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: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
_,(*a,),*r=(map(int,s.split())for s in open(0))
b=[[0],[0]]
for x in b:
for # TODO: Your code here:x+=x[-1]+max(0,u-v),
max=min
for s,t in r:l=b[s>t];print(l[t]-l[s])
|
_,(*a,),*r=(map(int,s.split())for s in open(0))
b=[[0],[0]]
for x in b:
for {{completion}}:x+=x[-1]+max(0,u-v),
max=min
for s,t in r:l=b[s>t];print(l[t]-l[s])
|
u,v in zip([0]+a,a)
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
control_completion_002890
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
n,m=map(int,input().split())
c=[int(i) for i in input().split()]
f=[0]*n
g=[0]*n
for i in range(1,n):
f[i]=f[i-1]+max(0,c[i-1]-c[i])
g[-i-1]=g[-i]+max(0,c[-i]-c[-i-1])
for i in range(m):
x,y=map(int,input().split())
if # TODO: Your code here:
print(f[y-1]-f[x-1])
else:
print(g[y-1]-g[x-1])
|
n,m=map(int,input().split())
c=[int(i) for i in input().split()]
f=[0]*n
g=[0]*n
for i in range(1,n):
f[i]=f[i-1]+max(0,c[i-1]-c[i])
g[-i-1]=g[-i]+max(0,c[-i]-c[-i-1])
for i in range(m):
x,y=map(int,input().split())
if {{completion}}:
print(f[y-1]-f[x-1])
else:
print(g[y-1]-g[x-1])
|
x<y
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
control_completion_002891
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
R=lambda:map(int,input().split())
n,m=R()
*a,=R()
b=[[0],[0]]
f=max
for x in b:
for # TODO: Your code here:x+=x[-1]+f(0,u-v),
f=min
for _ in[0]*m:s,t=R();l=b[s>t];print(l[t]-l[s])
|
R=lambda:map(int,input().split())
n,m=R()
*a,=R()
b=[[0],[0]]
f=max
for x in b:
for {{completion}}:x+=x[-1]+f(0,u-v),
f=min
for _ in[0]*m:s,t=R();l=b[s>t];print(l[t]-l[s])
|
u,v in zip([0]+a,a)
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
control_completion_002892
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
n,m=map(int,input().split())
world=['x']+list(map(int,input().split()))
L1=[0]
L2=[0]
for i in range(1,n):
L1.append(L1[i-1]+max(world[i]-world[i+1],0))
L2.append(L2[i-1]+max(world[i+1]-world[i],0))
for i in range(m):
s,t=map(int,input().split())
if # TODO: Your code here:
print(L1[t-1]-L1[s-1])
else:
print(L2[s-1]-L2[t-1])
|
n,m=map(int,input().split())
world=['x']+list(map(int,input().split()))
L1=[0]
L2=[0]
for i in range(1,n):
L1.append(L1[i-1]+max(world[i]-world[i+1],0))
L2.append(L2[i-1]+max(world[i+1]-world[i],0))
for i in range(m):
s,t=map(int,input().split())
if {{completion}}:
print(L1[t-1]-L1[s-1])
else:
print(L2[s-1]-L2[t-1])
|
s<t
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
control_completion_002893
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
n,m=[int(x) for x in input().split()]
a=[int(x) for x in input().split()]
ltr,rtl=[0],[0]
for i in range(1,n):
ltr.append(max(0,a[i-1]-a[i])+ltr[-1])
rtl.append(max(0,a[i]-a[i-1])+rtl[-1])
for i in range(m):
s,t=[int(x) for x in input().split()]
if # TODO: Your code here:
print(ltr[t-1]-ltr[s-1])
else:
print(rtl[s-1]-rtl[t-1])
|
n,m=[int(x) for x in input().split()]
a=[int(x) for x in input().split()]
ltr,rtl=[0],[0]
for i in range(1,n):
ltr.append(max(0,a[i-1]-a[i])+ltr[-1])
rtl.append(max(0,a[i]-a[i-1])+rtl[-1])
for i in range(m):
s,t=[int(x) for x in input().split()]
if {{completion}}:
print(ltr[t-1]-ltr[s-1])
else:
print(rtl[s-1]-rtl[t-1])
|
s<=t
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
control_completion_002894
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
(n,m),(*a,),*r=(map(int,s.split())for s in open(0))
b=[[0],[0]]
for x in b:
for # TODO: Your code here:x+=x[-1]+max(0,u-v),
max=min
for s,t in r:l=b[s>t];print(l[t]-l[s])
|
(n,m),(*a,),*r=(map(int,s.split())for s in open(0))
b=[[0],[0]]
for x in b:
for {{completion}}:x+=x[-1]+max(0,u-v),
max=min
for s,t in r:l=b[s>t];print(l[t]-l[s])
|
u,v in zip([0]+a,a)
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
control_completion_002895
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
n, m = map(int, input().split())
a = list(map(int, input().split()))
inc = [0]
dec = [0]
for i in range(n-1):
inc.append(inc[i] + max(0, a[i]-a[i+1]))
dec.append(dec[i] + max(0, a[i+1] - a[i]))
for i in range(m):
x, y = map(int, input().split())
#print(x, y)
#ans = 0
if # TODO: Your code here:
ans = inc[y-1] - inc[x-1]
else:
ans = dec[x-1] - dec[y-1]
print(ans)
|
n, m = map(int, input().split())
a = list(map(int, input().split()))
inc = [0]
dec = [0]
for i in range(n-1):
inc.append(inc[i] + max(0, a[i]-a[i+1]))
dec.append(dec[i] + max(0, a[i+1] - a[i]))
for i in range(m):
x, y = map(int, input().split())
#print(x, y)
#ans = 0
if {{completion}}:
ans = inc[y-1] - inc[x-1]
else:
ans = dec[x-1] - dec[y-1]
print(ans)
|
x < y
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
control_completion_002896
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
R=lambda:map(int,input().split())
n,m=R()
*a,=R()
b=[[0],[0]]
for x in b:
for # TODO: Your code here:x+=x[-1]+max(0,u-v),
a=a[::-1]
b[1]=[0]+b[1][::-1]
for _ in[0]*m:s,t=R();l=b[s>t];print(abs(l[s]-l[t]))
|
R=lambda:map(int,input().split())
n,m=R()
*a,=R()
b=[[0],[0]]
for x in b:
for {{completion}}:x+=x[-1]+max(0,u-v),
a=a[::-1]
b[1]=[0]+b[1][::-1]
for _ in[0]*m:s,t=R();l=b[s>t];print(abs(l[s]-l[t]))
|
u,v in zip([0]+a,a)
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
control_completion_002897
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
n, m = map(int, input().split());a = list(map(int, input().split()));l = [0] + [max(0, a[i] - a[i + 1]) for i in range(n - 1)];r = [0] + [max(0, a[i] - a[i - 1]) for i in range(1, n)]
for # TODO: Your code here: l[i + 1] += l[i];r[i + 1] += r[i]
for _ in range(m): s, t = map(int, input().split());print(l[t - 1] - l[s - 1]) if(s < t) else print(r[s - 1] - r[t - 1])
|
n, m = map(int, input().split());a = list(map(int, input().split()));l = [0] + [max(0, a[i] - a[i + 1]) for i in range(n - 1)];r = [0] + [max(0, a[i] - a[i - 1]) for i in range(1, n)]
for {{completion}}: l[i + 1] += l[i];r[i + 1] += r[i]
for _ in range(m): s, t = map(int, input().split());print(l[t - 1] - l[s - 1]) if(s < t) else print(r[s - 1] - r[t - 1])
|
i in range(n - 1)
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
control_completion_002898
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
n, m = map(int, input().split());a = list(map(int, input().split()));l = [0] + [max(0, a[i] - a[i + 1]) for i in range(n - 1)];r = [0] + [max(0, a[i] - a[i - 1]) for i in range(1, n)]
for i in range(n - 1): l[i + 1] += l[i];r[i + 1] += r[i]
for # TODO: Your code here: s, t = map(int, input().split());print(l[t - 1] - l[s - 1]) if(s < t) else print(r[s - 1] - r[t - 1])
|
n, m = map(int, input().split());a = list(map(int, input().split()));l = [0] + [max(0, a[i] - a[i + 1]) for i in range(n - 1)];r = [0] + [max(0, a[i] - a[i - 1]) for i in range(1, n)]
for i in range(n - 1): l[i + 1] += l[i];r[i + 1] += r[i]
for {{completion}}: s, t = map(int, input().split());print(l[t - 1] - l[s - 1]) if(s < t) else print(r[s - 1] - r[t - 1])
|
_ in range(m)
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
control_completion_002899
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$.
Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion.
Code:
n,m=(map(int,input().split()))
l=list(map(int,input().split()))
f=[0]*n
b=[0]*n
d=0
for j in range(1,n):
d=d+max(0,l[j-1]-l[j])
f[j]=d
l=l[::-1]
d=0
for k in range(1,n):
d=d+max(0,l[k-1]-l[k])
b[k]=d
b=b[::-1]
for i in range(m):
s,t=(map(int,input().split()))
if # TODO: Your code here:
print(f[t-1]-f[s-1])
else:
print(b[t-1]-b[s-1])
|
n,m=(map(int,input().split()))
l=list(map(int,input().split()))
f=[0]*n
b=[0]*n
d=0
for j in range(1,n):
d=d+max(0,l[j-1]-l[j])
f[j]=d
l=l[::-1]
d=0
for k in range(1,n):
d=d+max(0,l[k-1]-l[k])
b[k]=d
b=b[::-1]
for i in range(m):
s,t=(map(int,input().split()))
if {{completion}}:
print(f[t-1]-f[s-1])
else:
print(b[t-1]-b[s-1])
|
s<t
|
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
|
control_completion_002900
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
import sys, collections, math, itertools
input = lambda: sys.stdin.readline().strip()
ints = lambda: list(map(int, input().split()))
Int = lambda: int(input())
n, m = ints()
a = ints()
s = int(m ** .5 + 1)
maxs = [max(a[i:i + s]) for i in range(0, m, s)]
for i in range(Int()):
ys, xs, yf, xf, k = ints()
yes = (xs - xf) % k == 0 and (ys - yf) % k == 0
if # TODO: Your code here:
print('no')
continue
mi, ma = min(xs, xf), max(xs, xf)
high = max([0] + a[mi:min(ma, (mi // s + 1) * s)] + a[max(mi, ma // s * s):ma])
for j in range(min(xs, xf) // s + 1, max(xs, xf) // s):
high = max(high, maxs[j])
if high < ys:
print('yes')
continue
print('yes' if ((high - ys) // k + 1) * k + ys <= n else 'no')
|
import sys, collections, math, itertools
input = lambda: sys.stdin.readline().strip()
ints = lambda: list(map(int, input().split()))
Int = lambda: int(input())
n, m = ints()
a = ints()
s = int(m ** .5 + 1)
maxs = [max(a[i:i + s]) for i in range(0, m, s)]
for i in range(Int()):
ys, xs, yf, xf, k = ints()
yes = (xs - xf) % k == 0 and (ys - yf) % k == 0
if {{completion}}:
print('no')
continue
mi, ma = min(xs, xf), max(xs, xf)
high = max([0] + a[mi:min(ma, (mi // s + 1) * s)] + a[max(mi, ma // s * s):ma])
for j in range(min(xs, xf) // s + 1, max(xs, xf) // s):
high = max(high, maxs[j])
if high < ys:
print('yes')
continue
print('yes' if ((high - ys) // k + 1) * k + ys <= n else 'no')
|
not yes
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
control_completion_002939
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
import math
import sys
input = sys.stdin.buffer.readline
y, x_ = input().split()
block = [int(x) for x in input().split()]
list2 = [[0] *20 for i in range( len(block))]
for a in range(len(block)):
list2[a][0] = block[a]
z, x = 0, 1
while (1 << x) <= len(block):
z = 0
while (z + (1 << x) - 1) < len(block):
list2[z][x] = (max(list2[z][x - 1], list2[z + (1 << (x - 1))][x - 1]))
z += 1
x += 1
for i in range(int(input())):
s = [int(x) for x in input().split()]
if abs(s[0] - s[2]) % s[-1] != 0 or abs(s[1] - s[3]) % s[-1] != 0:
print("NO")
else:
smaller = min(s[1], s[3])
bigger = max(s[1], s[3])
k = 0
while # TODO: Your code here:
k += 1
highest = max(list2[smaller - 1][k], list2[bigger - (1 << k)][k])
if highest < s[-1] * ((int(y) - s[0]) // s[-1]) + s[0]:
print("YES")
else:
print("NO")
|
import math
import sys
input = sys.stdin.buffer.readline
y, x_ = input().split()
block = [int(x) for x in input().split()]
list2 = [[0] *20 for i in range( len(block))]
for a in range(len(block)):
list2[a][0] = block[a]
z, x = 0, 1
while (1 << x) <= len(block):
z = 0
while (z + (1 << x) - 1) < len(block):
list2[z][x] = (max(list2[z][x - 1], list2[z + (1 << (x - 1))][x - 1]))
z += 1
x += 1
for i in range(int(input())):
s = [int(x) for x in input().split()]
if abs(s[0] - s[2]) % s[-1] != 0 or abs(s[1] - s[3]) % s[-1] != 0:
print("NO")
else:
smaller = min(s[1], s[3])
bigger = max(s[1], s[3])
k = 0
while {{completion}}:
k += 1
highest = max(list2[smaller - 1][k], list2[bigger - (1 << k)][k])
if highest < s[-1] * ((int(y) - s[0]) // s[-1]) + s[0]:
print("YES")
else:
print("NO")
|
(1 << (k + 1)) <= bigger - smaller + 1
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
control_completion_002940
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
import math
import sys
input = sys.stdin.buffer.readline
y, x_ = input().split()
block = [int(x) for x in input().split()]
list2 = [[0] *20 for i in range( len(block))]
for a in range(len(block)):
list2[a][0] = block[a]
z, x = 0, 1
while (1 << x) <= len(block):
z = 0
while (z + (1 << x) - 1) < len(block):
list2[z][x] = (max(list2[z][x - 1], list2[z + (1 << (x - 1))][x - 1]))
z += 1
x += 1
for i in range(int(input())):
s = [int(x) for x in input().split()]
if abs(s[0] - s[2]) % s[-1] != 0 or abs(s[1] - s[3]) % s[-1] != 0:
print("NO")
else:
smaller = min(s[1], s[3])
bigger = max(s[1], s[3])
k = 0
while (1 << (k + 1)) <= bigger - smaller + 1:
k += 1
highest = max(list2[smaller - 1][k], list2[bigger - (1 << k)][k])
if # TODO: Your code here:
print("YES")
else:
print("NO")
|
import math
import sys
input = sys.stdin.buffer.readline
y, x_ = input().split()
block = [int(x) for x in input().split()]
list2 = [[0] *20 for i in range( len(block))]
for a in range(len(block)):
list2[a][0] = block[a]
z, x = 0, 1
while (1 << x) <= len(block):
z = 0
while (z + (1 << x) - 1) < len(block):
list2[z][x] = (max(list2[z][x - 1], list2[z + (1 << (x - 1))][x - 1]))
z += 1
x += 1
for i in range(int(input())):
s = [int(x) for x in input().split()]
if abs(s[0] - s[2]) % s[-1] != 0 or abs(s[1] - s[3]) % s[-1] != 0:
print("NO")
else:
smaller = min(s[1], s[3])
bigger = max(s[1], s[3])
k = 0
while (1 << (k + 1)) <= bigger - smaller + 1:
k += 1
highest = max(list2[smaller - 1][k], list2[bigger - (1 << k)][k])
if {{completion}}:
print("YES")
else:
print("NO")
|
highest < s[-1] * ((int(y) - s[0]) // s[-1]) + s[0]
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
control_completion_002941
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
from sys import stdin, stdout
from math import floor, ceil, log
input, print = stdin.readline, stdout.write
def main():
n, m = map(int, input().split())
arr = [0] + [int(x) for x in input().split()]
st = construct(arr, m)
for _ in range(int(input())):
x1, y1, x2, y2, k = map(int, input().split())
if # TODO: Your code here:
print("NO\n")
continue
if (x1 <= arr[y1] or x2 <= arr[y2]):
print("NO\n")
continue
max_x = ((n-x1)//k)*k + x1
if (max_x <= getMax(st, m, min(y1, y2), max(y1, y2))):
print("NO\n")
continue
print("YES\n")
def construct(arr, n):
x = ceil(log(n, 2))
max_size = 2 * pow(2, x) - 1
st = [0]*max_size
construct2(arr, 0, n-1, st, 0)
return st
def construct2(arr, ss, se, st, si):
if (ss == se):
st[si] = arr[ss]
return arr[ss]
mid = getMid(ss, se)
st[si] = max(construct2(arr, ss, mid, st, si * 2 + 1), construct2(arr, mid + 1, se, st, si * 2 + 2))
return st[si]
def getMid(s, e):
return s + (e - s) // 2
def getMax(st, n, l, r):
return MaxUtil(st, 0, n - 1, l, r, 0)
def MaxUtil(st, ss, se, l, r, node):
if (l <= ss and r >= se):
return st[node]
if (se < l or ss > r):
return -1
mid = getMid(ss, se)
return max(MaxUtil(st, ss, mid, l, r, 2 * node + 1), MaxUtil(st, mid + 1, se, l, r, 2 * node + 2))
main()
|
from sys import stdin, stdout
from math import floor, ceil, log
input, print = stdin.readline, stdout.write
def main():
n, m = map(int, input().split())
arr = [0] + [int(x) for x in input().split()]
st = construct(arr, m)
for _ in range(int(input())):
x1, y1, x2, y2, k = map(int, input().split())
if {{completion}}:
print("NO\n")
continue
if (x1 <= arr[y1] or x2 <= arr[y2]):
print("NO\n")
continue
max_x = ((n-x1)//k)*k + x1
if (max_x <= getMax(st, m, min(y1, y2), max(y1, y2))):
print("NO\n")
continue
print("YES\n")
def construct(arr, n):
x = ceil(log(n, 2))
max_size = 2 * pow(2, x) - 1
st = [0]*max_size
construct2(arr, 0, n-1, st, 0)
return st
def construct2(arr, ss, se, st, si):
if (ss == se):
st[si] = arr[ss]
return arr[ss]
mid = getMid(ss, se)
st[si] = max(construct2(arr, ss, mid, st, si * 2 + 1), construct2(arr, mid + 1, se, st, si * 2 + 2))
return st[si]
def getMid(s, e):
return s + (e - s) // 2
def getMax(st, n, l, r):
return MaxUtil(st, 0, n - 1, l, r, 0)
def MaxUtil(st, ss, se, l, r, node):
if (l <= ss and r >= se):
return st[node]
if (se < l or ss > r):
return -1
mid = getMid(ss, se)
return max(MaxUtil(st, ss, mid, l, r, 2 * node + 1), MaxUtil(st, mid + 1, se, l, r, 2 * node + 2))
main()
|
((y2 - y1) % k != 0 or (x2 - x1) % k != 0)
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
control_completion_002942
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
from sys import stdin, stdout
from math import floor, ceil, log
input, print = stdin.readline, stdout.write
def main():
n, m = map(int, input().split())
arr = [0] + [int(x) for x in input().split()]
st = construct(arr, m)
for _ in range(int(input())):
x1, y1, x2, y2, k = map(int, input().split())
if ((y2 - y1) % k != 0 or (x2 - x1) % k != 0):
print("NO\n")
continue
if (x1 <= arr[y1] or x2 <= arr[y2]):
print("NO\n")
continue
max_x = ((n-x1)//k)*k + x1
if # TODO: Your code here:
print("NO\n")
continue
print("YES\n")
def construct(arr, n):
x = ceil(log(n, 2))
max_size = 2 * pow(2, x) - 1
st = [0]*max_size
construct2(arr, 0, n-1, st, 0)
return st
def construct2(arr, ss, se, st, si):
if (ss == se):
st[si] = arr[ss]
return arr[ss]
mid = getMid(ss, se)
st[si] = max(construct2(arr, ss, mid, st, si * 2 + 1), construct2(arr, mid + 1, se, st, si * 2 + 2))
return st[si]
def getMid(s, e):
return s + (e - s) // 2
def getMax(st, n, l, r):
return MaxUtil(st, 0, n - 1, l, r, 0)
def MaxUtil(st, ss, se, l, r, node):
if (l <= ss and r >= se):
return st[node]
if (se < l or ss > r):
return -1
mid = getMid(ss, se)
return max(MaxUtil(st, ss, mid, l, r, 2 * node + 1), MaxUtil(st, mid + 1, se, l, r, 2 * node + 2))
main()
|
from sys import stdin, stdout
from math import floor, ceil, log
input, print = stdin.readline, stdout.write
def main():
n, m = map(int, input().split())
arr = [0] + [int(x) for x in input().split()]
st = construct(arr, m)
for _ in range(int(input())):
x1, y1, x2, y2, k = map(int, input().split())
if ((y2 - y1) % k != 0 or (x2 - x1) % k != 0):
print("NO\n")
continue
if (x1 <= arr[y1] or x2 <= arr[y2]):
print("NO\n")
continue
max_x = ((n-x1)//k)*k + x1
if {{completion}}:
print("NO\n")
continue
print("YES\n")
def construct(arr, n):
x = ceil(log(n, 2))
max_size = 2 * pow(2, x) - 1
st = [0]*max_size
construct2(arr, 0, n-1, st, 0)
return st
def construct2(arr, ss, se, st, si):
if (ss == se):
st[si] = arr[ss]
return arr[ss]
mid = getMid(ss, se)
st[si] = max(construct2(arr, ss, mid, st, si * 2 + 1), construct2(arr, mid + 1, se, st, si * 2 + 2))
return st[si]
def getMid(s, e):
return s + (e - s) // 2
def getMax(st, n, l, r):
return MaxUtil(st, 0, n - 1, l, r, 0)
def MaxUtil(st, ss, se, l, r, node):
if (l <= ss and r >= se):
return st[node]
if (se < l or ss > r):
return -1
mid = getMid(ss, se)
return max(MaxUtil(st, ss, mid, l, r, 2 * node + 1), MaxUtil(st, mid + 1, se, l, r, 2 * node + 2))
main()
|
(max_x <= getMax(st, m, min(y1, y2), max(y1, y2)))
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
control_completion_002943
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
import sys
from array import array
input = lambda: sys.stdin.buffer.readline().decode().strip()
inp = lambda dtype: [dtype(x) for x in input().split()]
inp_2d = lambda dtype, n: [dtype(input()) for _ in range(n)]
inp_2ds = lambda dtype, n: [inp(dtype) for _ in range(n)]
ceil1 = lambda a, b: (a + b - 1) // b
debug = lambda *x: print(*x, file=sys.stderr)
out = []
class sparse_table:
def __init__(self, n, a, default=0, func=max):
self.n, self.lg = n, 20
self.func = func
self.table = [array('i', [default] * n) for _ in range(self.lg)]
self.table[0] = a
self.preprocess()
def preprocess(self):
for j in range(1, self.lg):
i = 0
while # TODO: Your code here:
ix = i + (1 << (j - 1))
self.table[j][i] = self.func(self.table[j - 1][i], self.table[j - 1][ix])
i += 1
def quary(self, l, r):
'''[l,r]'''
l1 = (r - l + 1).bit_length() - 1
r1 = r - (1 << l1) + 1
return self.func(self.table[l1][l], self.table[l1][r1])
n, m = inp(int)
a = array('i', inp(int))
mem = sparse_table(m, a)
for _ in range(int(input())):
xs, ys, xf, yf, k = inp(int)
div, mod = divmod(n - xs, k)
xma = n - mod
qur = mem.quary(min(ys, yf) - 1, max(ys, yf) - 1)
if qur >= xma or abs(xma - xf) % k or abs(ys - yf) % k:
out.append('no')
else:
out.append('yes')
print('\n'.join(out))
|
import sys
from array import array
input = lambda: sys.stdin.buffer.readline().decode().strip()
inp = lambda dtype: [dtype(x) for x in input().split()]
inp_2d = lambda dtype, n: [dtype(input()) for _ in range(n)]
inp_2ds = lambda dtype, n: [inp(dtype) for _ in range(n)]
ceil1 = lambda a, b: (a + b - 1) // b
debug = lambda *x: print(*x, file=sys.stderr)
out = []
class sparse_table:
def __init__(self, n, a, default=0, func=max):
self.n, self.lg = n, 20
self.func = func
self.table = [array('i', [default] * n) for _ in range(self.lg)]
self.table[0] = a
self.preprocess()
def preprocess(self):
for j in range(1, self.lg):
i = 0
while {{completion}}:
ix = i + (1 << (j - 1))
self.table[j][i] = self.func(self.table[j - 1][i], self.table[j - 1][ix])
i += 1
def quary(self, l, r):
'''[l,r]'''
l1 = (r - l + 1).bit_length() - 1
r1 = r - (1 << l1) + 1
return self.func(self.table[l1][l], self.table[l1][r1])
n, m = inp(int)
a = array('i', inp(int))
mem = sparse_table(m, a)
for _ in range(int(input())):
xs, ys, xf, yf, k = inp(int)
div, mod = divmod(n - xs, k)
xma = n - mod
qur = mem.quary(min(ys, yf) - 1, max(ys, yf) - 1)
if qur >= xma or abs(xma - xf) % k or abs(ys - yf) % k:
out.append('no')
else:
out.append('yes')
print('\n'.join(out))
|
i + (1 << j) - 1 < self.n
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
control_completion_002944
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
from sys import setrecursionlimit, stdin
readline = stdin.readline
# def I(): return int(readline())
# def ST(): return readline()[:-1]
# def LI(): return list(map(int, readline().split()))
def solve():
N, m = map(int, readline().split())
A = [0] + list(map(int, readline().split()))
n = len(A)
num = 1 << (n - 1).bit_length()
tree = [0] * 2 * num
for i in range(n):
tree[num + i] = A[i]
for i in range(num - 1, 0, -1):
tree[i] = max(tree[2 * i], tree[2 * i + 1])
def query(l, r):
ret = 0
l += num
r += num
while l < r:
if # TODO: Your code here:
ret = max(ret, tree[l])
l += 1
if r & 1:
ret = max(ret, tree[r - 1])
l >>= 1
r >>= 1
return ret
for _ in range(int(input())):
sx, sy, gx, gy, k = map(int, readline().split())
dx = abs(sx - gx)
dy = abs(sy - gy)
if dx % k or dy % k:
print("NO")
continue
top = sx + k * ((N - sx) // k)
if sy > gy: sy, gy = gy, sy
if query(sy, gy + 1) < top: print("YES")
else: print("NO")
solve()
|
from sys import setrecursionlimit, stdin
readline = stdin.readline
# def I(): return int(readline())
# def ST(): return readline()[:-1]
# def LI(): return list(map(int, readline().split()))
def solve():
N, m = map(int, readline().split())
A = [0] + list(map(int, readline().split()))
n = len(A)
num = 1 << (n - 1).bit_length()
tree = [0] * 2 * num
for i in range(n):
tree[num + i] = A[i]
for i in range(num - 1, 0, -1):
tree[i] = max(tree[2 * i], tree[2 * i + 1])
def query(l, r):
ret = 0
l += num
r += num
while l < r:
if {{completion}}:
ret = max(ret, tree[l])
l += 1
if r & 1:
ret = max(ret, tree[r - 1])
l >>= 1
r >>= 1
return ret
for _ in range(int(input())):
sx, sy, gx, gy, k = map(int, readline().split())
dx = abs(sx - gx)
dy = abs(sy - gy)
if dx % k or dy % k:
print("NO")
continue
top = sx + k * ((N - sx) // k)
if sy > gy: sy, gy = gy, sy
if query(sy, gy + 1) < top: print("YES")
else: print("NO")
solve()
|
l & 1
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
control_completion_002945
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
from sys import setrecursionlimit, stdin
readline = stdin.readline
# def I(): return int(readline())
# def ST(): return readline()[:-1]
# def LI(): return list(map(int, readline().split()))
def solve():
N, m = map(int, readline().split())
A = [0] + list(map(int, readline().split()))
n = len(A)
num = 1 << (n - 1).bit_length()
tree = [0] * 2 * num
for i in range(n):
tree[num + i] = A[i]
for i in range(num - 1, 0, -1):
tree[i] = max(tree[2 * i], tree[2 * i + 1])
def query(l, r):
ret = 0
l += num
r += num
while l < r:
if l & 1:
ret = max(ret, tree[l])
l += 1
if # TODO: Your code here:
ret = max(ret, tree[r - 1])
l >>= 1
r >>= 1
return ret
for _ in range(int(input())):
sx, sy, gx, gy, k = map(int, readline().split())
dx = abs(sx - gx)
dy = abs(sy - gy)
if dx % k or dy % k:
print("NO")
continue
top = sx + k * ((N - sx) // k)
if sy > gy: sy, gy = gy, sy
if query(sy, gy + 1) < top: print("YES")
else: print("NO")
solve()
|
from sys import setrecursionlimit, stdin
readline = stdin.readline
# def I(): return int(readline())
# def ST(): return readline()[:-1]
# def LI(): return list(map(int, readline().split()))
def solve():
N, m = map(int, readline().split())
A = [0] + list(map(int, readline().split()))
n = len(A)
num = 1 << (n - 1).bit_length()
tree = [0] * 2 * num
for i in range(n):
tree[num + i] = A[i]
for i in range(num - 1, 0, -1):
tree[i] = max(tree[2 * i], tree[2 * i + 1])
def query(l, r):
ret = 0
l += num
r += num
while l < r:
if l & 1:
ret = max(ret, tree[l])
l += 1
if {{completion}}:
ret = max(ret, tree[r - 1])
l >>= 1
r >>= 1
return ret
for _ in range(int(input())):
sx, sy, gx, gy, k = map(int, readline().split())
dx = abs(sx - gx)
dy = abs(sy - gy)
if dx % k or dy % k:
print("NO")
continue
top = sx + k * ((N - sx) // k)
if sy > gy: sy, gy = gy, sy
if query(sy, gy + 1) < top: print("YES")
else: print("NO")
solve()
|
r & 1
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
control_completion_002946
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
input = __import__('sys').stdin.readline
log2s = [0, 0]
for i in range(2, 200005):
log2s.append(log2s[i // 2] + 1)
n, m = map(int, input().split())
sparse = [[0] + list(map(int, input().split()))]
for j in range(20):
sparse.append([max(sparse[j][min(i + (1 << j), m)], sparse[j][i]) for i in range(m+1)])
def getmax(L, R):
j = log2s[R - L + 1]
return max(sparse[j][R - (1 << j) + 1], sparse[j][L])
for _ in range(int(input())):
x1, y1, x2, y2, k = map(int, input().split())
if # TODO: Your code here:
print('NO')
continue
# i * k + x1 <= n
i = (n - x1) // k
h = i * k + x1
while h > n:
h -= k
print('YES' if getmax(min(y1, y2), max(y1, y2)) < h else 'NO')
|
input = __import__('sys').stdin.readline
log2s = [0, 0]
for i in range(2, 200005):
log2s.append(log2s[i // 2] + 1)
n, m = map(int, input().split())
sparse = [[0] + list(map(int, input().split()))]
for j in range(20):
sparse.append([max(sparse[j][min(i + (1 << j), m)], sparse[j][i]) for i in range(m+1)])
def getmax(L, R):
j = log2s[R - L + 1]
return max(sparse[j][R - (1 << j) + 1], sparse[j][L])
for _ in range(int(input())):
x1, y1, x2, y2, k = map(int, input().split())
if {{completion}}:
print('NO')
continue
# i * k + x1 <= n
i = (n - x1) // k
h = i * k + x1
while h > n:
h -= k
print('YES' if getmax(min(y1, y2), max(y1, y2)) < h else 'NO')
|
(x1 - x2) % k != 0 or (y1 - y2) % k != 0
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
control_completion_002947
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
import math,sys;input=sys.stdin.readline;S=lambda:input().rstrip();I=lambda:int(S());M=lambda:map(int,S().split());L=lambda:list(M());mod1=1000000007;mod2=998244353
n,m=M();a=L();q=I()
t=[0]*(2*m)
# a is the array of initial values
def build(t,n,a):
for i in range(n):t[i+n]=a[i]
for i in range(n-1,0,-1):t[i]=max(t[i<<1],t[(i<<1)|1])
# change value at position p to v
def modify(t,n,p,v):
p+=n
t[p]=v
while p>1:
t[p>>1]=max(t[p],t[p^1])
p>>=1
# find the combined value of range [l,r)
def query(t,n,l,r):
resl=resr=0
l+=n;r+=n
while l<r:
if # TODO: Your code here:resl=max(resl,t[l]);l+=1
if (r&1):r-=1;resr=max(t[r],resr)
l>>=1;r>>=1
return max(resl,resr)
build(t,m,a)
for i in range(q):
xs,ys,xf,yf,k=M()
if abs(xs-xf)%k or abs(ys-yf)%k:print("NO");continue
p=query(t,m,min(ys,yf)-1,max(ys,yf))
z=min(xs,xf)+((n-min(xs,xf))//k)*k
if z<=p:print("NO")
else:print("YES")
|
import math,sys;input=sys.stdin.readline;S=lambda:input().rstrip();I=lambda:int(S());M=lambda:map(int,S().split());L=lambda:list(M());mod1=1000000007;mod2=998244353
n,m=M();a=L();q=I()
t=[0]*(2*m)
# a is the array of initial values
def build(t,n,a):
for i in range(n):t[i+n]=a[i]
for i in range(n-1,0,-1):t[i]=max(t[i<<1],t[(i<<1)|1])
# change value at position p to v
def modify(t,n,p,v):
p+=n
t[p]=v
while p>1:
t[p>>1]=max(t[p],t[p^1])
p>>=1
# find the combined value of range [l,r)
def query(t,n,l,r):
resl=resr=0
l+=n;r+=n
while l<r:
if {{completion}}:resl=max(resl,t[l]);l+=1
if (r&1):r-=1;resr=max(t[r],resr)
l>>=1;r>>=1
return max(resl,resr)
build(t,m,a)
for i in range(q):
xs,ys,xf,yf,k=M()
if abs(xs-xf)%k or abs(ys-yf)%k:print("NO");continue
p=query(t,m,min(ys,yf)-1,max(ys,yf))
z=min(xs,xf)+((n-min(xs,xf))//k)*k
if z<=p:print("NO")
else:print("YES")
|
(l&1)
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
control_completion_002948
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count.
Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] < x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] < x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked.
Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO".
Code:
import math,sys;input=sys.stdin.readline;S=lambda:input().rstrip();I=lambda:int(S());M=lambda:map(int,S().split());L=lambda:list(M());mod1=1000000007;mod2=998244353
n,m=M();a=L();q=I()
t=[0]*(2*m)
# a is the array of initial values
def build(t,n,a):
for i in range(n):t[i+n]=a[i]
for i in range(n-1,0,-1):t[i]=max(t[i<<1],t[(i<<1)|1])
# change value at position p to v
def modify(t,n,p,v):
p+=n
t[p]=v
while p>1:
t[p>>1]=max(t[p],t[p^1])
p>>=1
# find the combined value of range [l,r)
def query(t,n,l,r):
resl=resr=0
l+=n;r+=n
while l<r:
if (l&1):resl=max(resl,t[l]);l+=1
if # TODO: Your code here:r-=1;resr=max(t[r],resr)
l>>=1;r>>=1
return max(resl,resr)
build(t,m,a)
for i in range(q):
xs,ys,xf,yf,k=M()
if abs(xs-xf)%k or abs(ys-yf)%k:print("NO");continue
p=query(t,m,min(ys,yf)-1,max(ys,yf))
z=min(xs,xf)+((n-min(xs,xf))//k)*k
if z<=p:print("NO")
else:print("YES")
|
import math,sys;input=sys.stdin.readline;S=lambda:input().rstrip();I=lambda:int(S());M=lambda:map(int,S().split());L=lambda:list(M());mod1=1000000007;mod2=998244353
n,m=M();a=L();q=I()
t=[0]*(2*m)
# a is the array of initial values
def build(t,n,a):
for i in range(n):t[i+n]=a[i]
for i in range(n-1,0,-1):t[i]=max(t[i<<1],t[(i<<1)|1])
# change value at position p to v
def modify(t,n,p,v):
p+=n
t[p]=v
while p>1:
t[p>>1]=max(t[p],t[p^1])
p>>=1
# find the combined value of range [l,r)
def query(t,n,l,r):
resl=resr=0
l+=n;r+=n
while l<r:
if (l&1):resl=max(resl,t[l]);l+=1
if {{completion}}:r-=1;resr=max(t[r],resr)
l>>=1;r>>=1
return max(resl,resr)
build(t,m,a)
for i in range(q):
xs,ys,xf,yf,k=M()
if abs(xs-xf)%k or abs(ys-yf)%k:print("NO");continue
p=query(t,m,min(ys,yf)-1,max(ys,yf))
z=min(xs,xf)+((n-min(xs,xf))//k)*k
if z<=p:print("NO")
else:print("YES")
|
(r&1)
|
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
|
control_completion_002949
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a tree consisting of $$$n$$$ vertices. A number is written on each vertex; the number on vertex $$$i$$$ is equal to $$$a_i$$$.Recall that a simple path is a path that visits each vertex at most once. Let the weight of the path be the bitwise XOR of the values written on vertices it consists of. Let's say that a tree is good if no simple path has weight $$$0$$$.You can apply the following operation any number of times (possibly, zero): select a vertex of the tree and replace the value written on it with an arbitrary positive integer. What is the minimum number of times you have to apply this operation in order to make the tree good?
Input Specification: The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of vertices. The second line contains $$$n$$$ integers $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \le a_i < 2^{30}$$$) — the numbers written on vertices. Then $$$n - 1$$$ lines follow, each containing two integers $$$x$$$ and $$$y$$$ ($$$1 \le x, y \le n; x \ne y$$$) denoting an edge connecting vertex $$$x$$$ with vertex $$$y$$$. It is guaranteed that these edges form a tree.
Output Specification: Print a single integer — the minimum number of times you have to apply the operation in order to make the tree good.
Notes: NoteIn the first example, it is enough to replace the value on the vertex $$$1$$$ with $$$13$$$, and the value on the vertex $$$4$$$ with $$$42$$$.
Code:
import sys
input = sys.stdin.buffer.readline
N = int(input())
A = list(map(int, input().split()))
adj = [[] for _ in range(N)]
for _ in range(N - 1):
a, b = map(lambda s: int(s) - 1, input().split())
adj[a].append(b)
adj[b].append(a)
O = [0]
for i in O:
for j in adj[i]:
adj[j].remove(i)
O.append(j)
class XORSet:
def __init__(self, el=None):
self.s = set()
self.xor = 0
if el is not None:
self.s.add(el)
def add(self, el: int):
self.s.add(el ^ self.xor)
def update(self, xor: int):
self.xor ^= xor
def __len__(self) -> int:
return len(self.s)
def __iter__(self):
return (x ^ self.xor for x in self.s)
def __contains__(self, el: int) -> bool:
return (el ^ self.xor) in self.s
r = 0
D = [XORSet(a) for a in A]
for i in reversed(O):
for j in adj[i]:
if len(D[j]) > len(D[i]):
D[i], D[j] = D[j], D[i]
D[i].update(A[i])
D[j].update(A[i])
#print('merging', i, j, list(D[i]), list(D[j]))
if any(x in D[i] for x in D[j]):
r += 1
D[i].s.clear()
break
else:
for # TODO: Your code here:
D[i].add(x ^ A[i])
#assert 0 not in D[i]
#print(i, A[i], adj[i], list(D[i]))
print(r)
|
import sys
input = sys.stdin.buffer.readline
N = int(input())
A = list(map(int, input().split()))
adj = [[] for _ in range(N)]
for _ in range(N - 1):
a, b = map(lambda s: int(s) - 1, input().split())
adj[a].append(b)
adj[b].append(a)
O = [0]
for i in O:
for j in adj[i]:
adj[j].remove(i)
O.append(j)
class XORSet:
def __init__(self, el=None):
self.s = set()
self.xor = 0
if el is not None:
self.s.add(el)
def add(self, el: int):
self.s.add(el ^ self.xor)
def update(self, xor: int):
self.xor ^= xor
def __len__(self) -> int:
return len(self.s)
def __iter__(self):
return (x ^ self.xor for x in self.s)
def __contains__(self, el: int) -> bool:
return (el ^ self.xor) in self.s
r = 0
D = [XORSet(a) for a in A]
for i in reversed(O):
for j in adj[i]:
if len(D[j]) > len(D[i]):
D[i], D[j] = D[j], D[i]
D[i].update(A[i])
D[j].update(A[i])
#print('merging', i, j, list(D[i]), list(D[j]))
if any(x in D[i] for x in D[j]):
r += 1
D[i].s.clear()
break
else:
for {{completion}}:
D[i].add(x ^ A[i])
#assert 0 not in D[i]
#print(i, A[i], adj[i], list(D[i]))
print(r)
|
x in D[j]
|
[{"input": "6\n3 2 1 3 2 1\n4 5\n3 4\n1 4\n2 1\n6 1", "output": ["2"]}, {"input": "4\n2 1 1 1\n1 2\n1 3\n1 4", "output": ["0"]}, {"input": "5\n2 2 2 2 2\n1 2\n2 3\n3 4\n4 5", "output": ["2"]}]
|
control_completion_002986
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a tree consisting of $$$n$$$ vertices. A number is written on each vertex; the number on vertex $$$i$$$ is equal to $$$a_i$$$.Recall that a simple path is a path that visits each vertex at most once. Let the weight of the path be the bitwise XOR of the values written on vertices it consists of. Let's say that a tree is good if no simple path has weight $$$0$$$.You can apply the following operation any number of times (possibly, zero): select a vertex of the tree and replace the value written on it with an arbitrary positive integer. What is the minimum number of times you have to apply this operation in order to make the tree good?
Input Specification: The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of vertices. The second line contains $$$n$$$ integers $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \le a_i < 2^{30}$$$) — the numbers written on vertices. Then $$$n - 1$$$ lines follow, each containing two integers $$$x$$$ and $$$y$$$ ($$$1 \le x, y \le n; x \ne y$$$) denoting an edge connecting vertex $$$x$$$ with vertex $$$y$$$. It is guaranteed that these edges form a tree.
Output Specification: Print a single integer — the minimum number of times you have to apply the operation in order to make the tree good.
Notes: NoteIn the first example, it is enough to replace the value on the vertex $$$1$$$ with $$$13$$$, and the value on the vertex $$$4$$$ with $$$42$$$.
Code:
import sys
from types import GeneratorType
from collections import defaultdict
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if # TODO: Your code here:
break
to = stack[-1].send(to)
return to
return wrappedfunc
input = lambda: sys.stdin.readline().rstrip()
def solve():
N = int(input())
A = [int(x) for x in input().split()]
G = [[] for _ in range(N)]
for _ in range(N - 1):
x, y = [int(x) - 1 for x in input().split()]
G[x].append(y)
G[y].append(x)
B = [0] * N
vals = defaultdict(set)
res = [0]
@bootstrap
def fill_dfs(v, p):
B[v] = A[v]
if p != -1:
B[v] ^= B[p]
for u in G[v]:
if u != p:
yield fill_dfs(u, v)
yield
@bootstrap
def calc_dfs(v, p):
zero = False
vals[v].add(B[v])
for u in G[v]:
if u != p:
yield calc_dfs(u, v)
if len(vals[v]) < len(vals[u]):
vals[v], vals[u] = vals[u], vals[v]
for x in vals[u]:
zero |= (x ^ A[v]) in vals[v]
for x in vals[u]:
vals[v].add(x)
vals[u].clear()
if zero:
res[0] += 1
vals[v].clear()
yield
fill_dfs(0, -1)
calc_dfs(0, -1)
return res[0]
print(solve())
|
import sys
from types import GeneratorType
from collections import defaultdict
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if {{completion}}:
break
to = stack[-1].send(to)
return to
return wrappedfunc
input = lambda: sys.stdin.readline().rstrip()
def solve():
N = int(input())
A = [int(x) for x in input().split()]
G = [[] for _ in range(N)]
for _ in range(N - 1):
x, y = [int(x) - 1 for x in input().split()]
G[x].append(y)
G[y].append(x)
B = [0] * N
vals = defaultdict(set)
res = [0]
@bootstrap
def fill_dfs(v, p):
B[v] = A[v]
if p != -1:
B[v] ^= B[p]
for u in G[v]:
if u != p:
yield fill_dfs(u, v)
yield
@bootstrap
def calc_dfs(v, p):
zero = False
vals[v].add(B[v])
for u in G[v]:
if u != p:
yield calc_dfs(u, v)
if len(vals[v]) < len(vals[u]):
vals[v], vals[u] = vals[u], vals[v]
for x in vals[u]:
zero |= (x ^ A[v]) in vals[v]
for x in vals[u]:
vals[v].add(x)
vals[u].clear()
if zero:
res[0] += 1
vals[v].clear()
yield
fill_dfs(0, -1)
calc_dfs(0, -1)
return res[0]
print(solve())
|
not stack
|
[{"input": "6\n3 2 1 3 2 1\n4 5\n3 4\n1 4\n2 1\n6 1", "output": ["2"]}, {"input": "4\n2 1 1 1\n1 2\n1 3\n1 4", "output": ["0"]}, {"input": "5\n2 2 2 2 2\n1 2\n2 3\n3 4\n4 5", "output": ["2"]}]
|
control_completion_002987
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a tree consisting of $$$n$$$ vertices. A number is written on each vertex; the number on vertex $$$i$$$ is equal to $$$a_i$$$.Recall that a simple path is a path that visits each vertex at most once. Let the weight of the path be the bitwise XOR of the values written on vertices it consists of. Let's say that a tree is good if no simple path has weight $$$0$$$.You can apply the following operation any number of times (possibly, zero): select a vertex of the tree and replace the value written on it with an arbitrary positive integer. What is the minimum number of times you have to apply this operation in order to make the tree good?
Input Specification: The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of vertices. The second line contains $$$n$$$ integers $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \le a_i < 2^{30}$$$) — the numbers written on vertices. Then $$$n - 1$$$ lines follow, each containing two integers $$$x$$$ and $$$y$$$ ($$$1 \le x, y \le n; x \ne y$$$) denoting an edge connecting vertex $$$x$$$ with vertex $$$y$$$. It is guaranteed that these edges form a tree.
Output Specification: Print a single integer — the minimum number of times you have to apply the operation in order to make the tree good.
Notes: NoteIn the first example, it is enough to replace the value on the vertex $$$1$$$ with $$$13$$$, and the value on the vertex $$$4$$$ with $$$42$$$.
Code:
import sys
from types import GeneratorType
from collections import defaultdict
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if # TODO: Your code here:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
input = lambda: sys.stdin.readline().rstrip()
def solve():
N = int(input())
A = [int(x) for x in input().split()]
G = [[] for _ in range(N)]
for _ in range(N - 1):
x, y = [int(x) - 1 for x in input().split()]
G[x].append(y)
G[y].append(x)
B = [0] * N
vals = defaultdict(set)
res = [0]
@bootstrap
def fill_dfs(v, p):
B[v] = A[v]
if p != -1:
B[v] ^= B[p]
for u in G[v]:
if u != p:
yield fill_dfs(u, v)
yield
@bootstrap
def calc_dfs(v, p):
zero = False
vals[v].add(B[v])
for u in G[v]:
if u != p:
yield calc_dfs(u, v)
if len(vals[v]) < len(vals[u]):
vals[v], vals[u] = vals[u], vals[v]
for x in vals[u]:
zero |= (x ^ A[v]) in vals[v]
for x in vals[u]:
vals[v].add(x)
vals[u].clear()
if zero:
res[0] += 1
vals[v].clear()
yield
fill_dfs(0, -1)
calc_dfs(0, -1)
return res[0]
print(solve())
|
import sys
from types import GeneratorType
from collections import defaultdict
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if {{completion}}:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
input = lambda: sys.stdin.readline().rstrip()
def solve():
N = int(input())
A = [int(x) for x in input().split()]
G = [[] for _ in range(N)]
for _ in range(N - 1):
x, y = [int(x) - 1 for x in input().split()]
G[x].append(y)
G[y].append(x)
B = [0] * N
vals = defaultdict(set)
res = [0]
@bootstrap
def fill_dfs(v, p):
B[v] = A[v]
if p != -1:
B[v] ^= B[p]
for u in G[v]:
if u != p:
yield fill_dfs(u, v)
yield
@bootstrap
def calc_dfs(v, p):
zero = False
vals[v].add(B[v])
for u in G[v]:
if u != p:
yield calc_dfs(u, v)
if len(vals[v]) < len(vals[u]):
vals[v], vals[u] = vals[u], vals[v]
for x in vals[u]:
zero |= (x ^ A[v]) in vals[v]
for x in vals[u]:
vals[v].add(x)
vals[u].clear()
if zero:
res[0] += 1
vals[v].clear()
yield
fill_dfs(0, -1)
calc_dfs(0, -1)
return res[0]
print(solve())
|
type(to) is GeneratorType
|
[{"input": "6\n3 2 1 3 2 1\n4 5\n3 4\n1 4\n2 1\n6 1", "output": ["2"]}, {"input": "4\n2 1 1 1\n1 2\n1 3\n1 4", "output": ["0"]}, {"input": "5\n2 2 2 2 2\n1 2\n2 3\n3 4\n4 5", "output": ["2"]}]
|
control_completion_002988
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given a tree consisting of $$$n$$$ vertices. A number is written on each vertex; the number on vertex $$$i$$$ is equal to $$$a_i$$$.Recall that a simple path is a path that visits each vertex at most once. Let the weight of the path be the bitwise XOR of the values written on vertices it consists of. Let's say that a tree is good if no simple path has weight $$$0$$$.You can apply the following operation any number of times (possibly, zero): select a vertex of the tree and replace the value written on it with an arbitrary positive integer. What is the minimum number of times you have to apply this operation in order to make the tree good?
Input Specification: The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of vertices. The second line contains $$$n$$$ integers $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \le a_i < 2^{30}$$$) — the numbers written on vertices. Then $$$n - 1$$$ lines follow, each containing two integers $$$x$$$ and $$$y$$$ ($$$1 \le x, y \le n; x \ne y$$$) denoting an edge connecting vertex $$$x$$$ with vertex $$$y$$$. It is guaranteed that these edges form a tree.
Output Specification: Print a single integer — the minimum number of times you have to apply the operation in order to make the tree good.
Notes: NoteIn the first example, it is enough to replace the value on the vertex $$$1$$$ with $$$13$$$, and the value on the vertex $$$4$$$ with $$$42$$$.
Code:
import sys
input = sys.stdin.buffer.readline
N = int(input())
A = list(map(int, input().split()))
adj = [[] for _ in range(N)]
for _ in range(N - 1):
a, b = map(lambda s: int(s) - 1, input().split())
adj[a].append(b)
adj[b].append(a)
O = [0]
for i in O:
for j in adj[i]:
adj[j].remove(i)
O.append(j)
class XORSet:
def __init__(self, el=None):
self.s = set()
self.xor = 0
if el is not None:
self.s.add(el)
def add(self, el: int):
self.s.add(el ^ self.xor)
def update(self, xor: int):
self.xor ^= xor
def __len__(self) -> int:
return len(self.s)
def __iter__(self):
return (x ^ self.xor for x in self.s)
def __contains__(self, el: int) -> bool:
return (el ^ self.xor) in self.s
r = 0
D = [XORSet(a) for a in A]
for i in reversed(O):
for j in adj[i]:
if len(D[j]) > len(D[i]):
D[i], D[j] = D[j], D[i]
D[i].update(A[i])
D[j].update(A[i])
l = list(D[j])
for x in l:
if x in D[i]:
r += 1
D[i].s.clear()
break
else:
for # TODO: Your code here:
D[i].add(x ^ A[i])
continue
break
#print(i, A[i], adj[i], list(D[i]))
print(r)
|
import sys
input = sys.stdin.buffer.readline
N = int(input())
A = list(map(int, input().split()))
adj = [[] for _ in range(N)]
for _ in range(N - 1):
a, b = map(lambda s: int(s) - 1, input().split())
adj[a].append(b)
adj[b].append(a)
O = [0]
for i in O:
for j in adj[i]:
adj[j].remove(i)
O.append(j)
class XORSet:
def __init__(self, el=None):
self.s = set()
self.xor = 0
if el is not None:
self.s.add(el)
def add(self, el: int):
self.s.add(el ^ self.xor)
def update(self, xor: int):
self.xor ^= xor
def __len__(self) -> int:
return len(self.s)
def __iter__(self):
return (x ^ self.xor for x in self.s)
def __contains__(self, el: int) -> bool:
return (el ^ self.xor) in self.s
r = 0
D = [XORSet(a) for a in A]
for i in reversed(O):
for j in adj[i]:
if len(D[j]) > len(D[i]):
D[i], D[j] = D[j], D[i]
D[i].update(A[i])
D[j].update(A[i])
l = list(D[j])
for x in l:
if x in D[i]:
r += 1
D[i].s.clear()
break
else:
for {{completion}}:
D[i].add(x ^ A[i])
continue
break
#print(i, A[i], adj[i], list(D[i]))
print(r)
|
x in l
|
[{"input": "6\n3 2 1 3 2 1\n4 5\n3 4\n1 4\n2 1\n6 1", "output": ["2"]}, {"input": "4\n2 1 1 1\n1 2\n1 3\n1 4", "output": ["0"]}, {"input": "5\n2 2 2 2 2\n1 2\n2 3\n3 4\n4 5", "output": ["2"]}]
|
control_completion_002989
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Leslie and Leon entered a labyrinth. The labyrinth consists of $$$n$$$ halls and $$$m$$$ one-way passages between them. The halls are numbered from $$$1$$$ to $$$n$$$.Leslie and Leon start their journey in the hall $$$s$$$. Right away, they quarrel and decide to explore the labyrinth separately. However, they want to meet again at the end of their journey.To help Leslie and Leon, your task is to find two different paths from the given hall $$$s$$$ to some other hall $$$t$$$, such that these two paths do not share halls other than the staring hall $$$s$$$ and the ending hall $$$t$$$. The hall $$$t$$$ has not been determined yet, so you can choose any of the labyrinth's halls as $$$t$$$ except $$$s$$$.Leslie's and Leon's paths do not have to be the shortest ones, but their paths must be simple, visiting any hall at most once. Also, they cannot visit any common halls except $$$s$$$ and $$$t$$$ during their journey, even at different times.
Input Specification: The first line contains three integers $$$n$$$, $$$m$$$, and $$$s$$$, where $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) is the number of vertices, $$$m$$$ ($$$0 \le m \le 2 \cdot 10^5$$$) is the number of edges in the labyrinth, and $$$s$$$ ($$$1 \le s \le n$$$) is the starting hall. Then $$$m$$$ lines with descriptions of passages follow. Each description contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$; $$$u_i \neq v_i$$$), denoting a passage from the hall $$$u_i$$$ to the hall $$$v_i$$$. The passages are one-way. Each tuple $$$(u_i, v_i)$$$ is present in the input at most once. The labyrinth can contain cycles and is not necessarily connected in any way.
Output Specification: If it is possible to find the desired two paths, output "Possible", otherwise output "Impossible". If the answer exists, output two path descriptions. Each description occupies two lines. The first line of the description contains an integer $$$h$$$ ($$$2 \le h \le n$$$) — the number of halls in a path, and the second line contains distinct integers $$$w_1, w_2, \dots, w_h$$$ ($$$w_1 = s$$$; $$$1 \le w_j \le n$$$; $$$w_h = t$$$) — the halls in the path in the order of passing. Both paths must end at the same vertex $$$t$$$. The paths must be different, and all intermediate halls in these paths must be distinct.
Code:
import collections
def caminho(parent, sala):
resp = []
while sala is not None:
resp.append(sala)
sala = parent[sala]
return list(reversed(resp))
def solve(grafo, total, inicio):
if len(grafo[inicio]) < 2:
return
globalParent = collections.defaultdict(lambda: None)
for sala1 in grafo[inicio]:
currentParent = collections.defaultdict(lambda: None)
currentParent[sala1] = inicio
fila = collections.deque()
fila.append(sala1)
while len(fila) > 0:
y = fila.popleft()
for x in grafo[y]:
if # TODO: Your code here:
currentParent[x] = y
fila.append(x)
for x in currentParent:
if x in globalParent:
#Deu bom
return (caminho(globalParent, x), caminho(currentParent, x))
for x,y in currentParent.items():
globalParent[x] = y
n, m, s = map(int, input().split())
g = collections.defaultdict(list)
for i in range(m):
x,y = map(int, input().split())
g[x].append(y)
paths = solve(g,n,s)
if paths is None:
print("Impossible")
else:
print("Possible")
for i in paths:
print(len(i))
print(" ".join(map(str,i)))
|
import collections
def caminho(parent, sala):
resp = []
while sala is not None:
resp.append(sala)
sala = parent[sala]
return list(reversed(resp))
def solve(grafo, total, inicio):
if len(grafo[inicio]) < 2:
return
globalParent = collections.defaultdict(lambda: None)
for sala1 in grafo[inicio]:
currentParent = collections.defaultdict(lambda: None)
currentParent[sala1] = inicio
fila = collections.deque()
fila.append(sala1)
while len(fila) > 0:
y = fila.popleft()
for x in grafo[y]:
if {{completion}}:
currentParent[x] = y
fila.append(x)
for x in currentParent:
if x in globalParent:
#Deu bom
return (caminho(globalParent, x), caminho(currentParent, x))
for x,y in currentParent.items():
globalParent[x] = y
n, m, s = map(int, input().split())
g = collections.defaultdict(list)
for i in range(m):
x,y = map(int, input().split())
g[x].append(y)
paths = solve(g,n,s)
if paths is None:
print("Impossible")
else:
print("Possible")
for i in paths:
print(len(i))
print(" ".join(map(str,i)))
|
x != inicio and currentParent[x] is None
|
[{"input": "5 5 1\n1 2\n2 3\n1 4\n4 3\n3 5", "output": ["Possible\n3\n1 2 3\n3\n1 4 3"]}, {"input": "5 5 1\n1 2\n2 3\n3 4\n2 5\n5 4", "output": ["Impossible"]}, {"input": "3 3 2\n1 2\n2 3\n3 1", "output": ["Impossible"]}]
|
control_completion_003108
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Leslie and Leon entered a labyrinth. The labyrinth consists of $$$n$$$ halls and $$$m$$$ one-way passages between them. The halls are numbered from $$$1$$$ to $$$n$$$.Leslie and Leon start their journey in the hall $$$s$$$. Right away, they quarrel and decide to explore the labyrinth separately. However, they want to meet again at the end of their journey.To help Leslie and Leon, your task is to find two different paths from the given hall $$$s$$$ to some other hall $$$t$$$, such that these two paths do not share halls other than the staring hall $$$s$$$ and the ending hall $$$t$$$. The hall $$$t$$$ has not been determined yet, so you can choose any of the labyrinth's halls as $$$t$$$ except $$$s$$$.Leslie's and Leon's paths do not have to be the shortest ones, but their paths must be simple, visiting any hall at most once. Also, they cannot visit any common halls except $$$s$$$ and $$$t$$$ during their journey, even at different times.
Input Specification: The first line contains three integers $$$n$$$, $$$m$$$, and $$$s$$$, where $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) is the number of vertices, $$$m$$$ ($$$0 \le m \le 2 \cdot 10^5$$$) is the number of edges in the labyrinth, and $$$s$$$ ($$$1 \le s \le n$$$) is the starting hall. Then $$$m$$$ lines with descriptions of passages follow. Each description contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$; $$$u_i \neq v_i$$$), denoting a passage from the hall $$$u_i$$$ to the hall $$$v_i$$$. The passages are one-way. Each tuple $$$(u_i, v_i)$$$ is present in the input at most once. The labyrinth can contain cycles and is not necessarily connected in any way.
Output Specification: If it is possible to find the desired two paths, output "Possible", otherwise output "Impossible". If the answer exists, output two path descriptions. Each description occupies two lines. The first line of the description contains an integer $$$h$$$ ($$$2 \le h \le n$$$) — the number of halls in a path, and the second line contains distinct integers $$$w_1, w_2, \dots, w_h$$$ ($$$w_1 = s$$$; $$$1 \le w_j \le n$$$; $$$w_h = t$$$) — the halls in the path in the order of passing. Both paths must end at the same vertex $$$t$$$. The paths must be different, and all intermediate halls in these paths must be distinct.
Code:
import collections
def caminho(parent, sala):
resp = []
while sala is not None:
resp.append(sala)
sala = parent[sala]
return list(reversed(resp))
def solve(grafo, total, inicio):
if len(grafo[inicio]) < 2:
return
globalParent = collections.defaultdict(lambda: None)
for sala1 in grafo[inicio]:
currentParent = collections.defaultdict(lambda: None)
currentParent[sala1] = inicio
fila = collections.deque()
fila.append(sala1)
while len(fila) > 0:
y = fila.popleft()
for # TODO: Your code here:
if x != inicio and currentParent[x] is None:
currentParent[x] = y
fila.append(x)
for x in currentParent:
if x in globalParent:
#Deu bom
return (caminho(globalParent, x), caminho(currentParent, x))
for x,y in currentParent.items():
globalParent[x] = y
n, m, s = map(int, input().split())
g = collections.defaultdict(list)
for i in range(m):
x,y = map(int, input().split())
g[x].append(y)
paths = solve(g,n,s)
if paths is None:
print("Impossible")
else:
print("Possible")
for i in paths:
print(len(i))
print(" ".join(map(str,i)))
|
import collections
def caminho(parent, sala):
resp = []
while sala is not None:
resp.append(sala)
sala = parent[sala]
return list(reversed(resp))
def solve(grafo, total, inicio):
if len(grafo[inicio]) < 2:
return
globalParent = collections.defaultdict(lambda: None)
for sala1 in grafo[inicio]:
currentParent = collections.defaultdict(lambda: None)
currentParent[sala1] = inicio
fila = collections.deque()
fila.append(sala1)
while len(fila) > 0:
y = fila.popleft()
for {{completion}}:
if x != inicio and currentParent[x] is None:
currentParent[x] = y
fila.append(x)
for x in currentParent:
if x in globalParent:
#Deu bom
return (caminho(globalParent, x), caminho(currentParent, x))
for x,y in currentParent.items():
globalParent[x] = y
n, m, s = map(int, input().split())
g = collections.defaultdict(list)
for i in range(m):
x,y = map(int, input().split())
g[x].append(y)
paths = solve(g,n,s)
if paths is None:
print("Impossible")
else:
print("Possible")
for i in paths:
print(len(i))
print(" ".join(map(str,i)))
|
x in grafo[y]
|
[{"input": "5 5 1\n1 2\n2 3\n1 4\n4 3\n3 5", "output": ["Possible\n3\n1 2 3\n3\n1 4 3"]}, {"input": "5 5 1\n1 2\n2 3\n3 4\n2 5\n5 4", "output": ["Impossible"]}, {"input": "3 3 2\n1 2\n2 3\n3 1", "output": ["Impossible"]}]
|
control_completion_003109
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Leslie and Leon entered a labyrinth. The labyrinth consists of $$$n$$$ halls and $$$m$$$ one-way passages between them. The halls are numbered from $$$1$$$ to $$$n$$$.Leslie and Leon start their journey in the hall $$$s$$$. Right away, they quarrel and decide to explore the labyrinth separately. However, they want to meet again at the end of their journey.To help Leslie and Leon, your task is to find two different paths from the given hall $$$s$$$ to some other hall $$$t$$$, such that these two paths do not share halls other than the staring hall $$$s$$$ and the ending hall $$$t$$$. The hall $$$t$$$ has not been determined yet, so you can choose any of the labyrinth's halls as $$$t$$$ except $$$s$$$.Leslie's and Leon's paths do not have to be the shortest ones, but their paths must be simple, visiting any hall at most once. Also, they cannot visit any common halls except $$$s$$$ and $$$t$$$ during their journey, even at different times.
Input Specification: The first line contains three integers $$$n$$$, $$$m$$$, and $$$s$$$, where $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) is the number of vertices, $$$m$$$ ($$$0 \le m \le 2 \cdot 10^5$$$) is the number of edges in the labyrinth, and $$$s$$$ ($$$1 \le s \le n$$$) is the starting hall. Then $$$m$$$ lines with descriptions of passages follow. Each description contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$; $$$u_i \neq v_i$$$), denoting a passage from the hall $$$u_i$$$ to the hall $$$v_i$$$. The passages are one-way. Each tuple $$$(u_i, v_i)$$$ is present in the input at most once. The labyrinth can contain cycles and is not necessarily connected in any way.
Output Specification: If it is possible to find the desired two paths, output "Possible", otherwise output "Impossible". If the answer exists, output two path descriptions. Each description occupies two lines. The first line of the description contains an integer $$$h$$$ ($$$2 \le h \le n$$$) — the number of halls in a path, and the second line contains distinct integers $$$w_1, w_2, \dots, w_h$$$ ($$$w_1 = s$$$; $$$1 \le w_j \le n$$$; $$$w_h = t$$$) — the halls in the path in the order of passing. Both paths must end at the same vertex $$$t$$$. The paths must be different, and all intermediate halls in these paths must be distinct.
Code:
n, m, s = [int(x) for x in input().split()]
labyrinth = {x: set() for x in range(1, n+1)}
for _ in range(m):
pt1, pt2 = [int(x) for x in input().split()]
labyrinth[pt1].add(pt2)
d_father = {}
for pt in labyrinth[s]:
d_father[pt] = s
if len(d_father) < 2: print('Impossible')
else:
for pt in labyrinth[s]:
visited = {pt, s}
to_visit = {pt}
while to_visit:
new_visit = set()
for origin in to_visit:
for new_pt in labyrinth[origin]:
if new_pt not in visited:
if new_pt in d_father:
path1 = [new_pt]
path2 = [new_pt, origin]
while # TODO: Your code here:
path1.append(d_father[path1[-1]])
while path2[-1] in d_father:
path2.append(d_father[path2[-1]])
path1.reverse()
path2.reverse()
print('Possible')
print(len(path1))
print(' '.join(str(x) for x in path1))
print(len(path2))
print(' '.join(str(x) for x in path2))
break
else:
d_father[new_pt] = origin
new_visit.add(new_pt)
visited.add(new_pt)
else: continue
break
else:
to_visit = new_visit
continue
break
else: continue
break
else: print('Impossible')
|
n, m, s = [int(x) for x in input().split()]
labyrinth = {x: set() for x in range(1, n+1)}
for _ in range(m):
pt1, pt2 = [int(x) for x in input().split()]
labyrinth[pt1].add(pt2)
d_father = {}
for pt in labyrinth[s]:
d_father[pt] = s
if len(d_father) < 2: print('Impossible')
else:
for pt in labyrinth[s]:
visited = {pt, s}
to_visit = {pt}
while to_visit:
new_visit = set()
for origin in to_visit:
for new_pt in labyrinth[origin]:
if new_pt not in visited:
if new_pt in d_father:
path1 = [new_pt]
path2 = [new_pt, origin]
while {{completion}}:
path1.append(d_father[path1[-1]])
while path2[-1] in d_father:
path2.append(d_father[path2[-1]])
path1.reverse()
path2.reverse()
print('Possible')
print(len(path1))
print(' '.join(str(x) for x in path1))
print(len(path2))
print(' '.join(str(x) for x in path2))
break
else:
d_father[new_pt] = origin
new_visit.add(new_pt)
visited.add(new_pt)
else: continue
break
else:
to_visit = new_visit
continue
break
else: continue
break
else: print('Impossible')
|
path1[-1] in d_father
|
[{"input": "5 5 1\n1 2\n2 3\n1 4\n4 3\n3 5", "output": ["Possible\n3\n1 2 3\n3\n1 4 3"]}, {"input": "5 5 1\n1 2\n2 3\n3 4\n2 5\n5 4", "output": ["Impossible"]}, {"input": "3 3 2\n1 2\n2 3\n3 1", "output": ["Impossible"]}]
|
control_completion_003110
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Leslie and Leon entered a labyrinth. The labyrinth consists of $$$n$$$ halls and $$$m$$$ one-way passages between them. The halls are numbered from $$$1$$$ to $$$n$$$.Leslie and Leon start their journey in the hall $$$s$$$. Right away, they quarrel and decide to explore the labyrinth separately. However, they want to meet again at the end of their journey.To help Leslie and Leon, your task is to find two different paths from the given hall $$$s$$$ to some other hall $$$t$$$, such that these two paths do not share halls other than the staring hall $$$s$$$ and the ending hall $$$t$$$. The hall $$$t$$$ has not been determined yet, so you can choose any of the labyrinth's halls as $$$t$$$ except $$$s$$$.Leslie's and Leon's paths do not have to be the shortest ones, but their paths must be simple, visiting any hall at most once. Also, they cannot visit any common halls except $$$s$$$ and $$$t$$$ during their journey, even at different times.
Input Specification: The first line contains three integers $$$n$$$, $$$m$$$, and $$$s$$$, where $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) is the number of vertices, $$$m$$$ ($$$0 \le m \le 2 \cdot 10^5$$$) is the number of edges in the labyrinth, and $$$s$$$ ($$$1 \le s \le n$$$) is the starting hall. Then $$$m$$$ lines with descriptions of passages follow. Each description contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$; $$$u_i \neq v_i$$$), denoting a passage from the hall $$$u_i$$$ to the hall $$$v_i$$$. The passages are one-way. Each tuple $$$(u_i, v_i)$$$ is present in the input at most once. The labyrinth can contain cycles and is not necessarily connected in any way.
Output Specification: If it is possible to find the desired two paths, output "Possible", otherwise output "Impossible". If the answer exists, output two path descriptions. Each description occupies two lines. The first line of the description contains an integer $$$h$$$ ($$$2 \le h \le n$$$) — the number of halls in a path, and the second line contains distinct integers $$$w_1, w_2, \dots, w_h$$$ ($$$w_1 = s$$$; $$$1 \le w_j \le n$$$; $$$w_h = t$$$) — the halls in the path in the order of passing. Both paths must end at the same vertex $$$t$$$. The paths must be different, and all intermediate halls in these paths must be distinct.
Code:
n, m, s = [int(x) for x in input().split()]
labyrinth = {x: set() for x in range(1, n+1)}
for _ in range(m):
pt1, pt2 = [int(x) for x in input().split()]
labyrinth[pt1].add(pt2)
d_father = {}
for pt in labyrinth[s]:
d_father[pt] = s
if len(d_father) < 2: print('Impossible')
else:
for pt in labyrinth[s]:
visited = {pt, s}
to_visit = {pt}
while to_visit:
new_visit = set()
for origin in to_visit:
for new_pt in labyrinth[origin]:
if new_pt not in visited:
if new_pt in d_father:
path1 = [new_pt]
path2 = [new_pt, origin]
while path1[-1] in d_father:
path1.append(d_father[path1[-1]])
while # TODO: Your code here:
path2.append(d_father[path2[-1]])
path1.reverse()
path2.reverse()
print('Possible')
print(len(path1))
print(' '.join(str(x) for x in path1))
print(len(path2))
print(' '.join(str(x) for x in path2))
break
else:
d_father[new_pt] = origin
new_visit.add(new_pt)
visited.add(new_pt)
else: continue
break
else:
to_visit = new_visit
continue
break
else: continue
break
else: print('Impossible')
|
n, m, s = [int(x) for x in input().split()]
labyrinth = {x: set() for x in range(1, n+1)}
for _ in range(m):
pt1, pt2 = [int(x) for x in input().split()]
labyrinth[pt1].add(pt2)
d_father = {}
for pt in labyrinth[s]:
d_father[pt] = s
if len(d_father) < 2: print('Impossible')
else:
for pt in labyrinth[s]:
visited = {pt, s}
to_visit = {pt}
while to_visit:
new_visit = set()
for origin in to_visit:
for new_pt in labyrinth[origin]:
if new_pt not in visited:
if new_pt in d_father:
path1 = [new_pt]
path2 = [new_pt, origin]
while path1[-1] in d_father:
path1.append(d_father[path1[-1]])
while {{completion}}:
path2.append(d_father[path2[-1]])
path1.reverse()
path2.reverse()
print('Possible')
print(len(path1))
print(' '.join(str(x) for x in path1))
print(len(path2))
print(' '.join(str(x) for x in path2))
break
else:
d_father[new_pt] = origin
new_visit.add(new_pt)
visited.add(new_pt)
else: continue
break
else:
to_visit = new_visit
continue
break
else: continue
break
else: print('Impossible')
|
path2[-1] in d_father
|
[{"input": "5 5 1\n1 2\n2 3\n1 4\n4 3\n3 5", "output": ["Possible\n3\n1 2 3\n3\n1 4 3"]}, {"input": "5 5 1\n1 2\n2 3\n3 4\n2 5\n5 4", "output": ["Impossible"]}, {"input": "3 3 2\n1 2\n2 3\n3 1", "output": ["Impossible"]}]
|
control_completion_003111
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Leslie and Leon entered a labyrinth. The labyrinth consists of $$$n$$$ halls and $$$m$$$ one-way passages between them. The halls are numbered from $$$1$$$ to $$$n$$$.Leslie and Leon start their journey in the hall $$$s$$$. Right away, they quarrel and decide to explore the labyrinth separately. However, they want to meet again at the end of their journey.To help Leslie and Leon, your task is to find two different paths from the given hall $$$s$$$ to some other hall $$$t$$$, such that these two paths do not share halls other than the staring hall $$$s$$$ and the ending hall $$$t$$$. The hall $$$t$$$ has not been determined yet, so you can choose any of the labyrinth's halls as $$$t$$$ except $$$s$$$.Leslie's and Leon's paths do not have to be the shortest ones, but their paths must be simple, visiting any hall at most once. Also, they cannot visit any common halls except $$$s$$$ and $$$t$$$ during their journey, even at different times.
Input Specification: The first line contains three integers $$$n$$$, $$$m$$$, and $$$s$$$, where $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) is the number of vertices, $$$m$$$ ($$$0 \le m \le 2 \cdot 10^5$$$) is the number of edges in the labyrinth, and $$$s$$$ ($$$1 \le s \le n$$$) is the starting hall. Then $$$m$$$ lines with descriptions of passages follow. Each description contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$; $$$u_i \neq v_i$$$), denoting a passage from the hall $$$u_i$$$ to the hall $$$v_i$$$. The passages are one-way. Each tuple $$$(u_i, v_i)$$$ is present in the input at most once. The labyrinth can contain cycles and is not necessarily connected in any way.
Output Specification: If it is possible to find the desired two paths, output "Possible", otherwise output "Impossible". If the answer exists, output two path descriptions. Each description occupies two lines. The first line of the description contains an integer $$$h$$$ ($$$2 \le h \le n$$$) — the number of halls in a path, and the second line contains distinct integers $$$w_1, w_2, \dots, w_h$$$ ($$$w_1 = s$$$; $$$1 \le w_j \le n$$$; $$$w_h = t$$$) — the halls in the path in the order of passing. Both paths must end at the same vertex $$$t$$$. The paths must be different, and all intermediate halls in these paths must be distinct.
Code:
import re
import sys
exit=sys.exit
from bisect import *
from collections import *
ddict=defaultdict
from functools import lru_cache
cache=lru_cache(None)
from heapq import *
from itertools import *
from math import inf
from pprint import pprint as pp
enum=enumerate
rb=lambda:list(map(int,rl()))
rfs=lambda:rln().split()
ri=lambda:int(rln())
ris=lambda:list(map(int,rfs()))
rl=lambda:rln().rstrip('\n')
rln=sys.stdin.readline
cat=''.join
catn='\n'.join
mod=1000000007
d4=[(0,-1),(1,0),(0,1),(-1,0)]
d8=[(-1,-1),(0,-1),(1,-1),(-1,0),(1,0),(-1,1),(0,1),(1,1)]
YES,NO,Yes,No,yes,no='YES','NO','Yes','No','yes','no'
########################################################################
n,m,s=ris()
adj=[[] for _ in range(n+1)]
for _ in range(m):
u,v=ris()
adj[u].append(v)
vis=[{} for _ in range(n+1)]
for i in adj[s]:
stk=[i]
vis[s][i]=0
vis[i][i]=s
while stk:
u=stk.pop()
if 1<len(vis[u]):
print('Possible')
for j in vis[u]:
x,path=u,[]
while # TODO: Your code here:
path.append(x)
x=vis[x][j]
print(len(path))
print(*reversed(path))
exit()
for v in adj[u]:
if i in vis[v] or v==s:
continue
stk.append(v)
vis[v][i]=u
print('Impossible')
|
import re
import sys
exit=sys.exit
from bisect import *
from collections import *
ddict=defaultdict
from functools import lru_cache
cache=lru_cache(None)
from heapq import *
from itertools import *
from math import inf
from pprint import pprint as pp
enum=enumerate
rb=lambda:list(map(int,rl()))
rfs=lambda:rln().split()
ri=lambda:int(rln())
ris=lambda:list(map(int,rfs()))
rl=lambda:rln().rstrip('\n')
rln=sys.stdin.readline
cat=''.join
catn='\n'.join
mod=1000000007
d4=[(0,-1),(1,0),(0,1),(-1,0)]
d8=[(-1,-1),(0,-1),(1,-1),(-1,0),(1,0),(-1,1),(0,1),(1,1)]
YES,NO,Yes,No,yes,no='YES','NO','Yes','No','yes','no'
########################################################################
n,m,s=ris()
adj=[[] for _ in range(n+1)]
for _ in range(m):
u,v=ris()
adj[u].append(v)
vis=[{} for _ in range(n+1)]
for i in adj[s]:
stk=[i]
vis[s][i]=0
vis[i][i]=s
while stk:
u=stk.pop()
if 1<len(vis[u]):
print('Possible')
for j in vis[u]:
x,path=u,[]
while {{completion}}:
path.append(x)
x=vis[x][j]
print(len(path))
print(*reversed(path))
exit()
for v in adj[u]:
if i in vis[v] or v==s:
continue
stk.append(v)
vis[v][i]=u
print('Impossible')
|
j in vis[x]
|
[{"input": "5 5 1\n1 2\n2 3\n1 4\n4 3\n3 5", "output": ["Possible\n3\n1 2 3\n3\n1 4 3"]}, {"input": "5 5 1\n1 2\n2 3\n3 4\n2 5\n5 4", "output": ["Impossible"]}, {"input": "3 3 2\n1 2\n2 3\n3 1", "output": ["Impossible"]}]
|
control_completion_003112
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Leslie and Leon entered a labyrinth. The labyrinth consists of $$$n$$$ halls and $$$m$$$ one-way passages between them. The halls are numbered from $$$1$$$ to $$$n$$$.Leslie and Leon start their journey in the hall $$$s$$$. Right away, they quarrel and decide to explore the labyrinth separately. However, they want to meet again at the end of their journey.To help Leslie and Leon, your task is to find two different paths from the given hall $$$s$$$ to some other hall $$$t$$$, such that these two paths do not share halls other than the staring hall $$$s$$$ and the ending hall $$$t$$$. The hall $$$t$$$ has not been determined yet, so you can choose any of the labyrinth's halls as $$$t$$$ except $$$s$$$.Leslie's and Leon's paths do not have to be the shortest ones, but their paths must be simple, visiting any hall at most once. Also, they cannot visit any common halls except $$$s$$$ and $$$t$$$ during their journey, even at different times.
Input Specification: The first line contains three integers $$$n$$$, $$$m$$$, and $$$s$$$, where $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) is the number of vertices, $$$m$$$ ($$$0 \le m \le 2 \cdot 10^5$$$) is the number of edges in the labyrinth, and $$$s$$$ ($$$1 \le s \le n$$$) is the starting hall. Then $$$m$$$ lines with descriptions of passages follow. Each description contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$; $$$u_i \neq v_i$$$), denoting a passage from the hall $$$u_i$$$ to the hall $$$v_i$$$. The passages are one-way. Each tuple $$$(u_i, v_i)$$$ is present in the input at most once. The labyrinth can contain cycles and is not necessarily connected in any way.
Output Specification: If it is possible to find the desired two paths, output "Possible", otherwise output "Impossible". If the answer exists, output two path descriptions. Each description occupies two lines. The first line of the description contains an integer $$$h$$$ ($$$2 \le h \le n$$$) — the number of halls in a path, and the second line contains distinct integers $$$w_1, w_2, \dots, w_h$$$ ($$$w_1 = s$$$; $$$1 \le w_j \le n$$$; $$$w_h = t$$$) — the halls in the path in the order of passing. Both paths must end at the same vertex $$$t$$$. The paths must be different, and all intermediate halls in these paths must be distinct.
Code:
def DFS(start):
nodes=set()
stack=[start]
while stack:
parent=stack.pop()
if(not visited[parent]):
nodes.add(parent)
visited[parent]=True
for child in graph[parent]:
if (not visited[child]):
stack.append(child)
else:
if # TODO: Your code here:
return child
else:
if parent not in nodes and parent != s:
return parent
return -1
def DFS_get_path(start):
stack=[start]
parent_list[start]=-1
while stack:
parent=stack.pop()
if parent==end:
visited[end]=False
return True
if(not visited[parent]):
visited[parent]=True
for child in graph[parent]:
if (not visited[child]):
stack.append(child)
parent_list[child]=parent
return False
def get_path(node):
path=[]
while node!=-1:
path.append(node)
node=parent_list[node]
path.reverse()
return path
n,m,s=map(int,input().split())
s-=1
graph=[[] for _ in range(n)]
for _ in range(m):
a,b=map(int,input().split())
a-=1
b-=1
graph[a].append(b)
visited=[False]*n
visited[s]=True
for child in graph[s]:
end=DFS(child)
if end!=-1:
visited = [False] * n
parent_list=[-1]*n
visited[s]=True
ans=[]
for child in graph[s]:
if DFS_get_path(child):
ans.append([s]+get_path(end))
if len(ans)==2:
break
print("Possible")
for i in ans:
print(len(i))
print(*[j+1 for j in i])
break
else:
print("Impossible")
# 3 3 1
# 1 2
# 2 1
# 1 3
|
def DFS(start):
nodes=set()
stack=[start]
while stack:
parent=stack.pop()
if(not visited[parent]):
nodes.add(parent)
visited[parent]=True
for child in graph[parent]:
if (not visited[child]):
stack.append(child)
else:
if {{completion}}:
return child
else:
if parent not in nodes and parent != s:
return parent
return -1
def DFS_get_path(start):
stack=[start]
parent_list[start]=-1
while stack:
parent=stack.pop()
if parent==end:
visited[end]=False
return True
if(not visited[parent]):
visited[parent]=True
for child in graph[parent]:
if (not visited[child]):
stack.append(child)
parent_list[child]=parent
return False
def get_path(node):
path=[]
while node!=-1:
path.append(node)
node=parent_list[node]
path.reverse()
return path
n,m,s=map(int,input().split())
s-=1
graph=[[] for _ in range(n)]
for _ in range(m):
a,b=map(int,input().split())
a-=1
b-=1
graph[a].append(b)
visited=[False]*n
visited[s]=True
for child in graph[s]:
end=DFS(child)
if end!=-1:
visited = [False] * n
parent_list=[-1]*n
visited[s]=True
ans=[]
for child in graph[s]:
if DFS_get_path(child):
ans.append([s]+get_path(end))
if len(ans)==2:
break
print("Possible")
for i in ans:
print(len(i))
print(*[j+1 for j in i])
break
else:
print("Impossible")
# 3 3 1
# 1 2
# 2 1
# 1 3
|
child not in nodes and child!=s
|
[{"input": "5 5 1\n1 2\n2 3\n1 4\n4 3\n3 5", "output": ["Possible\n3\n1 2 3\n3\n1 4 3"]}, {"input": "5 5 1\n1 2\n2 3\n3 4\n2 5\n5 4", "output": ["Impossible"]}, {"input": "3 3 2\n1 2\n2 3\n3 1", "output": ["Impossible"]}]
|
control_completion_003113
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Leslie and Leon entered a labyrinth. The labyrinth consists of $$$n$$$ halls and $$$m$$$ one-way passages between them. The halls are numbered from $$$1$$$ to $$$n$$$.Leslie and Leon start their journey in the hall $$$s$$$. Right away, they quarrel and decide to explore the labyrinth separately. However, they want to meet again at the end of their journey.To help Leslie and Leon, your task is to find two different paths from the given hall $$$s$$$ to some other hall $$$t$$$, such that these two paths do not share halls other than the staring hall $$$s$$$ and the ending hall $$$t$$$. The hall $$$t$$$ has not been determined yet, so you can choose any of the labyrinth's halls as $$$t$$$ except $$$s$$$.Leslie's and Leon's paths do not have to be the shortest ones, but their paths must be simple, visiting any hall at most once. Also, they cannot visit any common halls except $$$s$$$ and $$$t$$$ during their journey, even at different times.
Input Specification: The first line contains three integers $$$n$$$, $$$m$$$, and $$$s$$$, where $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) is the number of vertices, $$$m$$$ ($$$0 \le m \le 2 \cdot 10^5$$$) is the number of edges in the labyrinth, and $$$s$$$ ($$$1 \le s \le n$$$) is the starting hall. Then $$$m$$$ lines with descriptions of passages follow. Each description contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$; $$$u_i \neq v_i$$$), denoting a passage from the hall $$$u_i$$$ to the hall $$$v_i$$$. The passages are one-way. Each tuple $$$(u_i, v_i)$$$ is present in the input at most once. The labyrinth can contain cycles and is not necessarily connected in any way.
Output Specification: If it is possible to find the desired two paths, output "Possible", otherwise output "Impossible". If the answer exists, output two path descriptions. Each description occupies two lines. The first line of the description contains an integer $$$h$$$ ($$$2 \le h \le n$$$) — the number of halls in a path, and the second line contains distinct integers $$$w_1, w_2, \dots, w_h$$$ ($$$w_1 = s$$$; $$$1 \le w_j \le n$$$; $$$w_h = t$$$) — the halls in the path in the order of passing. Both paths must end at the same vertex $$$t$$$. The paths must be different, and all intermediate halls in these paths must be distinct.
Code:
def DFS(start):
nodes=set()
stack=[start]
while stack:
parent=stack.pop()
if(not visited[parent]):
nodes.add(parent)
visited[parent]=True
for child in graph[parent]:
if # TODO: Your code here:
stack.append(child)
else:
if child not in nodes and child!=s:
return child
else:
if parent not in nodes and parent != s:
return parent
return -1
def DFS_get_path(start):
stack=[start]
parent_list[start]=-1
while stack:
parent=stack.pop()
if parent==end:
visited[end]=False
return True
if(not visited[parent]):
visited[parent]=True
for child in graph[parent]:
if (not visited[child]):
stack.append(child)
parent_list[child]=parent
return False
def get_path(node):
path=[]
while node!=-1:
path.append(node)
node=parent_list[node]
path.reverse()
return path
n,m,s=map(int,input().split())
s-=1
graph=[[] for _ in range(n)]
for _ in range(m):
a,b=map(int,input().split())
a-=1
b-=1
graph[a].append(b)
visited=[False]*n
visited[s]=True
for child in graph[s]:
end=DFS(child)
if end!=-1:
visited = [False] * n
parent_list=[-1]*n
visited[s]=True
ans=[]
for child in graph[s]:
if DFS_get_path(child):
ans.append([s]+get_path(end))
if len(ans)==2:
break
print("Possible")
for i in ans:
print(len(i))
print(*[j+1 for j in i])
break
else:
print("Impossible")
# 3 3 1
# 1 2
# 2 1
# 1 3
|
def DFS(start):
nodes=set()
stack=[start]
while stack:
parent=stack.pop()
if(not visited[parent]):
nodes.add(parent)
visited[parent]=True
for child in graph[parent]:
if {{completion}}:
stack.append(child)
else:
if child not in nodes and child!=s:
return child
else:
if parent not in nodes and parent != s:
return parent
return -1
def DFS_get_path(start):
stack=[start]
parent_list[start]=-1
while stack:
parent=stack.pop()
if parent==end:
visited[end]=False
return True
if(not visited[parent]):
visited[parent]=True
for child in graph[parent]:
if (not visited[child]):
stack.append(child)
parent_list[child]=parent
return False
def get_path(node):
path=[]
while node!=-1:
path.append(node)
node=parent_list[node]
path.reverse()
return path
n,m,s=map(int,input().split())
s-=1
graph=[[] for _ in range(n)]
for _ in range(m):
a,b=map(int,input().split())
a-=1
b-=1
graph[a].append(b)
visited=[False]*n
visited[s]=True
for child in graph[s]:
end=DFS(child)
if end!=-1:
visited = [False] * n
parent_list=[-1]*n
visited[s]=True
ans=[]
for child in graph[s]:
if DFS_get_path(child):
ans.append([s]+get_path(end))
if len(ans)==2:
break
print("Possible")
for i in ans:
print(len(i))
print(*[j+1 for j in i])
break
else:
print("Impossible")
# 3 3 1
# 1 2
# 2 1
# 1 3
|
(not visited[child])
|
[{"input": "5 5 1\n1 2\n2 3\n1 4\n4 3\n3 5", "output": ["Possible\n3\n1 2 3\n3\n1 4 3"]}, {"input": "5 5 1\n1 2\n2 3\n3 4\n2 5\n5 4", "output": ["Impossible"]}, {"input": "3 3 2\n1 2\n2 3\n3 1", "output": ["Impossible"]}]
|
control_completion_003114
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given three points on a plane. You should choose some segments on the plane that are parallel to coordinate axes, so that all three points become connected. The total length of the chosen segments should be the minimal possible.Two points $$$a$$$ and $$$b$$$ are considered connected if there is a sequence of points $$$p_0 = a, p_1, \ldots, p_k = b$$$ such that points $$$p_i$$$ and $$$p_{i+1}$$$ lie on the same segment.
Input Specification: The input consists of three lines describing three points. Each line contains two integers $$$x$$$ and $$$y$$$ separated by a space — the coordinates of the point ($$$-10^9 \le x, y \le 10^9$$$). The points are pairwise distinct.
Output Specification: On the first line output $$$n$$$ — the number of segments, at most 100. The next $$$n$$$ lines should contain descriptions of segments. Output four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ on a line — the coordinates of the endpoints of the corresponding segment ($$$-10^9 \le x_1, y_1, x_2, y_2 \le 10^9$$$). Each segment should be either horizontal or vertical. It is guaranteed that the solution with the given constraints exists.
Notes: NoteThe points and the segments from the example are shown below.
Code:
l=[[*map(int,input().split())] for i in range(3)]
l=sorted(l,key=lambda x:x[1])
ans=[]
ans.append([l[0][0],l[0][1],l[0][0],l[1][1]])
l[0]=[l[0][0],l[1][1]]
if (l[1][0]<l[0][0] and l[2][0]>l[0][0]) or (l[1][0]>l[0][0] and l[2][0]<l[0][0]):
ans.append([*l[0],l[1][0],l[0][1]])
ans.append([*l[0],l[2][0],l[0][1]])
ans.append([l[2][0],l[0][1],l[2][0],l[2][1]])
else:
if # TODO: Your code here:leng=max(l[1][0],l[2][0])
else:leng=min(l[1][0],l[2][0])
ans.append([*l[0],leng,l[0][1]])
ans.append([l[2][0],l[0][1],l[2][0],l[2][1]])
print(len(ans))
for i in ans:
print(*i)
|
l=[[*map(int,input().split())] for i in range(3)]
l=sorted(l,key=lambda x:x[1])
ans=[]
ans.append([l[0][0],l[0][1],l[0][0],l[1][1]])
l[0]=[l[0][0],l[1][1]]
if (l[1][0]<l[0][0] and l[2][0]>l[0][0]) or (l[1][0]>l[0][0] and l[2][0]<l[0][0]):
ans.append([*l[0],l[1][0],l[0][1]])
ans.append([*l[0],l[2][0],l[0][1]])
ans.append([l[2][0],l[0][1],l[2][0],l[2][1]])
else:
if {{completion}}:leng=max(l[1][0],l[2][0])
else:leng=min(l[1][0],l[2][0])
ans.append([*l[0],leng,l[0][1]])
ans.append([l[2][0],l[0][1],l[2][0],l[2][1]])
print(len(ans))
for i in ans:
print(*i)
|
max(l[1][0],l[2][0])>l[0][0]
|
[{"input": "1 1\n3 5\n8 6", "output": ["3\n1 1 1 5\n1 5 8 5\n8 5 8 6"]}]
|
control_completion_003118
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Julia's $$$n$$$ friends want to organize a startup in a new country they moved to. They assigned each other numbers from 1 to $$$n$$$ according to the jobs they have, from the most front-end tasks to the most back-end ones. They also estimated a matrix $$$c$$$, where $$$c_{ij} = c_{ji}$$$ is the average number of messages per month between people doing jobs $$$i$$$ and $$$j$$$.Now they want to make a hierarchy tree. It will be a binary tree with each node containing one member of the team. Some member will be selected as a leader of the team and will be contained in the root node. In order for the leader to be able to easily reach any subordinate, for each node $$$v$$$ of the tree, the following should apply: all members in its left subtree must have smaller numbers than $$$v$$$, and all members in its right subtree must have larger numbers than $$$v$$$.After the hierarchy tree is settled, people doing jobs $$$i$$$ and $$$j$$$ will be communicating via the shortest path in the tree between their nodes. Let's denote the length of this path as $$$d_{ij}$$$. Thus, the cost of their communication is $$$c_{ij} \cdot d_{ij}$$$.Your task is to find a hierarchy tree that minimizes the total cost of communication over all pairs: $$$\sum_{1 \le i < j \le n} c_{ij} \cdot d_{ij}$$$.
Input Specification: The first line contains an integer $$$n$$$ ($$$1 \le n \le 200$$$) – the number of team members organizing a startup. The next $$$n$$$ lines contain $$$n$$$ integers each, $$$j$$$-th number in $$$i$$$-th line is $$$c_{ij}$$$ — the estimated number of messages per month between team members $$$i$$$ and $$$j$$$ ($$$0 \le c_{ij} \le 10^9; c_{ij} = c_{ji}; c_{ii} = 0$$$).
Output Specification: Output a description of a hierarchy tree that minimizes the total cost of communication. To do so, for each team member from 1 to $$$n$$$ output the number of the member in its parent node, or 0 for the leader. If there are many optimal trees, output a description of any one of them.
Notes: NoteThe minimal possible total cost is $$$566 \cdot 1+239 \cdot 1+30 \cdot 1+1 \cdot 2+1 \cdot 2=839$$$:
Code:
'''
题意：将n个人安排到二叉树中，每两人之间通过最短路径d[i,j]通信，消息数量c[i,j]通过矩阵给出，
要求构造的二叉树满足：任意节点u的左子树中节点序号<u、右子树…>，且所有的节点对之间的c[i,j]*d[i,j]的总和最小。
区间dp，对于f[l,r]，枚举根节点k，得到最大值时记住区间[l,r]的根节点k，
最后先序遍历为每个节点指定根节点。
'''
R,G=lambda:map(int,input().split()),range;n,=R();Fa=[0]*n;N=n+1
A=lambda:[[0]*N for _ in G(N)];S,Rt,f=A(),A(),A()
for i in G(n):
r=[*R()]
for j in G(n):S[i+1][j+1]=r[j]+S[i][j+1]+S[i+1][j]-S[i][j]
rec=lambda i,x:S[x][x]-S[x][i]-S[i][x]+S[i][i]
for c in G(1,N):
for l in G(N-c):
r=l+c;f[l][l+c]=float('inf')
for k in G(l,r):
C=f[l][k]+S[k][n]-S[l][n]-rec(l,k)+f[k+1][r]+S[r][n]-S[k+1][n]-rec(k+1,r)
if # TODO: Your code here:f[l][r]=C;Rt[l][r]=k
def F(l,r,fa):
if l==r:return
mid=Rt[l][r];Fa[mid]=fa+1;F(l,mid,mid);F(mid+1,r,mid)
F(0,n,-1);print(*Fa)
|
'''
题意：将n个人安排到二叉树中，每两人之间通过最短路径d[i,j]通信，消息数量c[i,j]通过矩阵给出，
要求构造的二叉树满足：任意节点u的左子树中节点序号<u、右子树…>，且所有的节点对之间的c[i,j]*d[i,j]的总和最小。
区间dp，对于f[l,r]，枚举根节点k，得到最大值时记住区间[l,r]的根节点k，
最后先序遍历为每个节点指定根节点。
'''
R,G=lambda:map(int,input().split()),range;n,=R();Fa=[0]*n;N=n+1
A=lambda:[[0]*N for _ in G(N)];S,Rt,f=A(),A(),A()
for i in G(n):
r=[*R()]
for j in G(n):S[i+1][j+1]=r[j]+S[i][j+1]+S[i+1][j]-S[i][j]
rec=lambda i,x:S[x][x]-S[x][i]-S[i][x]+S[i][i]
for c in G(1,N):
for l in G(N-c):
r=l+c;f[l][l+c]=float('inf')
for k in G(l,r):
C=f[l][k]+S[k][n]-S[l][n]-rec(l,k)+f[k+1][r]+S[r][n]-S[k+1][n]-rec(k+1,r)
if {{completion}}:f[l][r]=C;Rt[l][r]=k
def F(l,r,fa):
if l==r:return
mid=Rt[l][r];Fa[mid]=fa+1;F(l,mid,mid);F(mid+1,r,mid)
F(0,n,-1);print(*Fa)
|
C<f[l][r]
|
[{"input": "4\n0 566 1 0\n566 0 239 30\n1 239 0 1\n0 30 1 0", "output": ["2 4 2 0"]}]
|
control_completion_003154
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Julia's $$$n$$$ friends want to organize a startup in a new country they moved to. They assigned each other numbers from 1 to $$$n$$$ according to the jobs they have, from the most front-end tasks to the most back-end ones. They also estimated a matrix $$$c$$$, where $$$c_{ij} = c_{ji}$$$ is the average number of messages per month between people doing jobs $$$i$$$ and $$$j$$$.Now they want to make a hierarchy tree. It will be a binary tree with each node containing one member of the team. Some member will be selected as a leader of the team and will be contained in the root node. In order for the leader to be able to easily reach any subordinate, for each node $$$v$$$ of the tree, the following should apply: all members in its left subtree must have smaller numbers than $$$v$$$, and all members in its right subtree must have larger numbers than $$$v$$$.After the hierarchy tree is settled, people doing jobs $$$i$$$ and $$$j$$$ will be communicating via the shortest path in the tree between their nodes. Let's denote the length of this path as $$$d_{ij}$$$. Thus, the cost of their communication is $$$c_{ij} \cdot d_{ij}$$$.Your task is to find a hierarchy tree that minimizes the total cost of communication over all pairs: $$$\sum_{1 \le i < j \le n} c_{ij} \cdot d_{ij}$$$.
Input Specification: The first line contains an integer $$$n$$$ ($$$1 \le n \le 200$$$) – the number of team members organizing a startup. The next $$$n$$$ lines contain $$$n$$$ integers each, $$$j$$$-th number in $$$i$$$-th line is $$$c_{ij}$$$ — the estimated number of messages per month between team members $$$i$$$ and $$$j$$$ ($$$0 \le c_{ij} \le 10^9; c_{ij} = c_{ji}; c_{ii} = 0$$$).
Output Specification: Output a description of a hierarchy tree that minimizes the total cost of communication. To do so, for each team member from 1 to $$$n$$$ output the number of the member in its parent node, or 0 for the leader. If there are many optimal trees, output a description of any one of them.
Notes: NoteThe minimal possible total cost is $$$566 \cdot 1+239 \cdot 1+30 \cdot 1+1 \cdot 2+1 \cdot 2=839$$$:
Code:
n = int(input().strip())
S = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
for j, c in enumerate(map(int, input().strip().split())):
S[i][j] = c
for i in range(n):
for j in range(n):
if i > 0 and j > 0:
S[i][j] += S[i][j-1] + S[i-1][j] - S[i-1][j-1]
elif i > 0:
S[i][j] += S[i-1][j]
elif # TODO: Your code here:
S[i][j] += S[i][j-1]
def acc(i1, i2, j1, j2):
if i1 >= i2 or j1 >= j2:
return 0
a = S[i2-1][j2-1]
b = S[i2-1][j1-1] if j1 > 0 else 0
c = S[i1-1][j2-1] if i1 > 0 else 0
d = S[i1-1][j1-1] if i1 > 0 and j1 > 0 else 0
return a - b - c + d
M = [[-1 for i in range(n)] for j in range(n)]
P = [[-1 for i in range(n)] for j in range(n)]
def solve(b, e):
if e - b == 1:
M[b][e-1] = 0
return 0
if e - b == 0:
return 0
if M[b][e-1] != -1:
return M[b][e-1]
M[b][e-1] = 1e18
for i in range(b, e):
s = solve(b, i) + solve(i+1, e)
s += acc(0, b, b, i) + acc(b, i, i, n) + acc(0, i+1, i+1, e) + acc(i+1, e, e, n)
if s < M[b][e-1]:
M[b][e-1] = s
P[b][e-1] = i
return M[b][e-1]
solve(0, n)
sol = ["" for _ in range(n)]
def label(b, e, p):
if e - b == 1:
sol[b] = str(p)
return
elif e - b == 0:
return
i = P[b][e-1]
sol[i] = str(p)
label(b, i, i+1)
label(i+1, e, i+1)
label(0, n, 0)
print(" ".join(sol))
|
n = int(input().strip())
S = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
for j, c in enumerate(map(int, input().strip().split())):
S[i][j] = c
for i in range(n):
for j in range(n):
if i > 0 and j > 0:
S[i][j] += S[i][j-1] + S[i-1][j] - S[i-1][j-1]
elif i > 0:
S[i][j] += S[i-1][j]
elif {{completion}}:
S[i][j] += S[i][j-1]
def acc(i1, i2, j1, j2):
if i1 >= i2 or j1 >= j2:
return 0
a = S[i2-1][j2-1]
b = S[i2-1][j1-1] if j1 > 0 else 0
c = S[i1-1][j2-1] if i1 > 0 else 0
d = S[i1-1][j1-1] if i1 > 0 and j1 > 0 else 0
return a - b - c + d
M = [[-1 for i in range(n)] for j in range(n)]
P = [[-1 for i in range(n)] for j in range(n)]
def solve(b, e):
if e - b == 1:
M[b][e-1] = 0
return 0
if e - b == 0:
return 0
if M[b][e-1] != -1:
return M[b][e-1]
M[b][e-1] = 1e18
for i in range(b, e):
s = solve(b, i) + solve(i+1, e)
s += acc(0, b, b, i) + acc(b, i, i, n) + acc(0, i+1, i+1, e) + acc(i+1, e, e, n)
if s < M[b][e-1]:
M[b][e-1] = s
P[b][e-1] = i
return M[b][e-1]
solve(0, n)
sol = ["" for _ in range(n)]
def label(b, e, p):
if e - b == 1:
sol[b] = str(p)
return
elif e - b == 0:
return
i = P[b][e-1]
sol[i] = str(p)
label(b, i, i+1)
label(i+1, e, i+1)
label(0, n, 0)
print(" ".join(sol))
|
j > 0
|
[{"input": "4\n0 566 1 0\n566 0 239 30\n1 239 0 1\n0 30 1 0", "output": ["2 4 2 0"]}]
|
control_completion_003155
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Julia's $$$n$$$ friends want to organize a startup in a new country they moved to. They assigned each other numbers from 1 to $$$n$$$ according to the jobs they have, from the most front-end tasks to the most back-end ones. They also estimated a matrix $$$c$$$, where $$$c_{ij} = c_{ji}$$$ is the average number of messages per month between people doing jobs $$$i$$$ and $$$j$$$.Now they want to make a hierarchy tree. It will be a binary tree with each node containing one member of the team. Some member will be selected as a leader of the team and will be contained in the root node. In order for the leader to be able to easily reach any subordinate, for each node $$$v$$$ of the tree, the following should apply: all members in its left subtree must have smaller numbers than $$$v$$$, and all members in its right subtree must have larger numbers than $$$v$$$.After the hierarchy tree is settled, people doing jobs $$$i$$$ and $$$j$$$ will be communicating via the shortest path in the tree between their nodes. Let's denote the length of this path as $$$d_{ij}$$$. Thus, the cost of their communication is $$$c_{ij} \cdot d_{ij}$$$.Your task is to find a hierarchy tree that minimizes the total cost of communication over all pairs: $$$\sum_{1 \le i < j \le n} c_{ij} \cdot d_{ij}$$$.
Input Specification: The first line contains an integer $$$n$$$ ($$$1 \le n \le 200$$$) – the number of team members organizing a startup. The next $$$n$$$ lines contain $$$n$$$ integers each, $$$j$$$-th number in $$$i$$$-th line is $$$c_{ij}$$$ — the estimated number of messages per month between team members $$$i$$$ and $$$j$$$ ($$$0 \le c_{ij} \le 10^9; c_{ij} = c_{ji}; c_{ii} = 0$$$).
Output Specification: Output a description of a hierarchy tree that minimizes the total cost of communication. To do so, for each team member from 1 to $$$n$$$ output the number of the member in its parent node, or 0 for the leader. If there are many optimal trees, output a description of any one of them.
Notes: NoteThe minimal possible total cost is $$$566 \cdot 1+239 \cdot 1+30 \cdot 1+1 \cdot 2+1 \cdot 2=839$$$:
Code:
n=int(input())
c=[]
for _ in range(n):
c.append(tuple(map(int,input().split())))
prefix_sum=[[0]*(n+1) for _ in range(n+1)]
for i in range(1,n+1):
temp=0
for j in range(1,n+1):
temp+=c[i-1][j-1]
prefix_sum[i][j]+=prefix_sum[i-1][j]+temp
def get_rectangel_sum(x1,y1,x2,y2):
return prefix_sum[x2+1][y2+1]-prefix_sum[x1][y2+1]-prefix_sum[x2+1][y1]+prefix_sum[x1][y1]
def cost(x,y):
if x>y:
return 0
a=get_rectangel_sum(x,0,y,x-1) if x!=0 else 0
b=get_rectangel_sum(x,y+1,y,n-1) if y!=n-1 else 0
return a+b
dp=[[float("INF")]*n for _ in range(n)]
best_root_for_range=[[-1]*n for _ in range(n)]
for i in range(n):
dp[i][i]=0
best_root_for_range[i][i]=i
def get_dp_cost(x,y):
return dp[x][y] if x<=y else 0
for length in range(1,n):
# actual length is length+1
for i in range(n-length):
j=i+length
for root in range(i,j+1):
temp=cost(i,root-1)+cost(root+1,j)+get_dp_cost(i,root-1)+get_dp_cost(root+1,j)
if # TODO: Your code here:
dp[i][j]=temp
best_root_for_range[i][j]=root
ans=[-1]*n
def assign_ans(ansecstor,x,y):
if x>y:
return
root=best_root_for_range[x][y]
ans[root]=ansecstor
assign_ans(root,x,root-1)
assign_ans(root,root+1,y)
assign_ans(-1,0,n-1)
print(*[i+1 for i in ans])
# 3
# 0 1 2
# 1 0 3
# 2 3 0
# 4
# 0 1 2 3
# 1 0 5 7
# 2 5 0 4
# 3 7 4 0
# 6
# 0 100 20 37 14 73
# 100 0 17 13 20 2
# 20 17 0 1093 900 1
# 37 13 1093 0 2 4
# 14 20 900 2 0 1
# 73 2 1 4 1 0
|
n=int(input())
c=[]
for _ in range(n):
c.append(tuple(map(int,input().split())))
prefix_sum=[[0]*(n+1) for _ in range(n+1)]
for i in range(1,n+1):
temp=0
for j in range(1,n+1):
temp+=c[i-1][j-1]
prefix_sum[i][j]+=prefix_sum[i-1][j]+temp
def get_rectangel_sum(x1,y1,x2,y2):
return prefix_sum[x2+1][y2+1]-prefix_sum[x1][y2+1]-prefix_sum[x2+1][y1]+prefix_sum[x1][y1]
def cost(x,y):
if x>y:
return 0
a=get_rectangel_sum(x,0,y,x-1) if x!=0 else 0
b=get_rectangel_sum(x,y+1,y,n-1) if y!=n-1 else 0
return a+b
dp=[[float("INF")]*n for _ in range(n)]
best_root_for_range=[[-1]*n for _ in range(n)]
for i in range(n):
dp[i][i]=0
best_root_for_range[i][i]=i
def get_dp_cost(x,y):
return dp[x][y] if x<=y else 0
for length in range(1,n):
# actual length is length+1
for i in range(n-length):
j=i+length
for root in range(i,j+1):
temp=cost(i,root-1)+cost(root+1,j)+get_dp_cost(i,root-1)+get_dp_cost(root+1,j)
if {{completion}}:
dp[i][j]=temp
best_root_for_range[i][j]=root
ans=[-1]*n
def assign_ans(ansecstor,x,y):
if x>y:
return
root=best_root_for_range[x][y]
ans[root]=ansecstor
assign_ans(root,x,root-1)
assign_ans(root,root+1,y)
assign_ans(-1,0,n-1)
print(*[i+1 for i in ans])
# 3
# 0 1 2
# 1 0 3
# 2 3 0
# 4
# 0 1 2 3
# 1 0 5 7
# 2 5 0 4
# 3 7 4 0
# 6
# 0 100 20 37 14 73
# 100 0 17 13 20 2
# 20 17 0 1093 900 1
# 37 13 1093 0 2 4
# 14 20 900 2 0 1
# 73 2 1 4 1 0
|
temp<dp[i][j]
|
[{"input": "4\n0 566 1 0\n566 0 239 30\n1 239 0 1\n0 30 1 0", "output": ["2 4 2 0"]}]
|
control_completion_003156
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Suppose you have an integer $$$v$$$. In one operation, you can: either set $$$v = (v + 1) \bmod 32768$$$ or set $$$v = (2 \cdot v) \bmod 32768$$$. You are given $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. What is the minimum number of operations you need to make each $$$a_i$$$ equal to $$$0$$$?
Input Specification: The first line contains the single integer $$$n$$$ ($$$1 \le n \le 32768$$$) — the number of integers. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i < 32768$$$).
Output Specification: Print $$$n$$$ integers. The $$$i$$$-th integer should be equal to the minimum number of operations required to make $$$a_i$$$ equal to $$$0$$$.
Notes: NoteLet's consider each $$$a_i$$$: $$$a_1 = 19$$$. You can, firstly, increase it by one to get $$$20$$$ and then multiply it by two $$$13$$$ times. You'll get $$$0$$$ in $$$1 + 13 = 14$$$ steps. $$$a_2 = 32764$$$. You can increase it by one $$$4$$$ times: $$$32764 \rightarrow 32765 \rightarrow 32766 \rightarrow 32767 \rightarrow 0$$$. $$$a_3 = 10240$$$. You can multiply it by two $$$4$$$ times: $$$10240 \rightarrow 20480 \rightarrow 8192 \rightarrow 16384 \rightarrow 0$$$. $$$a_4 = 49$$$. You can multiply it by two $$$15$$$ times.
Code:
n,s=open(0)
for # TODO: Your code here:print(min(-int(x)%2**i-i+15for i in range(16)))
|
n,s=open(0)
for {{completion}}:print(min(-int(x)%2**i-i+15for i in range(16)))
|
x in s.split()
|
[{"input": "4\n19 32764 10240 49", "output": ["14 4 4 15"]}]
|
control_completion_003299
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Suppose you have an integer $$$v$$$. In one operation, you can: either set $$$v = (v + 1) \bmod 32768$$$ or set $$$v = (2 \cdot v) \bmod 32768$$$. You are given $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. What is the minimum number of operations you need to make each $$$a_i$$$ equal to $$$0$$$?
Input Specification: The first line contains the single integer $$$n$$$ ($$$1 \le n \le 32768$$$) — the number of integers. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i < 32768$$$).
Output Specification: Print $$$n$$$ integers. The $$$i$$$-th integer should be equal to the minimum number of operations required to make $$$a_i$$$ equal to $$$0$$$.
Notes: NoteLet's consider each $$$a_i$$$: $$$a_1 = 19$$$. You can, firstly, increase it by one to get $$$20$$$ and then multiply it by two $$$13$$$ times. You'll get $$$0$$$ in $$$1 + 13 = 14$$$ steps. $$$a_2 = 32764$$$. You can increase it by one $$$4$$$ times: $$$32764 \rightarrow 32765 \rightarrow 32766 \rightarrow 32767 \rightarrow 0$$$. $$$a_3 = 10240$$$. You can multiply it by two $$$4$$$ times: $$$10240 \rightarrow 20480 \rightarrow 8192 \rightarrow 16384 \rightarrow 0$$$. $$$a_4 = 49$$$. You can multiply it by two $$$15$$$ times.
Code:
n = int(input())
mod = 1 << 15
for x in map(int, input().split()):
res = 16
for a in range(15):
for b in range(15):
if # TODO: Your code here:
res = min(res, a + b)
print(res)
|
n = int(input())
mod = 1 << 15
for x in map(int, input().split()):
res = 16
for a in range(15):
for b in range(15):
if {{completion}}:
res = min(res, a + b)
print(res)
|
(x + a) * (1 << b) % mod == 0
|
[{"input": "4\n19 32764 10240 49", "output": ["14 4 4 15"]}]
|
control_completion_003300
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Suppose you have an integer $$$v$$$. In one operation, you can: either set $$$v = (v + 1) \bmod 32768$$$ or set $$$v = (2 \cdot v) \bmod 32768$$$. You are given $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. What is the minimum number of operations you need to make each $$$a_i$$$ equal to $$$0$$$?
Input Specification: The first line contains the single integer $$$n$$$ ($$$1 \le n \le 32768$$$) — the number of integers. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i < 32768$$$).
Output Specification: Print $$$n$$$ integers. The $$$i$$$-th integer should be equal to the minimum number of operations required to make $$$a_i$$$ equal to $$$0$$$.
Notes: NoteLet's consider each $$$a_i$$$: $$$a_1 = 19$$$. You can, firstly, increase it by one to get $$$20$$$ and then multiply it by two $$$13$$$ times. You'll get $$$0$$$ in $$$1 + 13 = 14$$$ steps. $$$a_2 = 32764$$$. You can increase it by one $$$4$$$ times: $$$32764 \rightarrow 32765 \rightarrow 32766 \rightarrow 32767 \rightarrow 0$$$. $$$a_3 = 10240$$$. You can multiply it by two $$$4$$$ times: $$$10240 \rightarrow 20480 \rightarrow 8192 \rightarrow 16384 \rightarrow 0$$$. $$$a_4 = 49$$$. You can multiply it by two $$$15$$$ times.
Code:
n, s = open(0)
for # TODO: Your code here:
print(min(15-i+-x % 2**i for i in range(16)))
|
n, s = open(0)
for {{completion}}:
print(min(15-i+-x % 2**i for i in range(16)))
|
x in map(int, s.split())
|
[{"input": "4\n19 32764 10240 49", "output": ["14 4 4 15"]}]
|
control_completion_003301
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Suppose you have an integer $$$v$$$. In one operation, you can: either set $$$v = (v + 1) \bmod 32768$$$ or set $$$v = (2 \cdot v) \bmod 32768$$$. You are given $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. What is the minimum number of operations you need to make each $$$a_i$$$ equal to $$$0$$$?
Input Specification: The first line contains the single integer $$$n$$$ ($$$1 \le n \le 32768$$$) — the number of integers. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i < 32768$$$).
Output Specification: Print $$$n$$$ integers. The $$$i$$$-th integer should be equal to the minimum number of operations required to make $$$a_i$$$ equal to $$$0$$$.
Notes: NoteLet's consider each $$$a_i$$$: $$$a_1 = 19$$$. You can, firstly, increase it by one to get $$$20$$$ and then multiply it by two $$$13$$$ times. You'll get $$$0$$$ in $$$1 + 13 = 14$$$ steps. $$$a_2 = 32764$$$. You can increase it by one $$$4$$$ times: $$$32764 \rightarrow 32765 \rightarrow 32766 \rightarrow 32767 \rightarrow 0$$$. $$$a_3 = 10240$$$. You can multiply it by two $$$4$$$ times: $$$10240 \rightarrow 20480 \rightarrow 8192 \rightarrow 16384 \rightarrow 0$$$. $$$a_4 = 49$$$. You can multiply it by two $$$15$$$ times.
Code:
n,s=open(0)
for # TODO: Your code here:print(min(-x%2**i-i+15for i in range(16)))
|
n,s=open(0)
for {{completion}}:print(min(-x%2**i-i+15for i in range(16)))
|
x in map(int,s.split())
|
[{"input": "4\n19 32764 10240 49", "output": ["14 4 4 15"]}]
|
control_completion_003302
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Suppose you have an integer $$$v$$$. In one operation, you can: either set $$$v = (v + 1) \bmod 32768$$$ or set $$$v = (2 \cdot v) \bmod 32768$$$. You are given $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. What is the minimum number of operations you need to make each $$$a_i$$$ equal to $$$0$$$?
Input Specification: The first line contains the single integer $$$n$$$ ($$$1 \le n \le 32768$$$) — the number of integers. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i < 32768$$$).
Output Specification: Print $$$n$$$ integers. The $$$i$$$-th integer should be equal to the minimum number of operations required to make $$$a_i$$$ equal to $$$0$$$.
Notes: NoteLet's consider each $$$a_i$$$: $$$a_1 = 19$$$. You can, firstly, increase it by one to get $$$20$$$ and then multiply it by two $$$13$$$ times. You'll get $$$0$$$ in $$$1 + 13 = 14$$$ steps. $$$a_2 = 32764$$$. You can increase it by one $$$4$$$ times: $$$32764 \rightarrow 32765 \rightarrow 32766 \rightarrow 32767 \rightarrow 0$$$. $$$a_3 = 10240$$$. You can multiply it by two $$$4$$$ times: $$$10240 \rightarrow 20480 \rightarrow 8192 \rightarrow 16384 \rightarrow 0$$$. $$$a_4 = 49$$$. You can multiply it by two $$$15$$$ times.
Code:
n,s=open(0)
for # TODO: Your code here:print(min(15-i+-x%2**i for i in range(16)))
|
n,s=open(0)
for {{completion}}:print(min(15-i+-x%2**i for i in range(16)))
|
x in map(int,s.split())
|
[{"input": "4\n19 32764 10240 49", "output": ["14 4 4 15"]}]
|
control_completion_003303
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Suppose you have an integer $$$v$$$. In one operation, you can: either set $$$v = (v + 1) \bmod 32768$$$ or set $$$v = (2 \cdot v) \bmod 32768$$$. You are given $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. What is the minimum number of operations you need to make each $$$a_i$$$ equal to $$$0$$$?
Input Specification: The first line contains the single integer $$$n$$$ ($$$1 \le n \le 32768$$$) — the number of integers. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i < 32768$$$).
Output Specification: Print $$$n$$$ integers. The $$$i$$$-th integer should be equal to the minimum number of operations required to make $$$a_i$$$ equal to $$$0$$$.
Notes: NoteLet's consider each $$$a_i$$$: $$$a_1 = 19$$$. You can, firstly, increase it by one to get $$$20$$$ and then multiply it by two $$$13$$$ times. You'll get $$$0$$$ in $$$1 + 13 = 14$$$ steps. $$$a_2 = 32764$$$. You can increase it by one $$$4$$$ times: $$$32764 \rightarrow 32765 \rightarrow 32766 \rightarrow 32767 \rightarrow 0$$$. $$$a_3 = 10240$$$. You can multiply it by two $$$4$$$ times: $$$10240 \rightarrow 20480 \rightarrow 8192 \rightarrow 16384 \rightarrow 0$$$. $$$a_4 = 49$$$. You can multiply it by two $$$15$$$ times.
Code:
n,s=open(0)
for # TODO: Your code here:print(min(-int(x)%2**i-i+15for i in range(16)))
|
n,s=open(0)
for {{completion}}:print(min(-int(x)%2**i-i+15for i in range(16)))
|
x in s.split()
|
[{"input": "4\n19 32764 10240 49", "output": ["14 4 4 15"]}]
|
control_completion_003304
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ trees in a park, numbered from $$$1$$$ to $$$n$$$. The initial height of the $$$i$$$-th tree is $$$h_i$$$.You want to water these trees, so they all grow to the same height.The watering process goes as follows. You start watering trees at day $$$1$$$. During the $$$j$$$-th day you can: Choose a tree and water it. If the day is odd (e.g. $$$1, 3, 5, 7, \dots$$$), then the height of the tree increases by $$$1$$$. If the day is even (e.g. $$$2, 4, 6, 8, \dots$$$), then the height of the tree increases by $$$2$$$. Or skip a day without watering any tree. Note that you can't water more than one tree in a day. Your task is to determine the minimum number of days required to water the trees so they grow to the same height.You have to answer $$$t$$$ independent test cases.
Input Specification: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$) — the number of trees. The second line of the test case contains $$$n$$$ integers $$$h_1, h_2, \ldots, h_n$$$ ($$$1 \le h_i \le 10^9$$$), where $$$h_i$$$ is the height of the $$$i$$$-th tree. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$ ($$$\sum n \le 3 \cdot 10^5$$$).
Output Specification: For each test case, print one integer — the minimum number of days required to water the trees, so they grow to the same height.
Notes: NoteConsider the first test case of the example. The initial state of the trees is $$$[1, 2, 4]$$$. During the first day, let's water the first tree, so the sequence of heights becomes $$$[2, 2, 4]$$$; during the second day, let's water the second tree, so the sequence of heights becomes $$$[2, 4, 4]$$$; let's skip the third day; during the fourth day, let's water the first tree, so the sequence of heights becomes $$$[4, 4, 4]$$$. Thus, the answer is $$$4$$$.
Code:
for test in range(int(input())):
n = int(input())
h = [int(i) for i in input().split()]
res = 2 << 69
for req in range(max(h), max(h)+3):
hm = req
d = 0
c = 0
for # TODO: Your code here:
d += req-i
c += (req-i) & 1
res = min(res, max((d//3)*2+d % 3, c*2-1))
print(res)
|
for test in range(int(input())):
n = int(input())
h = [int(i) for i in input().split()]
res = 2 << 69
for req in range(max(h), max(h)+3):
hm = req
d = 0
c = 0
for {{completion}}:
d += req-i
c += (req-i) & 1
res = min(res, max((d//3)*2+d % 3, c*2-1))
print(res)
|
i in h
|
[{"input": "3\n3\n1 2 4\n5\n4 4 3 5 5\n7\n2 5 4 8 3 7 4", "output": ["4\n3\n16"]}]
|
control_completion_003360
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ trees in a park, numbered from $$$1$$$ to $$$n$$$. The initial height of the $$$i$$$-th tree is $$$h_i$$$.You want to water these trees, so they all grow to the same height.The watering process goes as follows. You start watering trees at day $$$1$$$. During the $$$j$$$-th day you can: Choose a tree and water it. If the day is odd (e.g. $$$1, 3, 5, 7, \dots$$$), then the height of the tree increases by $$$1$$$. If the day is even (e.g. $$$2, 4, 6, 8, \dots$$$), then the height of the tree increases by $$$2$$$. Or skip a day without watering any tree. Note that you can't water more than one tree in a day. Your task is to determine the minimum number of days required to water the trees so they grow to the same height.You have to answer $$$t$$$ independent test cases.
Input Specification: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$) — the number of trees. The second line of the test case contains $$$n$$$ integers $$$h_1, h_2, \ldots, h_n$$$ ($$$1 \le h_i \le 10^9$$$), where $$$h_i$$$ is the height of the $$$i$$$-th tree. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$ ($$$\sum n \le 3 \cdot 10^5$$$).
Output Specification: For each test case, print one integer — the minimum number of days required to water the trees, so they grow to the same height.
Notes: NoteConsider the first test case of the example. The initial state of the trees is $$$[1, 2, 4]$$$. During the first day, let's water the first tree, so the sequence of heights becomes $$$[2, 2, 4]$$$; during the second day, let's water the second tree, so the sequence of heights becomes $$$[2, 4, 4]$$$; let's skip the third day; during the fourth day, let's water the first tree, so the sequence of heights becomes $$$[4, 4, 4]$$$. Thus, the answer is $$$4$$$.
Code:
from bisect import bisect
t = int(input().strip())
def solve(o, t):
if # TODO: Your code here:
d = (t - o) // 3 + ((t - o) % 3 == 2)
o, t = o + 2 * d, t - d
return (o + (t > o)) * 2 - (o > t)
out = []
for _ in range(t):
n = int(input().strip())
h = list(map(int, input().strip().split()))
mx = max(h)
o, t, e = 0, 0, 0
for x in h:
e += (mx - x + 1) % 2
o += (mx - x) % 2
t += (mx - x) // 2
out.append(str(min(solve(o, t), solve(e, t + o))))
print("\n".join(out))
|
from bisect import bisect
t = int(input().strip())
def solve(o, t):
if {{completion}}:
d = (t - o) // 3 + ((t - o) % 3 == 2)
o, t = o + 2 * d, t - d
return (o + (t > o)) * 2 - (o > t)
out = []
for _ in range(t):
n = int(input().strip())
h = list(map(int, input().strip().split()))
mx = max(h)
o, t, e = 0, 0, 0
for x in h:
e += (mx - x + 1) % 2
o += (mx - x) % 2
t += (mx - x) // 2
out.append(str(min(solve(o, t), solve(e, t + o))))
print("\n".join(out))
|
t - o > 1
|
[{"input": "3\n3\n1 2 4\n5\n4 4 3 5 5\n7\n2 5 4 8 3 7 4", "output": ["4\n3\n16"]}]
|
control_completion_003361
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ trees in a park, numbered from $$$1$$$ to $$$n$$$. The initial height of the $$$i$$$-th tree is $$$h_i$$$.You want to water these trees, so they all grow to the same height.The watering process goes as follows. You start watering trees at day $$$1$$$. During the $$$j$$$-th day you can: Choose a tree and water it. If the day is odd (e.g. $$$1, 3, 5, 7, \dots$$$), then the height of the tree increases by $$$1$$$. If the day is even (e.g. $$$2, 4, 6, 8, \dots$$$), then the height of the tree increases by $$$2$$$. Or skip a day without watering any tree. Note that you can't water more than one tree in a day. Your task is to determine the minimum number of days required to water the trees so they grow to the same height.You have to answer $$$t$$$ independent test cases.
Input Specification: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$) — the number of trees. The second line of the test case contains $$$n$$$ integers $$$h_1, h_2, \ldots, h_n$$$ ($$$1 \le h_i \le 10^9$$$), where $$$h_i$$$ is the height of the $$$i$$$-th tree. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$ ($$$\sum n \le 3 \cdot 10^5$$$).
Output Specification: For each test case, print one integer — the minimum number of days required to water the trees, so they grow to the same height.
Notes: NoteConsider the first test case of the example. The initial state of the trees is $$$[1, 2, 4]$$$. During the first day, let's water the first tree, so the sequence of heights becomes $$$[2, 2, 4]$$$; during the second day, let's water the second tree, so the sequence of heights becomes $$$[2, 4, 4]$$$; let's skip the third day; during the fourth day, let's water the first tree, so the sequence of heights becomes $$$[4, 4, 4]$$$. Thus, the answer is $$$4$$$.
Code:
from bisect import bisect
t = int(input().strip())
def solve(o, t):
if t - o > 1:
d = (t - o) // 3 + ((t - o) % 3 == 2)
o, t = o + 2 * d, t - d
return (o + (t > o)) * 2 - (o > t)
out = []
for _ in range(t):
n = int(input().strip())
h = list(map(int, input().strip().split()))
mx = max(h)
o, t, e = 0, 0, 0
for # TODO: Your code here:
e += (mx - x + 1) % 2
o += (mx - x) % 2
t += (mx - x) // 2
out.append(str(min(solve(o, t), solve(e, t + o))))
print("\n".join(out))
|
from bisect import bisect
t = int(input().strip())
def solve(o, t):
if t - o > 1:
d = (t - o) // 3 + ((t - o) % 3 == 2)
o, t = o + 2 * d, t - d
return (o + (t > o)) * 2 - (o > t)
out = []
for _ in range(t):
n = int(input().strip())
h = list(map(int, input().strip().split()))
mx = max(h)
o, t, e = 0, 0, 0
for {{completion}}:
e += (mx - x + 1) % 2
o += (mx - x) % 2
t += (mx - x) // 2
out.append(str(min(solve(o, t), solve(e, t + o))))
print("\n".join(out))
|
x in h
|
[{"input": "3\n3\n1 2 4\n5\n4 4 3 5 5\n7\n2 5 4 8 3 7 4", "output": ["4\n3\n16"]}]
|
control_completion_003362
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ trees in a park, numbered from $$$1$$$ to $$$n$$$. The initial height of the $$$i$$$-th tree is $$$h_i$$$.You want to water these trees, so they all grow to the same height.The watering process goes as follows. You start watering trees at day $$$1$$$. During the $$$j$$$-th day you can: Choose a tree and water it. If the day is odd (e.g. $$$1, 3, 5, 7, \dots$$$), then the height of the tree increases by $$$1$$$. If the day is even (e.g. $$$2, 4, 6, 8, \dots$$$), then the height of the tree increases by $$$2$$$. Or skip a day without watering any tree. Note that you can't water more than one tree in a day. Your task is to determine the minimum number of days required to water the trees so they grow to the same height.You have to answer $$$t$$$ independent test cases.
Input Specification: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$) — the number of trees. The second line of the test case contains $$$n$$$ integers $$$h_1, h_2, \ldots, h_n$$$ ($$$1 \le h_i \le 10^9$$$), where $$$h_i$$$ is the height of the $$$i$$$-th tree. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$ ($$$\sum n \le 3 \cdot 10^5$$$).
Output Specification: For each test case, print one integer — the minimum number of days required to water the trees, so they grow to the same height.
Notes: NoteConsider the first test case of the example. The initial state of the trees is $$$[1, 2, 4]$$$. During the first day, let's water the first tree, so the sequence of heights becomes $$$[2, 2, 4]$$$; during the second day, let's water the second tree, so the sequence of heights becomes $$$[2, 4, 4]$$$; let's skip the third day; during the fourth day, let's water the first tree, so the sequence of heights becomes $$$[4, 4, 4]$$$. Thus, the answer is $$$4$$$.
Code:
def solve(target,r):
k=len(r)
ones=twos=0
for # TODO: Your code here:
ones+=(target-r[i])%2
twos+=(target-r[i])//2
if ones>twos:return 2*ones-1
return (ones+twos*2)//3*2+(ones+twos*2)%3
for _ in [0]*int(input()):
input();r=[*map(int,input().split())]
print(min(solve(max(r),r),solve(max(r)+1,r)))
|
def solve(target,r):
k=len(r)
ones=twos=0
for {{completion}}:
ones+=(target-r[i])%2
twos+=(target-r[i])//2
if ones>twos:return 2*ones-1
return (ones+twos*2)//3*2+(ones+twos*2)%3
for _ in [0]*int(input()):
input();r=[*map(int,input().split())]
print(min(solve(max(r),r),solve(max(r)+1,r)))
|
i in range(k)
|
[{"input": "3\n3\n1 2 4\n5\n4 4 3 5 5\n7\n2 5 4 8 3 7 4", "output": ["4\n3\n16"]}]
|
control_completion_003363
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ trees in a park, numbered from $$$1$$$ to $$$n$$$. The initial height of the $$$i$$$-th tree is $$$h_i$$$.You want to water these trees, so they all grow to the same height.The watering process goes as follows. You start watering trees at day $$$1$$$. During the $$$j$$$-th day you can: Choose a tree and water it. If the day is odd (e.g. $$$1, 3, 5, 7, \dots$$$), then the height of the tree increases by $$$1$$$. If the day is even (e.g. $$$2, 4, 6, 8, \dots$$$), then the height of the tree increases by $$$2$$$. Or skip a day without watering any tree. Note that you can't water more than one tree in a day. Your task is to determine the minimum number of days required to water the trees so they grow to the same height.You have to answer $$$t$$$ independent test cases.
Input Specification: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$) — the number of trees. The second line of the test case contains $$$n$$$ integers $$$h_1, h_2, \ldots, h_n$$$ ($$$1 \le h_i \le 10^9$$$), where $$$h_i$$$ is the height of the $$$i$$$-th tree. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$ ($$$\sum n \le 3 \cdot 10^5$$$).
Output Specification: For each test case, print one integer — the minimum number of days required to water the trees, so they grow to the same height.
Notes: NoteConsider the first test case of the example. The initial state of the trees is $$$[1, 2, 4]$$$. During the first day, let's water the first tree, so the sequence of heights becomes $$$[2, 2, 4]$$$; during the second day, let's water the second tree, so the sequence of heights becomes $$$[2, 4, 4]$$$; let's skip the third day; during the fourth day, let's water the first tree, so the sequence of heights becomes $$$[4, 4, 4]$$$. Thus, the answer is $$$4$$$.
Code:
for _ in range(int(input())):
n=int(input())
l=list(map(int,input().strip().split()))
l.sort()
a=l.count(l[-1])
odd,even=0,0
for i in l:
if # TODO: Your code here:
even+=1
else:
odd+=1
su=sum(l[:n-a])
needed=l[-1]*(n-a)-su
if l[-1]%2==0:
p1,p2=odd,even
else:
p1,p2=even,odd
ans=max(2*(needed//3)+needed%3,2*p1-1)
needed+=n
ans2=max(2*(needed//3)+needed%3,2*p2-1)
print(min(ans,ans2))
|
for _ in range(int(input())):
n=int(input())
l=list(map(int,input().strip().split()))
l.sort()
a=l.count(l[-1])
odd,even=0,0
for i in l:
if {{completion}}:
even+=1
else:
odd+=1
su=sum(l[:n-a])
needed=l[-1]*(n-a)-su
if l[-1]%2==0:
p1,p2=odd,even
else:
p1,p2=even,odd
ans=max(2*(needed//3)+needed%3,2*p1-1)
needed+=n
ans2=max(2*(needed//3)+needed%3,2*p2-1)
print(min(ans,ans2))
|
i%2==0
|
[{"input": "3\n3\n1 2 4\n5\n4 4 3 5 5\n7\n2 5 4 8 3 7 4", "output": ["4\n3\n16"]}]
|
control_completion_003364
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ trees in a park, numbered from $$$1$$$ to $$$n$$$. The initial height of the $$$i$$$-th tree is $$$h_i$$$.You want to water these trees, so they all grow to the same height.The watering process goes as follows. You start watering trees at day $$$1$$$. During the $$$j$$$-th day you can: Choose a tree and water it. If the day is odd (e.g. $$$1, 3, 5, 7, \dots$$$), then the height of the tree increases by $$$1$$$. If the day is even (e.g. $$$2, 4, 6, 8, \dots$$$), then the height of the tree increases by $$$2$$$. Or skip a day without watering any tree. Note that you can't water more than one tree in a day. Your task is to determine the minimum number of days required to water the trees so they grow to the same height.You have to answer $$$t$$$ independent test cases.
Input Specification: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$) — the number of trees. The second line of the test case contains $$$n$$$ integers $$$h_1, h_2, \ldots, h_n$$$ ($$$1 \le h_i \le 10^9$$$), where $$$h_i$$$ is the height of the $$$i$$$-th tree. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$ ($$$\sum n \le 3 \cdot 10^5$$$).
Output Specification: For each test case, print one integer — the minimum number of days required to water the trees, so they grow to the same height.
Notes: NoteConsider the first test case of the example. The initial state of the trees is $$$[1, 2, 4]$$$. During the first day, let's water the first tree, so the sequence of heights becomes $$$[2, 2, 4]$$$; during the second day, let's water the second tree, so the sequence of heights becomes $$$[2, 4, 4]$$$; let's skip the third day; during the fourth day, let's water the first tree, so the sequence of heights becomes $$$[4, 4, 4]$$$. Thus, the answer is $$$4$$$.
Code:
def solve(m,a):
ev=od=0
for # TODO: Your code here:
ev += (m-i)//2; od += (m-i)%2
if(od>=ev):
return od*2-(od!=ev)
ev = (ev-od)*2
return od*2 + ev//3*2 + ev%3
I = lambda: map(int,input().split())
t,=I()
for _ in [1]*t:
n, = I()
b = [*I()]
mx = max(b)
print(min(solve(mx,b),solve(mx+1,b)))
|
def solve(m,a):
ev=od=0
for {{completion}}:
ev += (m-i)//2; od += (m-i)%2
if(od>=ev):
return od*2-(od!=ev)
ev = (ev-od)*2
return od*2 + ev//3*2 + ev%3
I = lambda: map(int,input().split())
t,=I()
for _ in [1]*t:
n, = I()
b = [*I()]
mx = max(b)
print(min(solve(mx,b),solve(mx+1,b)))
|
i in a
|
[{"input": "3\n3\n1 2 4\n5\n4 4 3 5 5\n7\n2 5 4 8 3 7 4", "output": ["4\n3\n16"]}]
|
control_completion_003365
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ trees in a park, numbered from $$$1$$$ to $$$n$$$. The initial height of the $$$i$$$-th tree is $$$h_i$$$.You want to water these trees, so they all grow to the same height.The watering process goes as follows. You start watering trees at day $$$1$$$. During the $$$j$$$-th day you can: Choose a tree and water it. If the day is odd (e.g. $$$1, 3, 5, 7, \dots$$$), then the height of the tree increases by $$$1$$$. If the day is even (e.g. $$$2, 4, 6, 8, \dots$$$), then the height of the tree increases by $$$2$$$. Or skip a day without watering any tree. Note that you can't water more than one tree in a day. Your task is to determine the minimum number of days required to water the trees so they grow to the same height.You have to answer $$$t$$$ independent test cases.
Input Specification: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$) — the number of trees. The second line of the test case contains $$$n$$$ integers $$$h_1, h_2, \ldots, h_n$$$ ($$$1 \le h_i \le 10^9$$$), where $$$h_i$$$ is the height of the $$$i$$$-th tree. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$ ($$$\sum n \le 3 \cdot 10^5$$$).
Output Specification: For each test case, print one integer — the minimum number of days required to water the trees, so they grow to the same height.
Notes: NoteConsider the first test case of the example. The initial state of the trees is $$$[1, 2, 4]$$$. During the first day, let's water the first tree, so the sequence of heights becomes $$$[2, 2, 4]$$$; during the second day, let's water the second tree, so the sequence of heights becomes $$$[2, 4, 4]$$$; let's skip the third day; during the fourth day, let's water the first tree, so the sequence of heights becomes $$$[4, 4, 4]$$$. Thus, the answer is $$$4$$$.
Code:
# by the authority of GOD author: Kritarth Sharma #
import sys
from collections import defaultdict,deque,Counter
from bisect import bisect_left
import math
input=sys.stdin.readline
def inp():
l=list(map(int,input().split()))
return l
for _ in range(int(input())):
n,=inp()
l=inp()
m=max(l)
a=float('inf')
for i in range(m,m+2):
x=0
y=0
for # TODO: Your code here:
x+=(i-l[j])//2
y+=(i-l[j])%2
a=min(a,max( 2*y -1, ((2*x+y)//3)*2 +((2*x+y)%3) ))
print(a)
|
# by the authority of GOD author: Kritarth Sharma #
import sys
from collections import defaultdict,deque,Counter
from bisect import bisect_left
import math
input=sys.stdin.readline
def inp():
l=list(map(int,input().split()))
return l
for _ in range(int(input())):
n,=inp()
l=inp()
m=max(l)
a=float('inf')
for i in range(m,m+2):
x=0
y=0
for {{completion}}:
x+=(i-l[j])//2
y+=(i-l[j])%2
a=min(a,max( 2*y -1, ((2*x+y)//3)*2 +((2*x+y)%3) ))
print(a)
|
j in range(n)
|
[{"input": "3\n3\n1 2 4\n5\n4 4 3 5 5\n7\n2 5 4 8 3 7 4", "output": ["4\n3\n16"]}]
|
control_completion_003366
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ trees in a park, numbered from $$$1$$$ to $$$n$$$. The initial height of the $$$i$$$-th tree is $$$h_i$$$.You want to water these trees, so they all grow to the same height.The watering process goes as follows. You start watering trees at day $$$1$$$. During the $$$j$$$-th day you can: Choose a tree and water it. If the day is odd (e.g. $$$1, 3, 5, 7, \dots$$$), then the height of the tree increases by $$$1$$$. If the day is even (e.g. $$$2, 4, 6, 8, \dots$$$), then the height of the tree increases by $$$2$$$. Or skip a day without watering any tree. Note that you can't water more than one tree in a day. Your task is to determine the minimum number of days required to water the trees so they grow to the same height.You have to answer $$$t$$$ independent test cases.
Input Specification: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$) — the number of trees. The second line of the test case contains $$$n$$$ integers $$$h_1, h_2, \ldots, h_n$$$ ($$$1 \le h_i \le 10^9$$$), where $$$h_i$$$ is the height of the $$$i$$$-th tree. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$ ($$$\sum n \le 3 \cdot 10^5$$$).
Output Specification: For each test case, print one integer — the minimum number of days required to water the trees, so they grow to the same height.
Notes: NoteConsider the first test case of the example. The initial state of the trees is $$$[1, 2, 4]$$$. During the first day, let's water the first tree, so the sequence of heights becomes $$$[2, 2, 4]$$$; during the second day, let's water the second tree, so the sequence of heights becomes $$$[2, 4, 4]$$$; let's skip the third day; during the fourth day, let's water the first tree, so the sequence of heights becomes $$$[4, 4, 4]$$$. Thus, the answer is $$$4$$$.
Code:
inp = lambda : list(map(int,input().split()))
for _ in range(int(input())):
n = int(input())
t = inp()
c = 0
m = max(t)
def mm(m):
n1 = n2 = 0
tot =0
for # TODO: Your code here:
n1+= (m-i)%2
n2+= (m-i)//2
return (n1*2-1) if n2<n1 else ((n2*2+n1)//3*2+(n2*2+n1)%3)
print(min(mm(m),mm(m+1)))
|
inp = lambda : list(map(int,input().split()))
for _ in range(int(input())):
n = int(input())
t = inp()
c = 0
m = max(t)
def mm(m):
n1 = n2 = 0
tot =0
for {{completion}}:
n1+= (m-i)%2
n2+= (m-i)//2
return (n1*2-1) if n2<n1 else ((n2*2+n1)//3*2+(n2*2+n1)%3)
print(min(mm(m),mm(m+1)))
|
i in t
|
[{"input": "3\n3\n1 2 4\n5\n4 4 3 5 5\n7\n2 5 4 8 3 7 4", "output": ["4\n3\n16"]}]
|
control_completion_003367
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n$$$ trees in a park, numbered from $$$1$$$ to $$$n$$$. The initial height of the $$$i$$$-th tree is $$$h_i$$$.You want to water these trees, so they all grow to the same height.The watering process goes as follows. You start watering trees at day $$$1$$$. During the $$$j$$$-th day you can: Choose a tree and water it. If the day is odd (e.g. $$$1, 3, 5, 7, \dots$$$), then the height of the tree increases by $$$1$$$. If the day is even (e.g. $$$2, 4, 6, 8, \dots$$$), then the height of the tree increases by $$$2$$$. Or skip a day without watering any tree. Note that you can't water more than one tree in a day. Your task is to determine the minimum number of days required to water the trees so they grow to the same height.You have to answer $$$t$$$ independent test cases.
Input Specification: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$) — the number of trees. The second line of the test case contains $$$n$$$ integers $$$h_1, h_2, \ldots, h_n$$$ ($$$1 \le h_i \le 10^9$$$), where $$$h_i$$$ is the height of the $$$i$$$-th tree. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$ ($$$\sum n \le 3 \cdot 10^5$$$).
Output Specification: For each test case, print one integer — the minimum number of days required to water the trees, so they grow to the same height.
Notes: NoteConsider the first test case of the example. The initial state of the trees is $$$[1, 2, 4]$$$. During the first day, let's water the first tree, so the sequence of heights becomes $$$[2, 2, 4]$$$; during the second day, let's water the second tree, so the sequence of heights becomes $$$[2, 4, 4]$$$; let's skip the third day; during the fourth day, let's water the first tree, so the sequence of heights becomes $$$[4, 4, 4]$$$. Thus, the answer is $$$4$$$.
Code:
for ii in range(int(input())):
n=int(input())
a = list(map(int, input().split()))
m=max(a)
ans=float("inf")
for jj in range(m,m+4):
x,y=0,0
for # TODO: Your code here:
x+=(jj-kk)%2
y+=(jj-kk)//2
ans=min(max(((x+y*2)//3*2)+(x+y*2)%3,x*2-1),ans)
print(ans)
|
for ii in range(int(input())):
n=int(input())
a = list(map(int, input().split()))
m=max(a)
ans=float("inf")
for jj in range(m,m+4):
x,y=0,0
for {{completion}}:
x+=(jj-kk)%2
y+=(jj-kk)//2
ans=min(max(((x+y*2)//3*2)+(x+y*2)%3,x*2-1),ans)
print(ans)
|
kk in a
|
[{"input": "3\n3\n1 2 4\n5\n4 4 3 5 5\n7\n2 5 4 8 3 7 4", "output": ["4\n3\n16"]}]
|
control_completion_003368
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given two arrays: an array $$$a$$$ consisting of $$$n$$$ zeros and an array $$$b$$$ consisting of $$$n$$$ integers.You can apply the following operation to the array $$$a$$$ an arbitrary number of times: choose some subsegment of $$$a$$$ of length $$$k$$$ and add the arithmetic progression $$$1, 2, \ldots, k$$$ to this subsegment — i. e. add $$$1$$$ to the first element of the subsegment, $$$2$$$ to the second element, and so on. The chosen subsegment should be inside the borders of the array $$$a$$$ (i.e., if the left border of the chosen subsegment is $$$l$$$, then the condition $$$1 \le l \le l + k - 1 \le n$$$ should be satisfied). Note that the progression added is always $$$1, 2, \ldots, k$$$ but not the $$$k, k - 1, \ldots, 1$$$ or anything else (i.e., the leftmost element of the subsegment always increases by $$$1$$$, the second element always increases by $$$2$$$ and so on).Your task is to find the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Note that the condition $$$a_i \ge b_i$$$ should be satisfied for all elements at once.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 3 \cdot 10^5$$$) — the number of elements in both arrays and the length of the subsegment, respectively. The second line of the input contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i \le 10^{12}$$$), where $$$b_i$$$ is the $$$i$$$-th element of the array $$$b$$$.
Output Specification: Print one integer — the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$.
Notes: NoteConsider the first example. In this test, we don't really have any choice, so we need to add at least five progressions to make the first element equals $$$5$$$. The array $$$a$$$ becomes $$$[5, 10, 15]$$$.Consider the second example. In this test, let's add one progression on the segment $$$[1; 3]$$$ and two progressions on the segment $$$[4; 6]$$$. Then, the array $$$a$$$ becomes $$$[1, 2, 3, 2, 4, 6]$$$.
Code:
n,k = map(int, input().split())
bb = list(map(int, input().split()))
ans = 0
sofar = 0
sumprog = 0
timeq = []
for ib,b in enumerate(bb[::-1]):
kk = min(k, n-ib)
time = (max(0,b-sofar)+kk-1)//kk
ans += time
timeq.append(time)
sumprog += time
if # TODO: Your code here:
sumprog -= timeq[ib-k]
sofar += kk*time
sofar -= sumprog
# print(time, sofar, timeq, sumprog)
print(ans)
|
n,k = map(int, input().split())
bb = list(map(int, input().split()))
ans = 0
sofar = 0
sumprog = 0
timeq = []
for ib,b in enumerate(bb[::-1]):
kk = min(k, n-ib)
time = (max(0,b-sofar)+kk-1)//kk
ans += time
timeq.append(time)
sumprog += time
if {{completion}}:
sumprog -= timeq[ib-k]
sofar += kk*time
sofar -= sumprog
# print(time, sofar, timeq, sumprog)
print(ans)
|
ib >= k
|
[{"input": "3 3\n5 4 6", "output": ["5"]}, {"input": "6 3\n1 2 3 2 2 3", "output": ["3"]}, {"input": "6 3\n1 2 4 1 2 3", "output": ["3"]}, {"input": "7 3\n50 17 81 25 42 39 96", "output": ["92"]}]
|
control_completion_003385
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given two arrays: an array $$$a$$$ consisting of $$$n$$$ zeros and an array $$$b$$$ consisting of $$$n$$$ integers.You can apply the following operation to the array $$$a$$$ an arbitrary number of times: choose some subsegment of $$$a$$$ of length $$$k$$$ and add the arithmetic progression $$$1, 2, \ldots, k$$$ to this subsegment — i. e. add $$$1$$$ to the first element of the subsegment, $$$2$$$ to the second element, and so on. The chosen subsegment should be inside the borders of the array $$$a$$$ (i.e., if the left border of the chosen subsegment is $$$l$$$, then the condition $$$1 \le l \le l + k - 1 \le n$$$ should be satisfied). Note that the progression added is always $$$1, 2, \ldots, k$$$ but not the $$$k, k - 1, \ldots, 1$$$ or anything else (i.e., the leftmost element of the subsegment always increases by $$$1$$$, the second element always increases by $$$2$$$ and so on).Your task is to find the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Note that the condition $$$a_i \ge b_i$$$ should be satisfied for all elements at once.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 3 \cdot 10^5$$$) — the number of elements in both arrays and the length of the subsegment, respectively. The second line of the input contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i \le 10^{12}$$$), where $$$b_i$$$ is the $$$i$$$-th element of the array $$$b$$$.
Output Specification: Print one integer — the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$.
Notes: NoteConsider the first example. In this test, we don't really have any choice, so we need to add at least five progressions to make the first element equals $$$5$$$. The array $$$a$$$ becomes $$$[5, 10, 15]$$$.Consider the second example. In this test, let's add one progression on the segment $$$[1; 3]$$$ and two progressions on the segment $$$[4; 6]$$$. Then, the array $$$a$$$ becomes $$$[1, 2, 3, 2, 4, 6]$$$.
Code:
I = lambda: [int(x) for x in input().split()]
n, k = I()
B, d = I() + [0]*k, [0] * (n + k)
s = total = 0
for i in range(n-1, -1, -1):
B[i] -= total
if # TODO: Your code here:
dd = min(k, i + 1)
d[i] = (B[i] + dd - 1)//dd
s += d[i] - d[i + k]
total += d[i] * dd - s
print(sum(d))
|
I = lambda: [int(x) for x in input().split()]
n, k = I()
B, d = I() + [0]*k, [0] * (n + k)
s = total = 0
for i in range(n-1, -1, -1):
B[i] -= total
if {{completion}}:
dd = min(k, i + 1)
d[i] = (B[i] + dd - 1)//dd
s += d[i] - d[i + k]
total += d[i] * dd - s
print(sum(d))
|
B[i] > 0
|
[{"input": "3 3\n5 4 6", "output": ["5"]}, {"input": "6 3\n1 2 3 2 2 3", "output": ["3"]}, {"input": "6 3\n1 2 4 1 2 3", "output": ["3"]}, {"input": "7 3\n50 17 81 25 42 39 96", "output": ["92"]}]
|
control_completion_003386
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given two arrays: an array $$$a$$$ consisting of $$$n$$$ zeros and an array $$$b$$$ consisting of $$$n$$$ integers.You can apply the following operation to the array $$$a$$$ an arbitrary number of times: choose some subsegment of $$$a$$$ of length $$$k$$$ and add the arithmetic progression $$$1, 2, \ldots, k$$$ to this subsegment — i. e. add $$$1$$$ to the first element of the subsegment, $$$2$$$ to the second element, and so on. The chosen subsegment should be inside the borders of the array $$$a$$$ (i.e., if the left border of the chosen subsegment is $$$l$$$, then the condition $$$1 \le l \le l + k - 1 \le n$$$ should be satisfied). Note that the progression added is always $$$1, 2, \ldots, k$$$ but not the $$$k, k - 1, \ldots, 1$$$ or anything else (i.e., the leftmost element of the subsegment always increases by $$$1$$$, the second element always increases by $$$2$$$ and so on).Your task is to find the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Note that the condition $$$a_i \ge b_i$$$ should be satisfied for all elements at once.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 3 \cdot 10^5$$$) — the number of elements in both arrays and the length of the subsegment, respectively. The second line of the input contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i \le 10^{12}$$$), where $$$b_i$$$ is the $$$i$$$-th element of the array $$$b$$$.
Output Specification: Print one integer — the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$.
Notes: NoteConsider the first example. In this test, we don't really have any choice, so we need to add at least five progressions to make the first element equals $$$5$$$. The array $$$a$$$ becomes $$$[5, 10, 15]$$$.Consider the second example. In this test, let's add one progression on the segment $$$[1; 3]$$$ and two progressions on the segment $$$[4; 6]$$$. Then, the array $$$a$$$ becomes $$$[1, 2, 3, 2, 4, 6]$$$.
Code:
#!/usr/bin/env PyPy3
from collections import Counter, defaultdict, deque
import itertools
import re
import math
from functools import reduce
import operator
import bisect
import heapq
import functools
mod=10**9+7
import sys
input=sys.stdin.readline
n,k=map(int,input().split())
b=list(map(int,input().split()))
ans = 0
dec = 0
cnt = [0] * n
tmp = 0
for i in range(k-1,n)[::-1]:
tmp -= cnt[i]
dec -= tmp
#print(tmp,dec)
if b[i] > dec:
#print(b[i]-dec)
x = -(-(b[i]-dec) // k)
ans += x
if # TODO: Your code here:
cnt[i-k-1] = x
dec += x * k
tmp += x
#print(ans)
#tmp -= cnt[i]
#print(cnt)
ma = 0
for i in range(k-1)[::-1]:
tmp -= cnt[i]
dec -= tmp
ma = max(ma,-(-(b[i]-dec) // (i+1)))
print(ans+ma)
|
#!/usr/bin/env PyPy3
from collections import Counter, defaultdict, deque
import itertools
import re
import math
from functools import reduce
import operator
import bisect
import heapq
import functools
mod=10**9+7
import sys
input=sys.stdin.readline
n,k=map(int,input().split())
b=list(map(int,input().split()))
ans = 0
dec = 0
cnt = [0] * n
tmp = 0
for i in range(k-1,n)[::-1]:
tmp -= cnt[i]
dec -= tmp
#print(tmp,dec)
if b[i] > dec:
#print(b[i]-dec)
x = -(-(b[i]-dec) // k)
ans += x
if {{completion}}:
cnt[i-k-1] = x
dec += x * k
tmp += x
#print(ans)
#tmp -= cnt[i]
#print(cnt)
ma = 0
for i in range(k-1)[::-1]:
tmp -= cnt[i]
dec -= tmp
ma = max(ma,-(-(b[i]-dec) // (i+1)))
print(ans+ma)
|
i - k - 1 >= 0
|
[{"input": "3 3\n5 4 6", "output": ["5"]}, {"input": "6 3\n1 2 3 2 2 3", "output": ["3"]}, {"input": "6 3\n1 2 4 1 2 3", "output": ["3"]}, {"input": "7 3\n50 17 81 25 42 39 96", "output": ["92"]}]
|
control_completion_003387
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given two arrays: an array $$$a$$$ consisting of $$$n$$$ zeros and an array $$$b$$$ consisting of $$$n$$$ integers.You can apply the following operation to the array $$$a$$$ an arbitrary number of times: choose some subsegment of $$$a$$$ of length $$$k$$$ and add the arithmetic progression $$$1, 2, \ldots, k$$$ to this subsegment — i. e. add $$$1$$$ to the first element of the subsegment, $$$2$$$ to the second element, and so on. The chosen subsegment should be inside the borders of the array $$$a$$$ (i.e., if the left border of the chosen subsegment is $$$l$$$, then the condition $$$1 \le l \le l + k - 1 \le n$$$ should be satisfied). Note that the progression added is always $$$1, 2, \ldots, k$$$ but not the $$$k, k - 1, \ldots, 1$$$ or anything else (i.e., the leftmost element of the subsegment always increases by $$$1$$$, the second element always increases by $$$2$$$ and so on).Your task is to find the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Note that the condition $$$a_i \ge b_i$$$ should be satisfied for all elements at once.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 3 \cdot 10^5$$$) — the number of elements in both arrays and the length of the subsegment, respectively. The second line of the input contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i \le 10^{12}$$$), where $$$b_i$$$ is the $$$i$$$-th element of the array $$$b$$$.
Output Specification: Print one integer — the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$.
Notes: NoteConsider the first example. In this test, we don't really have any choice, so we need to add at least five progressions to make the first element equals $$$5$$$. The array $$$a$$$ becomes $$$[5, 10, 15]$$$.Consider the second example. In this test, let's add one progression on the segment $$$[1; 3]$$$ and two progressions on the segment $$$[4; 6]$$$. Then, the array $$$a$$$ becomes $$$[1, 2, 3, 2, 4, 6]$$$.
Code:
n, k = [int(i) for i in input().split()]
a = [int(i) for i in input().split()]
dd = [0]*(len(a)+5)
add = 0
diff = 0
moves = 0
for key, i in reversed([*enumerate(a)]):
add += diff
i += add
diff += dd[-1]
dd.pop()
if # TODO: Your code here:
K = min(k, key+1)
dd[-K] -= (i+K-1)//K
diff += (i+K-1)//K
moves += (i+K-1)//K
add -= K*((i+K-1)//K)
print(moves)
|
n, k = [int(i) for i in input().split()]
a = [int(i) for i in input().split()]
dd = [0]*(len(a)+5)
add = 0
diff = 0
moves = 0
for key, i in reversed([*enumerate(a)]):
add += diff
i += add
diff += dd[-1]
dd.pop()
if {{completion}}:
K = min(k, key+1)
dd[-K] -= (i+K-1)//K
diff += (i+K-1)//K
moves += (i+K-1)//K
add -= K*((i+K-1)//K)
print(moves)
|
i > 0
|
[{"input": "3 3\n5 4 6", "output": ["5"]}, {"input": "6 3\n1 2 3 2 2 3", "output": ["3"]}, {"input": "6 3\n1 2 4 1 2 3", "output": ["3"]}, {"input": "7 3\n50 17 81 25 42 39 96", "output": ["92"]}]
|
control_completion_003388
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given two arrays: an array $$$a$$$ consisting of $$$n$$$ zeros and an array $$$b$$$ consisting of $$$n$$$ integers.You can apply the following operation to the array $$$a$$$ an arbitrary number of times: choose some subsegment of $$$a$$$ of length $$$k$$$ and add the arithmetic progression $$$1, 2, \ldots, k$$$ to this subsegment — i. e. add $$$1$$$ to the first element of the subsegment, $$$2$$$ to the second element, and so on. The chosen subsegment should be inside the borders of the array $$$a$$$ (i.e., if the left border of the chosen subsegment is $$$l$$$, then the condition $$$1 \le l \le l + k - 1 \le n$$$ should be satisfied). Note that the progression added is always $$$1, 2, \ldots, k$$$ but not the $$$k, k - 1, \ldots, 1$$$ or anything else (i.e., the leftmost element of the subsegment always increases by $$$1$$$, the second element always increases by $$$2$$$ and so on).Your task is to find the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Note that the condition $$$a_i \ge b_i$$$ should be satisfied for all elements at once.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 3 \cdot 10^5$$$) — the number of elements in both arrays and the length of the subsegment, respectively. The second line of the input contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i \le 10^{12}$$$), where $$$b_i$$$ is the $$$i$$$-th element of the array $$$b$$$.
Output Specification: Print one integer — the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$.
Notes: NoteConsider the first example. In this test, we don't really have any choice, so we need to add at least five progressions to make the first element equals $$$5$$$. The array $$$a$$$ becomes $$$[5, 10, 15]$$$.Consider the second example. In this test, let's add one progression on the segment $$$[1; 3]$$$ and two progressions on the segment $$$[4; 6]$$$. Then, the array $$$a$$$ becomes $$$[1, 2, 3, 2, 4, 6]$$$.
Code:
"""
take element as "the tail" will use more less operations,
use variables s,cnt and closed,
to avoid the inner iteration(update neighbor k elements every time),
complexity is O(n).
"""
row=lambda:map(int,input().split())
n,k=row()
a=list(row())
closed=[0]*n
s=cnt=res=0
for i in range(n-1,-1,-1):
# print(i,a[i],'s,cnt:',s,cnt,'closed:',closed)
s-=cnt
cnt-=closed[i]
a[i]-=s
if # TODO: Your code here:
continue
th=min(i+1,k)
need=(a[i]+th-1)//th#equals ceil()
s+=need*th
cnt+=need
res+=need
if i>=th:
closed[i-th]+=need
print(res)
|
"""
take element as "the tail" will use more less operations,
use variables s,cnt and closed,
to avoid the inner iteration(update neighbor k elements every time),
complexity is O(n).
"""
row=lambda:map(int,input().split())
n,k=row()
a=list(row())
closed=[0]*n
s=cnt=res=0
for i in range(n-1,-1,-1):
# print(i,a[i],'s,cnt:',s,cnt,'closed:',closed)
s-=cnt
cnt-=closed[i]
a[i]-=s
if {{completion}}:
continue
th=min(i+1,k)
need=(a[i]+th-1)//th#equals ceil()
s+=need*th
cnt+=need
res+=need
if i>=th:
closed[i-th]+=need
print(res)
|
a[i]<=0
|
[{"input": "3 3\n5 4 6", "output": ["5"]}, {"input": "6 3\n1 2 3 2 2 3", "output": ["3"]}, {"input": "6 3\n1 2 4 1 2 3", "output": ["3"]}, {"input": "7 3\n50 17 81 25 42 39 96", "output": ["92"]}]
|
control_completion_003389
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given two arrays: an array $$$a$$$ consisting of $$$n$$$ zeros and an array $$$b$$$ consisting of $$$n$$$ integers.You can apply the following operation to the array $$$a$$$ an arbitrary number of times: choose some subsegment of $$$a$$$ of length $$$k$$$ and add the arithmetic progression $$$1, 2, \ldots, k$$$ to this subsegment — i. e. add $$$1$$$ to the first element of the subsegment, $$$2$$$ to the second element, and so on. The chosen subsegment should be inside the borders of the array $$$a$$$ (i.e., if the left border of the chosen subsegment is $$$l$$$, then the condition $$$1 \le l \le l + k - 1 \le n$$$ should be satisfied). Note that the progression added is always $$$1, 2, \ldots, k$$$ but not the $$$k, k - 1, \ldots, 1$$$ or anything else (i.e., the leftmost element of the subsegment always increases by $$$1$$$, the second element always increases by $$$2$$$ and so on).Your task is to find the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Note that the condition $$$a_i \ge b_i$$$ should be satisfied for all elements at once.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 3 \cdot 10^5$$$) — the number of elements in both arrays and the length of the subsegment, respectively. The second line of the input contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i \le 10^{12}$$$), where $$$b_i$$$ is the $$$i$$$-th element of the array $$$b$$$.
Output Specification: Print one integer — the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$.
Notes: NoteConsider the first example. In this test, we don't really have any choice, so we need to add at least five progressions to make the first element equals $$$5$$$. The array $$$a$$$ becomes $$$[5, 10, 15]$$$.Consider the second example. In this test, let's add one progression on the segment $$$[1; 3]$$$ and two progressions on the segment $$$[4; 6]$$$. Then, the array $$$a$$$ becomes $$$[1, 2, 3, 2, 4, 6]$$$.
Code:
"""
take element as "the tail" will use more less operations,
use variables s,cnt and closed,
to avoid the inner iteration(update neighbor k elements every time),
complexity is O(n).
"""
row=lambda:map(int,input().split())
n,k=row()
a=list(row())
closed=[0]*n
s=cnt=res=0
for i in range(n-1,-1,-1):
# print(i,a[i],'s,cnt:',s,cnt,'closed:',closed)
s-=cnt
cnt-=closed[i]
a[i]-=s
if a[i]<=0:
continue
th=min(i+1,k)
need=(a[i]+th-1)//th#equals ceil()
s+=need*th
cnt+=need
res+=need
if # TODO: Your code here:
closed[i-th]+=need
print(res)
|
"""
take element as "the tail" will use more less operations,
use variables s,cnt and closed,
to avoid the inner iteration(update neighbor k elements every time),
complexity is O(n).
"""
row=lambda:map(int,input().split())
n,k=row()
a=list(row())
closed=[0]*n
s=cnt=res=0
for i in range(n-1,-1,-1):
# print(i,a[i],'s,cnt:',s,cnt,'closed:',closed)
s-=cnt
cnt-=closed[i]
a[i]-=s
if a[i]<=0:
continue
th=min(i+1,k)
need=(a[i]+th-1)//th#equals ceil()
s+=need*th
cnt+=need
res+=need
if {{completion}}:
closed[i-th]+=need
print(res)
|
i>=th
|
[{"input": "3 3\n5 4 6", "output": ["5"]}, {"input": "6 3\n1 2 3 2 2 3", "output": ["3"]}, {"input": "6 3\n1 2 4 1 2 3", "output": ["3"]}, {"input": "7 3\n50 17 81 25 42 39 96", "output": ["92"]}]
|
control_completion_003390
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given two arrays: an array $$$a$$$ consisting of $$$n$$$ zeros and an array $$$b$$$ consisting of $$$n$$$ integers.You can apply the following operation to the array $$$a$$$ an arbitrary number of times: choose some subsegment of $$$a$$$ of length $$$k$$$ and add the arithmetic progression $$$1, 2, \ldots, k$$$ to this subsegment — i. e. add $$$1$$$ to the first element of the subsegment, $$$2$$$ to the second element, and so on. The chosen subsegment should be inside the borders of the array $$$a$$$ (i.e., if the left border of the chosen subsegment is $$$l$$$, then the condition $$$1 \le l \le l + k - 1 \le n$$$ should be satisfied). Note that the progression added is always $$$1, 2, \ldots, k$$$ but not the $$$k, k - 1, \ldots, 1$$$ or anything else (i.e., the leftmost element of the subsegment always increases by $$$1$$$, the second element always increases by $$$2$$$ and so on).Your task is to find the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Note that the condition $$$a_i \ge b_i$$$ should be satisfied for all elements at once.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 3 \cdot 10^5$$$) — the number of elements in both arrays and the length of the subsegment, respectively. The second line of the input contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i \le 10^{12}$$$), where $$$b_i$$$ is the $$$i$$$-th element of the array $$$b$$$.
Output Specification: Print one integer — the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$.
Notes: NoteConsider the first example. In this test, we don't really have any choice, so we need to add at least five progressions to make the first element equals $$$5$$$. The array $$$a$$$ becomes $$$[5, 10, 15]$$$.Consider the second example. In this test, let's add one progression on the segment $$$[1; 3]$$$ and two progressions on the segment $$$[4; 6]$$$. Then, the array $$$a$$$ becomes $$$[1, 2, 3, 2, 4, 6]$$$.
Code:
import math
n, k = list(map(int, input().split(' ')))
nums = list(map(int, input().split(' ')))
ans = res = tot = minus = 0
pre = []
prefix = []
for i in range(n)[::-1]:
if # TODO: Your code here:
minus += k * pre[-1] - prefix[-1]
nums[i] -= minus
cur = max(0, math.ceil(nums[i] / k))
ans += (cur if i >= k else 0)
pre.append(cur if i >= k else 0)
tot += (cur if i >= k else 0)
if len(pre) > k:
tot -= pre[- k - 1]
prefix.append(tot)
for i in range(k):
res = max(res, math.ceil(nums[i] / (i + 1)))
print(ans + res)
|
import math
n, k = list(map(int, input().split(' ')))
nums = list(map(int, input().split(' ')))
ans = res = tot = minus = 0
pre = []
prefix = []
for i in range(n)[::-1]:
if {{completion}}:
minus += k * pre[-1] - prefix[-1]
nums[i] -= minus
cur = max(0, math.ceil(nums[i] / k))
ans += (cur if i >= k else 0)
pre.append(cur if i >= k else 0)
tot += (cur if i >= k else 0)
if len(pre) > k:
tot -= pre[- k - 1]
prefix.append(tot)
for i in range(k):
res = max(res, math.ceil(nums[i] / (i + 1)))
print(ans + res)
|
i < n - 1
|
[{"input": "3 3\n5 4 6", "output": ["5"]}, {"input": "6 3\n1 2 3 2 2 3", "output": ["3"]}, {"input": "6 3\n1 2 4 1 2 3", "output": ["3"]}, {"input": "7 3\n50 17 81 25 42 39 96", "output": ["92"]}]
|
control_completion_003391
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given two arrays: an array $$$a$$$ consisting of $$$n$$$ zeros and an array $$$b$$$ consisting of $$$n$$$ integers.You can apply the following operation to the array $$$a$$$ an arbitrary number of times: choose some subsegment of $$$a$$$ of length $$$k$$$ and add the arithmetic progression $$$1, 2, \ldots, k$$$ to this subsegment — i. e. add $$$1$$$ to the first element of the subsegment, $$$2$$$ to the second element, and so on. The chosen subsegment should be inside the borders of the array $$$a$$$ (i.e., if the left border of the chosen subsegment is $$$l$$$, then the condition $$$1 \le l \le l + k - 1 \le n$$$ should be satisfied). Note that the progression added is always $$$1, 2, \ldots, k$$$ but not the $$$k, k - 1, \ldots, 1$$$ or anything else (i.e., the leftmost element of the subsegment always increases by $$$1$$$, the second element always increases by $$$2$$$ and so on).Your task is to find the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Note that the condition $$$a_i \ge b_i$$$ should be satisfied for all elements at once.
Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 3 \cdot 10^5$$$) — the number of elements in both arrays and the length of the subsegment, respectively. The second line of the input contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i \le 10^{12}$$$), where $$$b_i$$$ is the $$$i$$$-th element of the array $$$b$$$.
Output Specification: Print one integer — the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$.
Notes: NoteConsider the first example. In this test, we don't really have any choice, so we need to add at least five progressions to make the first element equals $$$5$$$. The array $$$a$$$ becomes $$$[5, 10, 15]$$$.Consider the second example. In this test, let's add one progression on the segment $$$[1; 3]$$$ and two progressions on the segment $$$[4; 6]$$$. Then, the array $$$a$$$ becomes $$$[1, 2, 3, 2, 4, 6]$$$.
Code:
import math
n, k = list(map(int, input().split(' ')))
nums = list(map(int, input().split(' ')))
ans = res = tot = minus = 0
pre = []
prefix = []
for i in range(n)[::-1]:
if i < n - 1:
minus += k * pre[-1] - prefix[-1]
nums[i] -= minus
cur = max(0, math.ceil(nums[i] / k))
ans += (cur if i >= k else 0)
pre.append(cur if i >= k else 0)
tot += (cur if i >= k else 0)
if # TODO: Your code here:
tot -= pre[- k - 1]
prefix.append(tot)
for i in range(k):
res = max(res, math.ceil(nums[i] / (i + 1)))
print(ans + res)
|
import math
n, k = list(map(int, input().split(' ')))
nums = list(map(int, input().split(' ')))
ans = res = tot = minus = 0
pre = []
prefix = []
for i in range(n)[::-1]:
if i < n - 1:
minus += k * pre[-1] - prefix[-1]
nums[i] -= minus
cur = max(0, math.ceil(nums[i] / k))
ans += (cur if i >= k else 0)
pre.append(cur if i >= k else 0)
tot += (cur if i >= k else 0)
if {{completion}}:
tot -= pre[- k - 1]
prefix.append(tot)
for i in range(k):
res = max(res, math.ceil(nums[i] / (i + 1)))
print(ans + res)
|
len(pre) > k
|
[{"input": "3 3\n5 4 6", "output": ["5"]}, {"input": "6 3\n1 2 3 2 2 3", "output": ["3"]}, {"input": "6 3\n1 2 4 1 2 3", "output": ["3"]}, {"input": "7 3\n50 17 81 25 42 39 96", "output": ["92"]}]
|
control_completion_003392
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n+1$$$ teleporters on a straight line, located in points $$$0$$$, $$$a_1$$$, $$$a_2$$$, $$$a_3$$$, ..., $$$a_n$$$. It's possible to teleport from point $$$x$$$ to point $$$y$$$ if there are teleporters in both of those points, and it costs $$$(x-y)^2$$$ energy.You want to install some additional teleporters so that it is possible to get from the point $$$0$$$ to the point $$$a_n$$$ (possibly through some other teleporters) spending no more than $$$m$$$ energy in total. Each teleporter you install must be located in an integer point.What is the minimum number of teleporters you have to install?
Input Specification: The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_1 < a_2 < a_3 < \dots < a_n \le 10^9$$$). The third line contains one integer $$$m$$$ ($$$a_n \le m \le 10^{18}$$$).
Output Specification: Print one integer — the minimum number of teleporters you have to install so that it is possible to get from $$$0$$$ to $$$a_n$$$ spending at most $$$m$$$ energy. It can be shown that it's always possible under the constraints from the input format.
Code:
I=lambda:[*map(int,input().split())]
def F(g,i):s=g//i;b=g%i;return b*(s+1)**2+(i-b)*s**2
def f(g,c):
if c>g**2//2:return 0,g**2
s=0;b=g
while b-s>1:
m=(b+s)//2
if # TODO: Your code here:s=m
else:b=m
return s,F(g,b)
n,=I()
a=[0]+I()
m,=I()
G=[a[i+1]-a[i]for i in range(n)]
s=2
b=1<<59
while b-s>1:
C=0;M=(b+s)//2
for g in G:a,c=f(g,M);C+=c
if C>m:b=M
else:s=M
A=C=0
for g in G:a,c=f(g,b);A+=a;C+=c
print(A+max(0,(C-m-1)//s+1))
|
I=lambda:[*map(int,input().split())]
def F(g,i):s=g//i;b=g%i;return b*(s+1)**2+(i-b)*s**2
def f(g,c):
if c>g**2//2:return 0,g**2
s=0;b=g
while b-s>1:
m=(b+s)//2
if {{completion}}:s=m
else:b=m
return s,F(g,b)
n,=I()
a=[0]+I()
m,=I()
G=[a[i+1]-a[i]for i in range(n)]
s=2
b=1<<59
while b-s>1:
C=0;M=(b+s)//2
for g in G:a,c=f(g,M);C+=c
if C>m:b=M
else:s=M
A=C=0
for g in G:a,c=f(g,b);A+=a;C+=c
print(A+max(0,(C-m-1)//s+1))
|
F(g,m)-F(g,m+1)>=c
|
[{"input": "2\n1 5\n7", "output": ["2"]}, {"input": "2\n1 5\n6", "output": ["3"]}, {"input": "1\n5\n5", "output": ["4"]}, {"input": "1\n1000000000\n1000000043", "output": ["999999978"]}]
|
control_completion_003404
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n+1$$$ teleporters on a straight line, located in points $$$0$$$, $$$a_1$$$, $$$a_2$$$, $$$a_3$$$, ..., $$$a_n$$$. It's possible to teleport from point $$$x$$$ to point $$$y$$$ if there are teleporters in both of those points, and it costs $$$(x-y)^2$$$ energy.You want to install some additional teleporters so that it is possible to get from the point $$$0$$$ to the point $$$a_n$$$ (possibly through some other teleporters) spending no more than $$$m$$$ energy in total. Each teleporter you install must be located in an integer point.What is the minimum number of teleporters you have to install?
Input Specification: The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_1 < a_2 < a_3 < \dots < a_n \le 10^9$$$). The third line contains one integer $$$m$$$ ($$$a_n \le m \le 10^{18}$$$).
Output Specification: Print one integer — the minimum number of teleporters you have to install so that it is possible to get from $$$0$$$ to $$$a_n$$$ spending at most $$$m$$$ energy. It can be shown that it's always possible under the constraints from the input format.
Code:
from collections import Counter
n, a, m = int(input()), [*map(int, input().split())], int(input())
def energy(l, t):
x,y = divmod(l, t+1)
return x*x*(t+1-y)+(x+1)*(x+1)*y
def getdiff(l, diff):
lo, hi = 0, l
while lo < hi:
mid = lo + hi >> 1
if # TODO: Your code here: hi = mid
else: lo = mid + 1
return lo, energy(l, lo)
def getsum(d, c=0): a1,a2 = zip(*[getdiff(i, d) for i,_ in a]); return (0 if c else sum(a1[i]*x for i, (_, x) in enumerate(a)), sum(a2[i]*x for i, (_, x) in enumerate(a)))
a = [0] + a
a = Counter([a[i+1]-a[i] for i in range(n)]).items()
lo, hi = 1, m
while lo < hi:
mid = lo + hi >> 1
if getsum(mid, 1)[1] > m: hi = mid
else: lo = mid + 1
lo-=1
a1, a2 = getsum(lo)
print(a1-(m-a2)//lo if lo else a1)
|
from collections import Counter
n, a, m = int(input()), [*map(int, input().split())], int(input())
def energy(l, t):
x,y = divmod(l, t+1)
return x*x*(t+1-y)+(x+1)*(x+1)*y
def getdiff(l, diff):
lo, hi = 0, l
while lo < hi:
mid = lo + hi >> 1
if {{completion}}: hi = mid
else: lo = mid + 1
return lo, energy(l, lo)
def getsum(d, c=0): a1,a2 = zip(*[getdiff(i, d) for i,_ in a]); return (0 if c else sum(a1[i]*x for i, (_, x) in enumerate(a)), sum(a2[i]*x for i, (_, x) in enumerate(a)))
a = [0] + a
a = Counter([a[i+1]-a[i] for i in range(n)]).items()
lo, hi = 1, m
while lo < hi:
mid = lo + hi >> 1
if getsum(mid, 1)[1] > m: hi = mid
else: lo = mid + 1
lo-=1
a1, a2 = getsum(lo)
print(a1-(m-a2)//lo if lo else a1)
|
energy(l, mid) - energy(l, mid+1) < diff
|
[{"input": "2\n1 5\n7", "output": ["2"]}, {"input": "2\n1 5\n6", "output": ["3"]}, {"input": "1\n5\n5", "output": ["4"]}, {"input": "1\n1000000000\n1000000043", "output": ["999999978"]}]
|
control_completion_003405
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: There are $$$n+1$$$ teleporters on a straight line, located in points $$$0$$$, $$$a_1$$$, $$$a_2$$$, $$$a_3$$$, ..., $$$a_n$$$. It's possible to teleport from point $$$x$$$ to point $$$y$$$ if there are teleporters in both of those points, and it costs $$$(x-y)^2$$$ energy.You want to install some additional teleporters so that it is possible to get from the point $$$0$$$ to the point $$$a_n$$$ (possibly through some other teleporters) spending no more than $$$m$$$ energy in total. Each teleporter you install must be located in an integer point.What is the minimum number of teleporters you have to install?
Input Specification: The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_1 < a_2 < a_3 < \dots < a_n \le 10^9$$$). The third line contains one integer $$$m$$$ ($$$a_n \le m \le 10^{18}$$$).
Output Specification: Print one integer — the minimum number of teleporters you have to install so that it is possible to get from $$$0$$$ to $$$a_n$$$ spending at most $$$m$$$ energy. It can be shown that it's always possible under the constraints from the input format.
Code:
# import io,os
# read = io.BytesIO(os.read(0, os.fstat(0).st_size))
# I = lambda: [*map(int, read.readline().split())]
import sys
I=lambda:[*map(int,sys.stdin.readline().split())]
def ff(gap, ints):
sml = gap // ints
bigcount = gap % ints
return bigcount * (sml + 1) ** 2 + (ints - bigcount) * sml ** 2
def f(gap, c):
if c > gap ** 2 // 2:
return 0, gap ** 2
sml = 0
big = gap
while big - sml > 1:
mid = (big + sml) // 2
a = ff(gap, mid)
b = ff(gap, mid + 1)
if # TODO: Your code here:
sml = mid
else:
big = mid
return sml, ff(gap, big)
n, = I()
a = I()
m, = I()
gaps = [a[0]]
for i in range(n - 1):
gaps.append(a[i + 1] - a[i])
sml = 2
big = 1 << 59 + 2
while big - sml > 1:
cost = 0
mid = (big + sml) // 2
for g in gaps:
a, c = f(g, mid)
cost += c
if cost > m:
big = mid
else:
sml = mid
abig = 0
cbig = 0
for g in gaps:
a, c = f(g, big)
abig += a
cbig += c
print(abig + max(0, (cbig - m - 1) // sml + 1))
|
# import io,os
# read = io.BytesIO(os.read(0, os.fstat(0).st_size))
# I = lambda: [*map(int, read.readline().split())]
import sys
I=lambda:[*map(int,sys.stdin.readline().split())]
def ff(gap, ints):
sml = gap // ints
bigcount = gap % ints
return bigcount * (sml + 1) ** 2 + (ints - bigcount) * sml ** 2
def f(gap, c):
if c > gap ** 2 // 2:
return 0, gap ** 2
sml = 0
big = gap
while big - sml > 1:
mid = (big + sml) // 2
a = ff(gap, mid)
b = ff(gap, mid + 1)
if {{completion}}:
sml = mid
else:
big = mid
return sml, ff(gap, big)
n, = I()
a = I()
m, = I()
gaps = [a[0]]
for i in range(n - 1):
gaps.append(a[i + 1] - a[i])
sml = 2
big = 1 << 59 + 2
while big - sml > 1:
cost = 0
mid = (big + sml) // 2
for g in gaps:
a, c = f(g, mid)
cost += c
if cost > m:
big = mid
else:
sml = mid
abig = 0
cbig = 0
for g in gaps:
a, c = f(g, big)
abig += a
cbig += c
print(abig + max(0, (cbig - m - 1) // sml + 1))
|
a - b >= c
|
[{"input": "2\n1 5\n7", "output": ["2"]}, {"input": "2\n1 5\n6", "output": ["3"]}, {"input": "1\n5\n5", "output": ["4"]}, {"input": "1\n1000000000\n1000000043", "output": ["999999978"]}]
|
control_completion_003406
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Nastya baked $$$m$$$ pancakes and spread them on $$$n$$$ dishes. The dishes are in a row and numbered from left to right. She put $$$a_i$$$ pancakes on the dish with the index $$$i$$$.Seeing the dishes, Vlad decided to bring order to the stacks and move some pancakes. In one move, he can shift one pancake from any dish to the closest one, that is, select the dish $$$i$$$ ($$$a_i > 0$$$) and do one of the following: if $$$i > 1$$$, put the pancake on a dish with the previous index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i - 1} = a_{i - 1} + 1$$$; if $$$i < n$$$, put the pancake on a dish with the following index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i + 1} = a_{i + 1} + 1$$$.Vlad wants to make the array $$$a$$$non-increasing, after moving as few pancakes as possible. Help him find the minimum number of moves needed for this.The array $$$a=[a_1, a_2,\dots,a_n]$$$ is called non-increasing if $$$a_i \ge a_{i+1}$$$ for all $$$i$$$ from $$$1$$$ to $$$n-1$$$.
Input Specification: The first line of the input contains two numbers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 250$$$) — the number of dishes and the number of pancakes, respectively. The second line contains $$$n$$$ numbers $$$a_i$$$ ($$$0 \le a_i \le m$$$), the sum of all $$$a_i$$$ is equal to $$$m$$$.
Output Specification: Print a single number: the minimum number of moves required to make the array $$$a$$$ non-increasing.
Notes: NoteIn the first example, you first need to move the pancake from the dish $$$1$$$, after which $$$a = [4, 4, 2, 3, 3, 3]$$$. After that, you need to move the pancake from the dish $$$2$$$ to the dish $$$3$$$ and the array will become non-growing $$$a = [4, 3, 3, 3, 3, 3]$$$.
Code:
from functools import cache
from math import inf
from itertools import accumulate
def solve(A, m):
n = len(A)
A.reverse()
@cache
def dp(i, val, balance):
if abs(balance) > m:
return inf
if # TODO: Your code here:
return inf
if i == n:
return 0 if balance == 0 else inf
curr = A[i] + balance
take = abs(curr - val) + dp(i + 1, val, curr - val)
skip = dp(i, val + 1, balance)
return min(take, skip)
return dp(0, 0, 0)
n, m = map(int, input().split())
A = list(map(int, input().split()))
print(solve(A, m))
|
from functools import cache
from math import inf
from itertools import accumulate
def solve(A, m):
n = len(A)
A.reverse()
@cache
def dp(i, val, balance):
if abs(balance) > m:
return inf
if {{completion}}:
return inf
if i == n:
return 0 if balance == 0 else inf
curr = A[i] + balance
take = abs(curr - val) + dp(i + 1, val, curr - val)
skip = dp(i, val + 1, balance)
return min(take, skip)
return dp(0, 0, 0)
n, m = map(int, input().split())
A = list(map(int, input().split()))
print(solve(A, m))
|
(n - i) * val > m
|
[{"input": "6 19\n5 3 2 3 3 3", "output": ["2"]}, {"input": "3 6\n3 2 1", "output": ["0"]}, {"input": "3 6\n2 1 3", "output": ["1"]}, {"input": "6 19\n3 4 3 3 5 1", "output": ["3"]}, {"input": "10 1\n0 0 0 0 0 0 0 0 0 1", "output": ["9"]}]
|
control_completion_003521
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Nastya baked $$$m$$$ pancakes and spread them on $$$n$$$ dishes. The dishes are in a row and numbered from left to right. She put $$$a_i$$$ pancakes on the dish with the index $$$i$$$.Seeing the dishes, Vlad decided to bring order to the stacks and move some pancakes. In one move, he can shift one pancake from any dish to the closest one, that is, select the dish $$$i$$$ ($$$a_i > 0$$$) and do one of the following: if $$$i > 1$$$, put the pancake on a dish with the previous index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i - 1} = a_{i - 1} + 1$$$; if $$$i < n$$$, put the pancake on a dish with the following index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i + 1} = a_{i + 1} + 1$$$.Vlad wants to make the array $$$a$$$non-increasing, after moving as few pancakes as possible. Help him find the minimum number of moves needed for this.The array $$$a=[a_1, a_2,\dots,a_n]$$$ is called non-increasing if $$$a_i \ge a_{i+1}$$$ for all $$$i$$$ from $$$1$$$ to $$$n-1$$$.
Input Specification: The first line of the input contains two numbers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 250$$$) — the number of dishes and the number of pancakes, respectively. The second line contains $$$n$$$ numbers $$$a_i$$$ ($$$0 \le a_i \le m$$$), the sum of all $$$a_i$$$ is equal to $$$m$$$.
Output Specification: Print a single number: the minimum number of moves required to make the array $$$a$$$ non-increasing.
Notes: NoteIn the first example, you first need to move the pancake from the dish $$$1$$$, after which $$$a = [4, 4, 2, 3, 3, 3]$$$. After that, you need to move the pancake from the dish $$$2$$$ to the dish $$$3$$$ and the array will become non-growing $$$a = [4, 3, 3, 3, 3, 3]$$$.
Code:
from functools import cache
from math import inf
from itertools import accumulate
def solve(A, m):
n = len(A)
A.reverse()
@cache
def dp(i, val, balance):
if abs(balance) > m:
return inf
if (n - i) * val > m:
return inf
if # TODO: Your code here:
return 0 if balance == 0 else inf
curr = A[i] + balance
take = abs(curr - val) + dp(i + 1, val, curr - val)
skip = dp(i, val + 1, balance)
return min(take, skip)
return dp(0, 0, 0)
n, m = map(int, input().split())
A = list(map(int, input().split()))
print(solve(A, m))
|
from functools import cache
from math import inf
from itertools import accumulate
def solve(A, m):
n = len(A)
A.reverse()
@cache
def dp(i, val, balance):
if abs(balance) > m:
return inf
if (n - i) * val > m:
return inf
if {{completion}}:
return 0 if balance == 0 else inf
curr = A[i] + balance
take = abs(curr - val) + dp(i + 1, val, curr - val)
skip = dp(i, val + 1, balance)
return min(take, skip)
return dp(0, 0, 0)
n, m = map(int, input().split())
A = list(map(int, input().split()))
print(solve(A, m))
|
i == n
|
[{"input": "6 19\n5 3 2 3 3 3", "output": ["2"]}, {"input": "3 6\n3 2 1", "output": ["0"]}, {"input": "3 6\n2 1 3", "output": ["1"]}, {"input": "6 19\n3 4 3 3 5 1", "output": ["3"]}, {"input": "10 1\n0 0 0 0 0 0 0 0 0 1", "output": ["9"]}]
|
control_completion_003522
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Nastya baked $$$m$$$ pancakes and spread them on $$$n$$$ dishes. The dishes are in a row and numbered from left to right. She put $$$a_i$$$ pancakes on the dish with the index $$$i$$$.Seeing the dishes, Vlad decided to bring order to the stacks and move some pancakes. In one move, he can shift one pancake from any dish to the closest one, that is, select the dish $$$i$$$ ($$$a_i > 0$$$) and do one of the following: if $$$i > 1$$$, put the pancake on a dish with the previous index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i - 1} = a_{i - 1} + 1$$$; if $$$i < n$$$, put the pancake on a dish with the following index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i + 1} = a_{i + 1} + 1$$$.Vlad wants to make the array $$$a$$$non-increasing, after moving as few pancakes as possible. Help him find the minimum number of moves needed for this.The array $$$a=[a_1, a_2,\dots,a_n]$$$ is called non-increasing if $$$a_i \ge a_{i+1}$$$ for all $$$i$$$ from $$$1$$$ to $$$n-1$$$.
Input Specification: The first line of the input contains two numbers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 250$$$) — the number of dishes and the number of pancakes, respectively. The second line contains $$$n$$$ numbers $$$a_i$$$ ($$$0 \le a_i \le m$$$), the sum of all $$$a_i$$$ is equal to $$$m$$$.
Output Specification: Print a single number: the minimum number of moves required to make the array $$$a$$$ non-increasing.
Notes: NoteIn the first example, you first need to move the pancake from the dish $$$1$$$, after which $$$a = [4, 4, 2, 3, 3, 3]$$$. After that, you need to move the pancake from the dish $$$2$$$ to the dish $$$3$$$ and the array will become non-growing $$$a = [4, 3, 3, 3, 3, 3]$$$.
Code:
from itertools import accumulate
I=lambda:map(int,input().split())
n,m=I()
a,inf=[*I()],float('inf')
pre=[0]+[*accumulate(a)]
dp=[[inf]*(m+1) for _ in range(m+1)]
dp[m][0]=0
for i in range(n):
cur=[[inf]*(m+1) for _ in range(m+1)]
for lst in reversed(range(m+1)):
for sums in range(m+1):
if # TODO: Your code here:
dp[lst][sums]=min(dp[lst][sums],dp[lst+1][sums])
if sums+lst<=m:
cur[lst][sums+lst]=min(cur[lst][sums+lst], dp[lst][sums]+abs(pre[i+1]-(sums+lst)))
dp,cur=cur,dp
print(min(dp[lst][m] for lst in range(m+1)))
|
from itertools import accumulate
I=lambda:map(int,input().split())
n,m=I()
a,inf=[*I()],float('inf')
pre=[0]+[*accumulate(a)]
dp=[[inf]*(m+1) for _ in range(m+1)]
dp[m][0]=0
for i in range(n):
cur=[[inf]*(m+1) for _ in range(m+1)]
for lst in reversed(range(m+1)):
for sums in range(m+1):
if {{completion}}:
dp[lst][sums]=min(dp[lst][sums],dp[lst+1][sums])
if sums+lst<=m:
cur[lst][sums+lst]=min(cur[lst][sums+lst], dp[lst][sums]+abs(pre[i+1]-(sums+lst)))
dp,cur=cur,dp
print(min(dp[lst][m] for lst in range(m+1)))
|
lst<m
|
[{"input": "6 19\n5 3 2 3 3 3", "output": ["2"]}, {"input": "3 6\n3 2 1", "output": ["0"]}, {"input": "3 6\n2 1 3", "output": ["1"]}, {"input": "6 19\n3 4 3 3 5 1", "output": ["3"]}, {"input": "10 1\n0 0 0 0 0 0 0 0 0 1", "output": ["9"]}]
|
control_completion_003523
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Nastya baked $$$m$$$ pancakes and spread them on $$$n$$$ dishes. The dishes are in a row and numbered from left to right. She put $$$a_i$$$ pancakes on the dish with the index $$$i$$$.Seeing the dishes, Vlad decided to bring order to the stacks and move some pancakes. In one move, he can shift one pancake from any dish to the closest one, that is, select the dish $$$i$$$ ($$$a_i > 0$$$) and do one of the following: if $$$i > 1$$$, put the pancake on a dish with the previous index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i - 1} = a_{i - 1} + 1$$$; if $$$i < n$$$, put the pancake on a dish with the following index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i + 1} = a_{i + 1} + 1$$$.Vlad wants to make the array $$$a$$$non-increasing, after moving as few pancakes as possible. Help him find the minimum number of moves needed for this.The array $$$a=[a_1, a_2,\dots,a_n]$$$ is called non-increasing if $$$a_i \ge a_{i+1}$$$ for all $$$i$$$ from $$$1$$$ to $$$n-1$$$.
Input Specification: The first line of the input contains two numbers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 250$$$) — the number of dishes and the number of pancakes, respectively. The second line contains $$$n$$$ numbers $$$a_i$$$ ($$$0 \le a_i \le m$$$), the sum of all $$$a_i$$$ is equal to $$$m$$$.
Output Specification: Print a single number: the minimum number of moves required to make the array $$$a$$$ non-increasing.
Notes: NoteIn the first example, you first need to move the pancake from the dish $$$1$$$, after which $$$a = [4, 4, 2, 3, 3, 3]$$$. After that, you need to move the pancake from the dish $$$2$$$ to the dish $$$3$$$ and the array will become non-growing $$$a = [4, 3, 3, 3, 3, 3]$$$.
Code:
from itertools import accumulate
I=lambda:map(int,input().split())
n,m=I()
a,inf=[*I()],float('inf')
pre=[0]+[*accumulate(a)]
dp=[[inf]*(m+1) for _ in range(m+1)]
dp[m][0]=0
for i in range(n):
cur=[[inf]*(m+1) for _ in range(m+1)]
for lst in reversed(range(m+1)):
for sums in range(m+1):
if lst<m:
dp[lst][sums]=min(dp[lst][sums],dp[lst+1][sums])
if # TODO: Your code here:
cur[lst][sums+lst]=min(cur[lst][sums+lst], dp[lst][sums]+abs(pre[i+1]-(sums+lst)))
dp,cur=cur,dp
print(min(dp[lst][m] for lst in range(m+1)))
|
from itertools import accumulate
I=lambda:map(int,input().split())
n,m=I()
a,inf=[*I()],float('inf')
pre=[0]+[*accumulate(a)]
dp=[[inf]*(m+1) for _ in range(m+1)]
dp[m][0]=0
for i in range(n):
cur=[[inf]*(m+1) for _ in range(m+1)]
for lst in reversed(range(m+1)):
for sums in range(m+1):
if lst<m:
dp[lst][sums]=min(dp[lst][sums],dp[lst+1][sums])
if {{completion}}:
cur[lst][sums+lst]=min(cur[lst][sums+lst], dp[lst][sums]+abs(pre[i+1]-(sums+lst)))
dp,cur=cur,dp
print(min(dp[lst][m] for lst in range(m+1)))
|
sums+lst<=m
|
[{"input": "6 19\n5 3 2 3 3 3", "output": ["2"]}, {"input": "3 6\n3 2 1", "output": ["0"]}, {"input": "3 6\n2 1 3", "output": ["1"]}, {"input": "6 19\n3 4 3 3 5 1", "output": ["3"]}, {"input": "10 1\n0 0 0 0 0 0 0 0 0 1", "output": ["9"]}]
|
control_completion_003524
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Nastya baked $$$m$$$ pancakes and spread them on $$$n$$$ dishes. The dishes are in a row and numbered from left to right. She put $$$a_i$$$ pancakes on the dish with the index $$$i$$$.Seeing the dishes, Vlad decided to bring order to the stacks and move some pancakes. In one move, he can shift one pancake from any dish to the closest one, that is, select the dish $$$i$$$ ($$$a_i > 0$$$) and do one of the following: if $$$i > 1$$$, put the pancake on a dish with the previous index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i - 1} = a_{i - 1} + 1$$$; if $$$i < n$$$, put the pancake on a dish with the following index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i + 1} = a_{i + 1} + 1$$$.Vlad wants to make the array $$$a$$$non-increasing, after moving as few pancakes as possible. Help him find the minimum number of moves needed for this.The array $$$a=[a_1, a_2,\dots,a_n]$$$ is called non-increasing if $$$a_i \ge a_{i+1}$$$ for all $$$i$$$ from $$$1$$$ to $$$n-1$$$.
Input Specification: The first line of the input contains two numbers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 250$$$) — the number of dishes and the number of pancakes, respectively. The second line contains $$$n$$$ numbers $$$a_i$$$ ($$$0 \le a_i \le m$$$), the sum of all $$$a_i$$$ is equal to $$$m$$$.
Output Specification: Print a single number: the minimum number of moves required to make the array $$$a$$$ non-increasing.
Notes: NoteIn the first example, you first need to move the pancake from the dish $$$1$$$, after which $$$a = [4, 4, 2, 3, 3, 3]$$$. After that, you need to move the pancake from the dish $$$2$$$ to the dish $$$3$$$ and the array will become non-growing $$$a = [4, 3, 3, 3, 3, 3]$$$.
Code:
n,m=map(int,input().split())
a=list(map(int,input().split()))[::-1]
id=[]
for i in range(n):
for _ in range(a[i]):
id.append(i)
inf=10**5
dp=[[inf]*(m+1) for i in range(m+1)]
dp[0][0]=0
for i in range(n):
cost=[]
for j in id:
cost.append(abs(i-j))
cum=[0]
tmp=0
for j in cost:
tmp+=j
cum.append(tmp)
dp_new=[[inf]*(m+1) for i in range(m+1)]
for j in range(m+1):
mx=(m-j)//(n-i)
for k in range(mx+1):
if dp[j][k]==inf:continue
#print(i,j,k,mx)
for # TODO: Your code here:
#print(l)
c=cum[l+j]-cum[j]
dp_new[j+l][l]=min(dp_new[j+l][l],dp[j][k]+c)
dp=dp_new
print(min(dp[-1]))
|
n,m=map(int,input().split())
a=list(map(int,input().split()))[::-1]
id=[]
for i in range(n):
for _ in range(a[i]):
id.append(i)
inf=10**5
dp=[[inf]*(m+1) for i in range(m+1)]
dp[0][0]=0
for i in range(n):
cost=[]
for j in id:
cost.append(abs(i-j))
cum=[0]
tmp=0
for j in cost:
tmp+=j
cum.append(tmp)
dp_new=[[inf]*(m+1) for i in range(m+1)]
for j in range(m+1):
mx=(m-j)//(n-i)
for k in range(mx+1):
if dp[j][k]==inf:continue
#print(i,j,k,mx)
for {{completion}}:
#print(l)
c=cum[l+j]-cum[j]
dp_new[j+l][l]=min(dp_new[j+l][l],dp[j][k]+c)
dp=dp_new
print(min(dp[-1]))
|
l in range(k,mx+1)
|
[{"input": "6 19\n5 3 2 3 3 3", "output": ["2"]}, {"input": "3 6\n3 2 1", "output": ["0"]}, {"input": "3 6\n2 1 3", "output": ["1"]}, {"input": "6 19\n3 4 3 3 5 1", "output": ["3"]}, {"input": "10 1\n0 0 0 0 0 0 0 0 0 1", "output": ["9"]}]
|
control_completion_003525
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: This is the easy version of the problem. The only difference between the versions is the constraints on $$$n$$$, $$$k$$$, $$$a_i$$$, and the sum of $$$n$$$ over all test cases. You can make hacks only if both versions of the problem are solved.Note the unusual memory limit.You are given an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$, and an integer $$$k$$$.The cost of an array of integers $$$p_1, p_2, \ldots, p_n$$$ of length $$$n$$$ is $$$$$$\max\limits_{1 \le i \le n}\left(\left \lfloor \frac{a_i}{p_i} \right \rfloor \right) - \min\limits_{1 \le i \le n}\left(\left \lfloor \frac{a_i}{p_i} \right \rfloor \right).$$$$$$Here, $$$\lfloor \frac{x}{y} \rfloor$$$ denotes the integer part of the division of $$$x$$$ by $$$y$$$. Find the minimum cost of an array $$$p$$$ such that $$$1 \le p_i \le k$$$ for all $$$1 \le i \le n$$$.
Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n, k \le 3000$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_1 \le a_2 \le \ldots \le a_n \le 3000$$$). It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3000$$$.
Output Specification: For each test case, print a single integer — the minimum possible cost of an array $$$p$$$ satisfying the condition above.
Notes: NoteIn the first test case, the optimal array is $$$p = [1, 1, 1, 2, 2]$$$. The resulting array of values of $$$\lfloor \frac{a_i}{p_i} \rfloor$$$ is $$$[4, 5, 6, 4, 5]$$$. The cost of $$$p$$$ is $$$\max\limits_{1 \le i \le n}(\lfloor \frac{a_i}{p_i} \rfloor) - \min\limits_{1 \le i \le n}(\lfloor \frac{a_i}{p_i} \rfloor) = 6 - 4 = 2$$$. We can show that there is no array (satisfying the condition from the statement) with a smaller cost.In the second test case, one of the optimal arrays is $$$p = [12, 12, 12, 12, 12]$$$, which results in all $$$\lfloor \frac{a_i}{p_i} \rfloor$$$ being $$$0$$$.In the third test case, the only possible array is $$$p = [1, 1, 1]$$$.
Code:
import sys
tokens = ''.join(sys.stdin.readlines()).split()[::-1]
def next(): return tokens.pop()
def nextInt(): return int(next())
def nextFloat(): return float(next())
def getIntArray(n): return [nextInt() for _ in range(n)]
def getFloatArray(n): return [nextFloat() for _ in range(n)]
def getStringArray(n): return [next() for _ in range(n)]
testcase = True
def solve(testcase = 1):
N, K = map(int, getIntArray(2))
A = getIntArray(N)
s = [set() for i in range(3005)]
for i in range(N):
for k in range(1, K + 1):
s[A[i] // k].add(i)
ans = 999999999999999999999999999999999999999999999999999999999999999999999
r = 0
freq = {}
for l in range(len(s)):
while len(freq) < N and r < len(s):
for v in s[r]:
if # TODO: Your code here: freq[v] = 0
freq[v] += 1
r += 1
if len(freq) < N: break
ans = min(ans, r - l - 1)
for v in s[l]:
if freq[v] == 1: del freq[v]
else: freq[v] -= 1
print(ans)
if testcase is None:
testcaseCount = 1
while tokens:
solve(testcaseCount)
testcaseCount += 1
else:
testcaseCount = nextInt() if testcase else 1
for tc in range(testcaseCount):
solve(tc + 1)
assert not tokens
|
import sys
tokens = ''.join(sys.stdin.readlines()).split()[::-1]
def next(): return tokens.pop()
def nextInt(): return int(next())
def nextFloat(): return float(next())
def getIntArray(n): return [nextInt() for _ in range(n)]
def getFloatArray(n): return [nextFloat() for _ in range(n)]
def getStringArray(n): return [next() for _ in range(n)]
testcase = True
def solve(testcase = 1):
N, K = map(int, getIntArray(2))
A = getIntArray(N)
s = [set() for i in range(3005)]
for i in range(N):
for k in range(1, K + 1):
s[A[i] // k].add(i)
ans = 999999999999999999999999999999999999999999999999999999999999999999999
r = 0
freq = {}
for l in range(len(s)):
while len(freq) < N and r < len(s):
for v in s[r]:
if {{completion}}: freq[v] = 0
freq[v] += 1
r += 1
if len(freq) < N: break
ans = min(ans, r - l - 1)
for v in s[l]:
if freq[v] == 1: del freq[v]
else: freq[v] -= 1
print(ans)
if testcase is None:
testcaseCount = 1
while tokens:
solve(testcaseCount)
testcaseCount += 1
else:
testcaseCount = nextInt() if testcase else 1
for tc in range(testcaseCount):
solve(tc + 1)
assert not tokens
|
v not in freq
|
[{"input": "7\n\n5 2\n\n4 5 6 8 11\n\n5 12\n\n4 5 6 8 11\n\n3 1\n\n2 9 15\n\n7 3\n\n2 3 5 5 6 9 10\n\n6 56\n\n54 286 527 1436 2450 2681\n\n3 95\n\n16 340 2241\n\n2 2\n\n1 3", "output": ["2\n0\n13\n1\n4\n7\n0"]}]
|
control_completion_003594
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: This is the easy version of the problem. The only difference between the versions is the constraints on $$$n$$$, $$$k$$$, $$$a_i$$$, and the sum of $$$n$$$ over all test cases. You can make hacks only if both versions of the problem are solved.Note the unusual memory limit.You are given an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$, and an integer $$$k$$$.The cost of an array of integers $$$p_1, p_2, \ldots, p_n$$$ of length $$$n$$$ is $$$$$$\max\limits_{1 \le i \le n}\left(\left \lfloor \frac{a_i}{p_i} \right \rfloor \right) - \min\limits_{1 \le i \le n}\left(\left \lfloor \frac{a_i}{p_i} \right \rfloor \right).$$$$$$Here, $$$\lfloor \frac{x}{y} \rfloor$$$ denotes the integer part of the division of $$$x$$$ by $$$y$$$. Find the minimum cost of an array $$$p$$$ such that $$$1 \le p_i \le k$$$ for all $$$1 \le i \le n$$$.
Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n, k \le 3000$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_1 \le a_2 \le \ldots \le a_n \le 3000$$$). It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3000$$$.
Output Specification: For each test case, print a single integer — the minimum possible cost of an array $$$p$$$ satisfying the condition above.
Notes: NoteIn the first test case, the optimal array is $$$p = [1, 1, 1, 2, 2]$$$. The resulting array of values of $$$\lfloor \frac{a_i}{p_i} \rfloor$$$ is $$$[4, 5, 6, 4, 5]$$$. The cost of $$$p$$$ is $$$\max\limits_{1 \le i \le n}(\lfloor \frac{a_i}{p_i} \rfloor) - \min\limits_{1 \le i \le n}(\lfloor \frac{a_i}{p_i} \rfloor) = 6 - 4 = 2$$$. We can show that there is no array (satisfying the condition from the statement) with a smaller cost.In the second test case, one of the optimal arrays is $$$p = [12, 12, 12, 12, 12]$$$, which results in all $$$\lfloor \frac{a_i}{p_i} \rfloor$$$ being $$$0$$$.In the third test case, the only possible array is $$$p = [1, 1, 1]$$$.
Code:
import sys
tokens = ''.join(sys.stdin.readlines()).split()[::-1]
def next(): return tokens.pop()
def nextInt(): return int(next())
def nextFloat(): return float(next())
def getIntArray(n): return [nextInt() for _ in range(n)]
def getFloatArray(n): return [nextFloat() for _ in range(n)]
def getStringArray(n): return [next() for _ in range(n)]
testcase = True
def solve(testcase = 1):
N, K = map(int, getIntArray(2))
A = getIntArray(N)
s = [set() for i in range(3005)]
for i in range(N):
for k in range(1, K + 1):
s[A[i] // k].add(i)
ans = 999999999999999999999999999999999999999999999999999999999999999999999
r = 0
freq = {}
for l in range(len(s)):
while len(freq) < N and r < len(s):
for # TODO: Your code here:
if v not in freq: freq[v] = 0
freq[v] += 1
r += 1
if len(freq) < N: break
ans = min(ans, r - l - 1)
for v in s[l]:
if freq[v] == 1: del freq[v]
else: freq[v] -= 1
print(ans)
if testcase is None:
testcaseCount = 1
while tokens:
solve(testcaseCount)
testcaseCount += 1
else:
testcaseCount = nextInt() if testcase else 1
for tc in range(testcaseCount):
solve(tc + 1)
assert not tokens
|
import sys
tokens = ''.join(sys.stdin.readlines()).split()[::-1]
def next(): return tokens.pop()
def nextInt(): return int(next())
def nextFloat(): return float(next())
def getIntArray(n): return [nextInt() for _ in range(n)]
def getFloatArray(n): return [nextFloat() for _ in range(n)]
def getStringArray(n): return [next() for _ in range(n)]
testcase = True
def solve(testcase = 1):
N, K = map(int, getIntArray(2))
A = getIntArray(N)
s = [set() for i in range(3005)]
for i in range(N):
for k in range(1, K + 1):
s[A[i] // k].add(i)
ans = 999999999999999999999999999999999999999999999999999999999999999999999
r = 0
freq = {}
for l in range(len(s)):
while len(freq) < N and r < len(s):
for {{completion}}:
if v not in freq: freq[v] = 0
freq[v] += 1
r += 1
if len(freq) < N: break
ans = min(ans, r - l - 1)
for v in s[l]:
if freq[v] == 1: del freq[v]
else: freq[v] -= 1
print(ans)
if testcase is None:
testcaseCount = 1
while tokens:
solve(testcaseCount)
testcaseCount += 1
else:
testcaseCount = nextInt() if testcase else 1
for tc in range(testcaseCount):
solve(tc + 1)
assert not tokens
|
v in s[r]
|
[{"input": "7\n\n5 2\n\n4 5 6 8 11\n\n5 12\n\n4 5 6 8 11\n\n3 1\n\n2 9 15\n\n7 3\n\n2 3 5 5 6 9 10\n\n6 56\n\n54 286 527 1436 2450 2681\n\n3 95\n\n16 340 2241\n\n2 2\n\n1 3", "output": ["2\n0\n13\n1\n4\n7\n0"]}]
|
control_completion_003595
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You have a sequence of $$$n$$$ colored blocks. The color of the $$$i$$$-th block is $$$c_i$$$, an integer between $$$1$$$ and $$$n$$$.You will place the blocks down in sequence on an infinite coordinate grid in the following way. Initially, you place block $$$1$$$ at $$$(0, 0)$$$. For $$$2 \le i \le n$$$, if the $$$(i - 1)$$$-th block is placed at position $$$(x, y)$$$, then the $$$i$$$-th block can be placed at one of positions $$$(x + 1, y)$$$, $$$(x - 1, y)$$$, $$$(x, y + 1)$$$ (but not at position $$$(x, y - 1)$$$), as long no previous block was placed at that position. A tower is formed by $$$s$$$ blocks such that they are placed at positions $$$(x, y), (x, y + 1), \ldots, (x, y + s - 1)$$$ for some position $$$(x, y)$$$ and integer $$$s$$$. The size of the tower is $$$s$$$, the number of blocks in it. A tower of color $$$r$$$ is a tower such that all blocks in it have the color $$$r$$$.For each color $$$r$$$ from $$$1$$$ to $$$n$$$, solve the following problem independently: Find the maximum size of a tower of color $$$r$$$ that you can form by placing down the blocks according to the rules.
Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$1 \le c_i \le n$$$). It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.
Output Specification: For each test case, output $$$n$$$ integers. The $$$r$$$-th of them should be the maximum size of an tower of color $$$r$$$ you can form by following the given rules. If you cannot form any tower of color $$$r$$$, the $$$r$$$-th integer should be $$$0$$$.
Notes: NoteIn the first test case, one of the possible ways to form a tower of color $$$1$$$ and size $$$3$$$ is: place block $$$1$$$ at position $$$(0, 0)$$$; place block $$$2$$$ to the right of block $$$1$$$, at position $$$(1, 0)$$$; place block $$$3$$$ above block $$$2$$$, at position $$$(1, 1)$$$; place block $$$4$$$ to the left of block $$$3$$$, at position $$$(0, 1)$$$; place block $$$5$$$ to the left of block $$$4$$$, at position $$$(-1, 1)$$$; place block $$$6$$$ above block $$$5$$$, at position $$$(-1, 2)$$$; place block $$$7$$$ to the right of block $$$6$$$, at position $$$(0, 2)$$$. The blocks at positions $$$(0, 0)$$$, $$$(0, 1)$$$, and $$$(0, 2)$$$ all have color $$$1$$$, forming an tower of size $$$3$$$.In the second test case, note that the following placement is not valid, since you are not allowed to place block $$$6$$$ under block $$$5$$$: It can be shown that it is impossible to form a tower of color $$$4$$$ and size $$$3$$$.
Code:
import sys
tokens = ''.join(sys.stdin.readlines()).split()[::-1]
def next(): return tokens.pop()
def nextInt(): return int(next())
def nextFloat(): return float(next())
def getIntArray(n): return [nextInt() for _ in range(n)]
def getFloatArray(n): return [nextFloat() for _ in range(n)]
def getStringArray(n): return [next() for _ in range(n)]
testcase = True
def solve(testcase = 1):
N = nextInt()
A = getIntArray(N)
map = {}
for i in range(N):
if A[i] not in map: map[A[i]] = []
map[A[i]].append(i)
for color in range(1, N + 1):
if color not in map:
print(0, end=' ')
continue
ar = map[color]
oddCount = evenCount = 0
for i in ar:
if # TODO: Your code here:
evenCount = max(evenCount, oddCount + 1)
else:
oddCount = max(oddCount, evenCount + 1)
print(max(oddCount, evenCount), end=' ')
print()
if testcase is None:
testcaseCount = 1
while tokens:
solve(testcaseCount)
testcaseCount += 1
else:
testcaseCount = nextInt() if testcase else 1
for tc in range(testcaseCount):
solve(tc + 1)
assert not tokens
|
import sys
tokens = ''.join(sys.stdin.readlines()).split()[::-1]
def next(): return tokens.pop()
def nextInt(): return int(next())
def nextFloat(): return float(next())
def getIntArray(n): return [nextInt() for _ in range(n)]
def getFloatArray(n): return [nextFloat() for _ in range(n)]
def getStringArray(n): return [next() for _ in range(n)]
testcase = True
def solve(testcase = 1):
N = nextInt()
A = getIntArray(N)
map = {}
for i in range(N):
if A[i] not in map: map[A[i]] = []
map[A[i]].append(i)
for color in range(1, N + 1):
if color not in map:
print(0, end=' ')
continue
ar = map[color]
oddCount = evenCount = 0
for i in ar:
if {{completion}}:
evenCount = max(evenCount, oddCount + 1)
else:
oddCount = max(oddCount, evenCount + 1)
print(max(oddCount, evenCount), end=' ')
print()
if testcase is None:
testcaseCount = 1
while tokens:
solve(testcaseCount)
testcaseCount += 1
else:
testcaseCount = nextInt() if testcase else 1
for tc in range(testcaseCount):
solve(tc + 1)
assert not tokens
|
i % 2 == 0
|
[{"input": "6\n\n7\n\n1 2 3 1 2 3 1\n\n6\n\n4 2 2 2 4 4\n\n1\n\n1\n\n5\n\n5 4 5 3 5\n\n6\n\n3 3 3 1 3 3\n\n8\n\n1 2 3 4 4 3 2 1", "output": ["3 2 2 0 0 0 0 \n0 3 0 2 0 0 \n1 \n0 0 1 1 1 \n1 0 4 0 0 0 \n2 2 2 2 0 0 0 0"]}]
|
control_completion_003617
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: A basketball competition is held where the number of players in a team does not have a maximum or minimum limit (not necessarily $$$5$$$ players in one team for each match). There are $$$N$$$ candidate players in the competition that will be trained by Pak Chanek, the best basketball coach on earth. The $$$i$$$-th candidate player has a power of $$$P_i$$$.Pak Chanek will form zero or more teams from the $$$N$$$ candidate players on the condition that each candidate player may only join in at most one team. Each of Pak Chanek's teams will be sent to compete once with an enemy team that has a power of $$$D$$$. In each match, the team sent is said to defeat the enemy team if the sum of powers from the formed players is strictly greater than $$$D$$$.One of Pak Chanek's skills is that when a team that has been formed plays in a match, he can change the power of each player in the team to be equal to the biggest player power from the team.Determine the maximum number of wins that can be achieved by Pak Chanek.
Input Specification: The first line contains two integers $$$N$$$ and $$$D$$$ ($$$1 \le N \le 10^5$$$, $$$1 \le D \le 10^9$$$) — the number of candidate players and the power of the enemy team. The second line contains $$$N$$$ integers $$$P_1, P_2, \ldots, P_N$$$ ($$$1 \le P_i \le 10^9$$$) — the powers of all candidate players.
Output Specification: A line containing an integer representing the maximum number of wins that can be achieved by Pak Chanek.
Notes: NoteThe $$$1$$$-st team formed is a team containing players $$$4$$$ and $$$6$$$. The power of each player in the team becomes $$$100$$$. So the total power of the team is $$$100 + 100 = 200 > 180$$$.The $$$2$$$-nd team formed is a team containing players $$$1$$$, $$$2$$$, and $$$5$$$. The power of each player in the team becomes $$$90$$$. So the total power of the team is $$$90 + 90 + 90 = 270 > 180$$$.
Code:
n, d = map(int, input().split())
p = sorted(map(int, input().split()), reverse=True)
ans = 0
for num in p:
if # TODO: Your code here:
n -= d // num + 1
ans += 1
else:
break
print(ans)
|
n, d = map(int, input().split())
p = sorted(map(int, input().split()), reverse=True)
ans = 0
for num in p:
if {{completion}}:
n -= d // num + 1
ans += 1
else:
break
print(ans)
|
n >= d // num + 1
|
[{"input": "6 180\n90 80 70 60 50 100", "output": ["2"]}]
|
control_completion_003663
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: A basketball competition is held where the number of players in a team does not have a maximum or minimum limit (not necessarily $$$5$$$ players in one team for each match). There are $$$N$$$ candidate players in the competition that will be trained by Pak Chanek, the best basketball coach on earth. The $$$i$$$-th candidate player has a power of $$$P_i$$$.Pak Chanek will form zero or more teams from the $$$N$$$ candidate players on the condition that each candidate player may only join in at most one team. Each of Pak Chanek's teams will be sent to compete once with an enemy team that has a power of $$$D$$$. In each match, the team sent is said to defeat the enemy team if the sum of powers from the formed players is strictly greater than $$$D$$$.One of Pak Chanek's skills is that when a team that has been formed plays in a match, he can change the power of each player in the team to be equal to the biggest player power from the team.Determine the maximum number of wins that can be achieved by Pak Chanek.
Input Specification: The first line contains two integers $$$N$$$ and $$$D$$$ ($$$1 \le N \le 10^5$$$, $$$1 \le D \le 10^9$$$) — the number of candidate players and the power of the enemy team. The second line contains $$$N$$$ integers $$$P_1, P_2, \ldots, P_N$$$ ($$$1 \le P_i \le 10^9$$$) — the powers of all candidate players.
Output Specification: A line containing an integer representing the maximum number of wins that can be achieved by Pak Chanek.
Notes: NoteThe $$$1$$$-st team formed is a team containing players $$$4$$$ and $$$6$$$. The power of each player in the team becomes $$$100$$$. So the total power of the team is $$$100 + 100 = 200 > 180$$$.The $$$2$$$-nd team formed is a team containing players $$$1$$$, $$$2$$$, and $$$5$$$. The power of each player in the team becomes $$$90$$$. So the total power of the team is $$$90 + 90 + 90 = 270 > 180$$$.
Code:
def solve():
n, d = [int(i) for i in input().split(' ')]
power = [int(i) for i in input().split(' ')]
power.sort()
used = 0
w = 0
for i in range(len(power)-1, -1, -1):
min_players = -(d // -power[i])
p = power[i] * min_players
if(p > d):
used += min_players
elif# TODO: Your code here:
used += min_players + 1
if(used > n):
break
w += 1
print(w)
solve()
|
def solve():
n, d = [int(i) for i in input().split(' ')]
power = [int(i) for i in input().split(' ')]
power.sort()
used = 0
w = 0
for i in range(len(power)-1, -1, -1):
min_players = -(d // -power[i])
p = power[i] * min_players
if(p > d):
used += min_players
elif{{completion}}:
used += min_players + 1
if(used > n):
break
w += 1
print(w)
solve()
|
(p == d)
|
[{"input": "6 180\n90 80 70 60 50 100", "output": ["2"]}]
|
control_completion_003664
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: A basketball competition is held where the number of players in a team does not have a maximum or minimum limit (not necessarily $$$5$$$ players in one team for each match). There are $$$N$$$ candidate players in the competition that will be trained by Pak Chanek, the best basketball coach on earth. The $$$i$$$-th candidate player has a power of $$$P_i$$$.Pak Chanek will form zero or more teams from the $$$N$$$ candidate players on the condition that each candidate player may only join in at most one team. Each of Pak Chanek's teams will be sent to compete once with an enemy team that has a power of $$$D$$$. In each match, the team sent is said to defeat the enemy team if the sum of powers from the formed players is strictly greater than $$$D$$$.One of Pak Chanek's skills is that when a team that has been formed plays in a match, he can change the power of each player in the team to be equal to the biggest player power from the team.Determine the maximum number of wins that can be achieved by Pak Chanek.
Input Specification: The first line contains two integers $$$N$$$ and $$$D$$$ ($$$1 \le N \le 10^5$$$, $$$1 \le D \le 10^9$$$) — the number of candidate players and the power of the enemy team. The second line contains $$$N$$$ integers $$$P_1, P_2, \ldots, P_N$$$ ($$$1 \le P_i \le 10^9$$$) — the powers of all candidate players.
Output Specification: A line containing an integer representing the maximum number of wins that can be achieved by Pak Chanek.
Notes: NoteThe $$$1$$$-st team formed is a team containing players $$$4$$$ and $$$6$$$. The power of each player in the team becomes $$$100$$$. So the total power of the team is $$$100 + 100 = 200 > 180$$$.The $$$2$$$-nd team formed is a team containing players $$$1$$$, $$$2$$$, and $$$5$$$. The power of each player in the team becomes $$$90$$$. So the total power of the team is $$$90 + 90 + 90 = 270 > 180$$$.
Code:
import math
enemy_power=int(input().split()[1])
team=[int(i) for i in input().split()]
team.sort()
days=0
while len(team)>0:
num=enemy_power//team[-1]+1
if # TODO: Your code here:
break;
else:
del team[-1]
del team[0:num-1]
days+=1
print(days)
|
import math
enemy_power=int(input().split()[1])
team=[int(i) for i in input().split()]
team.sort()
days=0
while len(team)>0:
num=enemy_power//team[-1]+1
if {{completion}}:
break;
else:
del team[-1]
del team[0:num-1]
days+=1
print(days)
|
len(team)<num
|
[{"input": "6 180\n90 80 70 60 50 100", "output": ["2"]}]
|
control_completion_003665
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: A basketball competition is held where the number of players in a team does not have a maximum or minimum limit (not necessarily $$$5$$$ players in one team for each match). There are $$$N$$$ candidate players in the competition that will be trained by Pak Chanek, the best basketball coach on earth. The $$$i$$$-th candidate player has a power of $$$P_i$$$.Pak Chanek will form zero or more teams from the $$$N$$$ candidate players on the condition that each candidate player may only join in at most one team. Each of Pak Chanek's teams will be sent to compete once with an enemy team that has a power of $$$D$$$. In each match, the team sent is said to defeat the enemy team if the sum of powers from the formed players is strictly greater than $$$D$$$.One of Pak Chanek's skills is that when a team that has been formed plays in a match, he can change the power of each player in the team to be equal to the biggest player power from the team.Determine the maximum number of wins that can be achieved by Pak Chanek.
Input Specification: The first line contains two integers $$$N$$$ and $$$D$$$ ($$$1 \le N \le 10^5$$$, $$$1 \le D \le 10^9$$$) — the number of candidate players and the power of the enemy team. The second line contains $$$N$$$ integers $$$P_1, P_2, \ldots, P_N$$$ ($$$1 \le P_i \le 10^9$$$) — the powers of all candidate players.
Output Specification: A line containing an integer representing the maximum number of wins that can be achieved by Pak Chanek.
Notes: NoteThe $$$1$$$-st team formed is a team containing players $$$4$$$ and $$$6$$$. The power of each player in the team becomes $$$100$$$. So the total power of the team is $$$100 + 100 = 200 > 180$$$.The $$$2$$$-nd team formed is a team containing players $$$1$$$, $$$2$$$, and $$$5$$$. The power of each player in the team becomes $$$90$$$. So the total power of the team is $$$90 + 90 + 90 = 270 > 180$$$.
Code:
import sys,math
n,team=map(int,sys.stdin.readline().split())
arr=sorted(map(int,sys.stdin.readline().split()),reverse=True)
# print(arr)
all=n+1
count=0
for i in range(n):
sub=int(math.floor(team/arr[i])+1)
all-=sub
if # TODO: Your code here:
count+=1
else:break
print(count)
|
import sys,math
n,team=map(int,sys.stdin.readline().split())
arr=sorted(map(int,sys.stdin.readline().split()),reverse=True)
# print(arr)
all=n+1
count=0
for i in range(n):
sub=int(math.floor(team/arr[i])+1)
all-=sub
if {{completion}}:
count+=1
else:break
print(count)
|
all>0
|
[{"input": "6 180\n90 80 70 60 50 100", "output": ["2"]}]
|
control_completion_003666
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: A basketball competition is held where the number of players in a team does not have a maximum or minimum limit (not necessarily $$$5$$$ players in one team for each match). There are $$$N$$$ candidate players in the competition that will be trained by Pak Chanek, the best basketball coach on earth. The $$$i$$$-th candidate player has a power of $$$P_i$$$.Pak Chanek will form zero or more teams from the $$$N$$$ candidate players on the condition that each candidate player may only join in at most one team. Each of Pak Chanek's teams will be sent to compete once with an enemy team that has a power of $$$D$$$. In each match, the team sent is said to defeat the enemy team if the sum of powers from the formed players is strictly greater than $$$D$$$.One of Pak Chanek's skills is that when a team that has been formed plays in a match, he can change the power of each player in the team to be equal to the biggest player power from the team.Determine the maximum number of wins that can be achieved by Pak Chanek.
Input Specification: The first line contains two integers $$$N$$$ and $$$D$$$ ($$$1 \le N \le 10^5$$$, $$$1 \le D \le 10^9$$$) — the number of candidate players and the power of the enemy team. The second line contains $$$N$$$ integers $$$P_1, P_2, \ldots, P_N$$$ ($$$1 \le P_i \le 10^9$$$) — the powers of all candidate players.
Output Specification: A line containing an integer representing the maximum number of wins that can be achieved by Pak Chanek.
Notes: NoteThe $$$1$$$-st team formed is a team containing players $$$4$$$ and $$$6$$$. The power of each player in the team becomes $$$100$$$. So the total power of the team is $$$100 + 100 = 200 > 180$$$.The $$$2$$$-nd team formed is a team containing players $$$1$$$, $$$2$$$, and $$$5$$$. The power of each player in the team becomes $$$90$$$. So the total power of the team is $$$90 + 90 + 90 = 270 > 180$$$.
Code:
def solve():
n,d=map(int,input().split())
a=sorted([*map(int,input().split())])[::-1]
i,j,r=0,len(a),0
while i<j:
x=a[i]
while x<=d:
j-=1
if # TODO: Your code here:
x+=a[i]
else:
return r
else:
r+=1
i+=1
return r
print(solve())
|
def solve():
n,d=map(int,input().split())
a=sorted([*map(int,input().split())])[::-1]
i,j,r=0,len(a),0
while i<j:
x=a[i]
while x<=d:
j-=1
if {{completion}}:
x+=a[i]
else:
return r
else:
r+=1
i+=1
return r
print(solve())
|
i<j
|
[{"input": "6 180\n90 80 70 60 50 100", "output": ["2"]}]
|
control_completion_003667
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: A basketball competition is held where the number of players in a team does not have a maximum or minimum limit (not necessarily $$$5$$$ players in one team for each match). There are $$$N$$$ candidate players in the competition that will be trained by Pak Chanek, the best basketball coach on earth. The $$$i$$$-th candidate player has a power of $$$P_i$$$.Pak Chanek will form zero or more teams from the $$$N$$$ candidate players on the condition that each candidate player may only join in at most one team. Each of Pak Chanek's teams will be sent to compete once with an enemy team that has a power of $$$D$$$. In each match, the team sent is said to defeat the enemy team if the sum of powers from the formed players is strictly greater than $$$D$$$.One of Pak Chanek's skills is that when a team that has been formed plays in a match, he can change the power of each player in the team to be equal to the biggest player power from the team.Determine the maximum number of wins that can be achieved by Pak Chanek.
Input Specification: The first line contains two integers $$$N$$$ and $$$D$$$ ($$$1 \le N \le 10^5$$$, $$$1 \le D \le 10^9$$$) — the number of candidate players and the power of the enemy team. The second line contains $$$N$$$ integers $$$P_1, P_2, \ldots, P_N$$$ ($$$1 \le P_i \le 10^9$$$) — the powers of all candidate players.
Output Specification: A line containing an integer representing the maximum number of wins that can be achieved by Pak Chanek.
Notes: NoteThe $$$1$$$-st team formed is a team containing players $$$4$$$ and $$$6$$$. The power of each player in the team becomes $$$100$$$. So the total power of the team is $$$100 + 100 = 200 > 180$$$.The $$$2$$$-nd team formed is a team containing players $$$1$$$, $$$2$$$, and $$$5$$$. The power of each player in the team becomes $$$90$$$. So the total power of the team is $$$90 + 90 + 90 = 270 > 180$$$.
Code:
from sys import stdin,stdout
def ans():
n,d=map(int,stdin.readline().strip().split())
p=list(map(int,stdin.readline().strip().split()))
temp=int(n)
ans=0
for x in sorted(p,reverse=True):
if # TODO: Your code here:
temp-=((d//x)+1)
ans+=1
print(ans)
if __name__=='__main__':
ans()
|
from sys import stdin,stdout
def ans():
n,d=map(int,stdin.readline().strip().split())
p=list(map(int,stdin.readline().strip().split()))
temp=int(n)
ans=0
for x in sorted(p,reverse=True):
if {{completion}}:
temp-=((d//x)+1)
ans+=1
print(ans)
if __name__=='__main__':
ans()
|
temp>=((d//x)+1)
|
[{"input": "6 180\n90 80 70 60 50 100", "output": ["2"]}]
|
control_completion_003668
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: A basketball competition is held where the number of players in a team does not have a maximum or minimum limit (not necessarily $$$5$$$ players in one team for each match). There are $$$N$$$ candidate players in the competition that will be trained by Pak Chanek, the best basketball coach on earth. The $$$i$$$-th candidate player has a power of $$$P_i$$$.Pak Chanek will form zero or more teams from the $$$N$$$ candidate players on the condition that each candidate player may only join in at most one team. Each of Pak Chanek's teams will be sent to compete once with an enemy team that has a power of $$$D$$$. In each match, the team sent is said to defeat the enemy team if the sum of powers from the formed players is strictly greater than $$$D$$$.One of Pak Chanek's skills is that when a team that has been formed plays in a match, he can change the power of each player in the team to be equal to the biggest player power from the team.Determine the maximum number of wins that can be achieved by Pak Chanek.
Input Specification: The first line contains two integers $$$N$$$ and $$$D$$$ ($$$1 \le N \le 10^5$$$, $$$1 \le D \le 10^9$$$) — the number of candidate players and the power of the enemy team. The second line contains $$$N$$$ integers $$$P_1, P_2, \ldots, P_N$$$ ($$$1 \le P_i \le 10^9$$$) — the powers of all candidate players.
Output Specification: A line containing an integer representing the maximum number of wins that can be achieved by Pak Chanek.
Notes: NoteThe $$$1$$$-st team formed is a team containing players $$$4$$$ and $$$6$$$. The power of each player in the team becomes $$$100$$$. So the total power of the team is $$$100 + 100 = 200 > 180$$$.The $$$2$$$-nd team formed is a team containing players $$$1$$$, $$$2$$$, and $$$5$$$. The power of each player in the team becomes $$$90$$$. So the total power of the team is $$$90 + 90 + 90 = 270 > 180$$$.
Code:
d = int(input().split(" ")[1])
p = sorted(map(int, input().split(" ")))
c = 0
l = 0
r = len(p) - 1
s = p[r]
while r > l:
while # TODO: Your code here:
s += p[r]
l += 1
if l > r:
break
r -= 1
s = p[r]
c += 1
if p[0] > d:
c += 1
print(c)
|
d = int(input().split(" ")[1])
p = sorted(map(int, input().split(" ")))
c = 0
l = 0
r = len(p) - 1
s = p[r]
while r > l:
while {{completion}}:
s += p[r]
l += 1
if l > r:
break
r -= 1
s = p[r]
c += 1
if p[0] > d:
c += 1
print(c)
|
s <= d
|
[{"input": "6 180\n90 80 70 60 50 100", "output": ["2"]}]
|
control_completion_003669
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Pak Chanek is playing one of his favourite board games. In the game, there is a directed graph with $$$N$$$ vertices and $$$M$$$ edges. In the graph, edge $$$i$$$ connects two different vertices $$$U_i$$$ and $$$V_i$$$ with a length of $$$W_i$$$. By using the $$$i$$$-th edge, something can move from $$$U_i$$$ to $$$V_i$$$, but not from $$$V_i$$$ to $$$U_i$$$.To play this game, initially Pak Chanek must place both of his hands onto two different vertices. In one move, he can move one of his hands to another vertex using an edge. To move a hand from vertex $$$U_i$$$ to vertex $$$V_i$$$, Pak Chanek needs a time of $$$W_i$$$ seconds. Note that Pak Chanek can only move one hand at a time. This game ends when both of Pak Chanek's hands are on the same vertex.Pak Chanek has several questions. For each $$$p$$$ satisfying $$$2 \leq p \leq N$$$, you need to find the minimum time in seconds needed for Pak Chanek to end the game if initially Pak Chanek's left hand and right hand are placed on vertex $$$1$$$ and vertex $$$p$$$, or report if it is impossible.
Input Specification: The first line contains two integers $$$N$$$ and $$$M$$$ ($$$2 \leq N \leq 10^5$$$, $$$0 \leq M \leq 2 \cdot 10^5$$$) — the number of vertices and edges in the graph. The $$$i$$$-th of the next $$$M$$$ lines contains three integers $$$U_i$$$, $$$V_i$$$, and $$$W_i$$$ ($$$1 \le U_i, V_i \le N$$$, $$$U_i \neq V_i$$$, $$$1 \le W_i \le 10^9$$$) — a directed edge that connects two different vertices $$$U_i$$$ and $$$V_i$$$ with a length of $$$W_i$$$. There is no pair of different edges $$$i$$$ and $$$j$$$ such that $$$U_i = U_j$$$ and $$$V_i = V_j$$$.
Output Specification: Output a line containing $$$N-1$$$ integers. The $$$j$$$-th integer represents the minimum time in seconds needed by Pak Chanek to end the game if initially Pak Chanek's left hand and right hand are placed on vertex $$$1$$$ and vertex $$$j+1$$$, or $$$-1$$$ if it is impossible.
Notes: NoteIf initially Pak Chanek's left hand is on vertex $$$1$$$ and his right hand is on vertex $$$5$$$, Pak Chanek can do the following moves: Move his right hand to vertex $$$4$$$ in $$$1$$$ second. Move his left hand to vertex $$$2$$$ in $$$2$$$ seconds. Move his left hand to vertex $$$4$$$ in $$$1$$$ second. In total it needs $$$1+2+1=4$$$ seconds. It can be proven that there is no other way that is faster.
Code:
import sys
input=sys.stdin.readline
from heapq import heappush,heappop,heapify
from collections import defaultdict
def main():
n,m=map(int,input().split())
gf=defaultdict(list)
gb=defaultdict(list)
for _ in range(m):
u,v,w=map(int,input().split())
gf[u].append((v,w))
gb[v].append((u,w))
dis=[float('inf')]*(n+1)
dis[1]=0
h=[]
heapify(h)
heappush(h,(0,1))
while h:
cd,cn=heappop(h)
if dis[cn]==cd:
for nn,nw in gf[cn]:
if # TODO: Your code here:
dis[nn]=cd+nw
heappush(h,(nw+cd,nn))
res=[float('inf')]*(n+1)
res[1]=0
h=[]
for i in range(1,n+1):
if dis[i]!=float('inf'):
res[i]=dis[i]
h.append((dis[i],i))
heapify(h)
while h:
cd,cn=heappop(h)
if res[cn]==cd:
for nn,nw in gb[cn]:
if cd+nw<res[nn]:
res[nn]=cd+nw
heappush(h,(nw+cd,nn))
for i in range(1,len(res)):
if res[i]==float('inf'):res[i]=-1
print(*res[2:])
main()
|
import sys
input=sys.stdin.readline
from heapq import heappush,heappop,heapify
from collections import defaultdict
def main():
n,m=map(int,input().split())
gf=defaultdict(list)
gb=defaultdict(list)
for _ in range(m):
u,v,w=map(int,input().split())
gf[u].append((v,w))
gb[v].append((u,w))
dis=[float('inf')]*(n+1)
dis[1]=0
h=[]
heapify(h)
heappush(h,(0,1))
while h:
cd,cn=heappop(h)
if dis[cn]==cd:
for nn,nw in gf[cn]:
if {{completion}}:
dis[nn]=cd+nw
heappush(h,(nw+cd,nn))
res=[float('inf')]*(n+1)
res[1]=0
h=[]
for i in range(1,n+1):
if dis[i]!=float('inf'):
res[i]=dis[i]
h.append((dis[i],i))
heapify(h)
while h:
cd,cn=heappop(h)
if res[cn]==cd:
for nn,nw in gb[cn]:
if cd+nw<res[nn]:
res[nn]=cd+nw
heappush(h,(nw+cd,nn))
for i in range(1,len(res)):
if res[i]==float('inf'):res[i]=-1
print(*res[2:])
main()
|
cd+nw<dis[nn]
|
[{"input": "5 7\n1 2 2\n2 4 1\n4 1 4\n2 5 3\n5 4 1\n5 2 4\n2 1 1", "output": ["1 -1 3 4"]}]
|
control_completion_003685
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Pak Chanek is playing one of his favourite board games. In the game, there is a directed graph with $$$N$$$ vertices and $$$M$$$ edges. In the graph, edge $$$i$$$ connects two different vertices $$$U_i$$$ and $$$V_i$$$ with a length of $$$W_i$$$. By using the $$$i$$$-th edge, something can move from $$$U_i$$$ to $$$V_i$$$, but not from $$$V_i$$$ to $$$U_i$$$.To play this game, initially Pak Chanek must place both of his hands onto two different vertices. In one move, he can move one of his hands to another vertex using an edge. To move a hand from vertex $$$U_i$$$ to vertex $$$V_i$$$, Pak Chanek needs a time of $$$W_i$$$ seconds. Note that Pak Chanek can only move one hand at a time. This game ends when both of Pak Chanek's hands are on the same vertex.Pak Chanek has several questions. For each $$$p$$$ satisfying $$$2 \leq p \leq N$$$, you need to find the minimum time in seconds needed for Pak Chanek to end the game if initially Pak Chanek's left hand and right hand are placed on vertex $$$1$$$ and vertex $$$p$$$, or report if it is impossible.
Input Specification: The first line contains two integers $$$N$$$ and $$$M$$$ ($$$2 \leq N \leq 10^5$$$, $$$0 \leq M \leq 2 \cdot 10^5$$$) — the number of vertices and edges in the graph. The $$$i$$$-th of the next $$$M$$$ lines contains three integers $$$U_i$$$, $$$V_i$$$, and $$$W_i$$$ ($$$1 \le U_i, V_i \le N$$$, $$$U_i \neq V_i$$$, $$$1 \le W_i \le 10^9$$$) — a directed edge that connects two different vertices $$$U_i$$$ and $$$V_i$$$ with a length of $$$W_i$$$. There is no pair of different edges $$$i$$$ and $$$j$$$ such that $$$U_i = U_j$$$ and $$$V_i = V_j$$$.
Output Specification: Output a line containing $$$N-1$$$ integers. The $$$j$$$-th integer represents the minimum time in seconds needed by Pak Chanek to end the game if initially Pak Chanek's left hand and right hand are placed on vertex $$$1$$$ and vertex $$$j+1$$$, or $$$-1$$$ if it is impossible.
Notes: NoteIf initially Pak Chanek's left hand is on vertex $$$1$$$ and his right hand is on vertex $$$5$$$, Pak Chanek can do the following moves: Move his right hand to vertex $$$4$$$ in $$$1$$$ second. Move his left hand to vertex $$$2$$$ in $$$2$$$ seconds. Move his left hand to vertex $$$4$$$ in $$$1$$$ second. In total it needs $$$1+2+1=4$$$ seconds. It can be proven that there is no other way that is faster.
Code:
import sys
input=sys.stdin.readline
from heapq import heappush,heappop,heapify
from collections import defaultdict
def main():
n,m=map(int,input().split())
gf=defaultdict(list)
gb=defaultdict(list)
for _ in range(m):
u,v,w=map(int,input().split())
gf[u].append((v,w))
gb[v].append((u,w))
dis=[float('inf')]*(n+1)
dis[1]=0
h=[]
heapify(h)
heappush(h,(0,1))
while h:
cd,cn=heappop(h)
if dis[cn]==cd:
for nn,nw in gf[cn]:
if cd+nw<dis[nn]:
dis[nn]=cd+nw
heappush(h,(nw+cd,nn))
res=[float('inf')]*(n+1)
res[1]=0
h=[]
for i in range(1,n+1):
if dis[i]!=float('inf'):
res[i]=dis[i]
h.append((dis[i],i))
heapify(h)
while h:
cd,cn=heappop(h)
if res[cn]==cd:
for nn,nw in gb[cn]:
if # TODO: Your code here:
res[nn]=cd+nw
heappush(h,(nw+cd,nn))
for i in range(1,len(res)):
if res[i]==float('inf'):res[i]=-1
print(*res[2:])
main()
|
import sys
input=sys.stdin.readline
from heapq import heappush,heappop,heapify
from collections import defaultdict
def main():
n,m=map(int,input().split())
gf=defaultdict(list)
gb=defaultdict(list)
for _ in range(m):
u,v,w=map(int,input().split())
gf[u].append((v,w))
gb[v].append((u,w))
dis=[float('inf')]*(n+1)
dis[1]=0
h=[]
heapify(h)
heappush(h,(0,1))
while h:
cd,cn=heappop(h)
if dis[cn]==cd:
for nn,nw in gf[cn]:
if cd+nw<dis[nn]:
dis[nn]=cd+nw
heappush(h,(nw+cd,nn))
res=[float('inf')]*(n+1)
res[1]=0
h=[]
for i in range(1,n+1):
if dis[i]!=float('inf'):
res[i]=dis[i]
h.append((dis[i],i))
heapify(h)
while h:
cd,cn=heappop(h)
if res[cn]==cd:
for nn,nw in gb[cn]:
if {{completion}}:
res[nn]=cd+nw
heappush(h,(nw+cd,nn))
for i in range(1,len(res)):
if res[i]==float('inf'):res[i]=-1
print(*res[2:])
main()
|
cd+nw<res[nn]
|
[{"input": "5 7\n1 2 2\n2 4 1\n4 1 4\n2 5 3\n5 4 1\n5 2 4\n2 1 1", "output": ["1 -1 3 4"]}]
|
control_completion_003686
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Let's say Pak Chanek has an array $$$A$$$ consisting of $$$N$$$ positive integers. Pak Chanek will do a number of operations. In each operation, Pak Chanek will do the following: Choose an index $$$p$$$ ($$$1 \leq p \leq N$$$). Let $$$c$$$ be the number of operations that have been done on index $$$p$$$ before this operation. Decrease the value of $$$A_p$$$ by $$$2^c$$$. Multiply the value of $$$A_p$$$ by $$$2$$$. After each operation, all elements of $$$A$$$ must be positive integers.An array $$$A$$$ is said to be sortable if and only if Pak Chanek can do zero or more operations so that $$$A_1 < A_2 < A_3 < A_4 < \ldots < A_N$$$.Pak Chanek must find an array $$$A$$$ that is sortable with length $$$N$$$ such that $$$A_1 + A_2 + A_3 + A_4 + \ldots + A_N$$$ is the minimum possible. If there are more than one possibilities, Pak Chanek must choose the array that is lexicographically minimum among them.Pak Chanek must solve the following things: Pak Chanek must print the value of $$$A_1 + A_2 + A_3 + A_4 + \ldots + A_N$$$ for that array. $$$Q$$$ questions will be given. For the $$$i$$$-th question, an integer $$$P_i$$$ is given. Pak Chanek must print the value of $$$A_{P_i}$$$. Help Pak Chanek solve the problem.Note: an array $$$B$$$ of size $$$N$$$ is said to be lexicographically smaller than an array $$$C$$$ that is also of size $$$N$$$ if and only if there exists an index $$$i$$$ such that $$$B_i < C_i$$$ and for each $$$j < i$$$, $$$B_j = C_j$$$.
Input Specification: The first line contains two integers $$$N$$$ and $$$Q$$$ ($$$1 \leq N \leq 10^9$$$, $$$0 \leq Q \leq \min(N, 10^5)$$$) — the required length of array $$$A$$$ and the number of questions. The $$$i$$$-th of the next $$$Q$$$ lines contains a single integer $$$P_i$$$ ($$$1 \leq P_1 < P_2 < \ldots < P_Q \leq N$$$) — the index asked in the $$$i$$$-th question.
Output Specification: Print $$$Q+1$$$ lines. The $$$1$$$-st line contains an integer representing $$$A_1 + A_2 + A_3 + A_4 + \ldots + A_N$$$. For each $$$1 \leq i \leq Q$$$, the $$$(i+1)$$$-th line contains an integer representing $$$A_{P_i}$$$.
Notes: NoteIn the first example, the array $$$A$$$ obtained is $$$[1, 2, 3, 3, 4, 4]$$$. We can see that the array is sortable by doing the following operations: Choose index $$$5$$$, then $$$A = [1, 2, 3, 3, 6, 4]$$$. Choose index $$$6$$$, then $$$A = [1, 2, 3, 3, 6, 6]$$$. Choose index $$$4$$$, then $$$A = [1, 2, 3, 4, 6, 6]$$$. Choose index $$$6$$$, then $$$A = [1, 2, 3, 4, 6, 8]$$$.
Code:
def ev(x):
k = 0
while x%2 == 0:
x//=2
k+=1
return x+k
p,q = [int(i) for i in input().split()]
t,s,sum = 1,1,0
while True:
ss = s + (t+1)//2
if ss > p:
m = (p-s+1)
sum += m*t
break
sum += t*((t+1)//2)
s = ss
t += 1
mx = t-2*m+1
print(sum)
mul,ded,turn = 0,0,0
t -= 0 if mx==turn else 1
for i in range(q):
x = int(input())
if x == 1:
print(1)
continue
while True:
if # TODO: Your code here:
print(ev(x-ded)+mul)
break
ded += (t+1)//2
turn+=1
mul += 1
t -= 0 if mx==turn else 1
|
def ev(x):
k = 0
while x%2 == 0:
x//=2
k+=1
return x+k
p,q = [int(i) for i in input().split()]
t,s,sum = 1,1,0
while True:
ss = s + (t+1)//2
if ss > p:
m = (p-s+1)
sum += m*t
break
sum += t*((t+1)//2)
s = ss
t += 1
mx = t-2*m+1
print(sum)
mul,ded,turn = 0,0,0
t -= 0 if mx==turn else 1
for i in range(q):
x = int(input())
if x == 1:
print(1)
continue
while True:
if {{completion}}:
print(ev(x-ded)+mul)
break
ded += (t+1)//2
turn+=1
mul += 1
t -= 0 if mx==turn else 1
|
x-ded <= t
|
[{"input": "6 3\n1\n4\n5", "output": ["17\n1\n3\n4"]}, {"input": "1 0", "output": ["1"]}]
|
control_completion_003691
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Let's say Pak Chanek has an array $$$A$$$ consisting of $$$N$$$ positive integers. Pak Chanek will do a number of operations. In each operation, Pak Chanek will do the following: Choose an index $$$p$$$ ($$$1 \leq p \leq N$$$). Let $$$c$$$ be the number of operations that have been done on index $$$p$$$ before this operation. Decrease the value of $$$A_p$$$ by $$$2^c$$$. Multiply the value of $$$A_p$$$ by $$$2$$$. After each operation, all elements of $$$A$$$ must be positive integers.An array $$$A$$$ is said to be sortable if and only if Pak Chanek can do zero or more operations so that $$$A_1 < A_2 < A_3 < A_4 < \ldots < A_N$$$.Pak Chanek must find an array $$$A$$$ that is sortable with length $$$N$$$ such that $$$A_1 + A_2 + A_3 + A_4 + \ldots + A_N$$$ is the minimum possible. If there are more than one possibilities, Pak Chanek must choose the array that is lexicographically minimum among them.Pak Chanek must solve the following things: Pak Chanek must print the value of $$$A_1 + A_2 + A_3 + A_4 + \ldots + A_N$$$ for that array. $$$Q$$$ questions will be given. For the $$$i$$$-th question, an integer $$$P_i$$$ is given. Pak Chanek must print the value of $$$A_{P_i}$$$. Help Pak Chanek solve the problem.Note: an array $$$B$$$ of size $$$N$$$ is said to be lexicographically smaller than an array $$$C$$$ that is also of size $$$N$$$ if and only if there exists an index $$$i$$$ such that $$$B_i < C_i$$$ and for each $$$j < i$$$, $$$B_j = C_j$$$.
Input Specification: The first line contains two integers $$$N$$$ and $$$Q$$$ ($$$1 \leq N \leq 10^9$$$, $$$0 \leq Q \leq \min(N, 10^5)$$$) — the required length of array $$$A$$$ and the number of questions. The $$$i$$$-th of the next $$$Q$$$ lines contains a single integer $$$P_i$$$ ($$$1 \leq P_1 < P_2 < \ldots < P_Q \leq N$$$) — the index asked in the $$$i$$$-th question.
Output Specification: Print $$$Q+1$$$ lines. The $$$1$$$-st line contains an integer representing $$$A_1 + A_2 + A_3 + A_4 + \ldots + A_N$$$. For each $$$1 \leq i \leq Q$$$, the $$$(i+1)$$$-th line contains an integer representing $$$A_{P_i}$$$.
Notes: NoteIn the first example, the array $$$A$$$ obtained is $$$[1, 2, 3, 3, 4, 4]$$$. We can see that the array is sortable by doing the following operations: Choose index $$$5$$$, then $$$A = [1, 2, 3, 3, 6, 4]$$$. Choose index $$$6$$$, then $$$A = [1, 2, 3, 3, 6, 6]$$$. Choose index $$$4$$$, then $$$A = [1, 2, 3, 4, 6, 6]$$$. Choose index $$$6$$$, then $$$A = [1, 2, 3, 4, 6, 8]$$$.
Code:
def ev(x):
k = 0
while x%2 == 0:
x//=2
k+=1
return x+k
p,q = [int(i) for i in input().split()]
t = 1
s = 1
sum = 0
while True:
ss = s + (t+1)//2
if ss > p:
m = (p-s+1)
sum += m*t
break
sum += t*((t+1)//2)
s = ss
t += 1
mx = t-2*m+1
print(sum)
tt = t
mul = 0
ded = 0
turn = 0
tt -= 0 if mx==turn else 1
for i in range(q):
x = int(input())
if x == 1:
print(1)
continue
while True:
#print("A:",turn,tt,x-ded)
if # TODO: Your code here:
print(ev(x-ded)+mul)
break
#print("WUT")
if tt < 0:
print(ev(x-ded)+mul)
break
ded += (tt+1)//2
turn+=1
mul += 1
tt -= 0 if mx==turn else 1
|
def ev(x):
k = 0
while x%2 == 0:
x//=2
k+=1
return x+k
p,q = [int(i) for i in input().split()]
t = 1
s = 1
sum = 0
while True:
ss = s + (t+1)//2
if ss > p:
m = (p-s+1)
sum += m*t
break
sum += t*((t+1)//2)
s = ss
t += 1
mx = t-2*m+1
print(sum)
tt = t
mul = 0
ded = 0
turn = 0
tt -= 0 if mx==turn else 1
for i in range(q):
x = int(input())
if x == 1:
print(1)
continue
while True:
#print("A:",turn,tt,x-ded)
if {{completion}}:
print(ev(x-ded)+mul)
break
#print("WUT")
if tt < 0:
print(ev(x-ded)+mul)
break
ded += (tt+1)//2
turn+=1
mul += 1
tt -= 0 if mx==turn else 1
|
x-ded <= tt
|
[{"input": "6 3\n1\n4\n5", "output": ["17\n1\n3\n4"]}, {"input": "1 0", "output": ["1"]}]
|
control_completion_003692
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Pak Chanek has a mirror in the shape of a circle. There are $$$N$$$ lamps on the circumference numbered from $$$1$$$ to $$$N$$$ in clockwise order. The length of the arc from lamp $$$i$$$ to lamp $$$i+1$$$ is $$$D_i$$$ for $$$1 \leq i \leq N-1$$$. Meanwhile, the length of the arc between lamp $$$N$$$ and lamp $$$1$$$ is $$$D_N$$$.Pak Chanek wants to colour the lamps with $$$M$$$ different colours. Each lamp can be coloured with one of the $$$M$$$ colours. However, there cannot be three different lamps such that the colours of the three lamps are the same and the triangle made by considering the three lamps as vertices is a right triangle (triangle with one of its angles being exactly $$$90$$$ degrees).The following are examples of lamp colouring configurations on the circular mirror. Figure 1. an example of an incorrect colouring because lamps $$$1$$$, $$$2$$$, and $$$3$$$ form a right triangleFigure 2. an example of a correct colouringFigure 3. an example of a correct colouring Before colouring the lamps, Pak Chanek wants to know the number of distinct colouring configurations he can make. Count the number of distinct possible lamp colouring configurations, modulo $$$998\,244\,353$$$.
Input Specification: The first line contains two integers $$$N$$$ and $$$M$$$ ($$$1 \le N \le 3 \cdot 10^5$$$, $$$2 \le M \le 3 \cdot 10^5$$$) — the number of lamps in the mirror and the number of different colours used. The second line contains $$$N$$$ integers $$$D_1, D_2, \ldots, D_N$$$ ($$$1 \le D_i \le 10^9$$$) — the lengths of the arcs between the lamps in the mirror.
Output Specification: An integer representing the number of possible lamp colouring configurations, modulo $$$998\,244\,353$$$.
Notes: NoteIn the first example, all correct lamp colouring configurations are $$$[1, 1, 2, 1]$$$, $$$[1, 1, 2, 2]$$$, $$$[1, 2, 1, 2]$$$, $$$[1, 2, 2, 1]$$$, $$$[1, 2, 2, 2]$$$, $$$[2, 1, 1, 1]$$$, $$$[2, 1, 1, 2]$$$, $$$[2, 1, 2, 1]$$$, $$$[2, 2, 1, 1]$$$, and $$$[2, 2, 1, 2]$$$.
Code:
import sys
#sys.stdin = open("C:\\Users\\asawe\\Documents\\input.txt","r")
mx = 3 * (10 ** 5) + 10
mod = 998244353
fact = [1]
inv_fact = [1]
def modpow(a, b, m):
res = 1
x = a
y = b
while (y > 0):
if # TODO: Your code here:
res = (res * x) % m
y = y >> 1
x = ((x % m) * (x % m)) % m
return (res % m + m) % m
def inverse(a, m):
u,v = 0,1
while a != 0:
t = m // a
m -= t * a
a, m = m, a
u -= t * v
u, v = v, u
return u
for i in range(1, mx):
fact.append((fact[-1] * i) % mod)
inv_fact.append(inverse(fact[-1], mod)%mod)
def P(n, k):
if k < 0 or k > n:
return 0
return (fact[n] * inv_fact[n - k]) % mod
def C(n, k):
if k < 0 or k > n:
return 0
return (fact[n] * inv_fact[k] * inv_fact[n - k]) % mod
n,m = map(int,sys.stdin.readline().split())
a = list(map(int,sys.stdin.readline().split()))
s = sum(a)
psum = [0] * n
for i in range(n-1):
psum[i] = psum[i-1] + a[i]
di = {}
for i in psum:
di[i] = 0
count = 0
for i in psum:
if i + s/2 in di:
count+=1
v = n - 2 * count
ans = 0
for i in range(count+1):
ans = (ans + C(count,i) * P(m,count-i) * modpow(m-count+i,v,mod) * modpow(P(m-count+i,2),i,mod)) % mod
print(ans)
|
import sys
#sys.stdin = open("C:\\Users\\asawe\\Documents\\input.txt","r")
mx = 3 * (10 ** 5) + 10
mod = 998244353
fact = [1]
inv_fact = [1]
def modpow(a, b, m):
res = 1
x = a
y = b
while (y > 0):
if {{completion}}:
res = (res * x) % m
y = y >> 1
x = ((x % m) * (x % m)) % m
return (res % m + m) % m
def inverse(a, m):
u,v = 0,1
while a != 0:
t = m // a
m -= t * a
a, m = m, a
u -= t * v
u, v = v, u
return u
for i in range(1, mx):
fact.append((fact[-1] * i) % mod)
inv_fact.append(inverse(fact[-1], mod)%mod)
def P(n, k):
if k < 0 or k > n:
return 0
return (fact[n] * inv_fact[n - k]) % mod
def C(n, k):
if k < 0 or k > n:
return 0
return (fact[n] * inv_fact[k] * inv_fact[n - k]) % mod
n,m = map(int,sys.stdin.readline().split())
a = list(map(int,sys.stdin.readline().split()))
s = sum(a)
psum = [0] * n
for i in range(n-1):
psum[i] = psum[i-1] + a[i]
di = {}
for i in psum:
di[i] = 0
count = 0
for i in psum:
if i + s/2 in di:
count+=1
v = n - 2 * count
ans = 0
for i in range(count+1):
ans = (ans + C(count,i) * P(m,count-i) * modpow(m-count+i,v,mod) * modpow(P(m-count+i,2),i,mod)) % mod
print(ans)
|
(y & 1)
|
[{"input": "4 2\n10 10 6 14", "output": ["10"]}, {"input": "1 2\n10", "output": ["2"]}]
|
control_completion_003701
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Pak Chanek has a mirror in the shape of a circle. There are $$$N$$$ lamps on the circumference numbered from $$$1$$$ to $$$N$$$ in clockwise order. The length of the arc from lamp $$$i$$$ to lamp $$$i+1$$$ is $$$D_i$$$ for $$$1 \leq i \leq N-1$$$. Meanwhile, the length of the arc between lamp $$$N$$$ and lamp $$$1$$$ is $$$D_N$$$.Pak Chanek wants to colour the lamps with $$$M$$$ different colours. Each lamp can be coloured with one of the $$$M$$$ colours. However, there cannot be three different lamps such that the colours of the three lamps are the same and the triangle made by considering the three lamps as vertices is a right triangle (triangle with one of its angles being exactly $$$90$$$ degrees).The following are examples of lamp colouring configurations on the circular mirror. Figure 1. an example of an incorrect colouring because lamps $$$1$$$, $$$2$$$, and $$$3$$$ form a right triangleFigure 2. an example of a correct colouringFigure 3. an example of a correct colouring Before colouring the lamps, Pak Chanek wants to know the number of distinct colouring configurations he can make. Count the number of distinct possible lamp colouring configurations, modulo $$$998\,244\,353$$$.
Input Specification: The first line contains two integers $$$N$$$ and $$$M$$$ ($$$1 \le N \le 3 \cdot 10^5$$$, $$$2 \le M \le 3 \cdot 10^5$$$) — the number of lamps in the mirror and the number of different colours used. The second line contains $$$N$$$ integers $$$D_1, D_2, \ldots, D_N$$$ ($$$1 \le D_i \le 10^9$$$) — the lengths of the arcs between the lamps in the mirror.
Output Specification: An integer representing the number of possible lamp colouring configurations, modulo $$$998\,244\,353$$$.
Notes: NoteIn the first example, all correct lamp colouring configurations are $$$[1, 1, 2, 1]$$$, $$$[1, 1, 2, 2]$$$, $$$[1, 2, 1, 2]$$$, $$$[1, 2, 2, 1]$$$, $$$[1, 2, 2, 2]$$$, $$$[2, 1, 1, 1]$$$, $$$[2, 1, 1, 2]$$$, $$$[2, 1, 2, 1]$$$, $$$[2, 2, 1, 1]$$$, and $$$[2, 2, 1, 2]$$$.
Code:
import sys
#sys.stdin = open("C:\\Users\\asawe\\Documents\\input.txt","r")
mx = 3 * (10 ** 5) + 10
mod = 998244353
fact = [1]
inv_fact = [1]
def modpow(a, b, m):
res = 1
x = a
y = b
while # TODO: Your code here:
if (y & 1):
res = (res * x) % m
y = y >> 1
x = ((x % m) * (x % m)) % m
return (res % m + m) % m
def inverse(a, m):
u,v = 0,1
while a != 0:
t = m // a
m -= t * a
a, m = m, a
u -= t * v
u, v = v, u
return u
for i in range(1, mx):
fact.append((fact[-1] * i) % mod)
inv_fact.append(inverse(fact[-1], mod)%mod)
def P(n, k):
if k < 0 or k > n:
return 0
return (fact[n] * inv_fact[n - k]) % mod
def C(n, k):
if k < 0 or k > n:
return 0
return (fact[n] * inv_fact[k] * inv_fact[n - k]) % mod
n,m = map(int,sys.stdin.readline().split())
a = list(map(int,sys.stdin.readline().split()))
s = sum(a)
psum = [0] * n
for i in range(n-1):
psum[i] = psum[i-1] + a[i]
di = {}
for i in psum:
di[i] = 0
count = 0
for i in psum:
if i + s/2 in di:
count+=1
v = n - 2 * count
ans = 0
for i in range(count+1):
ans = (ans + C(count,i) * P(m,count-i) * modpow(m-count+i,v,mod) * modpow(P(m-count+i,2),i,mod)) % mod
print(ans)
|
import sys
#sys.stdin = open("C:\\Users\\asawe\\Documents\\input.txt","r")
mx = 3 * (10 ** 5) + 10
mod = 998244353
fact = [1]
inv_fact = [1]
def modpow(a, b, m):
res = 1
x = a
y = b
while {{completion}}:
if (y & 1):
res = (res * x) % m
y = y >> 1
x = ((x % m) * (x % m)) % m
return (res % m + m) % m
def inverse(a, m):
u,v = 0,1
while a != 0:
t = m // a
m -= t * a
a, m = m, a
u -= t * v
u, v = v, u
return u
for i in range(1, mx):
fact.append((fact[-1] * i) % mod)
inv_fact.append(inverse(fact[-1], mod)%mod)
def P(n, k):
if k < 0 or k > n:
return 0
return (fact[n] * inv_fact[n - k]) % mod
def C(n, k):
if k < 0 or k > n:
return 0
return (fact[n] * inv_fact[k] * inv_fact[n - k]) % mod
n,m = map(int,sys.stdin.readline().split())
a = list(map(int,sys.stdin.readline().split()))
s = sum(a)
psum = [0] * n
for i in range(n-1):
psum[i] = psum[i-1] + a[i]
di = {}
for i in psum:
di[i] = 0
count = 0
for i in psum:
if i + s/2 in di:
count+=1
v = n - 2 * count
ans = 0
for i in range(count+1):
ans = (ans + C(count,i) * P(m,count-i) * modpow(m-count+i,v,mod) * modpow(P(m-count+i,2),i,mod)) % mod
print(ans)
|
(y > 0)
|
[{"input": "4 2\n10 10 6 14", "output": ["10"]}, {"input": "1 2\n10", "output": ["2"]}]
|
control_completion_003702
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Pak Chanek has a mirror in the shape of a circle. There are $$$N$$$ lamps on the circumference numbered from $$$1$$$ to $$$N$$$ in clockwise order. The length of the arc from lamp $$$i$$$ to lamp $$$i+1$$$ is $$$D_i$$$ for $$$1 \leq i \leq N-1$$$. Meanwhile, the length of the arc between lamp $$$N$$$ and lamp $$$1$$$ is $$$D_N$$$.Pak Chanek wants to colour the lamps with $$$M$$$ different colours. Each lamp can be coloured with one of the $$$M$$$ colours. However, there cannot be three different lamps such that the colours of the three lamps are the same and the triangle made by considering the three lamps as vertices is a right triangle (triangle with one of its angles being exactly $$$90$$$ degrees).The following are examples of lamp colouring configurations on the circular mirror. Figure 1. an example of an incorrect colouring because lamps $$$1$$$, $$$2$$$, and $$$3$$$ form a right triangleFigure 2. an example of a correct colouringFigure 3. an example of a correct colouring Before colouring the lamps, Pak Chanek wants to know the number of distinct colouring configurations he can make. Count the number of distinct possible lamp colouring configurations, modulo $$$998\,244\,353$$$.
Input Specification: The first line contains two integers $$$N$$$ and $$$M$$$ ($$$1 \le N \le 3 \cdot 10^5$$$, $$$2 \le M \le 3 \cdot 10^5$$$) — the number of lamps in the mirror and the number of different colours used. The second line contains $$$N$$$ integers $$$D_1, D_2, \ldots, D_N$$$ ($$$1 \le D_i \le 10^9$$$) — the lengths of the arcs between the lamps in the mirror.
Output Specification: An integer representing the number of possible lamp colouring configurations, modulo $$$998\,244\,353$$$.
Notes: NoteIn the first example, all correct lamp colouring configurations are $$$[1, 1, 2, 1]$$$, $$$[1, 1, 2, 2]$$$, $$$[1, 2, 1, 2]$$$, $$$[1, 2, 2, 1]$$$, $$$[1, 2, 2, 2]$$$, $$$[2, 1, 1, 1]$$$, $$$[2, 1, 1, 2]$$$, $$$[2, 1, 2, 1]$$$, $$$[2, 2, 1, 1]$$$, and $$$[2, 2, 1, 2]$$$.
Code:
from math import comb
def bpow(a,n,p):
res = 1
while n:
if # TODO: Your code here:
res = (res*a)%p
n-=1
else:
a = (a*a)%p
n//=2
return res
N = 3 * 10**5 + 5
factorialNumInverse = [None] * (N + 1)
naturalNumInverse = [None] * (N + 1)
fact = [None] * (N + 1)
def InverseofNumber(p):
naturalNumInverse[0] = naturalNumInverse[1] = 1
for i in range(2, N + 1, 1):
naturalNumInverse[i] = (naturalNumInverse[p % i] *
(p - int(p / i)) % p)
def InverseofFactorial(p):
factorialNumInverse[0] = factorialNumInverse[1] = 1
for i in range(2, N + 1, 1):
factorialNumInverse[i] = (naturalNumInverse[i] *
factorialNumInverse[i - 1]) % p
def factorial(p):
fact[0] = 1
for i in range(1, N + 1):
fact[i] = (fact[i - 1] * i) % p
def Binomial(N, R, p):
ans = ((fact[N] * factorialNumInverse[R])% p *
factorialNumInverse[N - R])% p
return ans
pmod = 998244353
InverseofNumber(pmod)
InverseofFactorial(pmod)
factorial(pmod)
n,pp = map(int,input().split())
l = list(map(int,input().split()))
pref,a = 0,[]
for i in l:
pref+=i
a.append(pref)
qq = pref
qq = qq/2
q = 1
k = 0
po = 0
p = 0
while(q<n):
if(a[q]-a[po]>qq):
po+=1
elif(a[q]-a[po]<qq):
q+=1
else:
k+=1
po+=1
q+=1
p=pp
anss = 0
for i in range(k+1):
ans=1
ans*=Binomial(k,k-i,pmod)
#print(f'ans after step 1 is {ans}')
ans*=bpow(((p-(k-i))*(p-(k-i)-1)),i,pmod)
#print(f'ans after step 2 is {ans}')
ans*=fact[p]*factorialNumInverse[p-k+i]
#print(f'ans after step 3 is {ans}')
ans*=bpow(p-k+i,(n-k*2),pmod)
anss+=ans
print(anss%pmod)
|
from math import comb
def bpow(a,n,p):
res = 1
while n:
if {{completion}}:
res = (res*a)%p
n-=1
else:
a = (a*a)%p
n//=2
return res
N = 3 * 10**5 + 5
factorialNumInverse = [None] * (N + 1)
naturalNumInverse = [None] * (N + 1)
fact = [None] * (N + 1)
def InverseofNumber(p):
naturalNumInverse[0] = naturalNumInverse[1] = 1
for i in range(2, N + 1, 1):
naturalNumInverse[i] = (naturalNumInverse[p % i] *
(p - int(p / i)) % p)
def InverseofFactorial(p):
factorialNumInverse[0] = factorialNumInverse[1] = 1
for i in range(2, N + 1, 1):
factorialNumInverse[i] = (naturalNumInverse[i] *
factorialNumInverse[i - 1]) % p
def factorial(p):
fact[0] = 1
for i in range(1, N + 1):
fact[i] = (fact[i - 1] * i) % p
def Binomial(N, R, p):
ans = ((fact[N] * factorialNumInverse[R])% p *
factorialNumInverse[N - R])% p
return ans
pmod = 998244353
InverseofNumber(pmod)
InverseofFactorial(pmod)
factorial(pmod)
n,pp = map(int,input().split())
l = list(map(int,input().split()))
pref,a = 0,[]
for i in l:
pref+=i
a.append(pref)
qq = pref
qq = qq/2
q = 1
k = 0
po = 0
p = 0
while(q<n):
if(a[q]-a[po]>qq):
po+=1
elif(a[q]-a[po]<qq):
q+=1
else:
k+=1
po+=1
q+=1
p=pp
anss = 0
for i in range(k+1):
ans=1
ans*=Binomial(k,k-i,pmod)
#print(f'ans after step 1 is {ans}')
ans*=bpow(((p-(k-i))*(p-(k-i)-1)),i,pmod)
#print(f'ans after step 2 is {ans}')
ans*=fact[p]*factorialNumInverse[p-k+i]
#print(f'ans after step 3 is {ans}')
ans*=bpow(p-k+i,(n-k*2),pmod)
anss+=ans
print(anss%pmod)
|
n%2
|
[{"input": "4 2\n10 10 6 14", "output": ["10"]}, {"input": "1 2\n10", "output": ["2"]}]
|
control_completion_003703
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Pak Chanek has a mirror in the shape of a circle. There are $$$N$$$ lamps on the circumference numbered from $$$1$$$ to $$$N$$$ in clockwise order. The length of the arc from lamp $$$i$$$ to lamp $$$i+1$$$ is $$$D_i$$$ for $$$1 \leq i \leq N-1$$$. Meanwhile, the length of the arc between lamp $$$N$$$ and lamp $$$1$$$ is $$$D_N$$$.Pak Chanek wants to colour the lamps with $$$M$$$ different colours. Each lamp can be coloured with one of the $$$M$$$ colours. However, there cannot be three different lamps such that the colours of the three lamps are the same and the triangle made by considering the three lamps as vertices is a right triangle (triangle with one of its angles being exactly $$$90$$$ degrees).The following are examples of lamp colouring configurations on the circular mirror. Figure 1. an example of an incorrect colouring because lamps $$$1$$$, $$$2$$$, and $$$3$$$ form a right triangleFigure 2. an example of a correct colouringFigure 3. an example of a correct colouring Before colouring the lamps, Pak Chanek wants to know the number of distinct colouring configurations he can make. Count the number of distinct possible lamp colouring configurations, modulo $$$998\,244\,353$$$.
Input Specification: The first line contains two integers $$$N$$$ and $$$M$$$ ($$$1 \le N \le 3 \cdot 10^5$$$, $$$2 \le M \le 3 \cdot 10^5$$$) — the number of lamps in the mirror and the number of different colours used. The second line contains $$$N$$$ integers $$$D_1, D_2, \ldots, D_N$$$ ($$$1 \le D_i \le 10^9$$$) — the lengths of the arcs between the lamps in the mirror.
Output Specification: An integer representing the number of possible lamp colouring configurations, modulo $$$998\,244\,353$$$.
Notes: NoteIn the first example, all correct lamp colouring configurations are $$$[1, 1, 2, 1]$$$, $$$[1, 1, 2, 2]$$$, $$$[1, 2, 1, 2]$$$, $$$[1, 2, 2, 1]$$$, $$$[1, 2, 2, 2]$$$, $$$[2, 1, 1, 1]$$$, $$$[2, 1, 1, 2]$$$, $$$[2, 1, 2, 1]$$$, $$$[2, 2, 1, 1]$$$, and $$$[2, 2, 1, 2]$$$.
Code:
from math import comb
def bpow(a,n,p):
res = 1
while n:
if # TODO: Your code here:
res = (res*a)%p
n-=1
else:
a = (a*a)%p
n//=2
return res
N = 3 * 10**5 + 5
factorialNumInverse = [None] * (N + 1)
naturalNumInverse = [None] * (N + 1)
fact = [None] * (N + 1)
def InverseofNumber(p):
naturalNumInverse[0] = naturalNumInverse[1] = 1
for i in range(2, N + 1, 1):
naturalNumInverse[i] = (naturalNumInverse[p % i] *
(p - int(p / i)) % p)
def InverseofFactorial(p):
factorialNumInverse[0] = factorialNumInverse[1] = 1
for i in range(2, N + 1, 1):
factorialNumInverse[i] = (naturalNumInverse[i] *
factorialNumInverse[i - 1]) % p
def factorial(p):
fact[0] = 1
for i in range(1, N + 1):
fact[i] = (fact[i - 1] * i) % p
def Binomial(N, R, p):
ans = ((fact[N] * factorialNumInverse[R])% p *
factorialNumInverse[N - R])% p
return ans
pmod = 998244353
InverseofNumber(pmod)
InverseofFactorial(pmod)
factorial(pmod)
n,pp = map(int,input().split())
l = list(map(int,input().split()))
pref,a = 0,[]
for i in l:
pref+=i
a.append(pref)
qq = pref
qq = qq/2
q = 1
k = 0
po = 0
p = 0
while(q<n):
if(a[q]-a[po]>qq):
po+=1
elif(a[q]-a[po]<qq):
q+=1
else:
k+=1
po+=1
q+=1
p=pp
anss = 0
for i in range(k+1):
ans=1
ans*=Binomial(k,k-i,pmod)
ans%=pmod
#print(f'ans after step 1 is {ans}')
ans*=bpow(((p-(k-i))*(p-(k-i)-1)),i,pmod)
ans%=pmod
#print(f'ans after step 2 is {ans}')
ans*=fact[p]*factorialNumInverse[p-k+i]
ans%=pmod
#print(f'ans after step 3 is {ans}')
ans*=bpow(p-k+i,(n-k*2),pmod)
ans%=pmod
anss+=ans
print(anss%pmod)
|
from math import comb
def bpow(a,n,p):
res = 1
while n:
if {{completion}}:
res = (res*a)%p
n-=1
else:
a = (a*a)%p
n//=2
return res
N = 3 * 10**5 + 5
factorialNumInverse = [None] * (N + 1)
naturalNumInverse = [None] * (N + 1)
fact = [None] * (N + 1)
def InverseofNumber(p):
naturalNumInverse[0] = naturalNumInverse[1] = 1
for i in range(2, N + 1, 1):
naturalNumInverse[i] = (naturalNumInverse[p % i] *
(p - int(p / i)) % p)
def InverseofFactorial(p):
factorialNumInverse[0] = factorialNumInverse[1] = 1
for i in range(2, N + 1, 1):
factorialNumInverse[i] = (naturalNumInverse[i] *
factorialNumInverse[i - 1]) % p
def factorial(p):
fact[0] = 1
for i in range(1, N + 1):
fact[i] = (fact[i - 1] * i) % p
def Binomial(N, R, p):
ans = ((fact[N] * factorialNumInverse[R])% p *
factorialNumInverse[N - R])% p
return ans
pmod = 998244353
InverseofNumber(pmod)
InverseofFactorial(pmod)
factorial(pmod)
n,pp = map(int,input().split())
l = list(map(int,input().split()))
pref,a = 0,[]
for i in l:
pref+=i
a.append(pref)
qq = pref
qq = qq/2
q = 1
k = 0
po = 0
p = 0
while(q<n):
if(a[q]-a[po]>qq):
po+=1
elif(a[q]-a[po]<qq):
q+=1
else:
k+=1
po+=1
q+=1
p=pp
anss = 0
for i in range(k+1):
ans=1
ans*=Binomial(k,k-i,pmod)
ans%=pmod
#print(f'ans after step 1 is {ans}')
ans*=bpow(((p-(k-i))*(p-(k-i)-1)),i,pmod)
ans%=pmod
#print(f'ans after step 2 is {ans}')
ans*=fact[p]*factorialNumInverse[p-k+i]
ans%=pmod
#print(f'ans after step 3 is {ans}')
ans*=bpow(p-k+i,(n-k*2),pmod)
ans%=pmod
anss+=ans
print(anss%pmod)
|
n%2
|
[{"input": "4 2\n10 10 6 14", "output": ["10"]}, {"input": "1 2\n10", "output": ["2"]}]
|
control_completion_003704
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Kristina has two arrays $$$a$$$ and $$$b$$$, each containing $$$n$$$ non-negative integers. She can perform the following operation on array $$$a$$$ any number of times: apply a decrement to each non-zero element of the array, that is, replace the value of each element $$$a_i$$$ such that $$$a_i > 0$$$ with the value $$$a_i - 1$$$ ($$$1 \le i \le n$$$). If $$$a_i$$$ was $$$0$$$, its value does not change. Determine whether Kristina can get an array $$$b$$$ from an array $$$a$$$ in some number of operations (probably zero). In other words, can she make $$$a_i = b_i$$$ after some number of operations for each $$$1 \le i \le n$$$?For example, let $$$n = 4$$$, $$$a = [3, 5, 4, 1]$$$ and $$$b = [1, 3, 2, 0]$$$. In this case, she can apply the operation twice: after the first application of the operation she gets $$$a = [2, 4, 3, 0]$$$; after the second use of the operation she gets $$$a = [1, 3, 2, 0]$$$. Thus, in two operations, she can get an array $$$b$$$ from an array $$$a$$$.
Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \le t \le 10^4$$$) —the number of test cases in the test. The descriptions of the test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 5 \cdot 10^4$$$). The second line of each test case contains exactly $$$n$$$ non-negative integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 10^9$$$). The third line of each test case contains exactly $$$n$$$ non-negative integers $$$b_1, b_2, \dots, b_n$$$ ($$$0 \le b_i \le 10^9$$$). It is guaranteed that the sum of $$$n$$$ values over all test cases in the test does not exceed $$$2 \cdot 10^5$$$.
Output Specification: For each test case, output on a separate line: YES, if by doing some number of operations it is possible to get an array $$$b$$$ from an array $$$a$$$; NO otherwise. You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive response).
Notes: NoteThe first test case is analyzed in the statement.In the second test case, it is enough to apply the operation to array $$$a$$$ once.In the third test case, it is impossible to get array $$$b$$$ from array $$$a$$$.
Code:
def solve(a, b):
inf = 2 * 10 ** 6
d, n = inf, len(b)
for i in range(n):
if # TODO: Your code here:
d = min(d, a[i] - b[i])
# b[i] > a[i]
if d < 0:
print("NO")
return
# All elements of b are 0s
if d == inf:
print("YES")
return
for i in range(n):
if a[i] - b[i] > d:
print("NO")
return
if b[i] > 0 and a[i] - b[i] < d:
print("NO")
return
# all a[i] - b[i] == d
print("YES")
def main():
from sys import stdin
from itertools import islice
tkns = map(int, stdin.read().split())
t = next(tkns)
for T in range(t):
n = next(tkns)
a, b = list(islice(tkns, n)), list(islice(tkns, n))
solve(a, b)
main()
|
def solve(a, b):
inf = 2 * 10 ** 6
d, n = inf, len(b)
for i in range(n):
if {{completion}}:
d = min(d, a[i] - b[i])
# b[i] > a[i]
if d < 0:
print("NO")
return
# All elements of b are 0s
if d == inf:
print("YES")
return
for i in range(n):
if a[i] - b[i] > d:
print("NO")
return
if b[i] > 0 and a[i] - b[i] < d:
print("NO")
return
# all a[i] - b[i] == d
print("YES")
def main():
from sys import stdin
from itertools import islice
tkns = map(int, stdin.read().split())
t = next(tkns)
for T in range(t):
n = next(tkns)
a, b = list(islice(tkns, n)), list(islice(tkns, n))
solve(a, b)
main()
|
b[i] > 0
|
[{"input": "6\n\n4\n\n3 5 4 1\n\n1 3 2 0\n\n3\n\n1 2 1\n\n0 1 0\n\n4\n\n5 3 7 2\n\n1 1 1 1\n\n5\n\n1 2 3 4 5\n\n1 2 3 4 6\n\n1\n\n8\n\n0\n\n1\n\n4\n\n6", "output": ["YES\nYES\nNO\nNO\nYES\nNO"]}]
|
control_completion_003858
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Kristina has two arrays $$$a$$$ and $$$b$$$, each containing $$$n$$$ non-negative integers. She can perform the following operation on array $$$a$$$ any number of times: apply a decrement to each non-zero element of the array, that is, replace the value of each element $$$a_i$$$ such that $$$a_i > 0$$$ with the value $$$a_i - 1$$$ ($$$1 \le i \le n$$$). If $$$a_i$$$ was $$$0$$$, its value does not change. Determine whether Kristina can get an array $$$b$$$ from an array $$$a$$$ in some number of operations (probably zero). In other words, can she make $$$a_i = b_i$$$ after some number of operations for each $$$1 \le i \le n$$$?For example, let $$$n = 4$$$, $$$a = [3, 5, 4, 1]$$$ and $$$b = [1, 3, 2, 0]$$$. In this case, she can apply the operation twice: after the first application of the operation she gets $$$a = [2, 4, 3, 0]$$$; after the second use of the operation she gets $$$a = [1, 3, 2, 0]$$$. Thus, in two operations, she can get an array $$$b$$$ from an array $$$a$$$.
Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \le t \le 10^4$$$) —the number of test cases in the test. The descriptions of the test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 5 \cdot 10^4$$$). The second line of each test case contains exactly $$$n$$$ non-negative integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 10^9$$$). The third line of each test case contains exactly $$$n$$$ non-negative integers $$$b_1, b_2, \dots, b_n$$$ ($$$0 \le b_i \le 10^9$$$). It is guaranteed that the sum of $$$n$$$ values over all test cases in the test does not exceed $$$2 \cdot 10^5$$$.
Output Specification: For each test case, output on a separate line: YES, if by doing some number of operations it is possible to get an array $$$b$$$ from an array $$$a$$$; NO otherwise. You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive response).
Notes: NoteThe first test case is analyzed in the statement.In the second test case, it is enough to apply the operation to array $$$a$$$ once.In the third test case, it is impossible to get array $$$b$$$ from array $$$a$$$.
Code:
def solve(a, b):
inf = 2 * 10 ** 6
d, n = inf, len(b)
for i in range(n):
if b[i] > 0:
d = min(d, a[i] - b[i])
# b[i] > a[i]
if d < 0:
print("NO")
return
# All elements of b are 0s
if d == inf:
print("YES")
return
for i in range(n):
if # TODO: Your code here:
print("NO")
return
if b[i] > 0 and a[i] - b[i] < d:
print("NO")
return
# all a[i] - b[i] == d
print("YES")
def main():
from sys import stdin
from itertools import islice
tkns = map(int, stdin.read().split())
t = next(tkns)
for T in range(t):
n = next(tkns)
a, b = list(islice(tkns, n)), list(islice(tkns, n))
solve(a, b)
main()
|
def solve(a, b):
inf = 2 * 10 ** 6
d, n = inf, len(b)
for i in range(n):
if b[i] > 0:
d = min(d, a[i] - b[i])
# b[i] > a[i]
if d < 0:
print("NO")
return
# All elements of b are 0s
if d == inf:
print("YES")
return
for i in range(n):
if {{completion}}:
print("NO")
return
if b[i] > 0 and a[i] - b[i] < d:
print("NO")
return
# all a[i] - b[i] == d
print("YES")
def main():
from sys import stdin
from itertools import islice
tkns = map(int, stdin.read().split())
t = next(tkns)
for T in range(t):
n = next(tkns)
a, b = list(islice(tkns, n)), list(islice(tkns, n))
solve(a, b)
main()
|
a[i] - b[i] > d
|
[{"input": "6\n\n4\n\n3 5 4 1\n\n1 3 2 0\n\n3\n\n1 2 1\n\n0 1 0\n\n4\n\n5 3 7 2\n\n1 1 1 1\n\n5\n\n1 2 3 4 5\n\n1 2 3 4 6\n\n1\n\n8\n\n0\n\n1\n\n4\n\n6", "output": ["YES\nYES\nNO\nNO\nYES\nNO"]}]
|
control_completion_003859
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: An integer array $$$a_1, a_2, \ldots, a_n$$$ is being transformed into an array of lowercase English letters using the following prodecure:While there is at least one number in the array: Choose any number $$$x$$$ from the array $$$a$$$, and any letter of the English alphabet $$$y$$$. Replace all occurrences of number $$$x$$$ with the letter $$$y$$$. For example, if we initially had an array $$$a = [2, 3, 2, 4, 1]$$$, then we could transform it the following way: Choose the number $$$2$$$ and the letter c. After that $$$a = [c, 3, c, 4, 1]$$$. Choose the number $$$3$$$ and the letter a. After that $$$a = [c, a, c, 4, 1]$$$. Choose the number $$$4$$$ and the letter t. After that $$$a = [c, a, c, t, 1]$$$. Choose the number $$$1$$$ and the letter a. After that $$$a = [c, a, c, t, a]$$$. After the transformation all letters are united into a string, in our example we get the string "cacta".Having the array $$$a$$$ and the string $$$s$$$ determine if the string $$$s$$$ could be got from the array $$$a$$$ after the described transformation?
Input Specification: The first line contains a single integer $$$t$$$ $$$(1 \leq t \leq 10^3$$$) — the number of test cases. Then the description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 50$$$) — the length of the array $$$a$$$ and the string $$$s$$$. The second line of each test case contains exactly $$$n$$$ integers: $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 50$$$) — the elements of the array $$$a$$$. The third line of each test case contains a string $$$s$$$ of length $$$n$$$, consisting of lowercase English letters.
Output Specification: For each test case, output "YES", if we can get the string $$$s$$$ from the array $$$a$$$, and "NO" otherwise. You can output each letter in any case.
Notes: NoteThe first test case corresponds to the sample described in the statement.In the second test case we can choose the number $$$50$$$ and the letter a.In the third test case we can choose the number $$$11$$$ and the letter a, after that $$$a = [a, 22]$$$. Then we choose the number $$$22$$$ and the letter b and get $$$a = [a, b]$$$.In the fifth test case we can change all numbers one by one to the letter a.
Code:
from sys import stdin
from collections import deque
lst = stdin.read().split()
_s = 0
def inp(n=1):
global _s
ret = lst[_s:_s + n]
_s += n
return ret
def inp1():
return inp()[0]
t = int(inp1())
for _ in range(t):
n = int(inp1())
a = list(map(int, inp(n)))
s = list(inp1())
d = {}
ok = True
for i in range(n):
if a[i] not in d:
d[a[i]] = s[i]
elif # TODO: Your code here:
ok = not ok
break
print("YES" if ok else "NO")
|
from sys import stdin
from collections import deque
lst = stdin.read().split()
_s = 0
def inp(n=1):
global _s
ret = lst[_s:_s + n]
_s += n
return ret
def inp1():
return inp()[0]
t = int(inp1())
for _ in range(t):
n = int(inp1())
a = list(map(int, inp(n)))
s = list(inp1())
d = {}
ok = True
for i in range(n):
if a[i] not in d:
d[a[i]] = s[i]
elif {{completion}}:
ok = not ok
break
print("YES" if ok else "NO")
|
d[a[i]] != s[i]
|
[{"input": "7\n\n5\n\n2 3 2 4 1\n\ncacta\n\n1\n\n50\n\na\n\n2\n\n11 22\n\nab\n\n4\n\n1 2 2 1\n\naaab\n\n5\n\n1 2 3 2 1\n\naaaaa\n\n6\n\n1 10 2 9 3 8\n\nazzfdb\n\n7\n\n1 2 3 4 1 1 2\n\nabababb", "output": ["YES\nYES\nYES\nNO\nYES\nYES\nNO"]}]
|
control_completion_004006
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You find yourself on an unusual crossroad with a weird traffic light. That traffic light has three possible colors: red (r), yellow (y), green (g). It is known that the traffic light repeats its colors every $$$n$$$ seconds and at the $$$i$$$-th second the color $$$s_i$$$ is on.That way, the order of the colors is described by a string. For example, if $$$s=$$$"rggry", then the traffic light works as the following: red-green-green-red-yellow-red-green-green-red-yellow- ... and so on.More formally, you are given a string $$$s_1, s_2, \ldots, s_n$$$ of length $$$n$$$. At the first second the color $$$s_1$$$ is on, at the second — $$$s_2$$$, ..., at the $$$n$$$-th second the color $$$s_n$$$ is on, at the $$$n + 1$$$-st second the color $$$s_1$$$ is on and so on.You need to cross the road and that can only be done when the green color is on. You know which color is on the traffic light at the moment, but you don't know the current moment of time. You need to find the minimum amount of time in which you are guaranteed to cross the road.You can assume that you cross the road immediately. For example, with $$$s=$$$"rggry" and the current color r there are two options: either the green color will be on after $$$1$$$ second, or after $$$3$$$. That way, the answer is equal to $$$3$$$ — that is the number of seconds that we are guaranteed to cross the road, if the current color is r.
Input Specification: The first line contains a single integer $$$t$$$ $$$(1 \leq t \leq 10^4$$$) — the number of test cases. Then the description of the test cases follows. The first line of each test case contains an integer $$$n$$$ and a symbol $$$c$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$, $$$c$$$ is one of allowed traffic light colors r, y or g)— the length of the string $$$s$$$ and the current color of the traffic light. The second line of each test case contains a string $$$s$$$ of the length $$$n$$$, consisting of the letters r, y and g. It is guaranteed that the symbol g is in the string $$$s$$$ and the symbol $$$c$$$ is in the string $$$s$$$. It is guaranteed, that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case output the minimal number of second in which you are guaranteed to cross the road.
Notes: NoteThe first test case is explained in the statement.In the second test case the green color is on so you can cross the road immediately. In the third test case, if the red color was on at the second second, then we would wait for the green color for one second, and if the red light was on at the first second, then we would wait for the green light for two seconds.In the fourth test case the longest we would wait for the green color is if we wait for it starting from the fifth second.
Code:
from sys import stdin
from collections import deque
lst = stdin.read().split()
_s = 0
def inp(n=1):
global _s
ret = lst[_s:_s + n]
_s += n
return ret
def inp1():
return inp()[0]
t = int(inp1())
for _ in range(t):
n = int(inp1())
c = inp1()
s = inp1()
ret = 0
l = []
last = ''
for i in range(n):
cur = s[i]
if cur == last:
continue
if cur == c:
last = cur
l.append((c, i))
elif # TODO: Your code here:
last = cur
l.append(('g', i))
first_g = -1
for i in range(len(l)):
if l[i][0] == 'g' and first_g != -1:
continue
elif l[i][0] == 'g' and first_g == -1:
first_g = l[i][1]
elif i == len(l) - 1:
ret = max(ret, n - l[i][1] + first_g)
else:
ret = max(ret, l[i + 1][1] - l[i][1])
print(ret)
|
from sys import stdin
from collections import deque
lst = stdin.read().split()
_s = 0
def inp(n=1):
global _s
ret = lst[_s:_s + n]
_s += n
return ret
def inp1():
return inp()[0]
t = int(inp1())
for _ in range(t):
n = int(inp1())
c = inp1()
s = inp1()
ret = 0
l = []
last = ''
for i in range(n):
cur = s[i]
if cur == last:
continue
if cur == c:
last = cur
l.append((c, i))
elif {{completion}}:
last = cur
l.append(('g', i))
first_g = -1
for i in range(len(l)):
if l[i][0] == 'g' and first_g != -1:
continue
elif l[i][0] == 'g' and first_g == -1:
first_g = l[i][1]
elif i == len(l) - 1:
ret = max(ret, n - l[i][1] + first_g)
else:
ret = max(ret, l[i + 1][1] - l[i][1])
print(ret)
|
cur == 'g'
|
[{"input": "6\n\n5 r\n\nrggry\n\n1 g\n\ng\n\n3 r\n\nrrg\n\n5 y\n\nyrrgy\n\n7 r\n\nrgrgyrg\n\n9 y\n\nrrrgyyygy", "output": ["3\n0\n2\n4\n1\n4"]}]
|
control_completion_004067
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You find yourself on an unusual crossroad with a weird traffic light. That traffic light has three possible colors: red (r), yellow (y), green (g). It is known that the traffic light repeats its colors every $$$n$$$ seconds and at the $$$i$$$-th second the color $$$s_i$$$ is on.That way, the order of the colors is described by a string. For example, if $$$s=$$$"rggry", then the traffic light works as the following: red-green-green-red-yellow-red-green-green-red-yellow- ... and so on.More formally, you are given a string $$$s_1, s_2, \ldots, s_n$$$ of length $$$n$$$. At the first second the color $$$s_1$$$ is on, at the second — $$$s_2$$$, ..., at the $$$n$$$-th second the color $$$s_n$$$ is on, at the $$$n + 1$$$-st second the color $$$s_1$$$ is on and so on.You need to cross the road and that can only be done when the green color is on. You know which color is on the traffic light at the moment, but you don't know the current moment of time. You need to find the minimum amount of time in which you are guaranteed to cross the road.You can assume that you cross the road immediately. For example, with $$$s=$$$"rggry" and the current color r there are two options: either the green color will be on after $$$1$$$ second, or after $$$3$$$. That way, the answer is equal to $$$3$$$ — that is the number of seconds that we are guaranteed to cross the road, if the current color is r.
Input Specification: The first line contains a single integer $$$t$$$ $$$(1 \leq t \leq 10^4$$$) — the number of test cases. Then the description of the test cases follows. The first line of each test case contains an integer $$$n$$$ and a symbol $$$c$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$, $$$c$$$ is one of allowed traffic light colors r, y or g)— the length of the string $$$s$$$ and the current color of the traffic light. The second line of each test case contains a string $$$s$$$ of the length $$$n$$$, consisting of the letters r, y and g. It is guaranteed that the symbol g is in the string $$$s$$$ and the symbol $$$c$$$ is in the string $$$s$$$. It is guaranteed, that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.
Output Specification: For each test case output the minimal number of second in which you are guaranteed to cross the road.
Notes: NoteThe first test case is explained in the statement.In the second test case the green color is on so you can cross the road immediately. In the third test case, if the red color was on at the second second, then we would wait for the green color for one second, and if the red light was on at the first second, then we would wait for the green light for two seconds.In the fourth test case the longest we would wait for the green color is if we wait for it starting from the fifth second.
Code:
from sys import stdin
from collections import deque
lst = stdin.read().split()
_s = 0
def inp(n=1):
global _s
ret = lst[_s:_s + n]
_s += n
return ret
def inp1():
return inp()[0]
t = int(inp1())
for _ in range(t):
n = int(inp1())
c = inp1()
s = inp1()
ret = 0
l = []
last = ''
for i in range(n):
cur = s[i]
if cur == last:
continue
if cur == c:
last = cur
l.append((c, i))
elif cur == 'g':
last = cur
l.append(('g', i))
first_g = -1
for i in range(len(l)):
if l[i][0] == 'g' and first_g != -1:
continue
elif # TODO: Your code here:
first_g = l[i][1]
elif i == len(l) - 1:
ret = max(ret, n - l[i][1] + first_g)
else:
ret = max(ret, l[i + 1][1] - l[i][1])
print(ret)
|
from sys import stdin
from collections import deque
lst = stdin.read().split()
_s = 0
def inp(n=1):
global _s
ret = lst[_s:_s + n]
_s += n
return ret
def inp1():
return inp()[0]
t = int(inp1())
for _ in range(t):
n = int(inp1())
c = inp1()
s = inp1()
ret = 0
l = []
last = ''
for i in range(n):
cur = s[i]
if cur == last:
continue
if cur == c:
last = cur
l.append((c, i))
elif cur == 'g':
last = cur
l.append(('g', i))
first_g = -1
for i in range(len(l)):
if l[i][0] == 'g' and first_g != -1:
continue
elif {{completion}}:
first_g = l[i][1]
elif i == len(l) - 1:
ret = max(ret, n - l[i][1] + first_g)
else:
ret = max(ret, l[i + 1][1] - l[i][1])
print(ret)
|
l[i][0] == 'g' and first_g == -1
|
[{"input": "6\n\n5 r\n\nrggry\n\n1 g\n\ng\n\n3 r\n\nrrg\n\n5 y\n\nyrrgy\n\n7 r\n\nrgrgyrg\n\n9 y\n\nrrrgyyygy", "output": ["3\n0\n2\n4\n1\n4"]}]
|
control_completion_004068
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given $$$n$$$ of integers $$$a_1, a_2, \ldots, a_n$$$. Process $$$q$$$ queries of two types: query of the form "0 $$$x_j$$$": add the value $$$x_j$$$ to all even elements of the array $$$a$$$, query of the form "1 $$$x_j$$$": add the value $$$x_j$$$ to all odd elements of the array $$$a$$$.Note that when processing the query, we look specifically at the odd/even value of $$$a_i$$$, not its index.After processing each query, print the sum of the elements of the array $$$a$$$.Please note that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++).
Input Specification: The first line of the input contains an integer $$$t$$$ $$$(1 \leq t \leq 10^4$$$) — the number of test cases. The descriptions of the test cases follow. The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n$$$, $$$q \leq 10^5$$$) — the length of array $$$a$$$ and the number of queries. The second line of each test case contains exactly $$$n$$$ integers: $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — elements of the array $$$a$$$. The following $$$q$$$ lines contain queries as two integers $$$type_j$$$ and $$$x_j$$$ $$$(0 \leq type_j \leq 1$$$, $$$1 \leq x_j \leq 10^4$$$). It is guaranteed that the sum of values $$$n$$$ over all test cases in a test does not exceed $$$10^5$$$. Similarly, the sum of values $$$q$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: For each test case, print $$$q$$$ numbers: the sum of the elements of the array $$$a$$$ after processing a query.
Notes: NoteIn the first test case, the array $$$a = [2]$$$ after the first query.In the third test case, the array $$$a$$$ is modified as follows: $$$[1, 3, 2, 4, 10, 48]$$$ $$$\rightarrow$$$ $$$[7, 9, 2, 4, 10, 48]$$$ $$$\rightarrow$$$ $$$[7, 9, 7, 9, 15, 53]$$$ $$$\rightarrow$$$ $$$[7, 9, 7, 9, 15, 53]$$$ $$$\rightarrow$$$ $$$[10, 12, 10, 12, 18, 56]$$$ $$$\rightarrow$$$ $$$[22, 24, 22, 24, 30, 68]$$$ $$$\rightarrow$$$ $$$[23, 25, 23, 25, 31, 69]$$$.
Code:
from itertools import islice
from sys import stdin
data = iter([int(x) for x in stdin.read().split()[1:]])
res = []
while True:
try:
n = next(data)
except StopIteration:
break
q = next(data)
counts = [0, 0]
sums = [0, 0]
for v in islice(data, n):
counts[v % 2] += 1
sums[v % 2] += v
for _ in range(q):
mod = next(data)
x = next(data)
to_add = counts[mod] * x
if # TODO: Your code here:
counts[1 - mod] += counts[mod]
sums[1 - mod] += sums[mod] + to_add
counts[mod] = sums[mod] = 0
else:
sums[mod] += to_add
res.append(sum(sums))
print('\n'.join(str(x) for x in res))
|
from itertools import islice
from sys import stdin
data = iter([int(x) for x in stdin.read().split()[1:]])
res = []
while True:
try:
n = next(data)
except StopIteration:
break
q = next(data)
counts = [0, 0]
sums = [0, 0]
for v in islice(data, n):
counts[v % 2] += 1
sums[v % 2] += v
for _ in range(q):
mod = next(data)
x = next(data)
to_add = counts[mod] * x
if {{completion}}:
counts[1 - mod] += counts[mod]
sums[1 - mod] += sums[mod] + to_add
counts[mod] = sums[mod] = 0
else:
sums[mod] += to_add
res.append(sum(sums))
print('\n'.join(str(x) for x in res))
|
x % 2
|
[{"input": "4\n\n1 1\n\n1\n\n1 1\n\n3 3\n\n1 2 4\n\n0 2\n\n1 3\n\n0 5\n\n6 7\n\n1 3 2 4 10 48\n\n1 6\n\n0 5\n\n0 4\n\n0 5\n\n1 3\n\n0 12\n\n0 1\n\n6 7\n\n1000000000 1000000000 1000000000 11 15 17\n\n0 17\n\n1 10000\n\n1 51\n\n0 92\n\n0 53\n\n1 16\n\n0 1", "output": ["2\n11\n14\n29\n80\n100\n100\n100\n118\n190\n196\n3000000094\n3000060094\n3000060400\n3000060952\n3000061270\n3000061366\n3000061366"]}]
|
control_completion_004092
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: You are given $$$n$$$ of integers $$$a_1, a_2, \ldots, a_n$$$. Process $$$q$$$ queries of two types: query of the form "0 $$$x_j$$$": add the value $$$x_j$$$ to all even elements of the array $$$a$$$, query of the form "1 $$$x_j$$$": add the value $$$x_j$$$ to all odd elements of the array $$$a$$$.Note that when processing the query, we look specifically at the odd/even value of $$$a_i$$$, not its index.After processing each query, print the sum of the elements of the array $$$a$$$.Please note that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++).
Input Specification: The first line of the input contains an integer $$$t$$$ $$$(1 \leq t \leq 10^4$$$) — the number of test cases. The descriptions of the test cases follow. The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n$$$, $$$q \leq 10^5$$$) — the length of array $$$a$$$ and the number of queries. The second line of each test case contains exactly $$$n$$$ integers: $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — elements of the array $$$a$$$. The following $$$q$$$ lines contain queries as two integers $$$type_j$$$ and $$$x_j$$$ $$$(0 \leq type_j \leq 1$$$, $$$1 \leq x_j \leq 10^4$$$). It is guaranteed that the sum of values $$$n$$$ over all test cases in a test does not exceed $$$10^5$$$. Similarly, the sum of values $$$q$$$ over all test cases does not exceed $$$10^5$$$.
Output Specification: For each test case, print $$$q$$$ numbers: the sum of the elements of the array $$$a$$$ after processing a query.
Notes: NoteIn the first test case, the array $$$a = [2]$$$ after the first query.In the third test case, the array $$$a$$$ is modified as follows: $$$[1, 3, 2, 4, 10, 48]$$$ $$$\rightarrow$$$ $$$[7, 9, 2, 4, 10, 48]$$$ $$$\rightarrow$$$ $$$[7, 9, 7, 9, 15, 53]$$$ $$$\rightarrow$$$ $$$[7, 9, 7, 9, 15, 53]$$$ $$$\rightarrow$$$ $$$[10, 12, 10, 12, 18, 56]$$$ $$$\rightarrow$$$ $$$[22, 24, 22, 24, 30, 68]$$$ $$$\rightarrow$$$ $$$[23, 25, 23, 25, 31, 69]$$$.
Code:
from sys import stdin
from collections import deque
lst = list(map(int, stdin.read().split()))
_s = 0
def inp(n=1):
global _s
ret = lst[_s:_s + n]
_s += n
return ret
def inp1():
return inp()[0]
t = inp1()
for _ in range(t):
n = inp1()
q = inp1()
a = inp(n)
tx = [inp(2) for _ in range(q)]
odd = 0
even = 0
for i in a:
if i % 2 == 0:
even +=1
else:
odd +=1
ret = sum(a)
for i in tx:
if i[0] == 0:
ret += even * i[1]
if i[1] % 2 != 0:
odd = n
even = 0
else:
ret += odd * i[1]
if # TODO: Your code here:
even = n
odd = 0
print(ret)
|
from sys import stdin
from collections import deque
lst = list(map(int, stdin.read().split()))
_s = 0
def inp(n=1):
global _s
ret = lst[_s:_s + n]
_s += n
return ret
def inp1():
return inp()[0]
t = inp1()
for _ in range(t):
n = inp1()
q = inp1()
a = inp(n)
tx = [inp(2) for _ in range(q)]
odd = 0
even = 0
for i in a:
if i % 2 == 0:
even +=1
else:
odd +=1
ret = sum(a)
for i in tx:
if i[0] == 0:
ret += even * i[1]
if i[1] % 2 != 0:
odd = n
even = 0
else:
ret += odd * i[1]
if {{completion}}:
even = n
odd = 0
print(ret)
|
i[1] % 2 != 0
|
[{"input": "4\n\n1 1\n\n1\n\n1 1\n\n3 3\n\n1 2 4\n\n0 2\n\n1 3\n\n0 5\n\n6 7\n\n1 3 2 4 10 48\n\n1 6\n\n0 5\n\n0 4\n\n0 5\n\n1 3\n\n0 12\n\n0 1\n\n6 7\n\n1000000000 1000000000 1000000000 11 15 17\n\n0 17\n\n1 10000\n\n1 51\n\n0 92\n\n0 53\n\n1 16\n\n0 1", "output": ["2\n11\n14\n29\n80\n100\n100\n100\n118\n190\n196\n3000000094\n3000060094\n3000060400\n3000060952\n3000061270\n3000061366\n3000061366"]}]
|
control_completion_004093
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Little Leon lives in the forest. He has recently noticed that some trees near his favourite path are withering, while the other ones are overhydrated so he decided to learn how to control the level of the soil moisture to save the trees.There are $$$n$$$ trees growing near the path, the current levels of moisture of each tree are denoted by the array $$$a_1, a_2, \dots, a_n$$$. Leon has learned three abilities which will help him to dry and water the soil. Choose a position $$$i$$$ and decrease the level of moisture of the trees $$$1, 2, \dots, i$$$ by $$$1$$$. Choose a position $$$i$$$ and decrease the level of moisture of the trees $$$i, i + 1, \dots, n$$$ by $$$1$$$. Increase the level of moisture of all trees by $$$1$$$. Leon wants to know the minimum number of actions he needs to perform to make the moisture of each tree equal to $$$0$$$.
Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The description of $$$t$$$ test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 200\,000$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \leq a_i \leq 10^9$$$) — the initial levels of trees moisture. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$200\,000$$$.
Output Specification: For each test case output a single integer — the minimum number of actions. It can be shown that the answer exists.
Notes: NoteIn the first test case it's enough to apply the operation of adding $$$1$$$ to the whole array $$$2$$$ times. In the second test case you can apply the operation of decreasing $$$4$$$ times on the prefix of length $$$3$$$ and get an array $$$6, 0, 3$$$. After that apply the operation of decreasing $$$6$$$ times on the prefix of length $$$1$$$ and $$$3$$$ times on the suffix of length $$$1$$$. In total, the number of actions will be $$$4 + 6 + 3 = 13$$$. It can be shown that it's impossible to perform less actions to get the required array, so the answer is $$$13$$$.
Code:
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
prefix = suffix = 0
for i in range(n - 1):
if # TODO: Your code here:
prefix += d
else:
suffix -= d
print(abs(a[0] - prefix) + prefix + suffix)
|
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
prefix = suffix = 0
for i in range(n - 1):
if {{completion}}:
prefix += d
else:
suffix -= d
print(abs(a[0] - prefix) + prefix + suffix)
|
(d := a[i] - a[i + 1]) > 0
|
[{"input": "4\n3\n-2 -2 -2\n3\n10 4 7\n4\n4 -4 4 -4\n5\n1 -2 3 -4 5", "output": ["2\n13\n36\n33"]}]
|
control_completion_004116
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Little Leon lives in the forest. He has recently noticed that some trees near his favourite path are withering, while the other ones are overhydrated so he decided to learn how to control the level of the soil moisture to save the trees.There are $$$n$$$ trees growing near the path, the current levels of moisture of each tree are denoted by the array $$$a_1, a_2, \dots, a_n$$$. Leon has learned three abilities which will help him to dry and water the soil. Choose a position $$$i$$$ and decrease the level of moisture of the trees $$$1, 2, \dots, i$$$ by $$$1$$$. Choose a position $$$i$$$ and decrease the level of moisture of the trees $$$i, i + 1, \dots, n$$$ by $$$1$$$. Increase the level of moisture of all trees by $$$1$$$. Leon wants to know the minimum number of actions he needs to perform to make the moisture of each tree equal to $$$0$$$.
Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The description of $$$t$$$ test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 200\,000$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \leq a_i \leq 10^9$$$) — the initial levels of trees moisture. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$200\,000$$$.
Output Specification: For each test case output a single integer — the minimum number of actions. It can be shown that the answer exists.
Notes: NoteIn the first test case it's enough to apply the operation of adding $$$1$$$ to the whole array $$$2$$$ times. In the second test case you can apply the operation of decreasing $$$4$$$ times on the prefix of length $$$3$$$ and get an array $$$6, 0, 3$$$. After that apply the operation of decreasing $$$6$$$ times on the prefix of length $$$1$$$ and $$$3$$$ times on the suffix of length $$$1$$$. In total, the number of actions will be $$$4 + 6 + 3 = 13$$$. It can be shown that it's impossible to perform less actions to get the required array, so the answer is $$$13$$$.
Code:
for line in [*open(0)][2::2]:
tot = 0
list = line.split(' ')
list2 = [0] + list
minv = int(list[0])
for val in range(len(list)-1):
diff = int(list2[val+1]) - int(list[val+1])
if # TODO: Your code here:
tot += diff
minv -= diff
print(tot-minv+abs(minv)+int(list[len(list)-1]))
|
for line in [*open(0)][2::2]:
tot = 0
list = line.split(' ')
list2 = [0] + list
minv = int(list[0])
for val in range(len(list)-1):
diff = int(list2[val+1]) - int(list[val+1])
if {{completion}}:
tot += diff
minv -= diff
print(tot-minv+abs(minv)+int(list[len(list)-1]))
|
(diff >= 0)
|
[{"input": "4\n3\n-2 -2 -2\n3\n10 4 7\n4\n4 -4 4 -4\n5\n1 -2 3 -4 5", "output": ["2\n13\n36\n33"]}]
|
control_completion_004117
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Little Leon lives in the forest. He has recently noticed that some trees near his favourite path are withering, while the other ones are overhydrated so he decided to learn how to control the level of the soil moisture to save the trees.There are $$$n$$$ trees growing near the path, the current levels of moisture of each tree are denoted by the array $$$a_1, a_2, \dots, a_n$$$. Leon has learned three abilities which will help him to dry and water the soil. Choose a position $$$i$$$ and decrease the level of moisture of the trees $$$1, 2, \dots, i$$$ by $$$1$$$. Choose a position $$$i$$$ and decrease the level of moisture of the trees $$$i, i + 1, \dots, n$$$ by $$$1$$$. Increase the level of moisture of all trees by $$$1$$$. Leon wants to know the minimum number of actions he needs to perform to make the moisture of each tree equal to $$$0$$$.
Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The description of $$$t$$$ test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 200\,000$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \leq a_i \leq 10^9$$$) — the initial levels of trees moisture. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$200\,000$$$.
Output Specification: For each test case output a single integer — the minimum number of actions. It can be shown that the answer exists.
Notes: NoteIn the first test case it's enough to apply the operation of adding $$$1$$$ to the whole array $$$2$$$ times. In the second test case you can apply the operation of decreasing $$$4$$$ times on the prefix of length $$$3$$$ and get an array $$$6, 0, 3$$$. After that apply the operation of decreasing $$$6$$$ times on the prefix of length $$$1$$$ and $$$3$$$ times on the suffix of length $$$1$$$. In total, the number of actions will be $$$4 + 6 + 3 = 13$$$. It can be shown that it's impossible to perform less actions to get the required array, so the answer is $$$13$$$.
Code:
import math,sys;input=sys.stdin.readline;S=lambda:input().rstrip();I=lambda:int(S());M=lambda:map(int,S().split());L=lambda:list(M());mod1=1000000007;mod2=998244353
for _ in range(I()):
n=I();a=L();d=[a[0]]
for i in range(1,n):d.append(a[i]-a[i-1])
for i in range(1,n):
if # TODO: Your code here:d[0]+=d[i]
print(sum(abs(i) for i in d))
|
import math,sys;input=sys.stdin.readline;S=lambda:input().rstrip();I=lambda:int(S());M=lambda:map(int,S().split());L=lambda:list(M());mod1=1000000007;mod2=998244353
for _ in range(I()):
n=I();a=L();d=[a[0]]
for i in range(1,n):d.append(a[i]-a[i-1])
for i in range(1,n):
if {{completion}}:d[0]+=d[i]
print(sum(abs(i) for i in d))
|
d[i]<=0
|
[{"input": "4\n3\n-2 -2 -2\n3\n10 4 7\n4\n4 -4 4 -4\n5\n1 -2 3 -4 5", "output": ["2\n13\n36\n33"]}]
|
control_completion_004118
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Little Leon lives in the forest. He has recently noticed that some trees near his favourite path are withering, while the other ones are overhydrated so he decided to learn how to control the level of the soil moisture to save the trees.There are $$$n$$$ trees growing near the path, the current levels of moisture of each tree are denoted by the array $$$a_1, a_2, \dots, a_n$$$. Leon has learned three abilities which will help him to dry and water the soil. Choose a position $$$i$$$ and decrease the level of moisture of the trees $$$1, 2, \dots, i$$$ by $$$1$$$. Choose a position $$$i$$$ and decrease the level of moisture of the trees $$$i, i + 1, \dots, n$$$ by $$$1$$$. Increase the level of moisture of all trees by $$$1$$$. Leon wants to know the minimum number of actions he needs to perform to make the moisture of each tree equal to $$$0$$$.
Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The description of $$$t$$$ test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 200\,000$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \leq a_i \leq 10^9$$$) — the initial levels of trees moisture. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$200\,000$$$.
Output Specification: For each test case output a single integer — the minimum number of actions. It can be shown that the answer exists.
Notes: NoteIn the first test case it's enough to apply the operation of adding $$$1$$$ to the whole array $$$2$$$ times. In the second test case you can apply the operation of decreasing $$$4$$$ times on the prefix of length $$$3$$$ and get an array $$$6, 0, 3$$$. After that apply the operation of decreasing $$$6$$$ times on the prefix of length $$$1$$$ and $$$3$$$ times on the suffix of length $$$1$$$. In total, the number of actions will be $$$4 + 6 + 3 = 13$$$. It can be shown that it's impossible to perform less actions to get the required array, so the answer is $$$13$$$.
Code:
for _ in range(int(input())):
input()
a = [int(x) for x in input().split()]
prefix_value = a[0]
suffix_value = 0
steps = 0
for x, y in zip(a, a[1:]):
if x > y:
steps += x - y
prefix_value = y - suffix_value
elif # TODO: Your code here:
steps += y - x
suffix_value += y - x
print(steps + abs(prefix_value))
|
for _ in range(int(input())):
input()
a = [int(x) for x in input().split()]
prefix_value = a[0]
suffix_value = 0
steps = 0
for x, y in zip(a, a[1:]):
if x > y:
steps += x - y
prefix_value = y - suffix_value
elif {{completion}}:
steps += y - x
suffix_value += y - x
print(steps + abs(prefix_value))
|
y > x
|
[{"input": "4\n3\n-2 -2 -2\n3\n10 4 7\n4\n4 -4 4 -4\n5\n1 -2 3 -4 5", "output": ["2\n13\n36\n33"]}]
|
control_completion_004119
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Little Leon lives in the forest. He has recently noticed that some trees near his favourite path are withering, while the other ones are overhydrated so he decided to learn how to control the level of the soil moisture to save the trees.There are $$$n$$$ trees growing near the path, the current levels of moisture of each tree are denoted by the array $$$a_1, a_2, \dots, a_n$$$. Leon has learned three abilities which will help him to dry and water the soil. Choose a position $$$i$$$ and decrease the level of moisture of the trees $$$1, 2, \dots, i$$$ by $$$1$$$. Choose a position $$$i$$$ and decrease the level of moisture of the trees $$$i, i + 1, \dots, n$$$ by $$$1$$$. Increase the level of moisture of all trees by $$$1$$$. Leon wants to know the minimum number of actions he needs to perform to make the moisture of each tree equal to $$$0$$$.
Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The description of $$$t$$$ test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 200\,000$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \leq a_i \leq 10^9$$$) — the initial levels of trees moisture. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$200\,000$$$.
Output Specification: For each test case output a single integer — the minimum number of actions. It can be shown that the answer exists.
Notes: NoteIn the first test case it's enough to apply the operation of adding $$$1$$$ to the whole array $$$2$$$ times. In the second test case you can apply the operation of decreasing $$$4$$$ times on the prefix of length $$$3$$$ and get an array $$$6, 0, 3$$$. After that apply the operation of decreasing $$$6$$$ times on the prefix of length $$$1$$$ and $$$3$$$ times on the suffix of length $$$1$$$. In total, the number of actions will be $$$4 + 6 + 3 = 13$$$. It can be shown that it's impossible to perform less actions to get the required array, so the answer is $$$13$$$.
Code:
from sys import stdin
a = int(stdin.readline())
for t in range(0,a):
b = int(stdin.readline())
c = stdin.readline().split()
count = 0
current = int(c[0])
for u in range(0,b-1):
if int(c[u+1])>int(c[u]):
count+=(int(c[u+1])-int(c[u]))
elif # TODO: Your code here:
count+=(int(c[u]) - int(c[u+1]))
current = current - (int(c[u]) - int(c[u+1]))
print(abs(current)+count)
|
from sys import stdin
a = int(stdin.readline())
for t in range(0,a):
b = int(stdin.readline())
c = stdin.readline().split()
count = 0
current = int(c[0])
for u in range(0,b-1):
if int(c[u+1])>int(c[u]):
count+=(int(c[u+1])-int(c[u]))
elif {{completion}}:
count+=(int(c[u]) - int(c[u+1]))
current = current - (int(c[u]) - int(c[u+1]))
print(abs(current)+count)
|
int(c[u+1]) < int(c[u])
|
[{"input": "4\n3\n-2 -2 -2\n3\n10 4 7\n4\n4 -4 4 -4\n5\n1 -2 3 -4 5", "output": ["2\n13\n36\n33"]}]
|
control_completion_004120
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Little Leon lives in the forest. He has recently noticed that some trees near his favourite path are withering, while the other ones are overhydrated so he decided to learn how to control the level of the soil moisture to save the trees.There are $$$n$$$ trees growing near the path, the current levels of moisture of each tree are denoted by the array $$$a_1, a_2, \dots, a_n$$$. Leon has learned three abilities which will help him to dry and water the soil. Choose a position $$$i$$$ and decrease the level of moisture of the trees $$$1, 2, \dots, i$$$ by $$$1$$$. Choose a position $$$i$$$ and decrease the level of moisture of the trees $$$i, i + 1, \dots, n$$$ by $$$1$$$. Increase the level of moisture of all trees by $$$1$$$. Leon wants to know the minimum number of actions he needs to perform to make the moisture of each tree equal to $$$0$$$.
Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The description of $$$t$$$ test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 200\,000$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \leq a_i \leq 10^9$$$) — the initial levels of trees moisture. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$200\,000$$$.
Output Specification: For each test case output a single integer — the minimum number of actions. It can be shown that the answer exists.
Notes: NoteIn the first test case it's enough to apply the operation of adding $$$1$$$ to the whole array $$$2$$$ times. In the second test case you can apply the operation of decreasing $$$4$$$ times on the prefix of length $$$3$$$ and get an array $$$6, 0, 3$$$. After that apply the operation of decreasing $$$6$$$ times on the prefix of length $$$1$$$ and $$$3$$$ times on the suffix of length $$$1$$$. In total, the number of actions will be $$$4 + 6 + 3 = 13$$$. It can be shown that it's impossible to perform less actions to get the required array, so the answer is $$$13$$$.
Code:
t = int(input())
for _ in range(t):
n = int(input())
A = list(map(int,input().split()))
res,r = 0,0
for i in range(n-1):
x = A[i+1]-A[i]
if # TODO: Your code here:
r+=x
res+=abs(x)
res+=abs(r-A[n-1])
print(res)
|
t = int(input())
for _ in range(t):
n = int(input())
A = list(map(int,input().split()))
res,r = 0,0
for i in range(n-1):
x = A[i+1]-A[i]
if {{completion}}:
r+=x
res+=abs(x)
res+=abs(r-A[n-1])
print(res)
|
x>0
|
[{"input": "4\n3\n-2 -2 -2\n3\n10 4 7\n4\n4 -4 4 -4\n5\n1 -2 3 -4 5", "output": ["2\n13\n36\n33"]}]
|
control_completion_004121
|
control_fixed
|
python
|
Complete the code in python to solve this programming problem:
Description: Little Leon lives in the forest. He has recently noticed that some trees near his favourite path are withering, while the other ones are overhydrated so he decided to learn how to control the level of the soil moisture to save the trees.There are $$$n$$$ trees growing near the path, the current levels of moisture of each tree are denoted by the array $$$a_1, a_2, \dots, a_n$$$. Leon has learned three abilities which will help him to dry and water the soil. Choose a position $$$i$$$ and decrease the level of moisture of the trees $$$1, 2, \dots, i$$$ by $$$1$$$. Choose a position $$$i$$$ and decrease the level of moisture of the trees $$$i, i + 1, \dots, n$$$ by $$$1$$$. Increase the level of moisture of all trees by $$$1$$$. Leon wants to know the minimum number of actions he needs to perform to make the moisture of each tree equal to $$$0$$$.
Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. The description of $$$t$$$ test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 200\,000$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \leq a_i \leq 10^9$$$) — the initial levels of trees moisture. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$200\,000$$$.
Output Specification: For each test case output a single integer — the minimum number of actions. It can be shown that the answer exists.
Notes: NoteIn the first test case it's enough to apply the operation of adding $$$1$$$ to the whole array $$$2$$$ times. In the second test case you can apply the operation of decreasing $$$4$$$ times on the prefix of length $$$3$$$ and get an array $$$6, 0, 3$$$. After that apply the operation of decreasing $$$6$$$ times on the prefix of length $$$1$$$ and $$$3$$$ times on the suffix of length $$$1$$$. In total, the number of actions will be $$$4 + 6 + 3 = 13$$$. It can be shown that it's impossible to perform less actions to get the required array, so the answer is $$$13$$$.
Code:
def solve():
n = int(input())
a = [int(i) for i in input().split()]
dl, dr = 0, 0
for i in range(1, n):
if # TODO: Your code here:
dr += (a[i]-dr)-(a[0]-dl)
else:
dl += (a[0]-dl)-(a[i]-dr)
return dl+dr+abs(a[0]-dl)
for _ in range(int(input())):
print(solve())
|
def solve():
n = int(input())
a = [int(i) for i in input().split()]
dl, dr = 0, 0
for i in range(1, n):
if {{completion}}:
dr += (a[i]-dr)-(a[0]-dl)
else:
dl += (a[0]-dl)-(a[i]-dr)
return dl+dr+abs(a[0]-dl)
for _ in range(int(input())):
print(solve())
|
a[i]-dr >= a[0]-dl
|
[{"input": "4\n3\n-2 -2 -2\n3\n10 4 7\n4\n4 -4 4 -4\n5\n1 -2 3 -4 5", "output": ["2\n13\n36\n33"]}]
|
control_completion_004122
|
control_fixed
|
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