message stringlengths 2 30.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 237 109k | cluster float64 10 10 | __index_level_0__ int64 474 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
Submitted Solution:
```
N=int(input())
M=[0]*(N+1)
M[1]=1
for i in range(2,N+1):
L=[M[i-1]+1]
j=0
while i-6**j>=0:
L.append(M[i-6**j]+1)
j+=1
j=0
while i-9**j>=0:
L.append(M[i-9**j]+1)
j+=1
M[i]=min(L)
#print(M)
print(M[N])
``` | instruction | 0 | 107,711 | 10 | 215,422 |
Yes | output | 1 | 107,711 | 10 | 215,423 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
Submitted Solution:
```
N=int(input())
ans=N
for i in range(N+1):
cnt=0
t = i
while t>0:
cnt+=t%6
t//=6
t=N-i
while t>0:
cnt+=t%9
t//=9
ans=min(ans,cnt)
print(ans)
``` | instruction | 0 | 107,712 | 10 | 215,424 |
Yes | output | 1 | 107,712 | 10 | 215,425 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
Submitted Solution:
```
n=int(input())
a=n
for i in range(n+1):
x=i;y=n-i;c=0
while x>=6:
c+=x%6
x=x//6
c+=x
while y>=9:
c+=y%9
y=y//9
c+=y
a=min(a,c)
print(a)
``` | instruction | 0 | 107,713 | 10 | 215,426 |
Yes | output | 1 | 107,713 | 10 | 215,427 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
Submitted Solution:
```
N = int(input())
A = [1]
for i in range(1,7):
A.append(6 ** i)
for i in range(1, 6):
A.append(9 ** i)
A = sorted(A)
dp = [float("inf")] * (N+1)
dp[0] = 0
for i in range(1,N+1):
for k, v in enumerate(A):
if i >= v:
dp[i] = min(dp[i], dp[i-A[s]]+1)
print(dp[N])
``` | instruction | 0 | 107,714 | 10 | 215,428 |
No | output | 1 | 107,714 | 10 | 215,429 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
Submitted Solution:
```
N = int(input())
for i in range(6):
a = 9**(i+1)
if N<a:
nine = i
N = N-(9**i)
break
for i in range(7):
a = 6**(i+1)
if N<a:
six = i
N = N-(6**i)
one = N
break
out = one + six + nine
print(out)
``` | instruction | 0 | 107,715 | 10 | 215,430 |
No | output | 1 | 107,715 | 10 | 215,431 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
Submitted Solution:
```
# 引き出したいお金の額を入力する
N = int(input())
# 整数aの累乗のうち、整数N以下かつ最も大きいものを返す関数を定義
def func(N, a):
index = 0
while True:
if a ** (index + 1) <= N:
index += 1
else:
break
return a ** index
# 引き出し回数を再帰的に求める関数を定義
def search_times_re(N):
times = 0
if N < 6:
times = N
return times
elif N < 9:
times = 1 + (N - 6)
return times
else:
N1 = search_times_re(N - func(N, 6))
N2 = search_times_re(N - func(N, 9))
if N1 < N2:
return 1 + N1
else:
return 1 + N2
print(search_times_re(N))
``` | instruction | 0 | 107,716 | 10 | 215,432 |
No | output | 1 | 107,716 | 10 | 215,433 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
Submitted Solution:
```
#!/usr/bin/env python3
import math
# n進数の各位の和
def base_10_n(x,n):
a = int(math.log(x,n))
ans = 0
for i in range(a,-1,-1):
ans += x // n**i
x %= n**i
ans += x
return ans
def main():
N = int(input())
ans = N
for j in range(1,N//18 + 1):
a = base_10_n(N-18*j,6)
b = base_10_n(18*j,9)
ans = min(ans, a+b)
print(ans)
# print([i for i in range(2,-1,-1)])
if __name__ == "__main__":
main()
``` | instruction | 0 | 107,717 | 10 | 215,434 |
No | output | 1 | 107,717 | 10 | 215,435 |
Provide a correct Python 3 solution for this coding contest problem.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1 | instruction | 0 | 107,744 | 10 | 215,488 |
"Correct Solution:
```
N,A,B=map(int,input().split())
inf=float("inf")
dp=[[inf]*401 for _ in range(401)]
dp[0][0]=0
for i in range(N):
a,b,c=map(int,input().split())
for j in range(-1,-402,-1):
for k in range(-1,-402,-1):
if dp[j][k]!=inf:
dp[j+a][k+b]=min(dp[j+a][k+b],dp[j][k]+c)
ans=inf
for i in range(1,401):
for j in range(1,401):
if i/j==A/B:
ans=min(ans,dp[i][j])
if ans==inf:
ans=-1
print(ans)
``` | output | 1 | 107,744 | 10 | 215,489 |
Provide a correct Python 3 solution for this coding contest problem.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1 | instruction | 0 | 107,745 | 10 | 215,490 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
def inpl(): return map(int, input().split())
N, Ma, Mb = inpl()
INF = 4001
DP = [[INF]*(411) for _ in range(411)]
DP[0][0] = 0
for _ in range(N):
DP2 = [[INF]*(411) for _ in range(411)]
a, b, c = inpl()
for i in range(400):
for j in range(400):
DP2[i][j] = min(DP[i][j], DP2[i][j])
DP2[i+a][j+b] = min(DP[i+a][j+b], DP[i][j] + c)
DP = DP2
x = 1
ans = INF
while Ma*x <= 400 and Mb*x <= 400:
ans = min(ans, DP[Ma*x][Mb*x])
x += 1
if ans < INF:
print(ans)
else:
print(-1)
``` | output | 1 | 107,745 | 10 | 215,491 |
Provide a correct Python 3 solution for this coding contest problem.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1 | instruction | 0 | 107,746 | 10 | 215,492 |
"Correct Solution:
```
def inpl(): return [int(i) for i in input().split()]
N,Ma,Mb = inpl()
inf = float('inf')
H = {(0,0): 0}
for i in range(N):
a, b, c = inpl()
for (ia, ib), ic in H.copy().items():
H[(ia+a,ib+b)] = min(H.get((ia+a, ib+b), inf),ic+c)
ans = min(H.get((i*Ma, i*Mb), inf) for i in range(1,401))
print(-1 if ans == inf else ans)
``` | output | 1 | 107,746 | 10 | 215,493 |
Provide a correct Python 3 solution for this coding contest problem.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1 | instruction | 0 | 107,747 | 10 | 215,494 |
"Correct Solution:
```
n,ma,mb=map(int,input().split())
dp=[[[10**30 for j in range(401)]for i in range(401)]for k in range(n+1)]
dp[0][0][0]=0
l=[]
for i in range(n):
a,b,c=map(int,input().split())
l.append([a,b,c])
for i in range(1,n+1):
a,b,c=l[i-1][0],l[i-1][1],l[i-1][2]
for x in range(401):
for y in range(401):
if x<a or y<b:
dp[i][x][y]=dp[i-1][x][y]
continue
dp[i][x][y]=min(dp[i-1][x][y],dp[i-1][x-a][y-b]+c)
mini=10**30
for i in range(1,40):
mini=min(mini,dp[n][ma*i][mb*i])
if mini>=10**30 :
print(-1)
else:
print(mini)
``` | output | 1 | 107,747 | 10 | 215,495 |
Provide a correct Python 3 solution for this coding contest problem.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1 | instruction | 0 | 107,748 | 10 | 215,496 |
"Correct Solution:
```
inf = float('inf')
N,Ma,Mb = map(int,input().split())
abc = [list(map(int,input().split())) for _ in range(N)]
# dp[i][j][k]: i番目までみてj(g), k(g)となるような最小予算
dp = [[[inf]*401 for _ in range(401)] for _ in range(N+1)]
dp[0][0][0] = 0
for i in range(N):
a,b,c = abc[i]
for j in range(401):
for k in range(401):
if j - a >= 0 and k - b >= 0:
dp[i+1][j][k] = min(dp[i][j][k], dp[i][j-a][k-b] + c)
else:
dp[i+1][j][k] = dp[i][j][k]
ans = inf
for j in range(1,401):
for k in range(1,401):
if j * Mb == Ma * k:
ans = min(ans, dp[N][j][k])
if ans == inf:
ans = -1
print(ans)
``` | output | 1 | 107,748 | 10 | 215,497 |
Provide a correct Python 3 solution for this coding contest problem.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1 | instruction | 0 | 107,749 | 10 | 215,498 |
"Correct Solution:
```
INFTY = 10**4
N,Ma,Mb = map(int,input().split())
A = [list(map(int,input().split())) for _ in range(N)]
dp = [[[INFTY for _ in range(401)] for _ in range(401)] for _ in range(N+1)]
dp[0][0][0] = 0
for i in range(1,N+1):
for j in range(401):
for k in range(401):
dp[i][j][k] = dp[i-1][j][k]
a = A[i-1][0]
b = A[i-1][1]
c = A[i-1][2]
if j>=a and k>=b:
dp[i][j][k] = min(dp[i][j][k],dp[i-1][j-a][k-b]+c)
cmin = INFTY
for j in range(1,401):
for k in range(1,401):
if j*Mb==k*Ma:
cmin = min(cmin,dp[N][j][k])
if cmin>=INFTY:
print(-1)
else:
print(cmin)
``` | output | 1 | 107,749 | 10 | 215,499 |
Provide a correct Python 3 solution for this coding contest problem.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1 | instruction | 0 | 107,750 | 10 | 215,500 |
"Correct Solution:
```
N, Ma, Mb = map(int, input().split())
table = []
for i in range(N):
a,b,c = map(int, input().split())
table.append([a,b,c])
inf = 10**9
dp = [[[inf]*401 for i in range(401)] for i in range(N+1)]
dp[0][0][0] = 0
for i in range(1,N+1):
a,b,c = table[i-1]
for j in range(401):
for k in range(401):
dp[i][j][k] = dp[i-1][j][k]
if 400>=j-a>=0 and 400>=k-b>=0:
dp[i][j][k] = min(dp[i-1][j][k],dp[i-1][j-a][k-b]+c)
ans = 10**9
i = 1
while max(Ma,Mb)*i <= 400:
ans = min(ans,dp[N][Ma*i][Mb*i])
i += 1
if ans == 10**9:
ans = -1
print(ans)
``` | output | 1 | 107,750 | 10 | 215,501 |
Provide a correct Python 3 solution for this coding contest problem.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1 | instruction | 0 | 107,751 | 10 | 215,502 |
"Correct Solution:
```
INF = float('inf')
MAXAB = 10
n, ma, mb = map(int, input().split())
t = [[INF] * (n * MAXAB + 1) for _ in range(n * MAXAB + 1)]
t[0][0] = 0
for _ in range(n):
a, b, c = map(int, input().split())
for aa in range(n * MAXAB, -1, -1):
for bb in range(n * MAXAB, -1, -1):
if t[aa][bb] == INF:
continue
if t[a + aa][b + bb] > t[aa][bb] + c:
t[a + aa][b + bb] = t[aa][bb] + c
result = INF
for a in range(1, n * MAXAB + 1):
for b in range(1, n * MAXAB + 1):
if a * mb == b * ma and result > t[a][b]:
result = t[a][b]
if result == INF:
result = -1
print(result)
``` | output | 1 | 107,751 | 10 | 215,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
Submitted Solution:
```
# DP
INF = float('inf')
N, Ma, Mb = map(int, input().split())
cmax = N * 10
t = [[INF] * (cmax + 1) for _ in range(cmax + 1)]
for _ in range(N):
a, b, c = map(int, input().split())
for aa in range(cmax, 0, -1):
for bb in range(cmax, 0, -1):
if t[aa][bb] == INF:
continue
t[aa + a][bb + b] = min(t[aa + a][bb + b], t[aa][bb] + c)
t[a][b] = min(t[a][b], c)
result = INF
for a in range(1, cmax):
for b in range(1, cmax):
if a * Mb == b * Ma:
result = min(result, t[a][b])
if result == INF:
result = -1
print(result)
``` | instruction | 0 | 107,752 | 10 | 215,504 |
Yes | output | 1 | 107,752 | 10 | 215,505 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
Submitted Solution:
```
n, ma, mb = map(int, input().split())
abc = []
for _ in range(n):
a, b, c = map(int, input().split())
abc.append((a, b, c))
dp = {(0, 0): 0}
for a, b, c in abc:
newdp = dp.copy()
for (ka, kb), p in dp.items():
k = (a+ka, b+kb)
if k in newdp:
newdp[k] = min(newdp[k], p+c)
else:
newdp[k] = p+c
dp = newdp
INF = 10**10
ans = INF
for (a, b), c in dp.items():
if ma*b!=mb*a or a==b==0:
continue
ans = min(ans, c)
if ans==INF:
print(-1)
else:
print(ans)
``` | instruction | 0 | 107,753 | 10 | 215,506 |
Yes | output | 1 | 107,753 | 10 | 215,507 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
Submitted Solution:
```
N,Ma,Mb=map(int,input().split())
a,b,c=map(list,zip(*[list(map(int,input().split())) for i in range(N)]))
ma,mb=sum(a),sum(b)
MAX=5000
dp=[[[MAX]*(mb+1) for i in range(ma+1)] for j in range(N+1)]
dp[0][0][0]=0
for i in range(N):
for j in range(ma):
for k in range(mb):
dp[i+1][j][k]=min(dp[i][j][k],dp[i+1][j][k])
if j+a[i]<=ma and k+b[i]<=mb:
dp[i+1][j+a[i]][k+b[i]]=min(dp[i+1][j+a[i]][k+b[i]],dp[i][j][k]+c[i])
ans=MAX
x=1
while x*Ma<=ma and x*Mb<=mb:
ans=min(ans,dp[N][x*Ma][x*Mb])
x+=1
print(ans if ans<MAX else -1)
``` | instruction | 0 | 107,754 | 10 | 215,508 |
Yes | output | 1 | 107,754 | 10 | 215,509 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
Submitted Solution:
```
n, A, B = map(int, input().split())
z = [tuple(map(int, input().split())) for _ in range(n)]
inf = 10 ** 18
d = {(0, 0):0}
for a, b, cost in z:
newd = []
for (x, y), c in d.items():
new = (a+x, b+y)
if new not in d or d[new] > c + cost:
newd.append((new, c+cost))
for new, newc in newd:
d[new] = newc
ans = inf
for i in range(1, 10 * n):
t = (A * i, B * i)
if t in d:
ans = min(ans, d[t])
if ans == inf:
print(-1)
else:
print(ans)
``` | instruction | 0 | 107,755 | 10 | 215,510 |
Yes | output | 1 | 107,755 | 10 | 215,511 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
Submitted Solution:
```
N, Ma, Mb = map(int, input().split())
A = [0]*N
B = [0]*N
C = [0]*N
for i in range(N):
A[i], B[i], C[i] = map(int, input().split())
# print(N, Ma, Mb, A, B, C)
A = [Mb * a for a in A]
B = [Ma * b for b in B]
D = []
for i in range(N):
D.append((A[i] - B[i], C[i]))
# print(N, Ma, Mb, A, B, C, D)
Dp = []
Dn = []
Dz = []
for d in D:
if d[0] == 0:
Dz.append(d)
elif d[0] > 0:
Dp.append(d)
else:
Dn.append(d)
Dp.sort(key=lambda x: x[0])
Dn.sort(key=lambda x: x[0])
Dz.sort(key=lambda x: x[1])
min_c = 100*41
if len(Dz) > 0:
min_c = Dz[0][1]
# print(Dp, Dn, Dz, min_c)
Ds = []
Dl = []
if len(Dp) < len(Dn):
Ds, Dl = Dp, Dn
else:
Ds, Dl = Dn, Dp
def calc(D, l):
s = 0
c = 0
for i in range(len(D)):
if (l >> i) & 1 != 0:
s += D[i][0]
c += D[i][1]
return s, c
Ls = len(Ds)
Ll = len(Dl)
for ls in range(Ls):
s, c = calc(Ds, ls + 1)
# print('ls', ls, s, c)
if c >= min_c:
continue
for ll in range(Ll):
sl, cl = calc(Dl, ll + 1)
# print('ll', ll, sl, cl)
if c >= min_c:
continue
if s + sl == 0:
min_c = c + cl
if min_c == 100*41:
print(-1)
else:
print(min_c)
``` | instruction | 0 | 107,756 | 10 | 215,512 |
No | output | 1 | 107,756 | 10 | 215,513 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
Submitted Solution:
```
import math
import copy
from operator import mul
from functools import reduce
from collections import defaultdict
from collections import Counter
from collections import deque
# 直積 A={a, b, c}, B={d, e}:のとき，A×B={(a,d),(a,e),(b,d),(b,e),(c,d),(c,e)}: product(A, B)
from itertools import product
# 階乗 P!: permutations(seq), 順列 {}_len(seq) P_n: permutations(seq, n)
from itertools import permutations
# 組み合わせ {}_len(seq) C_n: combinations(seq, n)
from itertools import combinations
# 一次元累積和
from itertools import accumulate
from bisect import bisect_left, bisect_right
import re
# import numpy as np
# from scipy.misc import comb
# 再帰がやばいとき
# import sys
# sys.setrecursionlimit(10**9)
def inside(y, x, H, W):
return 0 <= y < H and 0 <= x < W
# 四方向: 右, 下, 左, 上
dy = [0, -1, 0, 1]
dx = [1, 0, -1, 0]
def i_inpl(): return int(input())
def l_inpl(): return list(map(int, input().split()))
def line_inpl(x): return [i_inpl() for _ in range(x)]
INF = int(1e50)
MOD = int(1e9)+7 # 10^9 + 7
# field[H][W]
def create_grid(H, W, value = 0):
return [[ value for _ in range(W)] for _ in range(H)]
########
def main():
N, Ma, Mb = l_inpl()
abc = []
for _ in range(N):
abci = l_inpl()
abc.append(abci)
max_ab = 40*10
# dp[i][j][k]: i番目までの薬品の組み合わせで，物質aがjグラム，物質bがkグラムのときのコスト
dp = [[[INF for _ in range(max_ab+1)] for _ in range(max_ab+1) ] for _ in range(N+1)]
dp[0][0][0] = 0
for i in range(N):
for j in range(max_ab+1):
for k in range(max_ab+1):
# ここでのINFはNoneと同じ意味なので
if dp[i][j][k] == INF:
continue
# i番目の薬品を加えない
dp[i+1][j][k] = min(dp[i+1][j][k], dp[i][j][k])
# i番目の薬品を加える
dp[i+1][j+abc[i][0]][k+abc[i][1]]
dp[i+1][j+abc[i][0]][k+abc[i][1]] = min(dp[i+1][j+abc[i][0]][k+abc[i][1]], dp[i+1][j][k] + abc[i][2])
ans = INF
for ca in range(1, max_ab+1):
for cb in range(1, max_ab+1):
if ca*Mb == cb*Ma:
ans = min(ans, dp[N][ca][cb])
if ans == INF:
print(-1)
else:
print(ans)
if __name__ == "__main__":
main()
``` | instruction | 0 | 107,757 | 10 | 215,514 |
No | output | 1 | 107,757 | 10 | 215,515 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
Submitted Solution:
```
n,ma,mb=map(int,input().split())
dp=[[[9999]*401for _ in range(401)]for _ in range(n+1)];dp[0][0][0]=0
sa=sb=0
for i in range(1,n+1):
a,b,c=map(int,input().split())
sa+=a;sb+=b
for j in range(401):
for k in range(401):
if j>=a and k>=b:dp[i][j][k]=min(dp[i-1][j-a][k-b]+c,dp[i-1][j][k])
else:dp[i][j][k]=dp[i-1][j][k]
a=9999
for i in range(1,min(sa//ma,sb//mb)+1):
if dp[n][ma*i][mb*i]!=9999:a=min(a,dp[n][ma*i][mb*i])
print(-1if a==9999else a)
``` | instruction | 0 | 107,758 | 10 | 215,516 |
No | output | 1 | 107,758 | 10 | 215,517 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
Submitted Solution:
```
N, Ma, Mb = map(int, input().split())
INF = 10 ** 8
lst = [[INF] * 401 for i in range(401)]
lst[0][0] = 0
for _ in range(N):
a, b, c = map(int, input().split())
for i in range(a, 401):
for j in range(b, 401):
lst[i][j] = min(lst[i][j], lst[i - a][j - b] + c)
ans = INF
n = 400 // max(Ma, Mb)
for i in range(1, n + 1):
ans = min(ans, lst[Ma * i][Mb * i])
if ans == INF:
print (-1)
else:
print (ans)
``` | instruction | 0 | 107,759 | 10 | 215,518 |
No | output | 1 | 107,759 | 10 | 215,519 |
Provide a correct Python 3 solution for this coding contest problem.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247 | instruction | 0 | 107,788 | 10 | 215,576 |
"Correct Solution:
```
if __name__ == '__main__':
A = ["A","B","C","D","E"]
while True:
try:
n,m = map(int,input().split())
max_num = n + m
number = 0
if n == 0 and m == 0:
break
for i in range(4):
n,m = map(int,input().split())
if max_num < n + m:
max_num = n + m
number = i + 1
print(A[number],max_num)
except EOFError:
break
``` | output | 1 | 107,788 | 10 | 215,577 |
Provide a correct Python 3 solution for this coding contest problem.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247 | instruction | 0 | 107,789 | 10 | 215,578 |
"Correct Solution:
```
while True :
a, p = map(int, input().split())
if(a == 0 and p== 0) :
break
else :
m = ["A", "B", "C", "D", "E"]
shop = list()
shop.append(a + p)
for i in range(4) :
a, p = map(int, input().split())
shop.append(a + p)
top = max(shop)
Shop = shop.index(top)
print(m[Shop], top)
``` | output | 1 | 107,789 | 10 | 215,579 |
Provide a correct Python 3 solution for this coding contest problem.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247 | instruction | 0 | 107,790 | 10 | 215,580 |
"Correct Solution:
```
# Aizu Problem 0195: What is the Most Popular Shop in Tokaichi?
#
import sys, math, os
# read input:
PYDEV = os.environ.get('PYDEV')
if PYDEV=="True":
sys.stdin = open("sample-input.txt", "rt")
while True:
most = 0
shop = 'A'
a, b = [int(_) for _ in input().split()]
if a == b == 0:
break
most = a + b
for s in range(4):
a = sum([int(_) for _ in input().split()])
if a > most:
most = a
shop = chr(66 + s)
print(shop, most)
``` | output | 1 | 107,790 | 10 | 215,581 |
Provide a correct Python 3 solution for this coding contest problem.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247 | instruction | 0 | 107,791 | 10 | 215,582 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0195
"""
import sys
from sys import stdin
input = stdin.readline
def main(args):
while True:
am, pm = map(int, input().split())
if am == 0 and pm == 0:
break
totals = [0] * 5
totals[0] = am + pm
for i in range(1, 5):
am, pm = map(int, input().split())
totals[i] = am + pm
top = max(totals)
shop = totals.index(top)
print('{} {}'.format(chr(ord('A')+shop), top))
if __name__ == '__main__':
main(sys.argv[1:])
``` | output | 1 | 107,791 | 10 | 215,583 |
Provide a correct Python 3 solution for this coding contest problem.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247 | instruction | 0 | 107,792 | 10 | 215,584 |
"Correct Solution:
```
S = ["A","B","C","D","E"]
while True:
L = []
a,p = map(int,input().split())
if a == 0:
break
L.append(a+p)
for i in range(4):
a,p = map(int,input().split())
L.append(a+p)
m = max(L)
s = L.index(m)
print(S[s],m)
``` | output | 1 | 107,792 | 10 | 215,585 |
Provide a correct Python 3 solution for this coding contest problem.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247 | instruction | 0 | 107,793 | 10 | 215,586 |
"Correct Solution:
```
while True:
try:
A=[]
for i in range(5):
s1,s2=map(int,input().split())
if s1==0 and s2==0:
break
s=s1+s2
A.append(s)
if s==0:
break
for i in range(len(A)):
if A[i]==max(A):
if i==0:
print("A",A[i])
if i==1:
print("B",A[i])
if i==2:
print("C",A[i])
if i==3:
print("D",A[i])
if i==4:
print("E",A[i])
except EOFError:
break
``` | output | 1 | 107,793 | 10 | 215,587 |
Provide a correct Python 3 solution for this coding contest problem.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247 | instruction | 0 | 107,794 | 10 | 215,588 |
"Correct Solution:
```
try:
while True:
num_list = []
num_di = {}
#リストに数値を格納して１番大きい数を出力する
for i in range(0,5):
line = list(map(int,input().split()))
num_list.append(sum(line))
num_di[sum(line)] = 65 + i
num_list.sort()
print(chr(num_di[num_list[4]]) + " " + str(num_list[4]))
#EOFErrorを検知する
except EOFError:
pass
``` | output | 1 | 107,794 | 10 | 215,589 |
Provide a correct Python 3 solution for this coding contest problem.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247 | instruction | 0 | 107,795 | 10 | 215,590 |
"Correct Solution:
```
# AOJ 0195 What is the Most Popular Shop in Tokaich
# Python3 2018.6.20 bal4u
import sys
while True:
tbl = {}
for i in range(5):
s = sum(list(map(int, input().split())))
if i == 0 and s == 0: sys.exit()
tbl[chr(ord('A')+i)] = s
ans = sorted(tbl.items(), key=lambda x: x[1], reverse = True)
print(*ans[0])
``` | output | 1 | 107,795 | 10 | 215,591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247
Submitted Solution:
```
end = 0
while end == 0 :
shop = -1
max_s = 0
for i in range(5) :
try :
s1, s2 = map(int, input().split())
except EOFError :
end = 1
break
if s1 + s2 > max_s :
max_s = s1 + s2
shop = i
if shop == 0 :
shop_name = "A"
elif shop == 1 :
shop_name = "B"
elif shop == 2 :
shop_name = "C"
elif shop == 3 :
shop_name = "D"
elif shop == 4 :
shop_name = "E"
if end == 0 :
print(shop_name, max_s)
``` | instruction | 0 | 107,796 | 10 | 215,592 |
Yes | output | 1 | 107,796 | 10 | 215,593 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247
Submitted Solution:
```
import sys
while True:
sells=[]
for i in range(5):
s1,s2=[int(j) for j in input().split(" ")]
if s1==0 and s2==0:
sys.exit()
sells.append(s1+s2)
print(chr(ord('A')+sells.index(max(sells))),max(sells))
``` | instruction | 0 | 107,797 | 10 | 215,594 |
Yes | output | 1 | 107,797 | 10 | 215,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247
Submitted Solution:
```
def solve():
from sys import stdin
f_i = stdin
while True:
s1A, s2A = map(int, f_i.readline().split())
if s1A == 0 and s2A == 0:
break
sales = [(s1A + s2A, 'A')]
for shop in 'BCDE':
cnt = sum(map(int, f_i.readline().split()))
sales.append((cnt, shop))
cnt, shop = max(sales)
print(f"{shop} {cnt}")
solve()
``` | instruction | 0 | 107,798 | 10 | 215,596 |
Yes | output | 1 | 107,798 | 10 | 215,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247
Submitted Solution:
```
while True :
a, b = map(int, input().split())
if(a == b and b == 0) :
break
else :
S = ["A", "B", "C", "D", "E"]
shop = list()
shop.append(a + b)
for i in range(4) :
a, b = map(int, input().split())
shop.append(a + b)
win = max(shop)
Shop = shop.index(win)
print(S[Shop], win)
``` | instruction | 0 | 107,799 | 10 | 215,598 |
Yes | output | 1 | 107,799 | 10 | 215,599 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You're given an array a of n integers, such that a_1 + a_2 + ⋅⋅⋅ + a_n = 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n), decrement a_i by one and increment a_j by one. If i < j this operation is free, otherwise it costs one coin.
How many coins do you have to spend in order to make all elements equal to 0?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 5000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 10^5) — the number of elements.
The next line contains n integers a_1, …, a_n (-10^9 ≤ a_i ≤ 10^9). It is given that ∑_{i=1}^n a_i = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the minimum number of coins we have to spend in order to make all elements equal to 0.
Example
Input
7
4
-3 5 -3 1
2
1 -1
4
-3 2 -3 4
4
-1 1 1 -1
7
-5 7 -6 -4 17 -13 4
6
-1000000000 -1000000000 -1000000000 1000000000 1000000000 1000000000
1
0
Output
3
0
4
1
8
3000000000
0
Note
Possible strategy for the first test case:
* Do (i=2, j=3) three times (free), a = [-3, 2, 0, 1].
* Do (i=2, j=1) two times (pay two coins), a = [-1, 0, 0, 1].
* Do (i=4, j=1) one time (pay one coin), a = [0, 0, 0, 0]. | instruction | 0 | 108,021 | 10 | 216,042 |
Tags: constructive algorithms, implementation
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
ans, pos, neg = 0, 0, 0
ans = 0
stack = 0
for i in range(n):
if arr[i] < 0 and stack == 0:
ans += abs(arr[i])
elif arr[i] > 0 and stack == 0:
stack += arr[i]
elif arr[i] < 0 and stack > 0:
stack += arr[i]
if stack < 0:
ans += abs(stack)
stack = 0
elif arr[i] > 0 and stack > 0:
stack += arr[i]
print(ans)
# if arr[0] > 0:
# pos += arr[0]
# if arr[-1] < 0:
# neg += arr[-1]
# for i in range(n):
# if arr[i] > 0 and pos = 0 and i != n-1:
# pos += arr[i]
# if arr[i] < 0 and neg = 0 and i != 0:
# neg += arr[i]
# if arr[i] > 0 and pos > 0:
# if pos > neg:
# pos += arr[i]
# pos -=
# ans = abs(pos - neg)
# pos
# for i in range(1, n):
# if arr[i] < 0 and arr[i-1] > 0:
# if abs(arr[i]) > abs(arr[i-1]):
# arr[i] = arr[i] + arr[i-1]
# arr[i-1] = 0
# else:
# arr[i-1] = arr[i] + arr[i-1]
# arr[i] = 0
``` | output | 1 | 108,021 | 10 | 216,043 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You're given an array a of n integers, such that a_1 + a_2 + ⋅⋅⋅ + a_n = 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n), decrement a_i by one and increment a_j by one. If i < j this operation is free, otherwise it costs one coin.
How many coins do you have to spend in order to make all elements equal to 0?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 5000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 10^5) — the number of elements.
The next line contains n integers a_1, …, a_n (-10^9 ≤ a_i ≤ 10^9). It is given that ∑_{i=1}^n a_i = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the minimum number of coins we have to spend in order to make all elements equal to 0.
Example
Input
7
4
-3 5 -3 1
2
1 -1
4
-3 2 -3 4
4
-1 1 1 -1
7
-5 7 -6 -4 17 -13 4
6
-1000000000 -1000000000 -1000000000 1000000000 1000000000 1000000000
1
0
Output
3
0
4
1
8
3000000000
0
Note
Possible strategy for the first test case:
* Do (i=2, j=3) three times (free), a = [-3, 2, 0, 1].
* Do (i=2, j=1) two times (pay two coins), a = [-1, 0, 0, 1].
* Do (i=4, j=1) one time (pay one coin), a = [0, 0, 0, 0]. | instruction | 0 | 108,022 | 10 | 216,044 |
Tags: constructive algorithms, implementation
Correct Solution:
```
from sys import stdin
input = stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
a = [int(x) for x in input().split()]
p = 0
ans = 0
for i in range(n):
if a[i] > 0:
p += a[i]
else:
ans += max(0, -a[i] - p)
p = max(0, p + a[i])
print(ans)
``` | output | 1 | 108,022 | 10 | 216,045 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You're given an array a of n integers, such that a_1 + a_2 + ⋅⋅⋅ + a_n = 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n), decrement a_i by one and increment a_j by one. If i < j this operation is free, otherwise it costs one coin.
How many coins do you have to spend in order to make all elements equal to 0?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 5000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 10^5) — the number of elements.
The next line contains n integers a_1, …, a_n (-10^9 ≤ a_i ≤ 10^9). It is given that ∑_{i=1}^n a_i = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the minimum number of coins we have to spend in order to make all elements equal to 0.
Example
Input
7
4
-3 5 -3 1
2
1 -1
4
-3 2 -3 4
4
-1 1 1 -1
7
-5 7 -6 -4 17 -13 4
6
-1000000000 -1000000000 -1000000000 1000000000 1000000000 1000000000
1
0
Output
3
0
4
1
8
3000000000
0
Note
Possible strategy for the first test case:
* Do (i=2, j=3) three times (free), a = [-3, 2, 0, 1].
* Do (i=2, j=1) two times (pay two coins), a = [-1, 0, 0, 1].
* Do (i=4, j=1) one time (pay one coin), a = [0, 0, 0, 0]. | instruction | 0 | 108,023 | 10 | 216,046 |
Tags: constructive algorithms, implementation
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
arr=list(map(int,input().split()))
c=0
pos=0
for i in range(n):
if arr[i]>0:
pos+=arr[i]
else:
arr[i]=arr[i]*-1
if pos>arr[i]:
pos-=arr[i]
else:
c+=arr[i]-pos
pos=0
print(c)
``` | output | 1 | 108,023 | 10 | 216,047 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You're given an array a of n integers, such that a_1 + a_2 + ⋅⋅⋅ + a_n = 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n), decrement a_i by one and increment a_j by one. If i < j this operation is free, otherwise it costs one coin.
How many coins do you have to spend in order to make all elements equal to 0?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 5000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 10^5) — the number of elements.
The next line contains n integers a_1, …, a_n (-10^9 ≤ a_i ≤ 10^9). It is given that ∑_{i=1}^n a_i = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the minimum number of coins we have to spend in order to make all elements equal to 0.
Example
Input
7
4
-3 5 -3 1
2
1 -1
4
-3 2 -3 4
4
-1 1 1 -1
7
-5 7 -6 -4 17 -13 4
6
-1000000000 -1000000000 -1000000000 1000000000 1000000000 1000000000
1
0
Output
3
0
4
1
8
3000000000
0
Note
Possible strategy for the first test case:
* Do (i=2, j=3) three times (free), a = [-3, 2, 0, 1].
* Do (i=2, j=1) two times (pay two coins), a = [-1, 0, 0, 1].
* Do (i=4, j=1) one time (pay one coin), a = [0, 0, 0, 0]. | instruction | 0 | 108,024 | 10 | 216,048 |
Tags: constructive algorithms, implementation
Correct Solution:
```
for _ in range(int(input())):
lens = int(input())
arrs = [int(x) for x in input().split()]
psum = 0
for i in arrs:
psum += i
psum = max(psum, 0)
print(psum)
``` | output | 1 | 108,024 | 10 | 216,049 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You're given an array a of n integers, such that a_1 + a_2 + ⋅⋅⋅ + a_n = 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n), decrement a_i by one and increment a_j by one. If i < j this operation is free, otherwise it costs one coin.
How many coins do you have to spend in order to make all elements equal to 0?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 5000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 10^5) — the number of elements.
The next line contains n integers a_1, …, a_n (-10^9 ≤ a_i ≤ 10^9). It is given that ∑_{i=1}^n a_i = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the minimum number of coins we have to spend in order to make all elements equal to 0.
Example
Input
7
4
-3 5 -3 1
2
1 -1
4
-3 2 -3 4
4
-1 1 1 -1
7
-5 7 -6 -4 17 -13 4
6
-1000000000 -1000000000 -1000000000 1000000000 1000000000 1000000000
1
0
Output
3
0
4
1
8
3000000000
0
Note
Possible strategy for the first test case:
* Do (i=2, j=3) three times (free), a = [-3, 2, 0, 1].
* Do (i=2, j=1) two times (pay two coins), a = [-1, 0, 0, 1].
* Do (i=4, j=1) one time (pay one coin), a = [0, 0, 0, 0]. | instruction | 0 | 108,025 | 10 | 216,050 |
Tags: constructive algorithms, implementation
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Thu Sep 17 07:50:00 2020
@author: Harshal
"""
test=int(input())
for _ in range(test):
n=int(input())
arr=list(map(int,input().split()))
ans=0
for i in arr:
ans=max(ans+i,0)
print(ans)
``` | output | 1 | 108,025 | 10 | 216,051 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You're given an array a of n integers, such that a_1 + a_2 + ⋅⋅⋅ + a_n = 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n), decrement a_i by one and increment a_j by one. If i < j this operation is free, otherwise it costs one coin.
How many coins do you have to spend in order to make all elements equal to 0?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 5000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 10^5) — the number of elements.
The next line contains n integers a_1, …, a_n (-10^9 ≤ a_i ≤ 10^9). It is given that ∑_{i=1}^n a_i = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the minimum number of coins we have to spend in order to make all elements equal to 0.
Example
Input
7
4
-3 5 -3 1
2
1 -1
4
-3 2 -3 4
4
-1 1 1 -1
7
-5 7 -6 -4 17 -13 4
6
-1000000000 -1000000000 -1000000000 1000000000 1000000000 1000000000
1
0
Output
3
0
4
1
8
3000000000
0
Note
Possible strategy for the first test case:
* Do (i=2, j=3) three times (free), a = [-3, 2, 0, 1].
* Do (i=2, j=1) two times (pay two coins), a = [-1, 0, 0, 1].
* Do (i=4, j=1) one time (pay one coin), a = [0, 0, 0, 0]. | instruction | 0 | 108,026 | 10 | 216,052 |
Tags: constructive algorithms, implementation
Correct Solution:
```
T = int(input())
for t in range(T):
n = int(input())
A = list(map(int, input().split()))
start = 1
for i in range(n):
if A[i] > 0:
temp = []
sum = 0
for j in range(max(i+1, start), n):
if A[j] < 0:
temp.append(j)
sum -= A[j]
if sum >= A[i]:
break
if sum >= A[i]:
# A[i] = 0
for k in temp:
if A[i] + A[k] > 0:
A[i] += A[k]
A[k] = 0
else:
A[k] += A[i]
A[i] = 0
start = k
# print(A[i], A[k], k,sum)
else:
A[i] -= sum
for k in temp:
A[k] = 0
# print(A[i], A[k], sum)
break
ans = 0
for i in range(n):
if A[i] > 0:
break
ans -= A[i]
print( ans)
``` | output | 1 | 108,026 | 10 | 216,053 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You're given an array a of n integers, such that a_1 + a_2 + ⋅⋅⋅ + a_n = 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n), decrement a_i by one and increment a_j by one. If i < j this operation is free, otherwise it costs one coin.
How many coins do you have to spend in order to make all elements equal to 0?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 5000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 10^5) — the number of elements.
The next line contains n integers a_1, …, a_n (-10^9 ≤ a_i ≤ 10^9). It is given that ∑_{i=1}^n a_i = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the minimum number of coins we have to spend in order to make all elements equal to 0.
Example
Input
7
4
-3 5 -3 1
2
1 -1
4
-3 2 -3 4
4
-1 1 1 -1
7
-5 7 -6 -4 17 -13 4
6
-1000000000 -1000000000 -1000000000 1000000000 1000000000 1000000000
1
0
Output
3
0
4
1
8
3000000000
0
Note
Possible strategy for the first test case:
* Do (i=2, j=3) three times (free), a = [-3, 2, 0, 1].
* Do (i=2, j=1) two times (pay two coins), a = [-1, 0, 0, 1].
* Do (i=4, j=1) one time (pay one coin), a = [0, 0, 0, 0]. | instruction | 0 | 108,027 | 10 | 216,054 |
Tags: constructive algorithms, implementation
Correct Solution:
```
import sys
input=sys.stdin.readline
for _ in range(int(input())):
n=int(input())
ar=list(map(int,input().split()))
ans=0
pos=0
neg=0
for i in range(n):
if(ar[i]<0):
pos+=ar[i]
if(pos<0):
ans+=abs(pos)
pos=0
elif(ar[i]>0):
pos+=ar[i]
print(ans)
``` | output | 1 | 108,027 | 10 | 216,055 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given an array a of n integers, such that a_1 + a_2 + ⋅⋅⋅ + a_n = 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n), decrement a_i by one and increment a_j by one. If i < j this operation is free, otherwise it costs one coin.
How many coins do you have to spend in order to make all elements equal to 0?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 5000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 10^5) — the number of elements.
The next line contains n integers a_1, …, a_n (-10^9 ≤ a_i ≤ 10^9). It is given that ∑_{i=1}^n a_i = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the minimum number of coins we have to spend in order to make all elements equal to 0.
Example
Input
7
4
-3 5 -3 1
2
1 -1
4
-3 2 -3 4
4
-1 1 1 -1
7
-5 7 -6 -4 17 -13 4
6
-1000000000 -1000000000 -1000000000 1000000000 1000000000 1000000000
1
0
Output
3
0
4
1
8
3000000000
0
Note
Possible strategy for the first test case:
* Do (i=2, j=3) three times (free), a = [-3, 2, 0, 1].
* Do (i=2, j=1) two times (pay two coins), a = [-1, 0, 0, 1].
* Do (i=4, j=1) one time (pay one coin), a = [0, 0, 0, 0].
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
minus=0
plus=0
for i in a:
if i==0:
continue
elif i>0:
plus+=i
else:
if plus>=abs(i):
plus+=i
else:
plus=0
hg=abs(i)-plus
minus-=hg
print(abs(plus))
``` | instruction | 0 | 108,028 | 10 | 216,056 |
Yes | output | 1 | 108,028 | 10 | 216,057 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given an array a of n integers, such that a_1 + a_2 + ⋅⋅⋅ + a_n = 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n), decrement a_i by one and increment a_j by one. If i < j this operation is free, otherwise it costs one coin.
How many coins do you have to spend in order to make all elements equal to 0?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 5000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 10^5) — the number of elements.
The next line contains n integers a_1, …, a_n (-10^9 ≤ a_i ≤ 10^9). It is given that ∑_{i=1}^n a_i = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the minimum number of coins we have to spend in order to make all elements equal to 0.
Example
Input
7
4
-3 5 -3 1
2
1 -1
4
-3 2 -3 4
4
-1 1 1 -1
7
-5 7 -6 -4 17 -13 4
6
-1000000000 -1000000000 -1000000000 1000000000 1000000000 1000000000
1
0
Output
3
0
4
1
8
3000000000
0
Note
Possible strategy for the first test case:
* Do (i=2, j=3) three times (free), a = [-3, 2, 0, 1].
* Do (i=2, j=1) two times (pay two coins), a = [-1, 0, 0, 1].
* Do (i=4, j=1) one time (pay one coin), a = [0, 0, 0, 0].
Submitted Solution:
```
t=int(input())
for i in range(t):
a=int(input())
b=[int(s) for s in input().split()]
s=0
t=0
for k in range(len(b)):
if s+b[k]>=0:
s+=b[k]
else:
t+=abs(b[k])-s
s-=min(s,abs(b[k]))
print(t)
``` | instruction | 0 | 108,029 | 10 | 216,058 |
Yes | output | 1 | 108,029 | 10 | 216,059 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given an array a of n integers, such that a_1 + a_2 + ⋅⋅⋅ + a_n = 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n), decrement a_i by one and increment a_j by one. If i < j this operation is free, otherwise it costs one coin.
How many coins do you have to spend in order to make all elements equal to 0?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 5000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 10^5) — the number of elements.
The next line contains n integers a_1, …, a_n (-10^9 ≤ a_i ≤ 10^9). It is given that ∑_{i=1}^n a_i = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the minimum number of coins we have to spend in order to make all elements equal to 0.
Example
Input
7
4
-3 5 -3 1
2
1 -1
4
-3 2 -3 4
4
-1 1 1 -1
7
-5 7 -6 -4 17 -13 4
6
-1000000000 -1000000000 -1000000000 1000000000 1000000000 1000000000
1
0
Output
3
0
4
1
8
3000000000
0
Note
Possible strategy for the first test case:
* Do (i=2, j=3) three times (free), a = [-3, 2, 0, 1].
* Do (i=2, j=1) two times (pay two coins), a = [-1, 0, 0, 1].
* Do (i=4, j=1) one time (pay one coin), a = [0, 0, 0, 0].
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
s = 0
ans = 0
for i in range(n):
s += arr[i]
ans = min(ans, s)
print(abs(ans))
``` | instruction | 0 | 108,030 | 10 | 216,060 |
Yes | output | 1 | 108,030 | 10 | 216,061 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given an array a of n integers, such that a_1 + a_2 + ⋅⋅⋅ + a_n = 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n), decrement a_i by one and increment a_j by one. If i < j this operation is free, otherwise it costs one coin.
How many coins do you have to spend in order to make all elements equal to 0?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 5000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 10^5) — the number of elements.
The next line contains n integers a_1, …, a_n (-10^9 ≤ a_i ≤ 10^9). It is given that ∑_{i=1}^n a_i = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the minimum number of coins we have to spend in order to make all elements equal to 0.
Example
Input
7
4
-3 5 -3 1
2
1 -1
4
-3 2 -3 4
4
-1 1 1 -1
7
-5 7 -6 -4 17 -13 4
6
-1000000000 -1000000000 -1000000000 1000000000 1000000000 1000000000
1
0
Output
3
0
4
1
8
3000000000
0
Note
Possible strategy for the first test case:
* Do (i=2, j=3) three times (free), a = [-3, 2, 0, 1].
* Do (i=2, j=1) two times (pay two coins), a = [-1, 0, 0, 1].
* Do (i=4, j=1) one time (pay one coin), a = [0, 0, 0, 0].
Submitted Solution:
```
# https://codeforces.com/contest/1405/problem/B
'''
Author @Subhajit Das (sdsubhajitdas.github.io)
SWE @Turbot HQ India PVT Ltd.
07/09/2020
'''
import math
def main():
n = int(input().strip())
arr = list(map(int,input().strip().split()))
cost = suffixSum = arr[-1]
for i in reversed(range(n-1)):
suffixSum += arr[i]
cost = max(cost,suffixSum)
print(cost)
if __name__ == "__main__":
for _ in range(int(input())):
main()
# main()
``` | instruction | 0 | 108,031 | 10 | 216,062 |
Yes | output | 1 | 108,031 | 10 | 216,063 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given an array a of n integers, such that a_1 + a_2 + ⋅⋅⋅ + a_n = 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n), decrement a_i by one and increment a_j by one. If i < j this operation is free, otherwise it costs one coin.
How many coins do you have to spend in order to make all elements equal to 0?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 5000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 10^5) — the number of elements.
The next line contains n integers a_1, …, a_n (-10^9 ≤ a_i ≤ 10^9). It is given that ∑_{i=1}^n a_i = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the minimum number of coins we have to spend in order to make all elements equal to 0.
Example
Input
7
4
-3 5 -3 1
2
1 -1
4
-3 2 -3 4
4
-1 1 1 -1
7
-5 7 -6 -4 17 -13 4
6
-1000000000 -1000000000 -1000000000 1000000000 1000000000 1000000000
1
0
Output
3
0
4
1
8
3000000000
0
Note
Possible strategy for the first test case:
* Do (i=2, j=3) three times (free), a = [-3, 2, 0, 1].
* Do (i=2, j=1) two times (pay two coins), a = [-1, 0, 0, 1].
* Do (i=4, j=1) one time (pay one coin), a = [0, 0, 0, 0].
Submitted Solution:
```
import math
test = int(input())
for t in range(test):
n = int(input())
A = list(map(int,input().split()))
if(n==2):
if(A[0]<=A[1]):
print(0)
continue
sump=0;sumn=0
B = []
for i in range(n):
if(A[i]>0):
sump+=A[i]
if(sumn!=0):
B.append(sumn)
sumn = 0
else:
sumn+=A[i]
if(sump!=0):
B.append(sump)
sump = 0
if(sumn!=0):
B.append(sumn)
if(sump!=0):
B.append(sump)
for i in range(1,len(B)-1):
if(B[i]<0):
continue
else:
a = min(B[i],abs(B[i+1]))
B[i+1] += a
B[i] -= a
ans = 0
for i in B:
if(i<0):
ans+=i
print(abs(ans))
``` | instruction | 0 | 108,032 | 10 | 216,064 |
No | output | 1 | 108,032 | 10 | 216,065 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given an array a of n integers, such that a_1 + a_2 + ⋅⋅⋅ + a_n = 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n), decrement a_i by one and increment a_j by one. If i < j this operation is free, otherwise it costs one coin.
How many coins do you have to spend in order to make all elements equal to 0?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 5000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 10^5) — the number of elements.
The next line contains n integers a_1, …, a_n (-10^9 ≤ a_i ≤ 10^9). It is given that ∑_{i=1}^n a_i = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the minimum number of coins we have to spend in order to make all elements equal to 0.
Example
Input
7
4
-3 5 -3 1
2
1 -1
4
-3 2 -3 4
4
-1 1 1 -1
7
-5 7 -6 -4 17 -13 4
6
-1000000000 -1000000000 -1000000000 1000000000 1000000000 1000000000
1
0
Output
3
0
4
1
8
3000000000
0
Note
Possible strategy for the first test case:
* Do (i=2, j=3) three times (free), a = [-3, 2, 0, 1].
* Do (i=2, j=1) two times (pay two coins), a = [-1, 0, 0, 1].
* Do (i=4, j=1) one time (pay one coin), a = [0, 0, 0, 0].
Submitted Solution:
```
t=int(input())
while (t>0):
n=int(input())
arr=[abs(int(i)) for i in input().split()]
print(sum(arr)//n)
t-=1
``` | instruction | 0 | 108,033 | 10 | 216,066 |
No | output | 1 | 108,033 | 10 | 216,067 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given an array a of n integers, such that a_1 + a_2 + ⋅⋅⋅ + a_n = 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n), decrement a_i by one and increment a_j by one. If i < j this operation is free, otherwise it costs one coin.
How many coins do you have to spend in order to make all elements equal to 0?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 5000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 10^5) — the number of elements.
The next line contains n integers a_1, …, a_n (-10^9 ≤ a_i ≤ 10^9). It is given that ∑_{i=1}^n a_i = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the minimum number of coins we have to spend in order to make all elements equal to 0.
Example
Input
7
4
-3 5 -3 1
2
1 -1
4
-3 2 -3 4
4
-1 1 1 -1
7
-5 7 -6 -4 17 -13 4
6
-1000000000 -1000000000 -1000000000 1000000000 1000000000 1000000000
1
0
Output
3
0
4
1
8
3000000000
0
Note
Possible strategy for the first test case:
* Do (i=2, j=3) three times (free), a = [-3, 2, 0, 1].
* Do (i=2, j=1) two times (pay two coins), a = [-1, 0, 0, 1].
* Do (i=4, j=1) one time (pay one coin), a = [0, 0, 0, 0].
Submitted Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
mod = 10 ** 9 + 7
mod1 = 998244353
# sys.setrecursionlimit(300000)
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
# sys.setrecursionlimit(300000)
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
# -----------------------------------------------binary seacrh tree---------------------------------------
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: max(a, b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] <= key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
for ik in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
ans=0
cur=0
for i in range(n-1,-1,-1):
if l[i]>=0:
ans+=max(0,l[i]+cur)
cur-=min(l[i],cur)
else:
cur+=l[i]
print(ans)
``` | instruction | 0 | 108,034 | 10 | 216,068 |
No | output | 1 | 108,034 | 10 | 216,069 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given an array a of n integers, such that a_1 + a_2 + ⋅⋅⋅ + a_n = 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n), decrement a_i by one and increment a_j by one. If i < j this operation is free, otherwise it costs one coin.
How many coins do you have to spend in order to make all elements equal to 0?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 5000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 10^5) — the number of elements.
The next line contains n integers a_1, …, a_n (-10^9 ≤ a_i ≤ 10^9). It is given that ∑_{i=1}^n a_i = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the minimum number of coins we have to spend in order to make all elements equal to 0.
Example
Input
7
4
-3 5 -3 1
2
1 -1
4
-3 2 -3 4
4
-1 1 1 -1
7
-5 7 -6 -4 17 -13 4
6
-1000000000 -1000000000 -1000000000 1000000000 1000000000 1000000000
1
0
Output
3
0
4
1
8
3000000000
0
Note
Possible strategy for the first test case:
* Do (i=2, j=3) three times (free), a = [-3, 2, 0, 1].
* Do (i=2, j=1) two times (pay two coins), a = [-1, 0, 0, 1].
* Do (i=4, j=1) one time (pay one coin), a = [0, 0, 0, 0].
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()));b=[]
for i in range(n):
if a[i]>0:
b.append(i)
j=0
if n==1:
print(0)
else:
for i in range(b[0]+1,n):
if a[i]>=0:
continue
if a[b[j]]>=-a[i]:
a[b[j]]+=a[i]
a[i]=0
else:
a[i]+=a[b[j]]
a[b[j]]=0
j+=1
s=0
a.sort()
for i in a:
if i>=0:
break
s-=i
print(s)
``` | instruction | 0 | 108,035 | 10 | 216,070 |
No | output | 1 | 108,035 | 10 | 216,071 |
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