input()
if s.count('A') != a+c+d:
print("NO")
continue
ult = 'X'
k = 0
z = []
for x in s:
if x == ult:
z.append((k, ult))
k = 1
else:
ult = x
k += 1
z.append((k, ult))
r = 0
z.sort()
for k,v in z:
if k % 2 == 0:
if v == 'A' and d >= k//2:
d -= k//2
elif # TODO: Your code here:
c -= k//2
else:
r += k//2 - 1
else:
r += k//2
print("YES" if r >= c+d else "NO")
|
t = int(input())
for _ in range(t):
a, b, c, d = [int(i) for i in input().split()]
s = input()
if s.count('A') != a+c+d:
print("NO")
continue
ult = 'X'
k = 0
z = []
for x in s:
if x == ult:
z.append((k, ult))
k = 1
else:
ult = x
k += 1
z.append((k, ult))
r = 0
z.sort()
for k,v in z:
if k % 2 == 0:
if v == 'A' and d >= k//2:
d -= k//2
elif {{completion}}:
c -= k//2
else:
r += k//2 - 1
else:
r += k//2
print("YES" if r >= c+d else "NO")
|
v == 'B' and c >= k//2
|
[{"input": "8\n1 0 0 0\nB\n0 0 1 0\nAB\n1 1 0 1\nABAB\n1 0 1 1\nABAAB\n1 1 2 2\nBAABBABBAA\n1 1 2 3\nABABABBAABAB\n2 3 5 4\nAABAABBABAAABABBABBBABB\n1 3 3 10\nBBABABABABBBABABABABABABAABABA", "output": ["NO\nYES\nYES\nYES\nYES\nYES\nNO\nYES"]}]
|
control_completion_001182
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Alina has discovered a weird language, which contains only $$$4$$$ words: $$$\texttt{A}$$$, $$$\texttt{B}$$$, $$$\texttt{AB}$$$, $$$\texttt{BA}$$$. It also turned out that there are no spaces in this language: a sentence is written by just concatenating its words into a single string.Alina has found one such sentence $$$s$$$ and she is curious: is it possible that it consists of precisely $$$a$$$ words $$$\texttt{A}$$$, $$$b$$$ words $$$\texttt{B}$$$, $$$c$$$ words $$$\texttt{AB}$$$, and $$$d$$$ words $$$\texttt{BA}$$$?In other words, determine, if it's possible to concatenate these $$$a+b+c+d$$$ words in some order so that the resulting string is $$$s$$$. Each of the $$$a+b+c+d$$$ words must be used exactly once in the concatenation, but you can choose the order in which they are concatenated.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains four integers $$$a$$$, $$$b$$$, $$$c$$$, $$$d$$$ ($$$0\le a,b,c,d\le 2\cdot 10^5$$$) — the number of times that words $$$\texttt{A}$$$, $$$\texttt{B}$$$, $$$\texttt{AB}$$$, $$$\texttt{BA}$$$ respectively must be used in the sentence. The second line contains the string $$$s$$$ ($$$s$$$ consists only of the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$, $$$1\le |s| \le 2\cdot 10^5$$$, $$$|s|=a+b+2c+2d$$$) — the sentence. Notice that the condition $$$|s|=a+b+2c+2d$$$ (here $$$|s|$$$ denotes the length of the string $$$s$$$) is equivalent to the fact that $$$s$$$ is as long as the concatenation of the $$$a+b+c+d$$$ words. The sum of the lengths of $$$s$$$ over all test cases doesn't exceed $$$2\cdot 10^5$$$.
Output Specification: For each test case output $$$\texttt{YES}$$$ if it is possible that the sentence $$$s$$$ consists of precisely $$$a$$$ words $$$\texttt{A}$$$, $$$b$$$ words $$$\texttt{B}$$$, $$$c$$$ words $$$\texttt{AB}$$$, and $$$d$$$ words $$$\texttt{BA}$$$, and $$$\texttt{NO}$$$ otherwise. You can output each letter in any case.
Notes: NoteIn the first test case, the sentence $$$s$$$ is $$$\texttt{B}$$$. Clearly, it can't consist of a single word $$$\texttt{A}$$$, so the answer is $$$\texttt{NO}$$$.In the second test case, the sentence $$$s$$$ is $$$\texttt{AB}$$$, and it's possible that it consists of a single word $$$\texttt{AB}$$$, so the answer is $$$\texttt{YES}$$$.In the third test case, the sentence $$$s$$$ is $$$\texttt{ABAB}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{B}$$$, and one word $$$\texttt{BA}$$$, as $$$\texttt{A} + \texttt{BA} + \texttt{B} = \texttt{ABAB}$$$.In the fourth test case, the sentence $$$s$$$ is $$$\texttt{ABAAB}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{AB}$$$, and one word $$$\texttt{BA}$$$, as $$$\texttt{A} + \texttt{BA} + \texttt{AB} = \texttt{ABAAB}$$$. In the fifth test case, the sentence $$$s$$$ is $$$\texttt{BAABBABBAA}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{B}$$$, two words $$$\texttt{AB}$$$, and two words $$$\texttt{BA}$$$, as $$$\texttt{BA} + \texttt{AB} + \texttt{B} + \texttt{AB} + \texttt{BA} + \texttt{A}= \texttt{BAABBABBAA}$$$.
Code:
def canmake(s,a,b,c,d):
anum = s.count('A')
bnum = s.count('B')
cnum = s.count('AB')
dnum = s.count('BA')
if cnum < c or dnum < d or anum != a + c + d or bnum != b + c + d:
return False
n=len(s)
ans=0
abls=[]
bals=[]
l=0
while l<n:
while l<n-1 and s[l]==s[l+1]:
l+=1
r=l
while # TODO: Your code here:
r+=1
if s[l]== s[r]=='B':
ans+=(r-l+1)//2
if s[l]==s[r]=='A':
ans+=(r-l+1)//2
if s[l]=='A' and s[r]=='B':
abls.append((r-l+1)//2)
if s[l]=='B' and s[r]=='A':
bals.append((r-l+1)//2)
l=r+1
abls.sort()
bals.sort()
for i in abls:
if i<=c:
c-=i
else:
d-=i-c-1
c = 0
for i in bals:
if i<=d:
d-=i
else:
c-=i-d-1
d = 0
return (c+d)<=ans
t=int(input())
for _ in range(t):
a,b,c,d=[int(x) for x in input().split()]
s=input()
res=canmake(s,a,b,c,d)
if res:
print('YES')
else:
print('NO')
|
def canmake(s,a,b,c,d):
anum = s.count('A')
bnum = s.count('B')
cnum = s.count('AB')
dnum = s.count('BA')
if cnum < c or dnum < d or anum != a + c + d or bnum != b + c + d:
return False
n=len(s)
ans=0
abls=[]
bals=[]
l=0
while l<n:
while l<n-1 and s[l]==s[l+1]:
l+=1
r=l
while {{completion}}:
r+=1
if s[l]== s[r]=='B':
ans+=(r-l+1)//2
if s[l]==s[r]=='A':
ans+=(r-l+1)//2
if s[l]=='A' and s[r]=='B':
abls.append((r-l+1)//2)
if s[l]=='B' and s[r]=='A':
bals.append((r-l+1)//2)
l=r+1
abls.sort()
bals.sort()
for i in abls:
if i<=c:
c-=i
else:
d-=i-c-1
c = 0
for i in bals:
if i<=d:
d-=i
else:
c-=i-d-1
d = 0
return (c+d)<=ans
t=int(input())
for _ in range(t):
a,b,c,d=[int(x) for x in input().split()]
s=input()
res=canmake(s,a,b,c,d)
if res:
print('YES')
else:
print('NO')
|
r<n-1 and s[r]!=s[r+1]
|
[{"input": "8\n1 0 0 0\nB\n0 0 1 0\nAB\n1 1 0 1\nABAB\n1 0 1 1\nABAAB\n1 1 2 2\nBAABBABBAA\n1 1 2 3\nABABABBAABAB\n2 3 5 4\nAABAABBABAAABABBABBBABB\n1 3 3 10\nBBABABABABBBABABABABABABAABABA", "output": ["NO\nYES\nYES\nYES\nYES\nYES\nNO\nYES"]}]
|
control_completion_001183
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Alina has discovered a weird language, which contains only $$$4$$$ words: $$$\texttt{A}$$$, $$$\texttt{B}$$$, $$$\texttt{AB}$$$, $$$\texttt{BA}$$$. It also turned out that there are no spaces in this language: a sentence is written by just concatenating its words into a single string.Alina has found one such sentence $$$s$$$ and she is curious: is it possible that it consists of precisely $$$a$$$ words $$$\texttt{A}$$$, $$$b$$$ words $$$\texttt{B}$$$, $$$c$$$ words $$$\texttt{AB}$$$, and $$$d$$$ words $$$\texttt{BA}$$$?In other words, determine, if it's possible to concatenate these $$$a+b+c+d$$$ words in some order so that the resulting string is $$$s$$$. Each of the $$$a+b+c+d$$$ words must be used exactly once in the concatenation, but you can choose the order in which they are concatenated.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains four integers $$$a$$$, $$$b$$$, $$$c$$$, $$$d$$$ ($$$0\le a,b,c,d\le 2\cdot 10^5$$$) — the number of times that words $$$\texttt{A}$$$, $$$\texttt{B}$$$, $$$\texttt{AB}$$$, $$$\texttt{BA}$$$ respectively must be used in the sentence. The second line contains the string $$$s$$$ ($$$s$$$ consists only of the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$, $$$1\le |s| \le 2\cdot 10^5$$$, $$$|s|=a+b+2c+2d$$$) — the sentence. Notice that the condition $$$|s|=a+b+2c+2d$$$ (here $$$|s|$$$ denotes the length of the string $$$s$$$) is equivalent to the fact that $$$s$$$ is as long as the concatenation of the $$$a+b+c+d$$$ words. The sum of the lengths of $$$s$$$ over all test cases doesn't exceed $$$2\cdot 10^5$$$.
Output Specification: For each test case output $$$\texttt{YES}$$$ if it is possible that the sentence $$$s$$$ consists of precisely $$$a$$$ words $$$\texttt{A}$$$, $$$b$$$ words $$$\texttt{B}$$$, $$$c$$$ words $$$\texttt{AB}$$$, and $$$d$$$ words $$$\texttt{BA}$$$, and $$$\texttt{NO}$$$ otherwise. You can output each letter in any case.
Notes: NoteIn the first test case, the sentence $$$s$$$ is $$$\texttt{B}$$$. Clearly, it can't consist of a single word $$$\texttt{A}$$$, so the answer is $$$\texttt{NO}$$$.In the second test case, the sentence $$$s$$$ is $$$\texttt{AB}$$$, and it's possible that it consists of a single word $$$\texttt{AB}$$$, so the answer is $$$\texttt{YES}$$$.In the third test case, the sentence $$$s$$$ is $$$\texttt{ABAB}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{B}$$$, and one word $$$\texttt{BA}$$$, as $$$\texttt{A} + \texttt{BA} + \texttt{B} = \texttt{ABAB}$$$.In the fourth test case, the sentence $$$s$$$ is $$$\texttt{ABAAB}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{AB}$$$, and one word $$$\texttt{BA}$$$, as $$$\texttt{A} + \texttt{BA} + \texttt{AB} = \texttt{ABAAB}$$$. In the fifth test case, the sentence $$$s$$$ is $$$\texttt{BAABBABBAA}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{B}$$$, two words $$$\texttt{AB}$$$, and two words $$$\texttt{BA}$$$, as $$$\texttt{BA} + \texttt{AB} + \texttt{B} + \texttt{AB} + \texttt{BA} + \texttt{A}= \texttt{BAABBABBAA}$$$.
Code:
def canmake(s,a,b,c,d):
anum = s.count('A')
bnum = s.count('B')
cnum = s.count('AB')
dnum = s.count('BA')
if cnum < c or dnum < d or anum != a + c + d or bnum != b + c + d:
return False
n=len(s)
ans=0
abls=[]
bals=[]
l=0
while l<n:
while l<n-1 and s[l]==s[l+1]:
l+=1
r=l
while r<n-1 and s[r]!=s[r+1]:
r+=1
if # TODO: Your code here:
ans+=(r-l+1)//2
if s[l]==s[r]=='A':
ans+=(r-l+1)//2
if s[l]=='A' and s[r]=='B':
abls.append((r-l+1)//2)
if s[l]=='B' and s[r]=='A':
bals.append((r-l+1)//2)
l=r+1
abls.sort()
bals.sort()
for i in abls:
if i<=c:
c-=i
else:
d-=i-c-1
c = 0
for i in bals:
if i<=d:
d-=i
else:
c-=i-d-1
d = 0
return (c+d)<=ans
t=int(input())
for _ in range(t):
a,b,c,d=[int(x) for x in input().split()]
s=input()
res=canmake(s,a,b,c,d)
if res:
print('YES')
else:
print('NO')
|
def canmake(s,a,b,c,d):
anum = s.count('A')
bnum = s.count('B')
cnum = s.count('AB')
dnum = s.count('BA')
if cnum < c or dnum < d or anum != a + c + d or bnum != b + c + d:
return False
n=len(s)
ans=0
abls=[]
bals=[]
l=0
while l<n:
while l<n-1 and s[l]==s[l+1]:
l+=1
r=l
while r<n-1 and s[r]!=s[r+1]:
r+=1
if {{completion}}:
ans+=(r-l+1)//2
if s[l]==s[r]=='A':
ans+=(r-l+1)//2
if s[l]=='A' and s[r]=='B':
abls.append((r-l+1)//2)
if s[l]=='B' and s[r]=='A':
bals.append((r-l+1)//2)
l=r+1
abls.sort()
bals.sort()
for i in abls:
if i<=c:
c-=i
else:
d-=i-c-1
c = 0
for i in bals:
if i<=d:
d-=i
else:
c-=i-d-1
d = 0
return (c+d)<=ans
t=int(input())
for _ in range(t):
a,b,c,d=[int(x) for x in input().split()]
s=input()
res=canmake(s,a,b,c,d)
if res:
print('YES')
else:
print('NO')
|
s[l]== s[r]=='B'
|
[{"input": "8\n1 0 0 0\nB\n0 0 1 0\nAB\n1 1 0 1\nABAB\n1 0 1 1\nABAAB\n1 1 2 2\nBAABBABBAA\n1 1 2 3\nABABABBAABAB\n2 3 5 4\nAABAABBABAAABABBABBBABB\n1 3 3 10\nBBABABABABBBABABABABABABAABABA", "output": ["NO\nYES\nYES\nYES\nYES\nYES\nNO\nYES"]}]
|
control_completion_001184
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Alina has discovered a weird language, which contains only $$$4$$$ words: $$$\texttt{A}$$$, $$$\texttt{B}$$$, $$$\texttt{AB}$$$, $$$\texttt{BA}$$$. It also turned out that there are no spaces in this language: a sentence is written by just concatenating its words into a single string.Alina has found one such sentence $$$s$$$ and she is curious: is it possible that it consists of precisely $$$a$$$ words $$$\texttt{A}$$$, $$$b$$$ words $$$\texttt{B}$$$, $$$c$$$ words $$$\texttt{AB}$$$, and $$$d$$$ words $$$\texttt{BA}$$$?In other words, determine, if it's possible to concatenate these $$$a+b+c+d$$$ words in some order so that the resulting string is $$$s$$$. Each of the $$$a+b+c+d$$$ words must be used exactly once in the concatenation, but you can choose the order in which they are concatenated.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains four integers $$$a$$$, $$$b$$$, $$$c$$$, $$$d$$$ ($$$0\le a,b,c,d\le 2\cdot 10^5$$$) — the number of times that words $$$\texttt{A}$$$, $$$\texttt{B}$$$, $$$\texttt{AB}$$$, $$$\texttt{BA}$$$ respectively must be used in the sentence. The second line contains the string $$$s$$$ ($$$s$$$ consists only of the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$, $$$1\le |s| \le 2\cdot 10^5$$$, $$$|s|=a+b+2c+2d$$$) — the sentence. Notice that the condition $$$|s|=a+b+2c+2d$$$ (here $$$|s|$$$ denotes the length of the string $$$s$$$) is equivalent to the fact that $$$s$$$ is as long as the concatenation of the $$$a+b+c+d$$$ words. The sum of the lengths of $$$s$$$ over all test cases doesn't exceed $$$2\cdot 10^5$$$.
Output Specification: For each test case output $$$\texttt{YES}$$$ if it is possible that the sentence $$$s$$$ consists of precisely $$$a$$$ words $$$\texttt{A}$$$, $$$b$$$ words $$$\texttt{B}$$$, $$$c$$$ words $$$\texttt{AB}$$$, and $$$d$$$ words $$$\texttt{BA}$$$, and $$$\texttt{NO}$$$ otherwise. You can output each letter in any case.
Notes: NoteIn the first test case, the sentence $$$s$$$ is $$$\texttt{B}$$$. Clearly, it can't consist of a single word $$$\texttt{A}$$$, so the answer is $$$\texttt{NO}$$$.In the second test case, the sentence $$$s$$$ is $$$\texttt{AB}$$$, and it's possible that it consists of a single word $$$\texttt{AB}$$$, so the answer is $$$\texttt{YES}$$$.In the third test case, the sentence $$$s$$$ is $$$\texttt{ABAB}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{B}$$$, and one word $$$\texttt{BA}$$$, as $$$\texttt{A} + \texttt{BA} + \texttt{B} = \texttt{ABAB}$$$.In the fourth test case, the sentence $$$s$$$ is $$$\texttt{ABAAB}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{AB}$$$, and one word $$$\texttt{BA}$$$, as $$$\texttt{A} + \texttt{BA} + \texttt{AB} = \texttt{ABAAB}$$$. In the fifth test case, the sentence $$$s$$$ is $$$\texttt{BAABBABBAA}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{B}$$$, two words $$$\texttt{AB}$$$, and two words $$$\texttt{BA}$$$, as $$$\texttt{BA} + \texttt{AB} + \texttt{B} + \texttt{AB} + \texttt{BA} + \texttt{A}= \texttt{BAABBABBAA}$$$.
Code:
for _ in range(int(input())):
a,b,ab,ba=map(int,input().split());s=input()
if s.count('A')!=a+ab+ba:print('NO');continue
stack=[[1,s[0]]]
for i in range(1,len(s)):
if stack[-1][1]!=s[i]:
x=stack.pop()
stack.append([x[0]+1,s[i]])
else: stack.append([1,s[i]])
stack.sort();trash=0
for val,ele in stack:
if not val%2:
if ele=='A' and ba>=val//2:ba-=(val//2)
elif # TODO: Your code here:ab-=(val//2)
else:trash+=(val//2-1)
else:
trash+=(val//2)
print('YES' if trash>=ab+ba else 'NO')
|
for _ in range(int(input())):
a,b,ab,ba=map(int,input().split());s=input()
if s.count('A')!=a+ab+ba:print('NO');continue
stack=[[1,s[0]]]
for i in range(1,len(s)):
if stack[-1][1]!=s[i]:
x=stack.pop()
stack.append([x[0]+1,s[i]])
else: stack.append([1,s[i]])
stack.sort();trash=0
for val,ele in stack:
if not val%2:
if ele=='A' and ba>=val//2:ba-=(val//2)
elif {{completion}}:ab-=(val//2)
else:trash+=(val//2-1)
else:
trash+=(val//2)
print('YES' if trash>=ab+ba else 'NO')
|
ele=='B' and ab>=val//2
|
[{"input": "8\n1 0 0 0\nB\n0 0 1 0\nAB\n1 1 0 1\nABAB\n1 0 1 1\nABAAB\n1 1 2 2\nBAABBABBAA\n1 1 2 3\nABABABBAABAB\n2 3 5 4\nAABAABBABAAABABBABBBABB\n1 3 3 10\nBBABABABABBBABABABABABABAABABA", "output": ["NO\nYES\nYES\nYES\nYES\nYES\nNO\nYES"]}]
|
control_completion_001185
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Alina has discovered a weird language, which contains only $$$4$$$ words: $$$\texttt{A}$$$, $$$\texttt{B}$$$, $$$\texttt{AB}$$$, $$$\texttt{BA}$$$. It also turned out that there are no spaces in this language: a sentence is written by just concatenating its words into a single string.Alina has found one such sentence $$$s$$$ and she is curious: is it possible that it consists of precisely $$$a$$$ words $$$\texttt{A}$$$, $$$b$$$ words $$$\texttt{B}$$$, $$$c$$$ words $$$\texttt{AB}$$$, and $$$d$$$ words $$$\texttt{BA}$$$?In other words, determine, if it's possible to concatenate these $$$a+b+c+d$$$ words in some order so that the resulting string is $$$s$$$. Each of the $$$a+b+c+d$$$ words must be used exactly once in the concatenation, but you can choose the order in which they are concatenated.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains four integers $$$a$$$, $$$b$$$, $$$c$$$, $$$d$$$ ($$$0\le a,b,c,d\le 2\cdot 10^5$$$) — the number of times that words $$$\texttt{A}$$$, $$$\texttt{B}$$$, $$$\texttt{AB}$$$, $$$\texttt{BA}$$$ respectively must be used in the sentence. The second line contains the string $$$s$$$ ($$$s$$$ consists only of the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$, $$$1\le |s| \le 2\cdot 10^5$$$, $$$|s|=a+b+2c+2d$$$) — the sentence. Notice that the condition $$$|s|=a+b+2c+2d$$$ (here $$$|s|$$$ denotes the length of the string $$$s$$$) is equivalent to the fact that $$$s$$$ is as long as the concatenation of the $$$a+b+c+d$$$ words. The sum of the lengths of $$$s$$$ over all test cases doesn't exceed $$$2\cdot 10^5$$$.
Output Specification: For each test case output $$$\texttt{YES}$$$ if it is possible that the sentence $$$s$$$ consists of precisely $$$a$$$ words $$$\texttt{A}$$$, $$$b$$$ words $$$\texttt{B}$$$, $$$c$$$ words $$$\texttt{AB}$$$, and $$$d$$$ words $$$\texttt{BA}$$$, and $$$\texttt{NO}$$$ otherwise. You can output each letter in any case.
Notes: NoteIn the first test case, the sentence $$$s$$$ is $$$\texttt{B}$$$. Clearly, it can't consist of a single word $$$\texttt{A}$$$, so the answer is $$$\texttt{NO}$$$.In the second test case, the sentence $$$s$$$ is $$$\texttt{AB}$$$, and it's possible that it consists of a single word $$$\texttt{AB}$$$, so the answer is $$$\texttt{YES}$$$.In the third test case, the sentence $$$s$$$ is $$$\texttt{ABAB}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{B}$$$, and one word $$$\texttt{BA}$$$, as $$$\texttt{A} + \texttt{BA} + \texttt{B} = \texttt{ABAB}$$$.In the fourth test case, the sentence $$$s$$$ is $$$\texttt{ABAAB}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{AB}$$$, and one word $$$\texttt{BA}$$$, as $$$\texttt{A} + \texttt{BA} + \texttt{AB} = \texttt{ABAAB}$$$. In the fifth test case, the sentence $$$s$$$ is $$$\texttt{BAABBABBAA}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{B}$$$, two words $$$\texttt{AB}$$$, and two words $$$\texttt{BA}$$$, as $$$\texttt{BA} + \texttt{AB} + \texttt{B} + \texttt{AB} + \texttt{BA} + \texttt{A}= \texttt{BAABBABBAA}$$$.
Code:
import sys,os,io
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
input = sys.stdin.readline
for _ in range (int(input())):
c = [int(i) for i in input().split()]
s = list(input().strip())
if s.count('A') != c[0] + c[2] + c[3] or s.count('B') != c[1] + c[2] + c[3]:
print("NO")
continue
n = len(s)
a = [[s[0]]]
for i in range (1,n):
if s[i]==s[i-1]:
a.append([s[i]])
else:
a[-1].append(s[i])
extra = 0
for i in a:
if len(i)%2:
c[ord(i[0]) - ord('A')] -= 1
extra += len(i)//2
a.sort(key = lambda x: len(x))
for i in a:
if len(i)%2==0:
cnt = len(i)//2
if # TODO: Your code here:
c[2 + ord(i[0]) - ord('A')]-=cnt
else:
extra += cnt - 1
if min(c)<0 or extra < c[2]+c[3]:
print("NO")
else:
print("YES")
|
import sys,os,io
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
input = sys.stdin.readline
for _ in range (int(input())):
c = [int(i) for i in input().split()]
s = list(input().strip())
if s.count('A') != c[0] + c[2] + c[3] or s.count('B') != c[1] + c[2] + c[3]:
print("NO")
continue
n = len(s)
a = [[s[0]]]
for i in range (1,n):
if s[i]==s[i-1]:
a.append([s[i]])
else:
a[-1].append(s[i])
extra = 0
for i in a:
if len(i)%2:
c[ord(i[0]) - ord('A')] -= 1
extra += len(i)//2
a.sort(key = lambda x: len(x))
for i in a:
if len(i)%2==0:
cnt = len(i)//2
if {{completion}}:
c[2 + ord(i[0]) - ord('A')]-=cnt
else:
extra += cnt - 1
if min(c)<0 or extra < c[2]+c[3]:
print("NO")
else:
print("YES")
|
cnt <= c[2 + ord(i[0])-ord('A')]
|
[{"input": "8\n1 0 0 0\nB\n0 0 1 0\nAB\n1 1 0 1\nABAB\n1 0 1 1\nABAAB\n1 1 2 2\nBAABBABBAA\n1 1 2 3\nABABABBAABAB\n2 3 5 4\nAABAABBABAAABABBABBBABB\n1 3 3 10\nBBABABABABBBABABABABABABAABABA", "output": ["NO\nYES\nYES\nYES\nYES\nYES\nNO\nYES"]}]
|
control_completion_001186
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Alina has discovered a weird language, which contains only $$$4$$$ words: $$$\texttt{A}$$$, $$$\texttt{B}$$$, $$$\texttt{AB}$$$, $$$\texttt{BA}$$$. It also turned out that there are no spaces in this language: a sentence is written by just concatenating its words into a single string.Alina has found one such sentence $$$s$$$ and she is curious: is it possible that it consists of precisely $$$a$$$ words $$$\texttt{A}$$$, $$$b$$$ words $$$\texttt{B}$$$, $$$c$$$ words $$$\texttt{AB}$$$, and $$$d$$$ words $$$\texttt{BA}$$$?In other words, determine, if it's possible to concatenate these $$$a+b+c+d$$$ words in some order so that the resulting string is $$$s$$$. Each of the $$$a+b+c+d$$$ words must be used exactly once in the concatenation, but you can choose the order in which they are concatenated.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains four integers $$$a$$$, $$$b$$$, $$$c$$$, $$$d$$$ ($$$0\le a,b,c,d\le 2\cdot 10^5$$$) — the number of times that words $$$\texttt{A}$$$, $$$\texttt{B}$$$, $$$\texttt{AB}$$$, $$$\texttt{BA}$$$ respectively must be used in the sentence. The second line contains the string $$$s$$$ ($$$s$$$ consists only of the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$, $$$1\le |s| \le 2\cdot 10^5$$$, $$$|s|=a+b+2c+2d$$$) — the sentence. Notice that the condition $$$|s|=a+b+2c+2d$$$ (here $$$|s|$$$ denotes the length of the string $$$s$$$) is equivalent to the fact that $$$s$$$ is as long as the concatenation of the $$$a+b+c+d$$$ words. The sum of the lengths of $$$s$$$ over all test cases doesn't exceed $$$2\cdot 10^5$$$.
Output Specification: For each test case output $$$\texttt{YES}$$$ if it is possible that the sentence $$$s$$$ consists of precisely $$$a$$$ words $$$\texttt{A}$$$, $$$b$$$ words $$$\texttt{B}$$$, $$$c$$$ words $$$\texttt{AB}$$$, and $$$d$$$ words $$$\texttt{BA}$$$, and $$$\texttt{NO}$$$ otherwise. You can output each letter in any case.
Notes: NoteIn the first test case, the sentence $$$s$$$ is $$$\texttt{B}$$$. Clearly, it can't consist of a single word $$$\texttt{A}$$$, so the answer is $$$\texttt{NO}$$$.In the second test case, the sentence $$$s$$$ is $$$\texttt{AB}$$$, and it's possible that it consists of a single word $$$\texttt{AB}$$$, so the answer is $$$\texttt{YES}$$$.In the third test case, the sentence $$$s$$$ is $$$\texttt{ABAB}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{B}$$$, and one word $$$\texttt{BA}$$$, as $$$\texttt{A} + \texttt{BA} + \texttt{B} = \texttt{ABAB}$$$.In the fourth test case, the sentence $$$s$$$ is $$$\texttt{ABAAB}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{AB}$$$, and one word $$$\texttt{BA}$$$, as $$$\texttt{A} + \texttt{BA} + \texttt{AB} = \texttt{ABAAB}$$$. In the fifth test case, the sentence $$$s$$$ is $$$\texttt{BAABBABBAA}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{B}$$$, two words $$$\texttt{AB}$$$, and two words $$$\texttt{BA}$$$, as $$$\texttt{BA} + \texttt{AB} + \texttt{B} + \texttt{AB} + \texttt{BA} + \texttt{A}= \texttt{BAABBABBAA}$$$.
Code:
import sys
import math
def do_test():
a,b,ab,ba = map(int, input().split())
S = input().strip()
n = len(S)
ac = 0
for i in S:
if i == 'A':
ac += 1
if (ac != a + ab + ba):
return "NO"
a_parts = []
b_parts = []
ab_total = 0
l = 0
f = -1
p = S[0]
S = S + S[n-1]
for i in S:
if i == p:
if l > 1:
if l % 2 == 1:
ab_total += l // 2
elif # TODO: Your code here:
a_parts.append(l // 2)
else:
b_parts.append(l // 2)
l = 1
f = i
else:
l += 1
p = i
a_parts.sort()
b_parts.sort()
for i in a_parts:
k = i
if ab >= k:
ab -= k
else:
k -= ab
ab = 0
if ba > 0:
ba -= k-1
for i in b_parts:
k = i
if ba >= k:
ba -= k
else:
k -= ba
ba = 0
if ab > 0:
ab -= k-1
if ab + ba > ab_total:
return "NO"
return "YES"
input = sys.stdin.readline
t = int(input())
for _test_ in range(t):
print(do_test())
|
import sys
import math
def do_test():
a,b,ab,ba = map(int, input().split())
S = input().strip()
n = len(S)
ac = 0
for i in S:
if i == 'A':
ac += 1
if (ac != a + ab + ba):
return "NO"
a_parts = []
b_parts = []
ab_total = 0
l = 0
f = -1
p = S[0]
S = S + S[n-1]
for i in S:
if i == p:
if l > 1:
if l % 2 == 1:
ab_total += l // 2
elif {{completion}}:
a_parts.append(l // 2)
else:
b_parts.append(l // 2)
l = 1
f = i
else:
l += 1
p = i
a_parts.sort()
b_parts.sort()
for i in a_parts:
k = i
if ab >= k:
ab -= k
else:
k -= ab
ab = 0
if ba > 0:
ba -= k-1
for i in b_parts:
k = i
if ba >= k:
ba -= k
else:
k -= ba
ba = 0
if ab > 0:
ab -= k-1
if ab + ba > ab_total:
return "NO"
return "YES"
input = sys.stdin.readline
t = int(input())
for _test_ in range(t):
print(do_test())
|
f == 'A'
|
[{"input": "8\n1 0 0 0\nB\n0 0 1 0\nAB\n1 1 0 1\nABAB\n1 0 1 1\nABAAB\n1 1 2 2\nBAABBABBAA\n1 1 2 3\nABABABBAABAB\n2 3 5 4\nAABAABBABAAABABBABBBABB\n1 3 3 10\nBBABABABABBBABABABABABABAABABA", "output": ["NO\nYES\nYES\nYES\nYES\nYES\nNO\nYES"]}]
|
control_completion_001187
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Alina has discovered a weird language, which contains only $$$4$$$ words: $$$\texttt{A}$$$, $$$\texttt{B}$$$, $$$\texttt{AB}$$$, $$$\texttt{BA}$$$. It also turned out that there are no spaces in this language: a sentence is written by just concatenating its words into a single string.Alina has found one such sentence $$$s$$$ and she is curious: is it possible that it consists of precisely $$$a$$$ words $$$\texttt{A}$$$, $$$b$$$ words $$$\texttt{B}$$$, $$$c$$$ words $$$\texttt{AB}$$$, and $$$d$$$ words $$$\texttt{BA}$$$?In other words, determine, if it's possible to concatenate these $$$a+b+c+d$$$ words in some order so that the resulting string is $$$s$$$. Each of the $$$a+b+c+d$$$ words must be used exactly once in the concatenation, but you can choose the order in which they are concatenated.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains four integers $$$a$$$, $$$b$$$, $$$c$$$, $$$d$$$ ($$$0\le a,b,c,d\le 2\cdot 10^5$$$) — the number of times that words $$$\texttt{A}$$$, $$$\texttt{B}$$$, $$$\texttt{AB}$$$, $$$\texttt{BA}$$$ respectively must be used in the sentence. The second line contains the string $$$s$$$ ($$$s$$$ consists only of the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$, $$$1\le |s| \le 2\cdot 10^5$$$, $$$|s|=a+b+2c+2d$$$) — the sentence. Notice that the condition $$$|s|=a+b+2c+2d$$$ (here $$$|s|$$$ denotes the length of the string $$$s$$$) is equivalent to the fact that $$$s$$$ is as long as the concatenation of the $$$a+b+c+d$$$ words. The sum of the lengths of $$$s$$$ over all test cases doesn't exceed $$$2\cdot 10^5$$$.
Output Specification: For each test case output $$$\texttt{YES}$$$ if it is possible that the sentence $$$s$$$ consists of precisely $$$a$$$ words $$$\texttt{A}$$$, $$$b$$$ words $$$\texttt{B}$$$, $$$c$$$ words $$$\texttt{AB}$$$, and $$$d$$$ words $$$\texttt{BA}$$$, and $$$\texttt{NO}$$$ otherwise. You can output each letter in any case.
Notes: NoteIn the first test case, the sentence $$$s$$$ is $$$\texttt{B}$$$. Clearly, it can't consist of a single word $$$\texttt{A}$$$, so the answer is $$$\texttt{NO}$$$.In the second test case, the sentence $$$s$$$ is $$$\texttt{AB}$$$, and it's possible that it consists of a single word $$$\texttt{AB}$$$, so the answer is $$$\texttt{YES}$$$.In the third test case, the sentence $$$s$$$ is $$$\texttt{ABAB}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{B}$$$, and one word $$$\texttt{BA}$$$, as $$$\texttt{A} + \texttt{BA} + \texttt{B} = \texttt{ABAB}$$$.In the fourth test case, the sentence $$$s$$$ is $$$\texttt{ABAAB}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{AB}$$$, and one word $$$\texttt{BA}$$$, as $$$\texttt{A} + \texttt{BA} + \texttt{AB} = \texttt{ABAAB}$$$. In the fifth test case, the sentence $$$s$$$ is $$$\texttt{BAABBABBAA}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{B}$$$, two words $$$\texttt{AB}$$$, and two words $$$\texttt{BA}$$$, as $$$\texttt{BA} + \texttt{AB} + \texttt{B} + \texttt{AB} + \texttt{BA} + \texttt{A}= \texttt{BAABBABBAA}$$$.
Code:
import sys
import math
def do_test():
a,b,ab,ba = map(int, input().split())
S = input().strip()
n = len(S)
ac = 0
for i in S:
if i == 'A':
ac += 1
if (ac != a + ab + ba):
return "NO"
a_parts = []
b_parts = []
ab_total = 0
l = 0
f = -1
p = S[0]
S = S + S[n-1]
for i in S:
if i == p:
if l > 1:
if # TODO: Your code here:
ab_total += l // 2
elif f == 'A':
a_parts.append(l // 2)
else:
b_parts.append(l // 2)
l = 1
f = i
else:
l += 1
p = i
a_parts.sort()
b_parts.sort()
for i in a_parts:
k = i
if ab >= k:
ab -= k
else:
k -= ab
ab = 0
if ba > 0:
ba -= k-1
for i in b_parts:
k = i
if ba >= k:
ba -= k
else:
k -= ba
ba = 0
if ab > 0:
ab -= k-1
if ab + ba > ab_total:
return "NO"
return "YES"
input = sys.stdin.readline
t = int(input())
for _test_ in range(t):
print(do_test())
|
import sys
import math
def do_test():
a,b,ab,ba = map(int, input().split())
S = input().strip()
n = len(S)
ac = 0
for i in S:
if i == 'A':
ac += 1
if (ac != a + ab + ba):
return "NO"
a_parts = []
b_parts = []
ab_total = 0
l = 0
f = -1
p = S[0]
S = S + S[n-1]
for i in S:
if i == p:
if l > 1:
if {{completion}}:
ab_total += l // 2
elif f == 'A':
a_parts.append(l // 2)
else:
b_parts.append(l // 2)
l = 1
f = i
else:
l += 1
p = i
a_parts.sort()
b_parts.sort()
for i in a_parts:
k = i
if ab >= k:
ab -= k
else:
k -= ab
ab = 0
if ba > 0:
ba -= k-1
for i in b_parts:
k = i
if ba >= k:
ba -= k
else:
k -= ba
ba = 0
if ab > 0:
ab -= k-1
if ab + ba > ab_total:
return "NO"
return "YES"
input = sys.stdin.readline
t = int(input())
for _test_ in range(t):
print(do_test())
|
l % 2 == 1
|
[{"input": "8\n1 0 0 0\nB\n0 0 1 0\nAB\n1 1 0 1\nABAB\n1 0 1 1\nABAAB\n1 1 2 2\nBAABBABBAA\n1 1 2 3\nABABABBAABAB\n2 3 5 4\nAABAABBABAAABABBABBBABB\n1 3 3 10\nBBABABABABBBABABABABABABAABABA", "output": ["NO\nYES\nYES\nYES\nYES\nYES\nNO\nYES"]}]
|
control_completion_001188
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Alina has discovered a weird language, which contains only $$$4$$$ words: $$$\texttt{A}$$$, $$$\texttt{B}$$$, $$$\texttt{AB}$$$, $$$\texttt{BA}$$$. It also turned out that there are no spaces in this language: a sentence is written by just concatenating its words into a single string.Alina has found one such sentence $$$s$$$ and she is curious: is it possible that it consists of precisely $$$a$$$ words $$$\texttt{A}$$$, $$$b$$$ words $$$\texttt{B}$$$, $$$c$$$ words $$$\texttt{AB}$$$, and $$$d$$$ words $$$\texttt{BA}$$$?In other words, determine, if it's possible to concatenate these $$$a+b+c+d$$$ words in some order so that the resulting string is $$$s$$$. Each of the $$$a+b+c+d$$$ words must be used exactly once in the concatenation, but you can choose the order in which they are concatenated.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains four integers $$$a$$$, $$$b$$$, $$$c$$$, $$$d$$$ ($$$0\le a,b,c,d\le 2\cdot 10^5$$$) — the number of times that words $$$\texttt{A}$$$, $$$\texttt{B}$$$, $$$\texttt{AB}$$$, $$$\texttt{BA}$$$ respectively must be used in the sentence. The second line contains the string $$$s$$$ ($$$s$$$ consists only of the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$, $$$1\le |s| \le 2\cdot 10^5$$$, $$$|s|=a+b+2c+2d$$$) — the sentence. Notice that the condition $$$|s|=a+b+2c+2d$$$ (here $$$|s|$$$ denotes the length of the string $$$s$$$) is equivalent to the fact that $$$s$$$ is as long as the concatenation of the $$$a+b+c+d$$$ words. The sum of the lengths of $$$s$$$ over all test cases doesn't exceed $$$2\cdot 10^5$$$.
Output Specification: For each test case output $$$\texttt{YES}$$$ if it is possible that the sentence $$$s$$$ consists of precisely $$$a$$$ words $$$\texttt{A}$$$, $$$b$$$ words $$$\texttt{B}$$$, $$$c$$$ words $$$\texttt{AB}$$$, and $$$d$$$ words $$$\texttt{BA}$$$, and $$$\texttt{NO}$$$ otherwise. You can output each letter in any case.
Notes: NoteIn the first test case, the sentence $$$s$$$ is $$$\texttt{B}$$$. Clearly, it can't consist of a single word $$$\texttt{A}$$$, so the answer is $$$\texttt{NO}$$$.In the second test case, the sentence $$$s$$$ is $$$\texttt{AB}$$$, and it's possible that it consists of a single word $$$\texttt{AB}$$$, so the answer is $$$\texttt{YES}$$$.In the third test case, the sentence $$$s$$$ is $$$\texttt{ABAB}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{B}$$$, and one word $$$\texttt{BA}$$$, as $$$\texttt{A} + \texttt{BA} + \texttt{B} = \texttt{ABAB}$$$.In the fourth test case, the sentence $$$s$$$ is $$$\texttt{ABAAB}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{AB}$$$, and one word $$$\texttt{BA}$$$, as $$$\texttt{A} + \texttt{BA} + \texttt{AB} = \texttt{ABAAB}$$$. In the fifth test case, the sentence $$$s$$$ is $$$\texttt{BAABBABBAA}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{B}$$$, two words $$$\texttt{AB}$$$, and two words $$$\texttt{BA}$$$, as $$$\texttt{BA} + \texttt{AB} + \texttt{B} + \texttt{AB} + \texttt{BA} + \texttt{A}= \texttt{BAABBABBAA}$$$.
Code:
def solve():
cnt_a, cnt_b, cnt_ab, cnt_ba = map(int, input().strip().split())
# print(cntA, cntB, cntAB, cntBA)
s = input()
if s.count('A') != cnt_a + cnt_ba + cnt_ab:
print("NO")
return
stk = [[1, s[0]]]
for i in range(1, len(s)):
if i == 0:
continue
c = s[i]
if c != stk[-1][1]:
x = stk.pop()
stk.append([x[0] + 1, c])
else:
stk.append([1, c])
stk.sort()
rest = 0
for cnt, last in stk:
# print(cnt, last)
if not cnt % 2:
if last == 'A' and cnt_ba >= (cnt >> 1):
cnt_ba -= cnt >> 1
elif # TODO: Your code here:
cnt_ab -= cnt >> 1
else:
rest += (cnt >> 1) - 1
else:
rest += cnt >> 1
# print(rest, cnt_ab, cnt_ba)
if rest >= cnt_ab + cnt_ba:
print("YES")
else:
print("NO")
if __name__ == '__main__':
t = int(input())
for _ in range(t):
solve()
|
def solve():
cnt_a, cnt_b, cnt_ab, cnt_ba = map(int, input().strip().split())
# print(cntA, cntB, cntAB, cntBA)
s = input()
if s.count('A') != cnt_a + cnt_ba + cnt_ab:
print("NO")
return
stk = [[1, s[0]]]
for i in range(1, len(s)):
if i == 0:
continue
c = s[i]
if c != stk[-1][1]:
x = stk.pop()
stk.append([x[0] + 1, c])
else:
stk.append([1, c])
stk.sort()
rest = 0
for cnt, last in stk:
# print(cnt, last)
if not cnt % 2:
if last == 'A' and cnt_ba >= (cnt >> 1):
cnt_ba -= cnt >> 1
elif {{completion}}:
cnt_ab -= cnt >> 1
else:
rest += (cnt >> 1) - 1
else:
rest += cnt >> 1
# print(rest, cnt_ab, cnt_ba)
if rest >= cnt_ab + cnt_ba:
print("YES")
else:
print("NO")
if __name__ == '__main__':
t = int(input())
for _ in range(t):
solve()
|
last == 'B' and cnt_ab >= (cnt >> 1)
|
[{"input": "8\n1 0 0 0\nB\n0 0 1 0\nAB\n1 1 0 1\nABAB\n1 0 1 1\nABAAB\n1 1 2 2\nBAABBABBAA\n1 1 2 3\nABABABBAABAB\n2 3 5 4\nAABAABBABAAABABBABBBABB\n1 3 3 10\nBBABABABABBBABABABABABABAABABA", "output": ["NO\nYES\nYES\nYES\nYES\nYES\nNO\nYES"]}]
|
control_completion_001189
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Alina has discovered a weird language, which contains only $$$4$$$ words: $$$\texttt{A}$$$, $$$\texttt{B}$$$, $$$\texttt{AB}$$$, $$$\texttt{BA}$$$. It also turned out that there are no spaces in this language: a sentence is written by just concatenating its words into a single string.Alina has found one such sentence $$$s$$$ and she is curious: is it possible that it consists of precisely $$$a$$$ words $$$\texttt{A}$$$, $$$b$$$ words $$$\texttt{B}$$$, $$$c$$$ words $$$\texttt{AB}$$$, and $$$d$$$ words $$$\texttt{BA}$$$?In other words, determine, if it's possible to concatenate these $$$a+b+c+d$$$ words in some order so that the resulting string is $$$s$$$. Each of the $$$a+b+c+d$$$ words must be used exactly once in the concatenation, but you can choose the order in which they are concatenated.
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains four integers $$$a$$$, $$$b$$$, $$$c$$$, $$$d$$$ ($$$0\le a,b,c,d\le 2\cdot 10^5$$$) — the number of times that words $$$\texttt{A}$$$, $$$\texttt{B}$$$, $$$\texttt{AB}$$$, $$$\texttt{BA}$$$ respectively must be used in the sentence. The second line contains the string $$$s$$$ ($$$s$$$ consists only of the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$, $$$1\le |s| \le 2\cdot 10^5$$$, $$$|s|=a+b+2c+2d$$$) — the sentence. Notice that the condition $$$|s|=a+b+2c+2d$$$ (here $$$|s|$$$ denotes the length of the string $$$s$$$) is equivalent to the fact that $$$s$$$ is as long as the concatenation of the $$$a+b+c+d$$$ words. The sum of the lengths of $$$s$$$ over all test cases doesn't exceed $$$2\cdot 10^5$$$.
Output Specification: For each test case output $$$\texttt{YES}$$$ if it is possible that the sentence $$$s$$$ consists of precisely $$$a$$$ words $$$\texttt{A}$$$, $$$b$$$ words $$$\texttt{B}$$$, $$$c$$$ words $$$\texttt{AB}$$$, and $$$d$$$ words $$$\texttt{BA}$$$, and $$$\texttt{NO}$$$ otherwise. You can output each letter in any case.
Notes: NoteIn the first test case, the sentence $$$s$$$ is $$$\texttt{B}$$$. Clearly, it can't consist of a single word $$$\texttt{A}$$$, so the answer is $$$\texttt{NO}$$$.In the second test case, the sentence $$$s$$$ is $$$\texttt{AB}$$$, and it's possible that it consists of a single word $$$\texttt{AB}$$$, so the answer is $$$\texttt{YES}$$$.In the third test case, the sentence $$$s$$$ is $$$\texttt{ABAB}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{B}$$$, and one word $$$\texttt{BA}$$$, as $$$\texttt{A} + \texttt{BA} + \texttt{B} = \texttt{ABAB}$$$.In the fourth test case, the sentence $$$s$$$ is $$$\texttt{ABAAB}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{AB}$$$, and one word $$$\texttt{BA}$$$, as $$$\texttt{A} + \texttt{BA} + \texttt{AB} = \texttt{ABAAB}$$$. In the fifth test case, the sentence $$$s$$$ is $$$\texttt{BAABBABBAA}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{B}$$$, two words $$$\texttt{AB}$$$, and two words $$$\texttt{BA}$$$, as $$$\texttt{BA} + \texttt{AB} + \texttt{B} + \texttt{AB} + \texttt{BA} + \texttt{A}= \texttt{BAABBABBAA}$$$.
Code:
def solve():
cnt_a, cnt_b, cnt_ab, cnt_ba = map(int, input().strip().split())
# print(cntA, cntB, cntAB, cntBA)
s = input()
if s.count('A') != cnt_a + cnt_ba + cnt_ab:
print("NO")
return
stk = [[1, s[0]]]
for i in range(1, len(s)):
if i == 0:
continue
c = s[i]
if c != stk[-1][1]:
x = stk.pop()
stk.append([x[0] + 1, c])
else:
stk.append([1, c])
stk.sort()
rest = 0
for cnt, last in stk:
# print(cnt, last)
if not cnt % 2:
if # TODO: Your code here:
cnt_ba -= cnt >> 1
elif last == 'B' and cnt_ab >= (cnt >> 1):
cnt_ab -= cnt >> 1
else:
rest += (cnt >> 1) - 1
else:
rest += cnt >> 1
# print(rest, cnt_ab, cnt_ba)
if rest >= cnt_ab + cnt_ba:
print("YES")
else:
print("NO")
if __name__ == '__main__':
t = int(input())
for _ in range(t):
solve()
|
def solve():
cnt_a, cnt_b, cnt_ab, cnt_ba = map(int, input().strip().split())
# print(cntA, cntB, cntAB, cntBA)
s = input()
if s.count('A') != cnt_a + cnt_ba + cnt_ab:
print("NO")
return
stk = [[1, s[0]]]
for i in range(1, len(s)):
if i == 0:
continue
c = s[i]
if c != stk[-1][1]:
x = stk.pop()
stk.append([x[0] + 1, c])
else:
stk.append([1, c])
stk.sort()
rest = 0
for cnt, last in stk:
# print(cnt, last)
if not cnt % 2:
if {{completion}}:
cnt_ba -= cnt >> 1
elif last == 'B' and cnt_ab >= (cnt >> 1):
cnt_ab -= cnt >> 1
else:
rest += (cnt >> 1) - 1
else:
rest += cnt >> 1
# print(rest, cnt_ab, cnt_ba)
if rest >= cnt_ab + cnt_ba:
print("YES")
else:
print("NO")
if __name__ == '__main__':
t = int(input())
for _ in range(t):
solve()
|
last == 'A' and cnt_ba >= (cnt >> 1)
|
[{"input": "8\n1 0 0 0\nB\n0 0 1 0\nAB\n1 1 0 1\nABAB\n1 0 1 1\nABAAB\n1 1 2 2\nBAABBABBAA\n1 1 2 3\nABABABBAABAB\n2 3 5 4\nAABAABBABAAABABBABBBABB\n1 3 3 10\nBBABABABABBBABABABABABABAABABA", "output": ["NO\nYES\nYES\nYES\nYES\nYES\nNO\nYES"]}]
|
control_completion_001190
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given an array of $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$. After you watched the amazing film "Everything Everywhere All At Once", you came up with the following operation.In one operation, you choose $$$n-1$$$ elements of the array and replace each of them with their arithmetic mean (which doesn't have to be an integer). For example, from the array $$$[1, 2, 3, 1]$$$ we can get the array $$$[2, 2, 2, 1]$$$, if we choose the first three elements, or we can get the array $$$[\frac{4}{3}, \frac{4}{3}, 3, \frac{4}{3}]$$$, if we choose all elements except the third.Is it possible to make all elements of the array equal by performing a finite number of such operations?
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 200$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$3 \le n \le 50$$$) — the number of integers. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i \le 100$$$).
Output Specification: For each test case, if it is possible to make all elements equal after some number of operations, output $$$\texttt{YES}$$$. Otherwise, output $$$\texttt{NO}$$$. You can output $$$\texttt{YES}$$$ and $$$\texttt{NO}$$$ in any case (for example, strings $$$\texttt{yEs}$$$, $$$\texttt{yes}$$$, $$$\texttt{Yes}$$$ will be recognized as a positive response).
Notes: NoteIn the first test case, all elements are already equal.In the second test case, you can choose all elements except the third, their average is $$$\frac{1 + 2 + 4 + 5}{4} = 3$$$, so the array will become $$$[3, 3, 3, 3, 3]$$$.It's possible to show that it's impossible to make all elements equal in the third and fourth test cases.
Code:
import fileinput
lines = []
for line in fileinput.input():
line_f = [int(x) for x in line.split()]
if len(line_f) > 0:
lines.append(line_f)
# print ffs
for i in range(1, len(lines), 2):
n = lines[i][0]
a = lines[i+1]
b = 123 + 23
c= b + 1
sm = 0
for elem in a:
sm += elem
found = False
for elem in a:
lhs = elem
rhs = ((sm - elem) / (n-1))
if # TODO: Your code here:
found = True
break
if found:
print("YES")
else:
print("NO")
|
import fileinput
lines = []
for line in fileinput.input():
line_f = [int(x) for x in line.split()]
if len(line_f) > 0:
lines.append(line_f)
# print ffs
for i in range(1, len(lines), 2):
n = lines[i][0]
a = lines[i+1]
b = 123 + 23
c= b + 1
sm = 0
for elem in a:
sm += elem
found = False
for elem in a:
lhs = elem
rhs = ((sm - elem) / (n-1))
if {{completion}}:
found = True
break
if found:
print("YES")
else:
print("NO")
|
lhs == rhs
|
[{"input": "4\n\n3\n\n42 42 42\n\n5\n\n1 2 3 4 5\n\n4\n\n4 3 2 1\n\n3\n\n24 2 22", "output": ["YES\nYES\nNO\nNO"]}]
|
control_completion_001218
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: For an array $$$[b_1, b_2, \ldots, b_m]$$$ define its number of inversions as the number of pairs $$$(i, j)$$$ of integers such that $$$1 \le i < j \le m$$$ and $$$b_i>b_j$$$. Let's call array $$$b$$$ odd if its number of inversions is odd. For example, array $$$[4, 2, 7]$$$ is odd, as its number of inversions is $$$1$$$, while array $$$[2, 1, 4, 3]$$$ isn't, as its number of inversions is $$$2$$$.You are given a permutation $$$[p_1, p_2, \ldots, p_n]$$$ of integers from $$$1$$$ to $$$n$$$ (each of them appears exactly once in the permutation). You want to split it into several consecutive subarrays (maybe just one), so that the number of the odd subarrays among them is as large as possible. What largest number of these subarrays may be odd?
Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the size of the permutation. The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$, all $$$p_i$$$ are distinct) — the elements of the permutation. The sum of $$$n$$$ over all test cases doesn't exceed $$$2\cdot 10^5$$$.
Output Specification: For each test case output a single integer — the largest possible number of odd subarrays that you can get after splitting the permutation into several consecutive subarrays.
Notes: NoteIn the first and third test cases, no matter how we split our permutation, there won't be any odd subarrays.In the second test case, we can split our permutation into subarrays $$$[4, 3], [2, 1]$$$, both of which are odd since their numbers of inversions are $$$1$$$.In the fourth test case, we can split our permutation into a single subarray $$$[2, 1]$$$, which is odd.In the fifth test case, we can split our permutation into subarrays $$$[4, 5], [6, 1, 2, 3]$$$. The first subarray has $$$0$$$ inversions, and the second has $$$3$$$, so it is odd.
Code:
import fileinput
lines = []
for line in fileinput.input():
line_f = [int(x) for x in line.split()]
if len(line_f) > 0:
lines.append(line_f)
# print ffs
for i in range(1, len(lines), 2):
n = lines[i][0]
a = lines[i+1]
numoddseg = 0
prev = -1
i = 0
while i < n:
if # TODO: Your code here:
numoddseg += 1
prev = -1
else:
prev = a[i]
i += 1
print(numoddseg)
|
import fileinput
lines = []
for line in fileinput.input():
line_f = [int(x) for x in line.split()]
if len(line_f) > 0:
lines.append(line_f)
# print ffs
for i in range(1, len(lines), 2):
n = lines[i][0]
a = lines[i+1]
numoddseg = 0
prev = -1
i = 0
while i < n:
if {{completion}}:
numoddseg += 1
prev = -1
else:
prev = a[i]
i += 1
print(numoddseg)
|
a[i] < prev
|
[{"input": "5\n\n3\n\n1 2 3\n\n4\n\n4 3 2 1\n\n2\n\n1 2\n\n2\n\n2 1\n\n6\n\n4 5 6 1 2 3", "output": ["0\n2\n0\n1\n1"]}]
|
control_completion_001260
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Inflation has occurred in Berlandia, so the store needs to change the price of goods.The current price of good $$$n$$$ is given. It is allowed to increase the price of the good by $$$k$$$ times, with $$$1 \le k \le m$$$, k is an integer. Output the roundest possible new price of the good. That is, the one that has the maximum number of zeros at the end.For example, the number 481000 is more round than the number 1000010 (three zeros at the end of 481000 and only one at the end of 1000010).If there are several possible variants, output the one in which the new price is maximal.If it is impossible to get a rounder price, output $$$n \cdot m$$$ (that is, the maximum possible price).
Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) —the number of test cases in the test. Each test case consists of one line. This line contains two integers: $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 10^9$$$). Where $$$n$$$ is the old price of the good, and the number $$$m$$$ means that you can increase the price $$$n$$$ no more than $$$m$$$ times.
Output Specification: For each test case, output on a separate line the roundest integer of the form $$$n \cdot k$$$ ($$$1 \le k \le m$$$, $$$k$$$ — an integer). If there are several possible variants, output the one in which the new price (value $$$n \cdot k$$$) is maximal. If it is impossible to get a more rounded price, output $$$n \cdot m$$$ (that is, the maximum possible price).
Notes: NoteIn the first case $$$n = 6$$$, $$$m = 11$$$. We cannot get a number with two zeros or more at the end, because we need to increase the price $$$50$$$ times, but $$$50 > m = 11$$$. The maximum price multiple of $$$10$$$ would be $$$6 \cdot 10 = 60$$$.In the second case $$$n = 5$$$, $$$m = 43$$$. The maximum price multiple of $$$100$$$ would be $$$5 \cdot 40 = 200$$$.In the third case, $$$n = 13$$$, $$$m = 5$$$. All possible new prices will not end in $$$0$$$, then you should output $$$n \cdot m = 65$$$.In the fourth case, you should increase the price $$$15$$$ times.In the fifth case, increase the price $$$12000$$$ times.
Code:
from sys import stdin, stderr
data = [int(x) for x in stdin.read().split()[1:]]
ns, ms = data[::2], data[1::2]
output = []
for n, m in zip(ns, ms):
# n = 2 ** a * 5 ** b * c
a = b = 0
c = n
while c % 2 == 0:
a += 1
c //= 2
while c % 5 == 0:
b += 1
c //= 5
t = 1 # our result should be a multiple of t
if a > b:
while a > b and 5 * t <= m:
t *= 5
b += 1
elif b > a:
while # TODO: Your code here:
t *= 2
a += 1
while 10 * t <= m:
t *= 10
#print(n, m, t, file=stderr)
output.append(n * (m - (m % t)))
print('\n'.join(str(x) for x in output))
|
from sys import stdin, stderr
data = [int(x) for x in stdin.read().split()[1:]]
ns, ms = data[::2], data[1::2]
output = []
for n, m in zip(ns, ms):
# n = 2 ** a * 5 ** b * c
a = b = 0
c = n
while c % 2 == 0:
a += 1
c //= 2
while c % 5 == 0:
b += 1
c //= 5
t = 1 # our result should be a multiple of t
if a > b:
while a > b and 5 * t <= m:
t *= 5
b += 1
elif b > a:
while {{completion}}:
t *= 2
a += 1
while 10 * t <= m:
t *= 10
#print(n, m, t, file=stderr)
output.append(n * (m - (m % t)))
print('\n'.join(str(x) for x in output))
|
b > a and 2 * t <= m
|
[{"input": "10\n\n6 11\n\n5 43\n\n13 5\n\n4 16\n\n10050 12345\n\n2 6\n\n4 30\n\n25 10\n\n2 81\n\n1 7", "output": ["60\n200\n65\n60\n120600000\n10\n100\n200\n100\n7"]}]
|
control_completion_001308
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a string $$$s$$$. You have to determine whether it is possible to build the string $$$s$$$ out of strings aa, aaa, bb and/or bbb by concatenating them. You can use the strings aa, aaa, bb and/or bbb any number of times and in any order.For example: aaaabbb can be built as aa $$$+$$$ aa $$$+$$$ bbb; bbaaaaabbb can be built as bb $$$+$$$ aaa $$$+$$$ aa $$$+$$$ bbb; aaaaaa can be built as aa $$$+$$$ aa $$$+$$$ aa; abab cannot be built from aa, aaa, bb and/or bbb.
Input Specification: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case consists of one line containing the string $$$s$$$ ($$$1 \le |s| \le 50$$$), consisting of characters a and/or b.
Output Specification: For each test case, print YES if it is possible to build the string $$$s$$$. Otherwise, print NO. You may print each letter in any case (for example, YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
Notes: NoteThe first four test cases of the example are described in the statement.
Code:
t=int(input())
while(t):
i=0
s=input()
if(len(s)==1):
print("NO")
t=t-1
continue
while(i<len(s)):
if(i==0):
if(s[0:2]=="ab" or s[0:2]=="ba"):
print("NO")
t=t-1
break
if(i>0 and i<len(s)-1):
if(s[i-1:i+2]=="bab" or s[i-1:i+2]=="aba"):
print("NO")
t=t-1
break
if(i==len(s)-1):
if# TODO: Your code here:
print("NO")
t=t-1
break
else:
print("YES")
t=t-1
break
i+=1
|
t=int(input())
while(t):
i=0
s=input()
if(len(s)==1):
print("NO")
t=t-1
continue
while(i<len(s)):
if(i==0):
if(s[0:2]=="ab" or s[0:2]=="ba"):
print("NO")
t=t-1
break
if(i>0 and i<len(s)-1):
if(s[i-1:i+2]=="bab" or s[i-1:i+2]=="aba"):
print("NO")
t=t-1
break
if(i==len(s)-1):
if{{completion}}:
print("NO")
t=t-1
break
else:
print("YES")
t=t-1
break
i+=1
|
(s[i-1:]=="ba" or s[i-1:]=="ab")
|
[{"input": "8\n\naaaabbb\n\nbbaaaaabbb\n\naaaaaa\n\nabab\n\na\n\nb\n\naaaab\n\nbbaaa", "output": ["YES\nYES\nYES\nNO\nNO\nNO\nNO\nYES"]}]
|
control_completion_001643
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a rooted tree of $$$2^n - 1$$$ vertices. Every vertex of this tree has either $$$0$$$ children, or $$$2$$$ children. All leaves of this tree have the same distance from the root, and for every non-leaf vertex, one of its children is the left one, and the other child is the right one. Formally, you are given a perfect binary tree.The vertices of the tree are numbered in the following order: the root has index $$$1$$$; if a vertex has index $$$x$$$, then its left child has index $$$2x$$$, and its right child has index $$$2x+1$$$. Every vertex of the tree has a letter written on it, either A or B. Let's define the character on the vertex $$$x$$$ as $$$s_x$$$.Let the preorder string of some vertex $$$x$$$ be defined in the following way: if the vertex $$$x$$$ is a leaf, then the preorder string of $$$x$$$ be consisting of only one character $$$s_x$$$; otherwise, the preorder string of $$$x$$$ is $$$s_x + f(l_x) + f(r_x)$$$, where $$$+$$$ operator defines concatenation of strings, $$$f(l_x)$$$ is the preorder string of the left child of $$$x$$$, and $$$f(r_x)$$$ is the preorder string of the right child of $$$x$$$. The preorder string of the tree is the preorder string of its root.Now, for the problem itself...You have to calculate the number of different strings that can be obtained as the preorder string of the given tree, if you are allowed to perform the following operation any number of times before constructing the preorder string of the tree: choose any non-leaf vertex $$$x$$$, and swap its children (so, the left child becomes the right one, and vice versa).
Input Specification: The first line contains one integer $$$n$$$ ($$$2 \le n \le 18$$$). The second line contains a sequence of $$$2^n-1$$$ characters $$$s_1, s_2, \dots, s_{2^n-1}$$$. Each character is either A or B. The characters are not separated by spaces or anything else.
Output Specification: Print one integer — the number of different strings that can be obtained as the preorder string of the given tree, if you can apply any number of operations described in the statement. Since it can be very large, print it modulo $$$998244353$$$.
Code:
n = int(input())
s = input()
c = 1
def dfs(i):
if # TODO: Your code here:
return s[i]
global c
l = dfs(2*i + 1)
r = dfs(2*i + 2)
if l != r:
c *= 2
if l > r:
l, r = r, l
return s[i] + l + r
dfs(0)
print(c % 998244353)
|
n = int(input())
s = input()
c = 1
def dfs(i):
if {{completion}}:
return s[i]
global c
l = dfs(2*i + 1)
r = dfs(2*i + 2)
if l != r:
c *= 2
if l > r:
l, r = r, l
return s[i] + l + r
dfs(0)
print(c % 998244353)
|
i >= 2**(n-1)-1
|
[{"input": "4\nBAAAAAAAABBABAB", "output": ["16"]}, {"input": "2\nBAA", "output": ["1"]}, {"input": "2\nABA", "output": ["2"]}, {"input": "2\nAAB", "output": ["2"]}, {"input": "2\nAAA", "output": ["1"]}]
|
control_completion_001661
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a rooted tree of $$$2^n - 1$$$ vertices. Every vertex of this tree has either $$$0$$$ children, or $$$2$$$ children. All leaves of this tree have the same distance from the root, and for every non-leaf vertex, one of its children is the left one, and the other child is the right one. Formally, you are given a perfect binary tree.The vertices of the tree are numbered in the following order: the root has index $$$1$$$; if a vertex has index $$$x$$$, then its left child has index $$$2x$$$, and its right child has index $$$2x+1$$$. Every vertex of the tree has a letter written on it, either A or B. Let's define the character on the vertex $$$x$$$ as $$$s_x$$$.Let the preorder string of some vertex $$$x$$$ be defined in the following way: if the vertex $$$x$$$ is a leaf, then the preorder string of $$$x$$$ be consisting of only one character $$$s_x$$$; otherwise, the preorder string of $$$x$$$ is $$$s_x + f(l_x) + f(r_x)$$$, where $$$+$$$ operator defines concatenation of strings, $$$f(l_x)$$$ is the preorder string of the left child of $$$x$$$, and $$$f(r_x)$$$ is the preorder string of the right child of $$$x$$$. The preorder string of the tree is the preorder string of its root.Now, for the problem itself...You have to calculate the number of different strings that can be obtained as the preorder string of the given tree, if you are allowed to perform the following operation any number of times before constructing the preorder string of the tree: choose any non-leaf vertex $$$x$$$, and swap its children (so, the left child becomes the right one, and vice versa).
Input Specification: The first line contains one integer $$$n$$$ ($$$2 \le n \le 18$$$). The second line contains a sequence of $$$2^n-1$$$ characters $$$s_1, s_2, \dots, s_{2^n-1}$$$. Each character is either A or B. The characters are not separated by spaces or anything else.
Output Specification: Print one integer — the number of different strings that can be obtained as the preorder string of the given tree, if you can apply any number of operations described in the statement. Since it can be very large, print it modulo $$$998244353$$$.
Code:
n = int(input())
s = input()
c = 1
def dfs(i):
if i >= 2**(n-1)-1:
return s[i]
global c
l = dfs(2*i + 1)
r = dfs(2*i + 2)
if # TODO: Your code here:
c *= 2
if l > r:
l, r = r, l
return s[i] + l + r
dfs(0)
print(c % 998244353)
|
n = int(input())
s = input()
c = 1
def dfs(i):
if i >= 2**(n-1)-1:
return s[i]
global c
l = dfs(2*i + 1)
r = dfs(2*i + 2)
if {{completion}}:
c *= 2
if l > r:
l, r = r, l
return s[i] + l + r
dfs(0)
print(c % 998244353)
|
l != r
|
[{"input": "4\nBAAAAAAAABBABAB", "output": ["16"]}, {"input": "2\nBAA", "output": ["1"]}, {"input": "2\nABA", "output": ["2"]}, {"input": "2\nAAB", "output": ["2"]}, {"input": "2\nAAA", "output": ["1"]}]
|
control_completion_001662
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a rooted tree of $$$2^n - 1$$$ vertices. Every vertex of this tree has either $$$0$$$ children, or $$$2$$$ children. All leaves of this tree have the same distance from the root, and for every non-leaf vertex, one of its children is the left one, and the other child is the right one. Formally, you are given a perfect binary tree.The vertices of the tree are numbered in the following order: the root has index $$$1$$$; if a vertex has index $$$x$$$, then its left child has index $$$2x$$$, and its right child has index $$$2x+1$$$. Every vertex of the tree has a letter written on it, either A or B. Let's define the character on the vertex $$$x$$$ as $$$s_x$$$.Let the preorder string of some vertex $$$x$$$ be defined in the following way: if the vertex $$$x$$$ is a leaf, then the preorder string of $$$x$$$ be consisting of only one character $$$s_x$$$; otherwise, the preorder string of $$$x$$$ is $$$s_x + f(l_x) + f(r_x)$$$, where $$$+$$$ operator defines concatenation of strings, $$$f(l_x)$$$ is the preorder string of the left child of $$$x$$$, and $$$f(r_x)$$$ is the preorder string of the right child of $$$x$$$. The preorder string of the tree is the preorder string of its root.Now, for the problem itself...You have to calculate the number of different strings that can be obtained as the preorder string of the given tree, if you are allowed to perform the following operation any number of times before constructing the preorder string of the tree: choose any non-leaf vertex $$$x$$$, and swap its children (so, the left child becomes the right one, and vice versa).
Input Specification: The first line contains one integer $$$n$$$ ($$$2 \le n \le 18$$$). The second line contains a sequence of $$$2^n-1$$$ characters $$$s_1, s_2, \dots, s_{2^n-1}$$$. Each character is either A or B. The characters are not separated by spaces or anything else.
Output Specification: Print one integer — the number of different strings that can be obtained as the preorder string of the given tree, if you can apply any number of operations described in the statement. Since it can be very large, print it modulo $$$998244353$$$.
Code:
mod=998244353
cnt=0
n=int(input())
s=input()
import random
q=random.randint(10**9,2*10**9)
p=random.randint(10**9,2*10**9)
r=10**9+7
a=[-1]
for i in s:
if # TODO: Your code here:
a.append(p)
else:
a.append(q)
for i in range(2**(n-1)-1,0,-1):
if a[2*i]!=a[2*i+1]:
cnt+=1
a[i]=a[i]^(2*a[2*i]+2*a[2*i+1])
a[i]%=r
print(pow(2,cnt,mod))
|
mod=998244353
cnt=0
n=int(input())
s=input()
import random
q=random.randint(10**9,2*10**9)
p=random.randint(10**9,2*10**9)
r=10**9+7
a=[-1]
for i in s:
if {{completion}}:
a.append(p)
else:
a.append(q)
for i in range(2**(n-1)-1,0,-1):
if a[2*i]!=a[2*i+1]:
cnt+=1
a[i]=a[i]^(2*a[2*i]+2*a[2*i+1])
a[i]%=r
print(pow(2,cnt,mod))
|
i=='A'
|
[{"input": "4\nBAAAAAAAABBABAB", "output": ["16"]}, {"input": "2\nBAA", "output": ["1"]}, {"input": "2\nABA", "output": ["2"]}, {"input": "2\nAAB", "output": ["2"]}, {"input": "2\nAAA", "output": ["1"]}]
|
control_completion_001663
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a rooted tree of $$$2^n - 1$$$ vertices. Every vertex of this tree has either $$$0$$$ children, or $$$2$$$ children. All leaves of this tree have the same distance from the root, and for every non-leaf vertex, one of its children is the left one, and the other child is the right one. Formally, you are given a perfect binary tree.The vertices of the tree are numbered in the following order: the root has index $$$1$$$; if a vertex has index $$$x$$$, then its left child has index $$$2x$$$, and its right child has index $$$2x+1$$$. Every vertex of the tree has a letter written on it, either A or B. Let's define the character on the vertex $$$x$$$ as $$$s_x$$$.Let the preorder string of some vertex $$$x$$$ be defined in the following way: if the vertex $$$x$$$ is a leaf, then the preorder string of $$$x$$$ be consisting of only one character $$$s_x$$$; otherwise, the preorder string of $$$x$$$ is $$$s_x + f(l_x) + f(r_x)$$$, where $$$+$$$ operator defines concatenation of strings, $$$f(l_x)$$$ is the preorder string of the left child of $$$x$$$, and $$$f(r_x)$$$ is the preorder string of the right child of $$$x$$$. The preorder string of the tree is the preorder string of its root.Now, for the problem itself...You have to calculate the number of different strings that can be obtained as the preorder string of the given tree, if you are allowed to perform the following operation any number of times before constructing the preorder string of the tree: choose any non-leaf vertex $$$x$$$, and swap its children (so, the left child becomes the right one, and vice versa).
Input Specification: The first line contains one integer $$$n$$$ ($$$2 \le n \le 18$$$). The second line contains a sequence of $$$2^n-1$$$ characters $$$s_1, s_2, \dots, s_{2^n-1}$$$. Each character is either A or B. The characters are not separated by spaces or anything else.
Output Specification: Print one integer — the number of different strings that can be obtained as the preorder string of the given tree, if you can apply any number of operations described in the statement. Since it can be very large, print it modulo $$$998244353$$$.
Code:
mod=998244353
cnt=0
n=int(input())
s=input()
import random
q=random.randint(10**9,2*10**9)
p=random.randint(10**9,2*10**9)
r=10**9+7
a=[-1]
for i in s:
if i=='A':
a.append(p)
else:
a.append(q)
for i in range(2**(n-1)-1,0,-1):
if # TODO: Your code here:
cnt+=1
a[i]=a[i]^(2*a[2*i]+2*a[2*i+1])
a[i]%=r
print(pow(2,cnt,mod))
|
mod=998244353
cnt=0
n=int(input())
s=input()
import random
q=random.randint(10**9,2*10**9)
p=random.randint(10**9,2*10**9)
r=10**9+7
a=[-1]
for i in s:
if i=='A':
a.append(p)
else:
a.append(q)
for i in range(2**(n-1)-1,0,-1):
if {{completion}}:
cnt+=1
a[i]=a[i]^(2*a[2*i]+2*a[2*i+1])
a[i]%=r
print(pow(2,cnt,mod))
|
a[2*i]!=a[2*i+1]
|
[{"input": "4\nBAAAAAAAABBABAB", "output": ["16"]}, {"input": "2\nBAA", "output": ["1"]}, {"input": "2\nABA", "output": ["2"]}, {"input": "2\nAAB", "output": ["2"]}, {"input": "2\nAAA", "output": ["1"]}]
|
control_completion_001664
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a rooted tree of $$$2^n - 1$$$ vertices. Every vertex of this tree has either $$$0$$$ children, or $$$2$$$ children. All leaves of this tree have the same distance from the root, and for every non-leaf vertex, one of its children is the left one, and the other child is the right one. Formally, you are given a perfect binary tree.The vertices of the tree are numbered in the following order: the root has index $$$1$$$; if a vertex has index $$$x$$$, then its left child has index $$$2x$$$, and its right child has index $$$2x+1$$$. Every vertex of the tree has a letter written on it, either A or B. Let's define the character on the vertex $$$x$$$ as $$$s_x$$$.Let the preorder string of some vertex $$$x$$$ be defined in the following way: if the vertex $$$x$$$ is a leaf, then the preorder string of $$$x$$$ be consisting of only one character $$$s_x$$$; otherwise, the preorder string of $$$x$$$ is $$$s_x + f(l_x) + f(r_x)$$$, where $$$+$$$ operator defines concatenation of strings, $$$f(l_x)$$$ is the preorder string of the left child of $$$x$$$, and $$$f(r_x)$$$ is the preorder string of the right child of $$$x$$$. The preorder string of the tree is the preorder string of its root.Now, for the problem itself...You have to calculate the number of different strings that can be obtained as the preorder string of the given tree, if you are allowed to perform the following operation any number of times before constructing the preorder string of the tree: choose any non-leaf vertex $$$x$$$, and swap its children (so, the left child becomes the right one, and vice versa).
Input Specification: The first line contains one integer $$$n$$$ ($$$2 \le n \le 18$$$). The second line contains a sequence of $$$2^n-1$$$ characters $$$s_1, s_2, \dots, s_{2^n-1}$$$. Each character is either A or B. The characters are not separated by spaces or anything else.
Output Specification: Print one integer — the number of different strings that can be obtained as the preorder string of the given tree, if you can apply any number of operations described in the statement. Since it can be very large, print it modulo $$$998244353$$$.
Code:
MOD = 998244353
n, s = int(input()), input()
def calc(u: int) -> tuple:
if # TODO: Your code here:
return (0, 0)
t1, t2 = calc(u * 2), calc(u * 2 + 1)
return (t1[0] + t2[0] + (t1[1] != t2[1]),
hash((min(t1[1], t2[1]), max(t1[1], t2[1]), s[u - 1])))
print(pow(2, calc(1)[0], MOD))
|
MOD = 998244353
n, s = int(input()), input()
def calc(u: int) -> tuple:
if {{completion}}:
return (0, 0)
t1, t2 = calc(u * 2), calc(u * 2 + 1)
return (t1[0] + t2[0] + (t1[1] != t2[1]),
hash((min(t1[1], t2[1]), max(t1[1], t2[1]), s[u - 1])))
print(pow(2, calc(1)[0], MOD))
|
u >= (1 << n)
|
[{"input": "4\nBAAAAAAAABBABAB", "output": ["16"]}, {"input": "2\nBAA", "output": ["1"]}, {"input": "2\nABA", "output": ["2"]}, {"input": "2\nAAB", "output": ["2"]}, {"input": "2\nAAA", "output": ["1"]}]
|
control_completion_001665
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a rooted tree of $$$2^n - 1$$$ vertices. Every vertex of this tree has either $$$0$$$ children, or $$$2$$$ children. All leaves of this tree have the same distance from the root, and for every non-leaf vertex, one of its children is the left one, and the other child is the right one. Formally, you are given a perfect binary tree.The vertices of the tree are numbered in the following order: the root has index $$$1$$$; if a vertex has index $$$x$$$, then its left child has index $$$2x$$$, and its right child has index $$$2x+1$$$. Every vertex of the tree has a letter written on it, either A or B. Let's define the character on the vertex $$$x$$$ as $$$s_x$$$.Let the preorder string of some vertex $$$x$$$ be defined in the following way: if the vertex $$$x$$$ is a leaf, then the preorder string of $$$x$$$ be consisting of only one character $$$s_x$$$; otherwise, the preorder string of $$$x$$$ is $$$s_x + f(l_x) + f(r_x)$$$, where $$$+$$$ operator defines concatenation of strings, $$$f(l_x)$$$ is the preorder string of the left child of $$$x$$$, and $$$f(r_x)$$$ is the preorder string of the right child of $$$x$$$. The preorder string of the tree is the preorder string of its root.Now, for the problem itself...You have to calculate the number of different strings that can be obtained as the preorder string of the given tree, if you are allowed to perform the following operation any number of times before constructing the preorder string of the tree: choose any non-leaf vertex $$$x$$$, and swap its children (so, the left child becomes the right one, and vice versa).
Input Specification: The first line contains one integer $$$n$$$ ($$$2 \le n \le 18$$$). The second line contains a sequence of $$$2^n-1$$$ characters $$$s_1, s_2, \dots, s_{2^n-1}$$$. Each character is either A or B. The characters are not separated by spaces or anything else.
Output Specification: Print one integer — the number of different strings that can be obtained as the preorder string of the given tree, if you can apply any number of operations described in the statement. Since it can be very large, print it modulo $$$998244353$$$.
Code:
def dfs(tree,i,h):
if # TODO: Your code here:
return [tree[i],1]
ls,li=dfs(tree,i*2+1,h)
rs,ri=dfs(tree,i*2+2,h)
res=li*ri
if ls!=rs:
res*=2
if ls>rs:
return [tree[i]+rs+ls,res]
else:
return [tree[i]+ls+rs,res]
h=int(input())
tree=input()
print(dfs(tree,0,h)[1]%998244353 )
|
def dfs(tree,i,h):
if {{completion}}:
return [tree[i],1]
ls,li=dfs(tree,i*2+1,h)
rs,ri=dfs(tree,i*2+2,h)
res=li*ri
if ls!=rs:
res*=2
if ls>rs:
return [tree[i]+rs+ls,res]
else:
return [tree[i]+ls+rs,res]
h=int(input())
tree=input()
print(dfs(tree,0,h)[1]%998244353 )
|
i>=2**(h-1)-1
|
[{"input": "4\nBAAAAAAAABBABAB", "output": ["16"]}, {"input": "2\nBAA", "output": ["1"]}, {"input": "2\nABA", "output": ["2"]}, {"input": "2\nAAB", "output": ["2"]}, {"input": "2\nAAA", "output": ["1"]}]
|
control_completion_001666
|
control_fixed
|
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