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.
Joisino is planning to open a shop in a shopping street.
Each of the five weekdays is divided into two periods, the morning and the evening. For each of those ten periods, a shop must be either open during the whole period, or closed during the whole period. Naturally, a shop must be open during at least one of those periods.
There are already N stores in the street, numbered 1 through N.
You are given information of the business hours of those shops, F_{i,j,k}. If F_{i,j,k}=1, Shop i is open during Period k on Day j (this notation is explained below); if F_{i,j,k}=0, Shop i is closed during that period. Here, the days of the week are denoted as follows. Monday: Day 1, Tuesday: Day 2, Wednesday: Day 3, Thursday: Day 4, Friday: Day 5. Also, the morning is denoted as Period 1, and the afternoon is denoted as Period 2.
Let c_i be the number of periods during which both Shop i and Joisino's shop are open. Then, the profit of Joisino's shop will be P_{1,c_1}+P_{2,c_2}+...+P_{N,c_N}.
Find the maximum possible profit of Joisino's shop when she decides whether her shop is open during each period, making sure that it is open during at least one period.
Constraints
* 1≤N≤100
* 0≤F_{i,j,k}≤1
* For every integer i such that 1≤i≤N, there exists at least one pair (j,k) such that F_{i,j,k}=1.
* -10^7≤P_{i,j}≤10^7
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
F_{1,1,1} F_{1,1,2} ... F_{1,5,1} F_{1,5,2}
:
F_{N,1,1} F_{N,1,2} ... F_{N,5,1} F_{N,5,2}
P_{1,0} ... P_{1,10}
:
P_{N,0} ... P_{N,10}
Output
Print the maximum possible profit of Joisino's shop.
Examples
Input
1
1 1 0 1 0 0 0 1 0 1
3 4 5 6 7 8 9 -2 -3 4 -2
Output
8
Input
2
1 1 1 1 1 0 0 0 0 0
0 0 0 0 0 1 1 1 1 1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1
Output
-2
Input
3
1 1 1 1 1 1 0 0 1 1
0 1 0 1 1 1 1 0 1 0
1 0 1 1 0 1 0 1 0 1
-8 6 -2 -8 -8 4 8 7 -6 2 2
-9 2 0 1 7 -5 0 -2 -6 5 5
6 -6 7 -9 6 -5 8 0 -9 -7 -7
Output
23
Submitted Solution:
```
import itertools
N = int(input())
F = [list(map(int,input().split())) for i in range(N)]
P = [list(map(int,input().split())) for i in range(N)]
ans =int(-10e15)
for k in range(1,11):
for j in itertools.combinations(range(10),k):
tmp = 0
for f,p in zip(F, P):
num = sum([f[cnt] for cnt in j])
tmp += p[num]
ans = max(ans,tmp)
print(ans)
``` | instruction | 0 | 57,859 | 10 | 115,718 |
Yes | output | 1 | 57,859 | 10 | 115,719 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Joisino is planning to open a shop in a shopping street.
Each of the five weekdays is divided into two periods, the morning and the evening. For each of those ten periods, a shop must be either open during the whole period, or closed during the whole period. Naturally, a shop must be open during at least one of those periods.
There are already N stores in the street, numbered 1 through N.
You are given information of the business hours of those shops, F_{i,j,k}. If F_{i,j,k}=1, Shop i is open during Period k on Day j (this notation is explained below); if F_{i,j,k}=0, Shop i is closed during that period. Here, the days of the week are denoted as follows. Monday: Day 1, Tuesday: Day 2, Wednesday: Day 3, Thursday: Day 4, Friday: Day 5. Also, the morning is denoted as Period 1, and the afternoon is denoted as Period 2.
Let c_i be the number of periods during which both Shop i and Joisino's shop are open. Then, the profit of Joisino's shop will be P_{1,c_1}+P_{2,c_2}+...+P_{N,c_N}.
Find the maximum possible profit of Joisino's shop when she decides whether her shop is open during each period, making sure that it is open during at least one period.
Constraints
* 1≤N≤100
* 0≤F_{i,j,k}≤1
* For every integer i such that 1≤i≤N, there exists at least one pair (j,k) such that F_{i,j,k}=1.
* -10^7≤P_{i,j}≤10^7
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
F_{1,1,1} F_{1,1,2} ... F_{1,5,1} F_{1,5,2}
:
F_{N,1,1} F_{N,1,2} ... F_{N,5,1} F_{N,5,2}
P_{1,0} ... P_{1,10}
:
P_{N,0} ... P_{N,10}
Output
Print the maximum possible profit of Joisino's shop.
Examples
Input
1
1 1 0 1 0 0 0 1 0 1
3 4 5 6 7 8 9 -2 -3 4 -2
Output
8
Input
2
1 1 1 1 1 0 0 0 0 0
0 0 0 0 0 1 1 1 1 1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1
Output
-2
Input
3
1 1 1 1 1 1 0 0 1 1
0 1 0 1 1 1 1 0 1 0
1 0 1 1 0 1 0 1 0 1
-8 6 -2 -8 -8 4 8 7 -6 2 2
-9 2 0 1 7 -5 0 -2 -6 5 5
6 -6 7 -9 6 -5 8 0 -9 -7 -7
Output
23
Submitted Solution:
```
N = int(input())
F = [list(map(int, input().split())) for _ in range(N)]
P = [list(map(int, input().split())) for _ in range(N)]
max_first, max_second = 0, 0
max_v = -float("inf")
for p in range(1, 1 << 10):
v = 0
for n in range(N):
c = 0
for m in range(10):
c += (p & 1 << m != 0) & (F[n][m])
v += P[n][c]
max_v = max(max_v, v)
print(max_v)
``` | instruction | 0 | 57,860 | 10 | 115,720 |
Yes | output | 1 | 57,860 | 10 | 115,721 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Joisino is planning to open a shop in a shopping street.
Each of the five weekdays is divided into two periods, the morning and the evening. For each of those ten periods, a shop must be either open during the whole period, or closed during the whole period. Naturally, a shop must be open during at least one of those periods.
There are already N stores in the street, numbered 1 through N.
You are given information of the business hours of those shops, F_{i,j,k}. If F_{i,j,k}=1, Shop i is open during Period k on Day j (this notation is explained below); if F_{i,j,k}=0, Shop i is closed during that period. Here, the days of the week are denoted as follows. Monday: Day 1, Tuesday: Day 2, Wednesday: Day 3, Thursday: Day 4, Friday: Day 5. Also, the morning is denoted as Period 1, and the afternoon is denoted as Period 2.
Let c_i be the number of periods during which both Shop i and Joisino's shop are open. Then, the profit of Joisino's shop will be P_{1,c_1}+P_{2,c_2}+...+P_{N,c_N}.
Find the maximum possible profit of Joisino's shop when she decides whether her shop is open during each period, making sure that it is open during at least one period.
Constraints
* 1≤N≤100
* 0≤F_{i,j,k}≤1
* For every integer i such that 1≤i≤N, there exists at least one pair (j,k) such that F_{i,j,k}=1.
* -10^7≤P_{i,j}≤10^7
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
F_{1,1,1} F_{1,1,2} ... F_{1,5,1} F_{1,5,2}
:
F_{N,1,1} F_{N,1,2} ... F_{N,5,1} F_{N,5,2}
P_{1,0} ... P_{1,10}
:
P_{N,0} ... P_{N,10}
Output
Print the maximum possible profit of Joisino's shop.
Examples
Input
1
1 1 0 1 0 0 0 1 0 1
3 4 5 6 7 8 9 -2 -3 4 -2
Output
8
Input
2
1 1 1 1 1 0 0 0 0 0
0 0 0 0 0 1 1 1 1 1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1
Output
-2
Input
3
1 1 1 1 1 1 0 0 1 1
0 1 0 1 1 1 1 0 1 0
1 0 1 1 0 1 0 1 0 1
-8 6 -2 -8 -8 4 8 7 -6 2 2
-9 2 0 1 7 -5 0 -2 -6 5 5
6 -6 7 -9 6 -5 8 0 -9 -7 -7
Output
23
Submitted Solution:
```
from itertools import product
from operator import and_
N = int(input())
F = [tuple(map(lambda s: s=="1", input().split())) for _ in range(N)]
P = [tuple(map(int, input().split())) for _ in range(N)]
print(max(sum(P[i][sum(map(and_, F[i], prod))] for i in range(N)) for prod in product((True, False), repeat=10) if sum(prod)))
``` | instruction | 0 | 57,861 | 10 | 115,722 |
Yes | output | 1 | 57,861 | 10 | 115,723 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Joisino is planning to open a shop in a shopping street.
Each of the five weekdays is divided into two periods, the morning and the evening. For each of those ten periods, a shop must be either open during the whole period, or closed during the whole period. Naturally, a shop must be open during at least one of those periods.
There are already N stores in the street, numbered 1 through N.
You are given information of the business hours of those shops, F_{i,j,k}. If F_{i,j,k}=1, Shop i is open during Period k on Day j (this notation is explained below); if F_{i,j,k}=0, Shop i is closed during that period. Here, the days of the week are denoted as follows. Monday: Day 1, Tuesday: Day 2, Wednesday: Day 3, Thursday: Day 4, Friday: Day 5. Also, the morning is denoted as Period 1, and the afternoon is denoted as Period 2.
Let c_i be the number of periods during which both Shop i and Joisino's shop are open. Then, the profit of Joisino's shop will be P_{1,c_1}+P_{2,c_2}+...+P_{N,c_N}.
Find the maximum possible profit of Joisino's shop when she decides whether her shop is open during each period, making sure that it is open during at least one period.
Constraints
* 1≤N≤100
* 0≤F_{i,j,k}≤1
* For every integer i such that 1≤i≤N, there exists at least one pair (j,k) such that F_{i,j,k}=1.
* -10^7≤P_{i,j}≤10^7
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
F_{1,1,1} F_{1,1,2} ... F_{1,5,1} F_{1,5,2}
:
F_{N,1,1} F_{N,1,2} ... F_{N,5,1} F_{N,5,2}
P_{1,0} ... P_{1,10}
:
P_{N,0} ... P_{N,10}
Output
Print the maximum possible profit of Joisino's shop.
Examples
Input
1
1 1 0 1 0 0 0 1 0 1
3 4 5 6 7 8 9 -2 -3 4 -2
Output
8
Input
2
1 1 1 1 1 0 0 0 0 0
0 0 0 0 0 1 1 1 1 1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1
Output
-2
Input
3
1 1 1 1 1 1 0 0 1 1
0 1 0 1 1 1 1 0 1 0
1 0 1 1 0 1 0 1 0 1
-8 6 -2 -8 -8 4 8 7 -6 2 2
-9 2 0 1 7 -5 0 -2 -6 5 5
6 -6 7 -9 6 -5 8 0 -9 -7 -7
Output
23
Submitted Solution:
```
N=int(input())
F=[]
for i in range(N):
f=list(map(int, input().split()))
F.append(f)
P=[]
for i in range(N):
p=list(map(int, input().split()))
P.append(p)
C=[0]*N
for i in range(N):
for j in range(10):
if F[i][j]==1:
C[i]+=1
t=0
T=[]
for i in range(N):
A=int(max(P[i][0:C[i]+1]))
T.append(A)
if A==P[i][0]:
t+=1
S=[]
if t==N:
for i in range(N):
S.append(P[i][1])
T.append(max(S))
print(sum(T))
``` | instruction | 0 | 57,862 | 10 | 115,724 |
No | output | 1 | 57,862 | 10 | 115,725 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Joisino is planning to open a shop in a shopping street.
Each of the five weekdays is divided into two periods, the morning and the evening. For each of those ten periods, a shop must be either open during the whole period, or closed during the whole period. Naturally, a shop must be open during at least one of those periods.
There are already N stores in the street, numbered 1 through N.
You are given information of the business hours of those shops, F_{i,j,k}. If F_{i,j,k}=1, Shop i is open during Period k on Day j (this notation is explained below); if F_{i,j,k}=0, Shop i is closed during that period. Here, the days of the week are denoted as follows. Monday: Day 1, Tuesday: Day 2, Wednesday: Day 3, Thursday: Day 4, Friday: Day 5. Also, the morning is denoted as Period 1, and the afternoon is denoted as Period 2.
Let c_i be the number of periods during which both Shop i and Joisino's shop are open. Then, the profit of Joisino's shop will be P_{1,c_1}+P_{2,c_2}+...+P_{N,c_N}.
Find the maximum possible profit of Joisino's shop when she decides whether her shop is open during each period, making sure that it is open during at least one period.
Constraints
* 1≤N≤100
* 0≤F_{i,j,k}≤1
* For every integer i such that 1≤i≤N, there exists at least one pair (j,k) such that F_{i,j,k}=1.
* -10^7≤P_{i,j}≤10^7
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
F_{1,1,1} F_{1,1,2} ... F_{1,5,1} F_{1,5,2}
:
F_{N,1,1} F_{N,1,2} ... F_{N,5,1} F_{N,5,2}
P_{1,0} ... P_{1,10}
:
P_{N,0} ... P_{N,10}
Output
Print the maximum possible profit of Joisino's shop.
Examples
Input
1
1 1 0 1 0 0 0 1 0 1
3 4 5 6 7 8 9 -2 -3 4 -2
Output
8
Input
2
1 1 1 1 1 0 0 0 0 0
0 0 0 0 0 1 1 1 1 1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1
Output
-2
Input
3
1 1 1 1 1 1 0 0 1 1
0 1 0 1 1 1 1 0 1 0
1 0 1 1 0 1 0 1 0 1
-8 6 -2 -8 -8 4 8 7 -6 2 2
-9 2 0 1 7 -5 0 -2 -6 5 5
6 -6 7 -9 6 -5 8 0 -9 -7 -7
Output
23
Submitted Solution:
```
import sys
N = int(input())
time_lis = [0] * N
profit = [0] * N
for n in range(N):
time_lis[n] = list(map(int, input().split()))
for n in range(N):
profit[n] = list(map(int, input().split()))
ans = 0
always_zero = True
for n in range(N):
count_1 = len([i for i in time_lis[n] if i == 1])
ans += max(profit[n][:count_1 + 1])
if profit[n].index(ans) != 0:
always_zero = False
if always_zero:
max_prof = sorted([item[0], item[1] for item in profit], key=lambda x: -x[1])
print(ans - max_prof[0][0] + max_prof[0][1])
sys.exit()
print(ans)
``` | instruction | 0 | 57,863 | 10 | 115,726 |
No | output | 1 | 57,863 | 10 | 115,727 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Joisino is planning to open a shop in a shopping street.
Each of the five weekdays is divided into two periods, the morning and the evening. For each of those ten periods, a shop must be either open during the whole period, or closed during the whole period. Naturally, a shop must be open during at least one of those periods.
There are already N stores in the street, numbered 1 through N.
You are given information of the business hours of those shops, F_{i,j,k}. If F_{i,j,k}=1, Shop i is open during Period k on Day j (this notation is explained below); if F_{i,j,k}=0, Shop i is closed during that period. Here, the days of the week are denoted as follows. Monday: Day 1, Tuesday: Day 2, Wednesday: Day 3, Thursday: Day 4, Friday: Day 5. Also, the morning is denoted as Period 1, and the afternoon is denoted as Period 2.
Let c_i be the number of periods during which both Shop i and Joisino's shop are open. Then, the profit of Joisino's shop will be P_{1,c_1}+P_{2,c_2}+...+P_{N,c_N}.
Find the maximum possible profit of Joisino's shop when she decides whether her shop is open during each period, making sure that it is open during at least one period.
Constraints
* 1≤N≤100
* 0≤F_{i,j,k}≤1
* For every integer i such that 1≤i≤N, there exists at least one pair (j,k) such that F_{i,j,k}=1.
* -10^7≤P_{i,j}≤10^7
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
F_{1,1,1} F_{1,1,2} ... F_{1,5,1} F_{1,5,2}
:
F_{N,1,1} F_{N,1,2} ... F_{N,5,1} F_{N,5,2}
P_{1,0} ... P_{1,10}
:
P_{N,0} ... P_{N,10}
Output
Print the maximum possible profit of Joisino's shop.
Examples
Input
1
1 1 0 1 0 0 0 1 0 1
3 4 5 6 7 8 9 -2 -3 4 -2
Output
8
Input
2
1 1 1 1 1 0 0 0 0 0
0 0 0 0 0 1 1 1 1 1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1
Output
-2
Input
3
1 1 1 1 1 1 0 0 1 1
0 1 0 1 1 1 1 0 1 0
1 0 1 1 0 1 0 1 0 1
-8 6 -2 -8 -8 4 8 7 -6 2 2
-9 2 0 1 7 -5 0 -2 -6 5 5
6 -6 7 -9 6 -5 8 0 -9 -7 -7
Output
23
Submitted Solution:
```
N = int(input())
F = [list(map(int, input().split())) for _ in range(N)]
P = [list(map(int, input().split())) for _ in range(N)]
ans = 0
for i in range(2 ** 10):
sum = 0
for j in range(N):
cnt = 0
for k in range(10):
if i >> k & 1 and F[j][k] == 1:
cnt += 1
sum += P[j][cnt]
ans = max(ans, sum)
print(ans)
``` | instruction | 0 | 57,864 | 10 | 115,728 |
No | output | 1 | 57,864 | 10 | 115,729 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Joisino is planning to open a shop in a shopping street.
Each of the five weekdays is divided into two periods, the morning and the evening. For each of those ten periods, a shop must be either open during the whole period, or closed during the whole period. Naturally, a shop must be open during at least one of those periods.
There are already N stores in the street, numbered 1 through N.
You are given information of the business hours of those shops, F_{i,j,k}. If F_{i,j,k}=1, Shop i is open during Period k on Day j (this notation is explained below); if F_{i,j,k}=0, Shop i is closed during that period. Here, the days of the week are denoted as follows. Monday: Day 1, Tuesday: Day 2, Wednesday: Day 3, Thursday: Day 4, Friday: Day 5. Also, the morning is denoted as Period 1, and the afternoon is denoted as Period 2.
Let c_i be the number of periods during which both Shop i and Joisino's shop are open. Then, the profit of Joisino's shop will be P_{1,c_1}+P_{2,c_2}+...+P_{N,c_N}.
Find the maximum possible profit of Joisino's shop when she decides whether her shop is open during each period, making sure that it is open during at least one period.
Constraints
* 1≤N≤100
* 0≤F_{i,j,k}≤1
* For every integer i such that 1≤i≤N, there exists at least one pair (j,k) such that F_{i,j,k}=1.
* -10^7≤P_{i,j}≤10^7
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
F_{1,1,1} F_{1,1,2} ... F_{1,5,1} F_{1,5,2}
:
F_{N,1,1} F_{N,1,2} ... F_{N,5,1} F_{N,5,2}
P_{1,0} ... P_{1,10}
:
P_{N,0} ... P_{N,10}
Output
Print the maximum possible profit of Joisino's shop.
Examples
Input
1
1 1 0 1 0 0 0 1 0 1
3 4 5 6 7 8 9 -2 -3 4 -2
Output
8
Input
2
1 1 1 1 1 0 0 0 0 0
0 0 0 0 0 1 1 1 1 1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1
Output
-2
Input
3
1 1 1 1 1 1 0 0 1 1
0 1 0 1 1 1 1 0 1 0
1 0 1 1 0 1 0 1 0 1
-8 6 -2 -8 -8 4 8 7 -6 2 2
-9 2 0 1 7 -5 0 -2 -6 5 5
6 -6 7 -9 6 -5 8 0 -9 -7 -7
Output
23
Submitted Solution:
```
# coding: utf-8
# hello worldと表示する
#dpでできないかな？
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10**7)
from collections import Counter, deque
from collections import defaultdict
from itertools import combinations, permutations, accumulate, groupby, product
from bisect import bisect_left,bisect_right
from heapq import heapify, heappop, heappush
from math import floor, ceil,pi
from operator import itemgetter
def I(): return int(input())
def MI(): return map(int, input().split())
def LI(): return list(map(int, input().split()))
def LI2(): return [int(input()) for i in range(n)]
def MXI(): return [[LI()]for i in range(n)]
def SI(): return input().rstrip()
def printns(x): print('\n'.join(x))
def printni(x): print('\n'.join(list(map(str,x))))
inf = 10**17
mod = 10**9 + 7
n=I()
op=[LI() for i in range(n)]
pro=[LI() for i in range(n)]
vals=[]
for j in range(1,2**10-1):
ans=0
common=[0 for i in range(n)]
for k in range(n):
for i in range(10):
x=(j>>i) %2
common[k]+=x*op[k][i]
ans+=pro[k][common[k]]
vals.append(ans)
print(max(vals))
``` | instruction | 0 | 57,865 | 10 | 115,730 |
No | output | 1 | 57,865 | 10 | 115,731 |
Provide a correct Python 3 solution for this coding contest problem.
You want to go on a trip with a friend. However, friends who have a habit of spending money cannot easily save travel expenses. I don't know when my friends will go on a trip if they continue their current lives. So, if you want to travel early, you decide to create a program to help your friends save in a planned manner.
If you have a friend's pocket money of M yen and the money you spend in that month is N yen, you will save (M --N) yen in that month. Create a program that inputs the monthly income and expenditure information M and N and outputs the number of months it takes for the savings amount to reach the travel cost L. However, if your savings do not reach your travel expenses after 12 months, print NA.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
L
M1 N1
M2 N2
::
M12 N12
The first line gives the travel cost L (1 ≤ L ≤ 1000000, integer). The next 12 lines are given the balance information for the i month, Mi, Ni (0 ≤ Mi, Ni ≤ 100000, Ni ≤ Mi, integer).
The number of datasets does not exceed 1000.
Output
For each input dataset, print the number of months it takes for your savings to reach your travel costs on a single line.
Example
Input
10000
5000 3150
5000 5000
0 0
5000 1050
5000 3980
5000 210
5000 5000
5000 5000
0 0
5000 2100
5000 2100
5000 2100
29170
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 70831
0
Output
6
NA | instruction | 0 | 57,915 | 10 | 115,830 |
"Correct Solution:
```
while True:
l = int(input())
if l == 0:
break
acc = 0
ans = 0
for i in range(1, 13):
m, n = map(int, input().split())
acc += (m - n)
if not ans and acc >= l:
ans = i
if not ans:
print("NA")
else:
print(ans)
``` | output | 1 | 57,915 | 10 | 115,831 |
Provide a correct Python 3 solution for this coding contest problem.
You want to go on a trip with a friend. However, friends who have a habit of spending money cannot easily save travel expenses. I don't know when my friends will go on a trip if they continue their current lives. So, if you want to travel early, you decide to create a program to help your friends save in a planned manner.
If you have a friend's pocket money of M yen and the money you spend in that month is N yen, you will save (M --N) yen in that month. Create a program that inputs the monthly income and expenditure information M and N and outputs the number of months it takes for the savings amount to reach the travel cost L. However, if your savings do not reach your travel expenses after 12 months, print NA.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
L
M1 N1
M2 N2
::
M12 N12
The first line gives the travel cost L (1 ≤ L ≤ 1000000, integer). The next 12 lines are given the balance information for the i month, Mi, Ni (0 ≤ Mi, Ni ≤ 100000, Ni ≤ Mi, integer).
The number of datasets does not exceed 1000.
Output
For each input dataset, print the number of months it takes for your savings to reach your travel costs on a single line.
Example
Input
10000
5000 3150
5000 5000
0 0
5000 1050
5000 3980
5000 210
5000 5000
5000 5000
0 0
5000 2100
5000 2100
5000 2100
29170
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 70831
0
Output
6
NA | instruction | 0 | 57,916 | 10 | 115,832 |
"Correct Solution:
```
while 1:
l = int(input())
if l == 0:
break
money = []
for _ in range(12):
m, n = map(int, input().split())
money.append(m-n)
total = 0
for i, p in enumerate(money):
total += p
if total >= l:
print(i+1)
break
else:
print("NA")
``` | output | 1 | 57,916 | 10 | 115,833 |
Provide a correct Python 3 solution for this coding contest problem.
You want to go on a trip with a friend. However, friends who have a habit of spending money cannot easily save travel expenses. I don't know when my friends will go on a trip if they continue their current lives. So, if you want to travel early, you decide to create a program to help your friends save in a planned manner.
If you have a friend's pocket money of M yen and the money you spend in that month is N yen, you will save (M --N) yen in that month. Create a program that inputs the monthly income and expenditure information M and N and outputs the number of months it takes for the savings amount to reach the travel cost L. However, if your savings do not reach your travel expenses after 12 months, print NA.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
L
M1 N1
M2 N2
::
M12 N12
The first line gives the travel cost L (1 ≤ L ≤ 1000000, integer). The next 12 lines are given the balance information for the i month, Mi, Ni (0 ≤ Mi, Ni ≤ 100000, Ni ≤ Mi, integer).
The number of datasets does not exceed 1000.
Output
For each input dataset, print the number of months it takes for your savings to reach your travel costs on a single line.
Example
Input
10000
5000 3150
5000 5000
0 0
5000 1050
5000 3980
5000 210
5000 5000
5000 5000
0 0
5000 2100
5000 2100
5000 2100
29170
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 70831
0
Output
6
NA | instruction | 0 | 57,917 | 10 | 115,834 |
"Correct Solution:
```
def jug(l,income):
assert len(income)==12, "income strange"
s = 0
for i in range(len(income)):
inc = income[i]
s += inc[0] - inc[1]
if s >= l:
return(str(i+1))
return("NA")
while True:
income=[]
l=int(input())
if l==0:
break
for i in range(12):
m,n = list(map(int,input().strip().split()))
income.append((m,n))
print(jug(l,income))
``` | output | 1 | 57,917 | 10 | 115,835 |
Provide a correct Python 3 solution for this coding contest problem.
You want to go on a trip with a friend. However, friends who have a habit of spending money cannot easily save travel expenses. I don't know when my friends will go on a trip if they continue their current lives. So, if you want to travel early, you decide to create a program to help your friends save in a planned manner.
If you have a friend's pocket money of M yen and the money you spend in that month is N yen, you will save (M --N) yen in that month. Create a program that inputs the monthly income and expenditure information M and N and outputs the number of months it takes for the savings amount to reach the travel cost L. However, if your savings do not reach your travel expenses after 12 months, print NA.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
L
M1 N1
M2 N2
::
M12 N12
The first line gives the travel cost L (1 ≤ L ≤ 1000000, integer). The next 12 lines are given the balance information for the i month, Mi, Ni (0 ≤ Mi, Ni ≤ 100000, Ni ≤ Mi, integer).
The number of datasets does not exceed 1000.
Output
For each input dataset, print the number of months it takes for your savings to reach your travel costs on a single line.
Example
Input
10000
5000 3150
5000 5000
0 0
5000 1050
5000 3980
5000 210
5000 5000
5000 5000
0 0
5000 2100
5000 2100
5000 2100
29170
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 70831
0
Output
6
NA | instruction | 0 | 57,918 | 10 | 115,836 |
"Correct Solution:
```
import sys
while True:
L = int(input())
if L == 0:
break
savings = 0
a = [map(int, sys.stdin.readline().split()) for _ in [0]*12]
for i, (M, N) in enumerate(a, start=1):
savings += M - N
if savings >= L:
print(i)
break
else:
print("NA")
``` | output | 1 | 57,918 | 10 | 115,837 |
Provide a correct Python 3 solution for this coding contest problem.
You want to go on a trip with a friend. However, friends who have a habit of spending money cannot easily save travel expenses. I don't know when my friends will go on a trip if they continue their current lives. So, if you want to travel early, you decide to create a program to help your friends save in a planned manner.
If you have a friend's pocket money of M yen and the money you spend in that month is N yen, you will save (M --N) yen in that month. Create a program that inputs the monthly income and expenditure information M and N and outputs the number of months it takes for the savings amount to reach the travel cost L. However, if your savings do not reach your travel expenses after 12 months, print NA.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
L
M1 N1
M2 N2
::
M12 N12
The first line gives the travel cost L (1 ≤ L ≤ 1000000, integer). The next 12 lines are given the balance information for the i month, Mi, Ni (0 ≤ Mi, Ni ≤ 100000, Ni ≤ Mi, integer).
The number of datasets does not exceed 1000.
Output
For each input dataset, print the number of months it takes for your savings to reach your travel costs on a single line.
Example
Input
10000
5000 3150
5000 5000
0 0
5000 1050
5000 3980
5000 210
5000 5000
5000 5000
0 0
5000 2100
5000 2100
5000 2100
29170
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 70831
0
Output
6
NA | instruction | 0 | 57,919 | 10 | 115,838 |
"Correct Solution:
```
while True:
L=int(input())
if L==0:
break
S=0
B=0
for i in range(1,13):
M,N=map(int,input().split())
A=M-N
S+=A
if S>=L and B==0:
B=i
if B>0:
print(B)
else:
print("NA")
``` | output | 1 | 57,919 | 10 | 115,839 |
Provide a correct Python 3 solution for this coding contest problem.
You want to go on a trip with a friend. However, friends who have a habit of spending money cannot easily save travel expenses. I don't know when my friends will go on a trip if they continue their current lives. So, if you want to travel early, you decide to create a program to help your friends save in a planned manner.
If you have a friend's pocket money of M yen and the money you spend in that month is N yen, you will save (M --N) yen in that month. Create a program that inputs the monthly income and expenditure information M and N and outputs the number of months it takes for the savings amount to reach the travel cost L. However, if your savings do not reach your travel expenses after 12 months, print NA.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
L
M1 N1
M2 N2
::
M12 N12
The first line gives the travel cost L (1 ≤ L ≤ 1000000, integer). The next 12 lines are given the balance information for the i month, Mi, Ni (0 ≤ Mi, Ni ≤ 100000, Ni ≤ Mi, integer).
The number of datasets does not exceed 1000.
Output
For each input dataset, print the number of months it takes for your savings to reach your travel costs on a single line.
Example
Input
10000
5000 3150
5000 5000
0 0
5000 1050
5000 3980
5000 210
5000 5000
5000 5000
0 0
5000 2100
5000 2100
5000 2100
29170
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 70831
0
Output
6
NA | instruction | 0 | 57,920 | 10 | 115,840 |
"Correct Solution:
```
# Aizu Problem 0206: The Next Trip
import sys, math, os
# read input:
PYDEV = os.environ.get('PYDEV')
if PYDEV=="True":
sys.stdin = open("sample-input.txt", "rt")
def next_trip(L, cashflow):
saved = 0
month = 0
for m, n in cashflow:
month += 1
saved += (m - n)
if saved >= L:
return month
return "NA"
while True:
L = int(input())
if L == 0:
break
cashflow = [[int(_) for _ in input().split()] for __ in range(12)]
print(next_trip(L, cashflow))
``` | output | 1 | 57,920 | 10 | 115,841 |
Provide a correct Python 3 solution for this coding contest problem.
You want to go on a trip with a friend. However, friends who have a habit of spending money cannot easily save travel expenses. I don't know when my friends will go on a trip if they continue their current lives. So, if you want to travel early, you decide to create a program to help your friends save in a planned manner.
If you have a friend's pocket money of M yen and the money you spend in that month is N yen, you will save (M --N) yen in that month. Create a program that inputs the monthly income and expenditure information M and N and outputs the number of months it takes for the savings amount to reach the travel cost L. However, if your savings do not reach your travel expenses after 12 months, print NA.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
L
M1 N1
M2 N2
::
M12 N12
The first line gives the travel cost L (1 ≤ L ≤ 1000000, integer). The next 12 lines are given the balance information for the i month, Mi, Ni (0 ≤ Mi, Ni ≤ 100000, Ni ≤ Mi, integer).
The number of datasets does not exceed 1000.
Output
For each input dataset, print the number of months it takes for your savings to reach your travel costs on a single line.
Example
Input
10000
5000 3150
5000 5000
0 0
5000 1050
5000 3980
5000 210
5000 5000
5000 5000
0 0
5000 2100
5000 2100
5000 2100
29170
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 70831
0
Output
6
NA | instruction | 0 | 57,921 | 10 | 115,842 |
"Correct Solution:
```
while True:
v = int(input())
if v == 0:
break
c = 0
f = 0
for i in range(1,13):
m, n = [int(x) for x in input().split()]
c += m - n
if f == 0 and c >= v:
f = i
if f > 0:
print(f)
else:
print("NA")
``` | output | 1 | 57,921 | 10 | 115,843 |
Provide a correct Python 3 solution for this coding contest problem.
You want to go on a trip with a friend. However, friends who have a habit of spending money cannot easily save travel expenses. I don't know when my friends will go on a trip if they continue their current lives. So, if you want to travel early, you decide to create a program to help your friends save in a planned manner.
If you have a friend's pocket money of M yen and the money you spend in that month is N yen, you will save (M --N) yen in that month. Create a program that inputs the monthly income and expenditure information M and N and outputs the number of months it takes for the savings amount to reach the travel cost L. However, if your savings do not reach your travel expenses after 12 months, print NA.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
L
M1 N1
M2 N2
::
M12 N12
The first line gives the travel cost L (1 ≤ L ≤ 1000000, integer). The next 12 lines are given the balance information for the i month, Mi, Ni (0 ≤ Mi, Ni ≤ 100000, Ni ≤ Mi, integer).
The number of datasets does not exceed 1000.
Output
For each input dataset, print the number of months it takes for your savings to reach your travel costs on a single line.
Example
Input
10000
5000 3150
5000 5000
0 0
5000 1050
5000 3980
5000 210
5000 5000
5000 5000
0 0
5000 2100
5000 2100
5000 2100
29170
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 70831
0
Output
6
NA | instruction | 0 | 57,922 | 10 | 115,844 |
"Correct Solution:
```
while True:
L= int(input())
if L== 0: break
mn= [list(map(int, input().split())) for _ in range(12)]
ans=c=0
b= True
for m, n in mn:
ans+= m-n
c+= 1
if ans>= L:
print(c)
b= False
break
if b: print("NA")
``` | output | 1 | 57,922 | 10 | 115,845 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You want to go on a trip with a friend. However, friends who have a habit of spending money cannot easily save travel expenses. I don't know when my friends will go on a trip if they continue their current lives. So, if you want to travel early, you decide to create a program to help your friends save in a planned manner.
If you have a friend's pocket money of M yen and the money you spend in that month is N yen, you will save (M --N) yen in that month. Create a program that inputs the monthly income and expenditure information M and N and outputs the number of months it takes for the savings amount to reach the travel cost L. However, if your savings do not reach your travel expenses after 12 months, print NA.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
L
M1 N1
M2 N2
::
M12 N12
The first line gives the travel cost L (1 ≤ L ≤ 1000000, integer). The next 12 lines are given the balance information for the i month, Mi, Ni (0 ≤ Mi, Ni ≤ 100000, Ni ≤ Mi, integer).
The number of datasets does not exceed 1000.
Output
For each input dataset, print the number of months it takes for your savings to reach your travel costs on a single line.
Example
Input
10000
5000 3150
5000 5000
0 0
5000 1050
5000 3980
5000 210
5000 5000
5000 5000
0 0
5000 2100
5000 2100
5000 2100
29170
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 70831
0
Output
6
NA
Submitted Solution:
```
while True:
L = int(input())
if L == 0:
break
flag = True
ans = -1
d = 0
for i in range(12):
M,N = [int(i) for i in input().split()]
d = d + M - N
if d >= L and flag:
ans = i+1
flag = False
if ans >= 0:
print(ans)
else:
print("NA")
``` | instruction | 0 | 57,923 | 10 | 115,846 |
Yes | output | 1 | 57,923 | 10 | 115,847 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You want to go on a trip with a friend. However, friends who have a habit of spending money cannot easily save travel expenses. I don't know when my friends will go on a trip if they continue their current lives. So, if you want to travel early, you decide to create a program to help your friends save in a planned manner.
If you have a friend's pocket money of M yen and the money you spend in that month is N yen, you will save (M --N) yen in that month. Create a program that inputs the monthly income and expenditure information M and N and outputs the number of months it takes for the savings amount to reach the travel cost L. However, if your savings do not reach your travel expenses after 12 months, print NA.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
L
M1 N1
M2 N2
::
M12 N12
The first line gives the travel cost L (1 ≤ L ≤ 1000000, integer). The next 12 lines are given the balance information for the i month, Mi, Ni (0 ≤ Mi, Ni ≤ 100000, Ni ≤ Mi, integer).
The number of datasets does not exceed 1000.
Output
For each input dataset, print the number of months it takes for your savings to reach your travel costs on a single line.
Example
Input
10000
5000 3150
5000 5000
0 0
5000 1050
5000 3980
5000 210
5000 5000
5000 5000
0 0
5000 2100
5000 2100
5000 2100
29170
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 70831
0
Output
6
NA
Submitted Solution:
```
while 1:
l = int(input())
if l == 0:
break;
s = 0
for i in range(1,13):
m, n = [int(_) for _ in input().split()]
l -= m - n
if s == 0 and l <= 0:
s = i
print(s) if s != 0 else print('NA')
``` | instruction | 0 | 57,924 | 10 | 115,848 |
Yes | output | 1 | 57,924 | 10 | 115,849 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You want to go on a trip with a friend. However, friends who have a habit of spending money cannot easily save travel expenses. I don't know when my friends will go on a trip if they continue their current lives. So, if you want to travel early, you decide to create a program to help your friends save in a planned manner.
If you have a friend's pocket money of M yen and the money you spend in that month is N yen, you will save (M --N) yen in that month. Create a program that inputs the monthly income and expenditure information M and N and outputs the number of months it takes for the savings amount to reach the travel cost L. However, if your savings do not reach your travel expenses after 12 months, print NA.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
L
M1 N1
M2 N2
::
M12 N12
The first line gives the travel cost L (1 ≤ L ≤ 1000000, integer). The next 12 lines are given the balance information for the i month, Mi, Ni (0 ≤ Mi, Ni ≤ 100000, Ni ≤ Mi, integer).
The number of datasets does not exceed 1000.
Output
For each input dataset, print the number of months it takes for your savings to reach your travel costs on a single line.
Example
Input
10000
5000 3150
5000 5000
0 0
5000 1050
5000 3980
5000 210
5000 5000
5000 5000
0 0
5000 2100
5000 2100
5000 2100
29170
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 70831
0
Output
6
NA
Submitted Solution:
```
while True:
L = int(input())
if not L: break
for i in range(1, 13):
m, n = map(int, input().split())
if L <= 0: continue
c = i
L -= m - n
if L > 0:
print('NA')
else:
print(c)
``` | instruction | 0 | 57,925 | 10 | 115,850 |
Yes | output | 1 | 57,925 | 10 | 115,851 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You want to go on a trip with a friend. However, friends who have a habit of spending money cannot easily save travel expenses. I don't know when my friends will go on a trip if they continue their current lives. So, if you want to travel early, you decide to create a program to help your friends save in a planned manner.
If you have a friend's pocket money of M yen and the money you spend in that month is N yen, you will save (M --N) yen in that month. Create a program that inputs the monthly income and expenditure information M and N and outputs the number of months it takes for the savings amount to reach the travel cost L. However, if your savings do not reach your travel expenses after 12 months, print NA.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
L
M1 N1
M2 N2
::
M12 N12
The first line gives the travel cost L (1 ≤ L ≤ 1000000, integer). The next 12 lines are given the balance information for the i month, Mi, Ni (0 ≤ Mi, Ni ≤ 100000, Ni ≤ Mi, integer).
The number of datasets does not exceed 1000.
Output
For each input dataset, print the number of months it takes for your savings to reach your travel costs on a single line.
Example
Input
10000
5000 3150
5000 5000
0 0
5000 1050
5000 3980
5000 210
5000 5000
5000 5000
0 0
5000 2100
5000 2100
5000 2100
29170
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 70831
0
Output
6
NA
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0206
"""
import sys
from sys import stdin
input = stdin.readline
def main(args):
while True:
L = int(input())
if L == 0:
break
ans = 'NA'
for i in range(1, 12+1):
if ans == 'NA':
M, N = map(int, input().split())
L -= (M - N)
if L <= 0:
ans = i
else:
_ = input()
print(ans)
if __name__ == '__main__':
main(sys.argv[1:])
``` | instruction | 0 | 57,926 | 10 | 115,852 |
Yes | output | 1 | 57,926 | 10 | 115,853 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You want to go on a trip with a friend. However, friends who have a habit of spending money cannot easily save travel expenses. I don't know when my friends will go on a trip if they continue their current lives. So, if you want to travel early, you decide to create a program to help your friends save in a planned manner.
If you have a friend's pocket money of M yen and the money you spend in that month is N yen, you will save (M --N) yen in that month. Create a program that inputs the monthly income and expenditure information M and N and outputs the number of months it takes for the savings amount to reach the travel cost L. However, if your savings do not reach your travel expenses after 12 months, print NA.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
L
M1 N1
M2 N2
::
M12 N12
The first line gives the travel cost L (1 ≤ L ≤ 1000000, integer). The next 12 lines are given the balance information for the i month, Mi, Ni (0 ≤ Mi, Ni ≤ 100000, Ni ≤ Mi, integer).
The number of datasets does not exceed 1000.
Output
For each input dataset, print the number of months it takes for your savings to reach your travel costs on a single line.
Example
Input
10000
5000 3150
5000 5000
0 0
5000 1050
5000 3980
5000 210
5000 5000
5000 5000
0 0
5000 2100
5000 2100
5000 2100
29170
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 70831
0
Output
6
NA
Submitted Solution:
```
def alogrithm():
while True:
budget = int(input())
if budget == 0:
break
total = 0
months = 0
for _ in range(12):
income, outcome = map(int, input().split())
total += income - outcome
months += 1 if income > outcome else 0
if total < budget:
print('NA')
else:
print(months)
def main():
alogrithm()
if __name__ == '__main__':
main()
``` | instruction | 0 | 57,927 | 10 | 115,854 |
No | output | 1 | 57,927 | 10 | 115,855 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You want to go on a trip with a friend. However, friends who have a habit of spending money cannot easily save travel expenses. I don't know when my friends will go on a trip if they continue their current lives. So, if you want to travel early, you decide to create a program to help your friends save in a planned manner.
If you have a friend's pocket money of M yen and the money you spend in that month is N yen, you will save (M --N) yen in that month. Create a program that inputs the monthly income and expenditure information M and N and outputs the number of months it takes for the savings amount to reach the travel cost L. However, if your savings do not reach your travel expenses after 12 months, print NA.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
L
M1 N1
M2 N2
::
M12 N12
The first line gives the travel cost L (1 ≤ L ≤ 1000000, integer). The next 12 lines are given the balance information for the i month, Mi, Ni (0 ≤ Mi, Ni ≤ 100000, Ni ≤ Mi, integer).
The number of datasets does not exceed 1000.
Output
For each input dataset, print the number of months it takes for your savings to reach your travel costs on a single line.
Example
Input
10000
5000 3150
5000 5000
0 0
5000 1050
5000 3980
5000 210
5000 5000
5000 5000
0 0
5000 2100
5000 2100
5000 2100
29170
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 70831
0
Output
6
NA
Submitted Solution:
```
while True:
L= int(input())
if L== 0: break
mn= [list(map(int, input().split())) for _ in range(12)]
ans=c=1
b= True
for m, n in mn:
ans+= m-n
if ans>= L:
print(c)
b= False
break
c+= 1
if b: print("NA")
``` | instruction | 0 | 57,928 | 10 | 115,856 |
No | output | 1 | 57,928 | 10 | 115,857 |
Provide a correct Python 3 solution for this coding contest problem.
KM country has N kinds of coins and each coin has its value a_i.
The king of the country, Kita_masa, thought that the current currency system is poor, and he decided to make it beautiful by changing the values of some (possibly no) coins.
A currency system is called beautiful if each coin has an integer value and the (i+1)-th smallest value is divisible by the i-th smallest value for all i (1 \leq i \leq N-1).
For example, the set {1, 5, 10, 50, 100, 500} is considered as a beautiful system, while the set {1, 5, 10, 25, 50, 100} is NOT, because 25 is not divisible by 10.
Since changing the currency system may confuse citizens, the king, Kita_masa, wants to minimize the maximum value of the confusion ratios. Here, the confusion ratio for the change in the i-th coin is defined as |a_i - b_i| / a_i, where a_i and b_i is the value of i-th coin before and after the structure changes, respectively.
Note that Kita_masa can change the value of each existing coin, but he cannot introduce new coins nor eliminate existing coins. After the modification, the values of two or more coins may coincide.
Input
Each dataset contains two lines. The first line contains a single integer, N, and the second line contains N integers, {a_i}.
You may assume the following constraints:
1 \leq N \leq 20
1 \leq a_1 \lt a_2 \lt... \lt a_N \lt 10^5
Output
Output one number that represents the minimum of the maximum value of the confusion ratios. The value may be printed with an arbitrary number of decimal digits, but may not contain an absolute error greater than or equal to 10^{-8}.
Examples
Input
3
6 11 12
Output
0.090909090909
Input
3
6 11 24
Output
0.090909090909
Input
3
6 11 30
Output
0.166666666667 | instruction | 0 | 57,951 | 10 | 115,902 |
"Correct Solution:
```
# coding:utf-8
import sys
input = sys.stdin.readline
INF = float('inf')
MOD = 10 ** 9 + 7
def inpl(): return list(map(int, input().split()))
def solve(N):
A = inpl()
dp = [INF] * (A[0] * 2)
for i in range(A[0]//2, A[0]*2):
dp[i] = abs(i - A[0]) / A[0] # A[0]の価格を変えたときのconfusion ratio
for i in range(N-1):
a1 = A[i + 1]
nn = a1 * 2
ndp = [INF] * nn # ndp[A[i+1]の変更後の価値] = confusion ratioの最小値
for j in range(1, len(dp)):
if dp[j] == INF:
continue
t = dp[j]
for k in range(j, nn, j):
u = abs(a1 - k) / a1
if u < t: # A[i]とA[i+1]でconfusion ratioの大きい方をndpと比較
u = t
if ndp[k] > u: # A[1]~A[i+1]まで帳尻を合わせたときのconfusion ratioの最小値を格納
ndp[k] = u
dp = ndp
return '{:0.12f}'.format(min(dp))
N = int(input())
if N != 0:
print(solve(N))
``` | output | 1 | 57,951 | 10 | 115,903 |
Provide a correct Python 3 solution for this coding contest problem.
KM country has N kinds of coins and each coin has its value a_i.
The king of the country, Kita_masa, thought that the current currency system is poor, and he decided to make it beautiful by changing the values of some (possibly no) coins.
A currency system is called beautiful if each coin has an integer value and the (i+1)-th smallest value is divisible by the i-th smallest value for all i (1 \leq i \leq N-1).
For example, the set {1, 5, 10, 50, 100, 500} is considered as a beautiful system, while the set {1, 5, 10, 25, 50, 100} is NOT, because 25 is not divisible by 10.
Since changing the currency system may confuse citizens, the king, Kita_masa, wants to minimize the maximum value of the confusion ratios. Here, the confusion ratio for the change in the i-th coin is defined as |a_i - b_i| / a_i, where a_i and b_i is the value of i-th coin before and after the structure changes, respectively.
Note that Kita_masa can change the value of each existing coin, but he cannot introduce new coins nor eliminate existing coins. After the modification, the values of two or more coins may coincide.
Input
Each dataset contains two lines. The first line contains a single integer, N, and the second line contains N integers, {a_i}.
You may assume the following constraints:
1 \leq N \leq 20
1 \leq a_1 \lt a_2 \lt... \lt a_N \lt 10^5
Output
Output one number that represents the minimum of the maximum value of the confusion ratios. The value may be printed with an arbitrary number of decimal digits, but may not contain an absolute error greater than or equal to 10^{-8}.
Examples
Input
3
6 11 12
Output
0.090909090909
Input
3
6 11 24
Output
0.090909090909
Input
3
6 11 30
Output
0.166666666667 | instruction | 0 | 57,952 | 10 | 115,904 |
"Correct Solution:
```
from heapq import heappush, heappop
def solve():
N = int(input())
*A, = map(int, input().split())
M = 2*10**5
INF = 10**18
dist = [[INF]*(M+1) for i in range(N+1)]
que = [(0, 0, 1)]
while que:
cost, i, p = heappop(que)
if i == N:
break
if dist[i][p] + 1e-10 < cost:
continue
a = A[i]
for k in range(p, M+1, p):
d = max(cost, abs(k - a)/a)
if d < dist[i+1][k]:
dist[i+1][k] = d
heappush(que, (d, i+1, k))
return min(dist[N])
print("%.16f" % solve())
``` | output | 1 | 57,952 | 10 | 115,905 |
Provide a correct Python 3 solution for this coding contest problem.
KM country has N kinds of coins and each coin has its value a_i.
The king of the country, Kita_masa, thought that the current currency system is poor, and he decided to make it beautiful by changing the values of some (possibly no) coins.
A currency system is called beautiful if each coin has an integer value and the (i+1)-th smallest value is divisible by the i-th smallest value for all i (1 \leq i \leq N-1).
For example, the set {1, 5, 10, 50, 100, 500} is considered as a beautiful system, while the set {1, 5, 10, 25, 50, 100} is NOT, because 25 is not divisible by 10.
Since changing the currency system may confuse citizens, the king, Kita_masa, wants to minimize the maximum value of the confusion ratios. Here, the confusion ratio for the change in the i-th coin is defined as |a_i - b_i| / a_i, where a_i and b_i is the value of i-th coin before and after the structure changes, respectively.
Note that Kita_masa can change the value of each existing coin, but he cannot introduce new coins nor eliminate existing coins. After the modification, the values of two or more coins may coincide.
Input
Each dataset contains two lines. The first line contains a single integer, N, and the second line contains N integers, {a_i}.
You may assume the following constraints:
1 \leq N \leq 20
1 \leq a_1 \lt a_2 \lt... \lt a_N \lt 10^5
Output
Output one number that represents the minimum of the maximum value of the confusion ratios. The value may be printed with an arbitrary number of decimal digits, but may not contain an absolute error greater than or equal to 10^{-8}.
Examples
Input
3
6 11 12
Output
0.090909090909
Input
3
6 11 24
Output
0.090909090909
Input
3
6 11 30
Output
0.166666666667 | instruction | 0 | 57,953 | 10 | 115,906 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**13
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
def f(n):
a = LI()
dp = [inf] * (a[0]*2)
for i in range(a[0]//2,a[0]*2):
dp[i] = abs(i-a[0]) / a[0]
for i in range(n-1):
a1 = a[i+1]
nn = a1 * 2
ndp = [inf] * nn
for j in range(1, len(dp)):
if dp[j] == inf:
continue
t = dp[j]
for k in range(j,nn,j):
u = abs(a1-k) / a1
if u < t:
u = t
if ndp[k] > u:
ndp[k] = u
dp = ndp
return '{:0.9f}'.format(min(dp))
while 1:
n = I()
if n == 0:
break
rr.append(f(n))
# print('rr', rr[-1])
break
return '\n'.join(map(str,rr))
print(main())
``` | output | 1 | 57,953 | 10 | 115,907 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya wants to buy himself a nice new car. Unfortunately, he lacks some money. Currently he has exactly 0 burles.
However, the local bank has n credit offers. Each offer can be described with three numbers a_i, b_i and k_i. Offers are numbered from 1 to n. If Vasya takes the i-th offer, then the bank gives him a_i burles at the beginning of the month and then Vasya pays bank b_i burles at the end of each month for the next k_i months (including the month he activated the offer). Vasya can take the offers any order he wants.
Each month Vasya can take no more than one credit offer. Also each credit offer can not be used more than once. Several credits can be active at the same time. It implies that Vasya pays bank the sum of b_i over all the i of active credits at the end of each month.
Vasya wants to buy a car in the middle of some month. He just takes all the money he currently has and buys the car of that exact price.
Vasya don't really care what he'll have to pay the bank back after he buys a car. He just goes out of the country on his car so that the bank can't find him anymore.
What is the maximum price that car can have?
Input
The first line contains one integer n (1 ≤ n ≤ 500) — the number of credit offers.
Each of the next n lines contains three integers a_i, b_i and k_i (1 ≤ a_i, b_i, k_i ≤ 10^9).
Output
Print one integer — the maximum price of the car.
Examples
Input
4
10 9 2
20 33 1
30 115 1
5 3 2
Output
32
Input
3
40 1 2
1000 1100 5
300 2 1
Output
1337
Note
In the first example, the following sequence of offers taken is optimal: 4 → 3.
The amount of burles Vasya has changes the following way: 5 → 32 → -86 → .... He takes the money he has in the middle of the second month (32 burles) and buys the car.
The negative amount of money means that Vasya has to pay the bank that amount of burles.
In the second example, the following sequence of offers taken is optimal: 3 → 1 → 2.
The amount of burles Vasya has changes the following way: 0 → 300 → 338 → 1337 → 236 → -866 → .... | instruction | 0 | 58,866 | 10 | 117,732 |
Tags: dp, flows, graph matchings, graphs, sortings
Correct Solution:
```
n = int(input())
a = [tuple(map(int, input().split())) for i in range(n)]
a = [(y, x, k) for x, y, k in a]
a.sort(reverse=True)
dp = [[-1] * (n + 1) for i in range(n)]
def f(i, j):
if i < 0 or j < -1: return 0
if dp[i][j] == -1:
y, x, k = a[i]
dp[i][j] = f(i - 1, j) + max(0, x - k * y)
if 0 <= j < k: dp[i][j] = max(dp[i][j], x - j * y + f(i - 1, j - 1))
return dp[i][j]
print(max(f(n - 1, j) for j in range(-1, n)))
``` | output | 1 | 58,866 | 10 | 117,733 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya wants to buy himself a nice new car. Unfortunately, he lacks some money. Currently he has exactly 0 burles.
However, the local bank has n credit offers. Each offer can be described with three numbers a_i, b_i and k_i. Offers are numbered from 1 to n. If Vasya takes the i-th offer, then the bank gives him a_i burles at the beginning of the month and then Vasya pays bank b_i burles at the end of each month for the next k_i months (including the month he activated the offer). Vasya can take the offers any order he wants.
Each month Vasya can take no more than one credit offer. Also each credit offer can not be used more than once. Several credits can be active at the same time. It implies that Vasya pays bank the sum of b_i over all the i of active credits at the end of each month.
Vasya wants to buy a car in the middle of some month. He just takes all the money he currently has and buys the car of that exact price.
Vasya don't really care what he'll have to pay the bank back after he buys a car. He just goes out of the country on his car so that the bank can't find him anymore.
What is the maximum price that car can have?
Input
The first line contains one integer n (1 ≤ n ≤ 500) — the number of credit offers.
Each of the next n lines contains three integers a_i, b_i and k_i (1 ≤ a_i, b_i, k_i ≤ 10^9).
Output
Print one integer — the maximum price of the car.
Examples
Input
4
10 9 2
20 33 1
30 115 1
5 3 2
Output
32
Input
3
40 1 2
1000 1100 5
300 2 1
Output
1337
Note
In the first example, the following sequence of offers taken is optimal: 4 → 3.
The amount of burles Vasya has changes the following way: 5 → 32 → -86 → .... He takes the money he has in the middle of the second month (32 burles) and buys the car.
The negative amount of money means that Vasya has to pay the bank that amount of burles.
In the second example, the following sequence of offers taken is optimal: 3 → 1 → 2.
The amount of burles Vasya has changes the following way: 0 → 300 → 338 → 1337 → 236 → -866 → .... | instruction | 0 | 58,867 | 10 | 117,734 |
Tags: dp, flows, graph matchings, graphs, sortings
Correct Solution:
```
n = int(input())
if n == 0:
print('0')
else:
data = []
ans = 0
for i in range(n):
line = list(map(lambda entry: int(entry), input().split()))
x = line[0]-line[1]*line[2]
if x > 0:
ans += x
line[0] -= x
data.append((line[0], line[1]))
data.sort(key = lambda entry: -entry[1])
dp = [[0 for i in range(n)]]
for j in range(n):
x = data[j][0]
nextline = [max(x, dp[-1][0])]
for i in range(1, n):
x -= data[j][1]
nextline.append(max(x + dp[-1][i - 1], nextline[-1], dp[-1][i]))
dp.append(nextline)
ans += dp[n][n-1]
print(ans)
``` | output | 1 | 58,867 | 10 | 117,735 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya wants to buy himself a nice new car. Unfortunately, he lacks some money. Currently he has exactly 0 burles.
However, the local bank has n credit offers. Each offer can be described with three numbers a_i, b_i and k_i. Offers are numbered from 1 to n. If Vasya takes the i-th offer, then the bank gives him a_i burles at the beginning of the month and then Vasya pays bank b_i burles at the end of each month for the next k_i months (including the month he activated the offer). Vasya can take the offers any order he wants.
Each month Vasya can take no more than one credit offer. Also each credit offer can not be used more than once. Several credits can be active at the same time. It implies that Vasya pays bank the sum of b_i over all the i of active credits at the end of each month.
Vasya wants to buy a car in the middle of some month. He just takes all the money he currently has and buys the car of that exact price.
Vasya don't really care what he'll have to pay the bank back after he buys a car. He just goes out of the country on his car so that the bank can't find him anymore.
What is the maximum price that car can have?
Input
The first line contains one integer n (1 ≤ n ≤ 500) — the number of credit offers.
Each of the next n lines contains three integers a_i, b_i and k_i (1 ≤ a_i, b_i, k_i ≤ 10^9).
Output
Print one integer — the maximum price of the car.
Examples
Input
4
10 9 2
20 33 1
30 115 1
5 3 2
Output
32
Input
3
40 1 2
1000 1100 5
300 2 1
Output
1337
Note
In the first example, the following sequence of offers taken is optimal: 4 → 3.
The amount of burles Vasya has changes the following way: 5 → 32 → -86 → .... He takes the money he has in the middle of the second month (32 burles) and buys the car.
The negative amount of money means that Vasya has to pay the bank that amount of burles.
In the second example, the following sequence of offers taken is optimal: 3 → 1 → 2.
The amount of burles Vasya has changes the following way: 0 → 300 → 338 → 1337 → 236 → -866 → .... | instruction | 0 | 58,868 | 10 | 117,736 |
Tags: dp, flows, graph matchings, graphs, sortings
Correct Solution:
```
from sys import stdin
n = int(input())
t = stdin.read().splitlines()
a = [list(map(int, i.split())) for i in t]
a.sort(reverse=True, key=lambda k: (k[1], k[0], k[2]))
# print(a)
dp = [[0] * (n + 1) for _ in range(n + 1)]
for i in range(n):
for l in range(n + 1):
dp[i + 1][l] = max(dp[i + 1][l], dp[i][l],
dp[i][l] + a[i][0] - a[i][1] * a[i][2])
if l < n:
dp[i + 1][l + 1] = max(dp[i + 1][l + 1],
dp[i][l] + a[i][0] - a[i][1] * l)
# print(dp)
print(max(dp[n]))
``` | output | 1 | 58,868 | 10 | 117,737 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya wants to buy himself a nice new car. Unfortunately, he lacks some money. Currently he has exactly 0 burles.
However, the local bank has n credit offers. Each offer can be described with three numbers a_i, b_i and k_i. Offers are numbered from 1 to n. If Vasya takes the i-th offer, then the bank gives him a_i burles at the beginning of the month and then Vasya pays bank b_i burles at the end of each month for the next k_i months (including the month he activated the offer). Vasya can take the offers any order he wants.
Each month Vasya can take no more than one credit offer. Also each credit offer can not be used more than once. Several credits can be active at the same time. It implies that Vasya pays bank the sum of b_i over all the i of active credits at the end of each month.
Vasya wants to buy a car in the middle of some month. He just takes all the money he currently has and buys the car of that exact price.
Vasya don't really care what he'll have to pay the bank back after he buys a car. He just goes out of the country on his car so that the bank can't find him anymore.
What is the maximum price that car can have?
Input
The first line contains one integer n (1 ≤ n ≤ 500) — the number of credit offers.
Each of the next n lines contains three integers a_i, b_i and k_i (1 ≤ a_i, b_i, k_i ≤ 10^9).
Output
Print one integer — the maximum price of the car.
Examples
Input
4
10 9 2
20 33 1
30 115 1
5 3 2
Output
32
Input
3
40 1 2
1000 1100 5
300 2 1
Output
1337
Note
In the first example, the following sequence of offers taken is optimal: 4 → 3.
The amount of burles Vasya has changes the following way: 5 → 32 → -86 → .... He takes the money he has in the middle of the second month (32 burles) and buys the car.
The negative amount of money means that Vasya has to pay the bank that amount of burles.
In the second example, the following sequence of offers taken is optimal: 3 → 1 → 2.
The amount of burles Vasya has changes the following way: 0 → 300 → 338 → 1337 → 236 → -866 → ....
Submitted Solution:
```
def maxindex(array, indices):
for j in range(len(array)):
if indices[j]==1: array[j]=-1;
return array.index(max(array));
n=int(input());
plus = list();
minus = list();
months = list();
results = list();
taken = list();
answer = 0;
for j in range(n):
x=input().split();
plus.append(int(x[0]));
minus.append(int(x[1]));
months.append(int(x[2]));
taken.append(0);
if j==0: results.append([int(x[0])])
else: results[0].append(int(x[0]))
for k in range(1,n):
results.append([plus[0]-minus[0]*min(months[0], k)])
for j in range(1,n):
results[k].append(plus[j]-minus[j]*min(months[j], k));
sum=0;
for j in range(n):
i = maxindex(results[n-j-1], taken);
if results[n-1-j][i]>=0:
answer = answer + results[n-j-1][i]; taken[i]=1;sum = sum+minus[n-j-1];
else: answer = answer+sum;
print(answer)
``` | instruction | 0 | 58,869 | 10 | 117,738 |
No | output | 1 | 58,869 | 10 | 117,739 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya wants to buy himself a nice new car. Unfortunately, he lacks some money. Currently he has exactly 0 burles.
However, the local bank has n credit offers. Each offer can be described with three numbers a_i, b_i and k_i. Offers are numbered from 1 to n. If Vasya takes the i-th offer, then the bank gives him a_i burles at the beginning of the month and then Vasya pays bank b_i burles at the end of each month for the next k_i months (including the month he activated the offer). Vasya can take the offers any order he wants.
Each month Vasya can take no more than one credit offer. Also each credit offer can not be used more than once. Several credits can be active at the same time. It implies that Vasya pays bank the sum of b_i over all the i of active credits at the end of each month.
Vasya wants to buy a car in the middle of some month. He just takes all the money he currently has and buys the car of that exact price.
Vasya don't really care what he'll have to pay the bank back after he buys a car. He just goes out of the country on his car so that the bank can't find him anymore.
What is the maximum price that car can have?
Input
The first line contains one integer n (1 ≤ n ≤ 500) — the number of credit offers.
Each of the next n lines contains three integers a_i, b_i and k_i (1 ≤ a_i, b_i, k_i ≤ 10^9).
Output
Print one integer — the maximum price of the car.
Examples
Input
4
10 9 2
20 33 1
30 115 1
5 3 2
Output
32
Input
3
40 1 2
1000 1100 5
300 2 1
Output
1337
Note
In the first example, the following sequence of offers taken is optimal: 4 → 3.
The amount of burles Vasya has changes the following way: 5 → 32 → -86 → .... He takes the money he has in the middle of the second month (32 burles) and buys the car.
The negative amount of money means that Vasya has to pay the bank that amount of burles.
In the second example, the following sequence of offers taken is optimal: 3 → 1 → 2.
The amount of burles Vasya has changes the following way: 0 → 300 → 338 → 1337 → 236 → -866 → ....
Submitted Solution:
```
n = int(input())
a = [tuple(map(int, input().split())) for i in range(n)]
a = [(y, x, k) for x, y, k in a]
a.sort(reverse=True)
dp = [[-1] * (n + 1) for i in range(n + 1)]
def f(i, j):
if i < 0 or j < 0: return 0
if dp[i][j] == -1:
y, x, k = a[i]
dp[i][j] = f(i - 1, j) + max(0, x - k * y)
if j < k: dp[i][j] = max(dp[i][j], x - y * j + f(i - 1, j - 1))
return dp[i][j]
print(max(f(n - 1, j) for j in range(n)))
``` | instruction | 0 | 58,870 | 10 | 117,740 |
No | output | 1 | 58,870 | 10 | 117,741 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya wants to buy himself a nice new car. Unfortunately, he lacks some money. Currently he has exactly 0 burles.
However, the local bank has n credit offers. Each offer can be described with three numbers a_i, b_i and k_i. Offers are numbered from 1 to n. If Vasya takes the i-th offer, then the bank gives him a_i burles at the beginning of the month and then Vasya pays bank b_i burles at the end of each month for the next k_i months (including the month he activated the offer). Vasya can take the offers any order he wants.
Each month Vasya can take no more than one credit offer. Also each credit offer can not be used more than once. Several credits can be active at the same time. It implies that Vasya pays bank the sum of b_i over all the i of active credits at the end of each month.
Vasya wants to buy a car in the middle of some month. He just takes all the money he currently has and buys the car of that exact price.
Vasya don't really care what he'll have to pay the bank back after he buys a car. He just goes out of the country on his car so that the bank can't find him anymore.
What is the maximum price that car can have?
Input
The first line contains one integer n (1 ≤ n ≤ 500) — the number of credit offers.
Each of the next n lines contains three integers a_i, b_i and k_i (1 ≤ a_i, b_i, k_i ≤ 10^9).
Output
Print one integer — the maximum price of the car.
Examples
Input
4
10 9 2
20 33 1
30 115 1
5 3 2
Output
32
Input
3
40 1 2
1000 1100 5
300 2 1
Output
1337
Note
In the first example, the following sequence of offers taken is optimal: 4 → 3.
The amount of burles Vasya has changes the following way: 5 → 32 → -86 → .... He takes the money he has in the middle of the second month (32 burles) and buys the car.
The negative amount of money means that Vasya has to pay the bank that amount of burles.
In the second example, the following sequence of offers taken is optimal: 3 → 1 → 2.
The amount of burles Vasya has changes the following way: 0 → 300 → 338 → 1337 → 236 → -866 → ....
Submitted Solution:
```
from sys import stdin
n = int(input())
t = stdin.read().splitlines()
a = [list(map(int, i.split())) for i in t]
a.sort(reverse=True, key=lambda k: (k[1], k[0], k[2]))
# print(a)
dp = [[0] * (n + 1) for _ in range(n + 1)]
for i in range(n):
for l in range(n + 1):
dp[i + 1][l] = max(dp[i + 1][l], dp[i][l],
dp[i][l] + a[i][0] - a[i][1] * a[i][2])
if l < n:
dp[i + 1][l + 1] = max(dp[i + 1][l + 1],
dp[i][l] + a[i][0] - a[i][1] * l)
print(dp)
print(max(dp[n]))
``` | instruction | 0 | 58,871 | 10 | 117,742 |
No | output | 1 | 58,871 | 10 | 117,743 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In a medieval kingdom, the economic crisis is raging. Milk drops fall, Economic indicators are deteriorating every day, money from the treasury disappear. To remedy the situation, King Charles Sunnyface decided make his n sons-princes marry the brides with as big dowry as possible.
In search of candidates, the king asked neighboring kingdoms, and after a while several delegations arrived with m unmarried princesses. Receiving guests, Karl learned that the dowry of the i th princess is wi of golden coins.
Although the action takes place in the Middle Ages, progressive ideas are widespread in society, according to which no one can force a princess to marry a prince whom she does not like. Therefore, each princess has an opportunity to choose two princes, for each of which she is ready to become a wife. The princes were less fortunate, they will obey the will of their father in the matter of choosing a bride.
Knowing the value of the dowry and the preferences of each princess, Charles wants to play weddings in such a way that the total dowry of the brides of all his sons would be as great as possible. At the same time to marry all the princes or princesses is not necessary. Each prince can marry no more than one princess, and vice versa, each princess can marry no more than one prince.
Help the king to organize the marriage of his sons in the most profitable way for the treasury.
Input
The first line contains two integers n, m (2 ≤ n ≤ 200 000, 1 ≤ m ≤ 200 000) — number of princes and princesses respectively.
Each of following m lines contains three integers ai, bi, wi (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ wi ≤ 10 000) — number of princes, which i-th princess is ready to marry and the value of her dowry.
Output
Print the only integer — the maximum number of gold coins that a king can get by playing the right weddings.
Examples
Input
2 3
1 2 5
1 2 1
2 1 10
Output
15
Input
3 2
1 2 10
3 2 20
Output
30
Submitted Solution:
```
n, m = map(int, input().split())
weights = []
princes = [[] for _ in range(n)]
for i in range(m):
a, b, w = map(int, input().split())
princes[a - 1].append(i)
princes[b - 1].append(i)
weights.append(w)
princes.sort(key=lambda x: len(x))
total = 0
for nums in princes:
if nums:
num = max(nums, key=lambda x: weights[x])
total += weights[num]
weights[num] = 0
print(total)
``` | instruction | 0 | 59,389 | 10 | 118,778 |
No | output | 1 | 59,389 | 10 | 118,779 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In a medieval kingdom, the economic crisis is raging. Milk drops fall, Economic indicators are deteriorating every day, money from the treasury disappear. To remedy the situation, King Charles Sunnyface decided make his n sons-princes marry the brides with as big dowry as possible.
In search of candidates, the king asked neighboring kingdoms, and after a while several delegations arrived with m unmarried princesses. Receiving guests, Karl learned that the dowry of the i th princess is wi of golden coins.
Although the action takes place in the Middle Ages, progressive ideas are widespread in society, according to which no one can force a princess to marry a prince whom she does not like. Therefore, each princess has an opportunity to choose two princes, for each of which she is ready to become a wife. The princes were less fortunate, they will obey the will of their father in the matter of choosing a bride.
Knowing the value of the dowry and the preferences of each princess, Charles wants to play weddings in such a way that the total dowry of the brides of all his sons would be as great as possible. At the same time to marry all the princes or princesses is not necessary. Each prince can marry no more than one princess, and vice versa, each princess can marry no more than one prince.
Help the king to organize the marriage of his sons in the most profitable way for the treasury.
Input
The first line contains two integers n, m (2 ≤ n ≤ 200 000, 1 ≤ m ≤ 200 000) — number of princes and princesses respectively.
Each of following m lines contains three integers ai, bi, wi (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ wi ≤ 10 000) — number of princes, which i-th princess is ready to marry and the value of her dowry.
Output
Print the only integer — the maximum number of gold coins that a king can get by playing the right weddings.
Examples
Input
2 3
1 2 5
1 2 1
2 1 10
Output
15
Input
3 2
1 2 10
3 2 20
Output
30
Submitted Solution:
```
prince, princess = map(int, input().split())
prince_data = {}
for i in range(1, prince+1):
prince_data[str(i)] = 0
for _ in range(princess):
prince_one, prince_two, money = map(int, input().split())
if prince_data[str(prince_one)] < money:
if prince_data[str(prince_one)] > prince_data[str(prince_two)]:
prince_data[str(prince_two)] = money
else:
prince_data[str(prince_one)] = money
elif prince_data[str(prince_two)] < money:
prince_data[str(prince_two)] = money
total_money = 0
for key, value in prince_data.items():
total_money += value
print(total_money)
``` | instruction | 0 | 59,390 | 10 | 118,780 |
No | output | 1 | 59,390 | 10 | 118,781 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In a medieval kingdom, the economic crisis is raging. Milk drops fall, Economic indicators are deteriorating every day, money from the treasury disappear. To remedy the situation, King Charles Sunnyface decided make his n sons-princes marry the brides with as big dowry as possible.
In search of candidates, the king asked neighboring kingdoms, and after a while several delegations arrived with m unmarried princesses. Receiving guests, Karl learned that the dowry of the i th princess is wi of golden coins.
Although the action takes place in the Middle Ages, progressive ideas are widespread in society, according to which no one can force a princess to marry a prince whom she does not like. Therefore, each princess has an opportunity to choose two princes, for each of which she is ready to become a wife. The princes were less fortunate, they will obey the will of their father in the matter of choosing a bride.
Knowing the value of the dowry and the preferences of each princess, Charles wants to play weddings in such a way that the total dowry of the brides of all his sons would be as great as possible. At the same time to marry all the princes or princesses is not necessary. Each prince can marry no more than one princess, and vice versa, each princess can marry no more than one prince.
Help the king to organize the marriage of his sons in the most profitable way for the treasury.
Input
The first line contains two integers n, m (2 ≤ n ≤ 200 000, 1 ≤ m ≤ 200 000) — number of princes and princesses respectively.
Each of following m lines contains three integers ai, bi, wi (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ wi ≤ 10 000) — number of princes, which i-th princess is ready to marry and the value of her dowry.
Output
Print the only integer — the maximum number of gold coins that a king can get by playing the right weddings.
Examples
Input
2 3
1 2 5
1 2 1
2 1 10
Output
15
Input
3 2
1 2 10
3 2 20
Output
30
Submitted Solution:
```
prince, princess = map(int, input().split())
prince_data = {}
for i in range(1, prince+1):
prince_data[str(i)] = 0
for _ in range(princess):
prince_one, prince_two, money = map(int, input().split())
if prince_data[str(prince_one)] < money:
prince_data[str(prince_one)] = money
continue
elif prince_data[str(prince_two)] < money:
prince_data[str(prince_two)] = money
total_money = 0
for key, value in prince_data.items():
total_money += value
print(total_money)
``` | instruction | 0 | 59,391 | 10 | 118,782 |
No | output | 1 | 59,391 | 10 | 118,783 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In a medieval kingdom, the economic crisis is raging. Milk drops fall, Economic indicators are deteriorating every day, money from the treasury disappear. To remedy the situation, King Charles Sunnyface decided make his n sons-princes marry the brides with as big dowry as possible.
In search of candidates, the king asked neighboring kingdoms, and after a while several delegations arrived with m unmarried princesses. Receiving guests, Karl learned that the dowry of the i th princess is wi of golden coins.
Although the action takes place in the Middle Ages, progressive ideas are widespread in society, according to which no one can force a princess to marry a prince whom she does not like. Therefore, each princess has an opportunity to choose two princes, for each of which she is ready to become a wife. The princes were less fortunate, they will obey the will of their father in the matter of choosing a bride.
Knowing the value of the dowry and the preferences of each princess, Charles wants to play weddings in such a way that the total dowry of the brides of all his sons would be as great as possible. At the same time to marry all the princes or princesses is not necessary. Each prince can marry no more than one princess, and vice versa, each princess can marry no more than one prince.
Help the king to organize the marriage of his sons in the most profitable way for the treasury.
Input
The first line contains two integers n, m (2 ≤ n ≤ 200 000, 1 ≤ m ≤ 200 000) — number of princes and princesses respectively.
Each of following m lines contains three integers ai, bi, wi (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ wi ≤ 10 000) — number of princes, which i-th princess is ready to marry and the value of her dowry.
Output
Print the only integer — the maximum number of gold coins that a king can get by playing the right weddings.
Examples
Input
2 3
1 2 5
1 2 1
2 1 10
Output
15
Input
3 2
1 2 10
3 2 20
Output
30
Submitted Solution:
```
n, m = map(int, input().split())
weights = []
princes = [[] for _ in range(n)]
for i in range(m):
a, b, w = map(int, input().split())
princes[a - 1].append(i)
princes[b - 1].append(i)
weights.append(w)
princes.sort(key=lambda x: len(x), reverse=True)
total = 0
for nums in princes:
if nums:
num = max(nums, key=lambda x: weights[x])
total += weights[num]
weights[num] = 0
print(total)
``` | instruction | 0 | 59,392 | 10 | 118,784 |
No | output | 1 | 59,392 | 10 | 118,785 |
Provide a correct Python 3 solution for this coding contest problem.
Problem statement
JOI decided to start a new social game from tomorrow.
In this social game, you can log in up to once a day, and you will get A coins each time you log in.
Also, if you log in for 7 consecutive days from Monday to Sunday, you will get an additional B coins each time.
No other coins will be given.
Tomorrow is Monday. Find the minimum number of times JOI must log in to get at least C coins.
Constraint
* 1 ≤ A ≤ 1000
* 0 ≤ B ≤ 1000
* 1 ≤ C ≤ 1000000 (= 10 ^ 6)
Input Output
input
Input is given from standard input in the following format.
A B C
output
Output the minimum number of times JOI must log in to get at least C coins.
<!-
Subtask
1. (40 points) B = 0
2. (60 points) There are no additional restrictions.
->
Input / output example
Input example 1
3 0 10
Output example 1
Four
* I want to get 3 coins per login and collect 10 coins.
* JOI can get 12 coins by logging in for 4 consecutive days from Monday.
* Since you cannot get more than 10 coins by logging in 3 times or less, the minimum number of times JOI must log in is 4. Therefore, 4 is output.
Input example 2
1 2 10
Output example 2
8
* You can get 1 coin for each login. Apart from that, you can get 2 coins by logging in for a week in a row. I want to collect 10 coins.
* If you log in consecutively from Monday to Sunday, you will get 2 coins in addition to 7 daily coins, so you will get a total of 9 coins. Therefore, if you log in one more time, you will get 10 coins.
* Since you cannot get more than 10 coins by logging in 7 times or less, the minimum number of times JOI must log in is 8. Therefore, 8 is output.
Creative Commons License
Information Olympics Japan Committee work "18th Japan Information Olympics JOI 2018/2019 Qualifying Competition Tasks"
Example
Input
3 0 10
Output
4 | instruction | 0 | 59,611 | 10 | 119,222 |
"Correct Solution:
```
a,b,c=(int(x) for x in input().split())
week=a*7+b
weeks=c//week
c=c-weeks*week
if c>a*7:
days=7
else:
days=-(-c // a)
total=days+weeks*7
print(total)
``` | output | 1 | 59,611 | 10 | 119,223 |
Provide a correct Python 3 solution for this coding contest problem.
Problem statement
JOI decided to start a new social game from tomorrow.
In this social game, you can log in up to once a day, and you will get A coins each time you log in.
Also, if you log in for 7 consecutive days from Monday to Sunday, you will get an additional B coins each time.
No other coins will be given.
Tomorrow is Monday. Find the minimum number of times JOI must log in to get at least C coins.
Constraint
* 1 ≤ A ≤ 1000
* 0 ≤ B ≤ 1000
* 1 ≤ C ≤ 1000000 (= 10 ^ 6)
Input Output
input
Input is given from standard input in the following format.
A B C
output
Output the minimum number of times JOI must log in to get at least C coins.
<!-
Subtask
1. (40 points) B = 0
2. (60 points) There are no additional restrictions.
->
Input / output example
Input example 1
3 0 10
Output example 1
Four
* I want to get 3 coins per login and collect 10 coins.
* JOI can get 12 coins by logging in for 4 consecutive days from Monday.
* Since you cannot get more than 10 coins by logging in 3 times or less, the minimum number of times JOI must log in is 4. Therefore, 4 is output.
Input example 2
1 2 10
Output example 2
8
* You can get 1 coin for each login. Apart from that, you can get 2 coins by logging in for a week in a row. I want to collect 10 coins.
* If you log in consecutively from Monday to Sunday, you will get 2 coins in addition to 7 daily coins, so you will get a total of 9 coins. Therefore, if you log in one more time, you will get 10 coins.
* Since you cannot get more than 10 coins by logging in 7 times or less, the minimum number of times JOI must log in is 8. Therefore, 8 is output.
Creative Commons License
Information Olympics Japan Committee work "18th Japan Information Olympics JOI 2018/2019 Qualifying Competition Tasks"
Example
Input
3 0 10
Output
4 | instruction | 0 | 59,612 | 10 | 119,224 |
"Correct Solution:
```
A,B,C=map(int,input().split())
x=0
for i in range(1,1+C):
x+=A
if i%7==0:
x+=B
if x>=C:
break
print(i)
``` | output | 1 | 59,612 | 10 | 119,225 |
Provide a correct Python 3 solution for this coding contest problem.
Problem statement
JOI decided to start a new social game from tomorrow.
In this social game, you can log in up to once a day, and you will get A coins each time you log in.
Also, if you log in for 7 consecutive days from Monday to Sunday, you will get an additional B coins each time.
No other coins will be given.
Tomorrow is Monday. Find the minimum number of times JOI must log in to get at least C coins.
Constraint
* 1 ≤ A ≤ 1000
* 0 ≤ B ≤ 1000
* 1 ≤ C ≤ 1000000 (= 10 ^ 6)
Input Output
input
Input is given from standard input in the following format.
A B C
output
Output the minimum number of times JOI must log in to get at least C coins.
<!-
Subtask
1. (40 points) B = 0
2. (60 points) There are no additional restrictions.
->
Input / output example
Input example 1
3 0 10
Output example 1
Four
* I want to get 3 coins per login and collect 10 coins.
* JOI can get 12 coins by logging in for 4 consecutive days from Monday.
* Since you cannot get more than 10 coins by logging in 3 times or less, the minimum number of times JOI must log in is 4. Therefore, 4 is output.
Input example 2
1 2 10
Output example 2
8
* You can get 1 coin for each login. Apart from that, you can get 2 coins by logging in for a week in a row. I want to collect 10 coins.
* If you log in consecutively from Monday to Sunday, you will get 2 coins in addition to 7 daily coins, so you will get a total of 9 coins. Therefore, if you log in one more time, you will get 10 coins.
* Since you cannot get more than 10 coins by logging in 7 times or less, the minimum number of times JOI must log in is 8. Therefore, 8 is output.
Creative Commons License
Information Olympics Japan Committee work "18th Japan Information Olympics JOI 2018/2019 Qualifying Competition Tasks"
Example
Input
3 0 10
Output
4 | instruction | 0 | 59,613 | 10 | 119,226 |
"Correct Solution:
```
a,b,c=map(int,input().split())
W_C=int((7*a)+b)
#1週間の総コイン数
W=c//W_C
#何週間必要か
C=W_C*W
#W週間で得られたコイン数
if 7*a<c-C<W_C:
#1日で貰えるコイン×７<
#必要なコイン - W週間で得られたコイン数<
#1週間の総コイン数
print((W+1)*7)
else:
day=0
coin=0
while coin<c-C:
coin+=a
day+=1
print(W*7+day)
``` | output | 1 | 59,613 | 10 | 119,227 |
Provide a correct Python 3 solution for this coding contest problem.
Problem statement
JOI decided to start a new social game from tomorrow.
In this social game, you can log in up to once a day, and you will get A coins each time you log in.
Also, if you log in for 7 consecutive days from Monday to Sunday, you will get an additional B coins each time.
No other coins will be given.
Tomorrow is Monday. Find the minimum number of times JOI must log in to get at least C coins.
Constraint
* 1 ≤ A ≤ 1000
* 0 ≤ B ≤ 1000
* 1 ≤ C ≤ 1000000 (= 10 ^ 6)
Input Output
input
Input is given from standard input in the following format.
A B C
output
Output the minimum number of times JOI must log in to get at least C coins.
<!-
Subtask
1. (40 points) B = 0
2. (60 points) There are no additional restrictions.
->
Input / output example
Input example 1
3 0 10
Output example 1
Four
* I want to get 3 coins per login and collect 10 coins.
* JOI can get 12 coins by logging in for 4 consecutive days from Monday.
* Since you cannot get more than 10 coins by logging in 3 times or less, the minimum number of times JOI must log in is 4. Therefore, 4 is output.
Input example 2
1 2 10
Output example 2
8
* You can get 1 coin for each login. Apart from that, you can get 2 coins by logging in for a week in a row. I want to collect 10 coins.
* If you log in consecutively from Monday to Sunday, you will get 2 coins in addition to 7 daily coins, so you will get a total of 9 coins. Therefore, if you log in one more time, you will get 10 coins.
* Since you cannot get more than 10 coins by logging in 7 times or less, the minimum number of times JOI must log in is 8. Therefore, 8 is output.
Creative Commons License
Information Olympics Japan Committee work "18th Japan Information Olympics JOI 2018/2019 Qualifying Competition Tasks"
Example
Input
3 0 10
Output
4 | instruction | 0 | 59,614 | 10 | 119,228 |
"Correct Solution:
```
#標準入力とリストの初期化
a,b,c = map(int,input().split())
count,coin = 0,0
#毎日ログインボーナスを加算し7日ログインしたらボーナスを追加する
while coin < c:
count += 1
coin += a
if count % 7 == 0:coin += b
#かかった日数を出力する
print(count)
``` | output | 1 | 59,614 | 10 | 119,229 |
Provide a correct Python 3 solution for this coding contest problem.
Problem statement
JOI decided to start a new social game from tomorrow.
In this social game, you can log in up to once a day, and you will get A coins each time you log in.
Also, if you log in for 7 consecutive days from Monday to Sunday, you will get an additional B coins each time.
No other coins will be given.
Tomorrow is Monday. Find the minimum number of times JOI must log in to get at least C coins.
Constraint
* 1 ≤ A ≤ 1000
* 0 ≤ B ≤ 1000
* 1 ≤ C ≤ 1000000 (= 10 ^ 6)
Input Output
input
Input is given from standard input in the following format.
A B C
output
Output the minimum number of times JOI must log in to get at least C coins.
<!-
Subtask
1. (40 points) B = 0
2. (60 points) There are no additional restrictions.
->
Input / output example
Input example 1
3 0 10
Output example 1
Four
* I want to get 3 coins per login and collect 10 coins.
* JOI can get 12 coins by logging in for 4 consecutive days from Monday.
* Since you cannot get more than 10 coins by logging in 3 times or less, the minimum number of times JOI must log in is 4. Therefore, 4 is output.
Input example 2
1 2 10
Output example 2
8
* You can get 1 coin for each login. Apart from that, you can get 2 coins by logging in for a week in a row. I want to collect 10 coins.
* If you log in consecutively from Monday to Sunday, you will get 2 coins in addition to 7 daily coins, so you will get a total of 9 coins. Therefore, if you log in one more time, you will get 10 coins.
* Since you cannot get more than 10 coins by logging in 7 times or less, the minimum number of times JOI must log in is 8. Therefore, 8 is output.
Creative Commons License
Information Olympics Japan Committee work "18th Japan Information Olympics JOI 2018/2019 Qualifying Competition Tasks"
Example
Input
3 0 10
Output
4 | instruction | 0 | 59,615 | 10 | 119,230 |
"Correct Solution:
```
a,b,c=map(int,input().split())
mon=0
day=0
while mon<c:
mon+=a
day+=1
if day%7==0:
mon+=b
print(day)
``` | output | 1 | 59,615 | 10 | 119,231 |
Provide a correct Python 3 solution for this coding contest problem.
Problem statement
JOI decided to start a new social game from tomorrow.
In this social game, you can log in up to once a day, and you will get A coins each time you log in.
Also, if you log in for 7 consecutive days from Monday to Sunday, you will get an additional B coins each time.
No other coins will be given.
Tomorrow is Monday. Find the minimum number of times JOI must log in to get at least C coins.
Constraint
* 1 ≤ A ≤ 1000
* 0 ≤ B ≤ 1000
* 1 ≤ C ≤ 1000000 (= 10 ^ 6)
Input Output
input
Input is given from standard input in the following format.
A B C
output
Output the minimum number of times JOI must log in to get at least C coins.
<!-
Subtask
1. (40 points) B = 0
2. (60 points) There are no additional restrictions.
->
Input / output example
Input example 1
3 0 10
Output example 1
Four
* I want to get 3 coins per login and collect 10 coins.
* JOI can get 12 coins by logging in for 4 consecutive days from Monday.
* Since you cannot get more than 10 coins by logging in 3 times or less, the minimum number of times JOI must log in is 4. Therefore, 4 is output.
Input example 2
1 2 10
Output example 2
8
* You can get 1 coin for each login. Apart from that, you can get 2 coins by logging in for a week in a row. I want to collect 10 coins.
* If you log in consecutively from Monday to Sunday, you will get 2 coins in addition to 7 daily coins, so you will get a total of 9 coins. Therefore, if you log in one more time, you will get 10 coins.
* Since you cannot get more than 10 coins by logging in 7 times or less, the minimum number of times JOI must log in is 8. Therefore, 8 is output.
Creative Commons License
Information Olympics Japan Committee work "18th Japan Information Olympics JOI 2018/2019 Qualifying Competition Tasks"
Example
Input
3 0 10
Output
4 | instruction | 0 | 59,616 | 10 | 119,232 |
"Correct Solution:
```
A,B,C = map(int, input().split())
ans = 0
while C > 0:
ans += 1
C -= A
if ans % 7 == 0:
C -= B
print(ans)
``` | output | 1 | 59,616 | 10 | 119,233 |
Provide a correct Python 3 solution for this coding contest problem.
Problem statement
JOI decided to start a new social game from tomorrow.
In this social game, you can log in up to once a day, and you will get A coins each time you log in.
Also, if you log in for 7 consecutive days from Monday to Sunday, you will get an additional B coins each time.
No other coins will be given.
Tomorrow is Monday. Find the minimum number of times JOI must log in to get at least C coins.
Constraint
* 1 ≤ A ≤ 1000
* 0 ≤ B ≤ 1000
* 1 ≤ C ≤ 1000000 (= 10 ^ 6)
Input Output
input
Input is given from standard input in the following format.
A B C
output
Output the minimum number of times JOI must log in to get at least C coins.
<!-
Subtask
1. (40 points) B = 0
2. (60 points) There are no additional restrictions.
->
Input / output example
Input example 1
3 0 10
Output example 1
Four
* I want to get 3 coins per login and collect 10 coins.
* JOI can get 12 coins by logging in for 4 consecutive days from Monday.
* Since you cannot get more than 10 coins by logging in 3 times or less, the minimum number of times JOI must log in is 4. Therefore, 4 is output.
Input example 2
1 2 10
Output example 2
8
* You can get 1 coin for each login. Apart from that, you can get 2 coins by logging in for a week in a row. I want to collect 10 coins.
* If you log in consecutively from Monday to Sunday, you will get 2 coins in addition to 7 daily coins, so you will get a total of 9 coins. Therefore, if you log in one more time, you will get 10 coins.
* Since you cannot get more than 10 coins by logging in 7 times or less, the minimum number of times JOI must log in is 8. Therefore, 8 is output.
Creative Commons License
Information Olympics Japan Committee work "18th Japan Information Olympics JOI 2018/2019 Qualifying Competition Tasks"
Example
Input
3 0 10
Output
4 | instruction | 0 | 59,617 | 10 | 119,234 |
"Correct Solution:
```
A,B,C=map(int,input().split())
x=C%(A*7+B)
y =C//(A*7+B)
if (C//A)<7:
if C%A==0:
print(C//A)
else:
print(C//A+1)
elif x==0:
print(y*7)
else:
z=x%A
w=x//A
if w>(A*7):
print(y*7+7)
elif w==7 and z<=B:
print(y*7+7)
elif z==0:
print(y*7+w)
else:
print(y*7+w+1)
``` | output | 1 | 59,617 | 10 | 119,235 |
Provide a correct Python 3 solution for this coding contest problem.
Problem statement
JOI decided to start a new social game from tomorrow.
In this social game, you can log in up to once a day, and you will get A coins each time you log in.
Also, if you log in for 7 consecutive days from Monday to Sunday, you will get an additional B coins each time.
No other coins will be given.
Tomorrow is Monday. Find the minimum number of times JOI must log in to get at least C coins.
Constraint
* 1 ≤ A ≤ 1000
* 0 ≤ B ≤ 1000
* 1 ≤ C ≤ 1000000 (= 10 ^ 6)
Input Output
input
Input is given from standard input in the following format.
A B C
output
Output the minimum number of times JOI must log in to get at least C coins.
<!-
Subtask
1. (40 points) B = 0
2. (60 points) There are no additional restrictions.
->
Input / output example
Input example 1
3 0 10
Output example 1
Four
* I want to get 3 coins per login and collect 10 coins.
* JOI can get 12 coins by logging in for 4 consecutive days from Monday.
* Since you cannot get more than 10 coins by logging in 3 times or less, the minimum number of times JOI must log in is 4. Therefore, 4 is output.
Input example 2
1 2 10
Output example 2
8
* You can get 1 coin for each login. Apart from that, you can get 2 coins by logging in for a week in a row. I want to collect 10 coins.
* If you log in consecutively from Monday to Sunday, you will get 2 coins in addition to 7 daily coins, so you will get a total of 9 coins. Therefore, if you log in one more time, you will get 10 coins.
* Since you cannot get more than 10 coins by logging in 7 times or less, the minimum number of times JOI must log in is 8. Therefore, 8 is output.
Creative Commons License
Information Olympics Japan Committee work "18th Japan Information Olympics JOI 2018/2019 Qualifying Competition Tasks"
Example
Input
3 0 10
Output
4 | instruction | 0 | 59,618 | 10 | 119,236 |
"Correct Solution:
```
a, b, x = map(int, input().split())
i = 0
while x>0:
i += 1
x -= a
if i%7==0: x-=b
print(i)
``` | output | 1 | 59,618 | 10 | 119,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem statement
JOI decided to start a new social game from tomorrow.
In this social game, you can log in up to once a day, and you will get A coins each time you log in.
Also, if you log in for 7 consecutive days from Monday to Sunday, you will get an additional B coins each time.
No other coins will be given.
Tomorrow is Monday. Find the minimum number of times JOI must log in to get at least C coins.
Constraint
* 1 ≤ A ≤ 1000
* 0 ≤ B ≤ 1000
* 1 ≤ C ≤ 1000000 (= 10 ^ 6)
Input Output
input
Input is given from standard input in the following format.
A B C
output
Output the minimum number of times JOI must log in to get at least C coins.
<!-
Subtask
1. (40 points) B = 0
2. (60 points) There are no additional restrictions.
->
Input / output example
Input example 1
3 0 10
Output example 1
Four
* I want to get 3 coins per login and collect 10 coins.
* JOI can get 12 coins by logging in for 4 consecutive days from Monday.
* Since you cannot get more than 10 coins by logging in 3 times or less, the minimum number of times JOI must log in is 4. Therefore, 4 is output.
Input example 2
1 2 10
Output example 2
8
* You can get 1 coin for each login. Apart from that, you can get 2 coins by logging in for a week in a row. I want to collect 10 coins.
* If you log in consecutively from Monday to Sunday, you will get 2 coins in addition to 7 daily coins, so you will get a total of 9 coins. Therefore, if you log in one more time, you will get 10 coins.
* Since you cannot get more than 10 coins by logging in 7 times or less, the minimum number of times JOI must log in is 8. Therefore, 8 is output.
Creative Commons License
Information Olympics Japan Committee work "18th Japan Information Olympics JOI 2018/2019 Qualifying Competition Tasks"
Example
Input
3 0 10
Output
4
Submitted Solution:
```
a,b,c=map(int,input().split())
d,e=0,0
while d<c:
d +=a
e +=1
if e%7 == 0:
d +=b
print(e)
``` | instruction | 0 | 59,619 | 10 | 119,238 |
Yes | output | 1 | 59,619 | 10 | 119,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem statement
JOI decided to start a new social game from tomorrow.
In this social game, you can log in up to once a day, and you will get A coins each time you log in.
Also, if you log in for 7 consecutive days from Monday to Sunday, you will get an additional B coins each time.
No other coins will be given.
Tomorrow is Monday. Find the minimum number of times JOI must log in to get at least C coins.
Constraint
* 1 ≤ A ≤ 1000
* 0 ≤ B ≤ 1000
* 1 ≤ C ≤ 1000000 (= 10 ^ 6)
Input Output
input
Input is given from standard input in the following format.
A B C
output
Output the minimum number of times JOI must log in to get at least C coins.
<!-
Subtask
1. (40 points) B = 0
2. (60 points) There are no additional restrictions.
->
Input / output example
Input example 1
3 0 10
Output example 1
Four
* I want to get 3 coins per login and collect 10 coins.
* JOI can get 12 coins by logging in for 4 consecutive days from Monday.
* Since you cannot get more than 10 coins by logging in 3 times or less, the minimum number of times JOI must log in is 4. Therefore, 4 is output.
Input example 2
1 2 10
Output example 2
8
* You can get 1 coin for each login. Apart from that, you can get 2 coins by logging in for a week in a row. I want to collect 10 coins.
* If you log in consecutively from Monday to Sunday, you will get 2 coins in addition to 7 daily coins, so you will get a total of 9 coins. Therefore, if you log in one more time, you will get 10 coins.
* Since you cannot get more than 10 coins by logging in 7 times or less, the minimum number of times JOI must log in is 8. Therefore, 8 is output.
Creative Commons License
Information Olympics Japan Committee work "18th Japan Information Olympics JOI 2018/2019 Qualifying Competition Tasks"
Example
Input
3 0 10
Output
4
Submitted Solution:
```
# AOJ Volume6 0652
# http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0652
A, B, C = list(map(int, input().split()))
coins = 0
day = 0
while coins < C:
day += 1
coins += A
if day % 7 == 0:
coins += B
print(day)
``` | instruction | 0 | 59,620 | 10 | 119,240 |
Yes | output | 1 | 59,620 | 10 | 119,241 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem statement
JOI decided to start a new social game from tomorrow.
In this social game, you can log in up to once a day, and you will get A coins each time you log in.
Also, if you log in for 7 consecutive days from Monday to Sunday, you will get an additional B coins each time.
No other coins will be given.
Tomorrow is Monday. Find the minimum number of times JOI must log in to get at least C coins.
Constraint
* 1 ≤ A ≤ 1000
* 0 ≤ B ≤ 1000
* 1 ≤ C ≤ 1000000 (= 10 ^ 6)
Input Output
input
Input is given from standard input in the following format.
A B C
output
Output the minimum number of times JOI must log in to get at least C coins.
<!-
Subtask
1. (40 points) B = 0
2. (60 points) There are no additional restrictions.
->
Input / output example
Input example 1
3 0 10
Output example 1
Four
* I want to get 3 coins per login and collect 10 coins.
* JOI can get 12 coins by logging in for 4 consecutive days from Monday.
* Since you cannot get more than 10 coins by logging in 3 times or less, the minimum number of times JOI must log in is 4. Therefore, 4 is output.
Input example 2
1 2 10
Output example 2
8
* You can get 1 coin for each login. Apart from that, you can get 2 coins by logging in for a week in a row. I want to collect 10 coins.
* If you log in consecutively from Monday to Sunday, you will get 2 coins in addition to 7 daily coins, so you will get a total of 9 coins. Therefore, if you log in one more time, you will get 10 coins.
* Since you cannot get more than 10 coins by logging in 7 times or less, the minimum number of times JOI must log in is 8. Therefore, 8 is output.
Creative Commons License
Information Olympics Japan Committee work "18th Japan Information Olympics JOI 2018/2019 Qualifying Competition Tasks"
Example
Input
3 0 10
Output
4
Submitted Solution:
```
A, B, C = map(int, input().split())
i = 0
sum = 0
cnt = 0
while True:
sum += A
i += 1
cnt += 1
if sum >= C:
break
if i%6 == 0:
sum += A+B
i = 0
cnt += 1
if sum >= C:
break
print(cnt)
``` | instruction | 0 | 59,621 | 10 | 119,242 |
Yes | output | 1 | 59,621 | 10 | 119,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem statement
JOI decided to start a new social game from tomorrow.
In this social game, you can log in up to once a day, and you will get A coins each time you log in.
Also, if you log in for 7 consecutive days from Monday to Sunday, you will get an additional B coins each time.
No other coins will be given.
Tomorrow is Monday. Find the minimum number of times JOI must log in to get at least C coins.
Constraint
* 1 ≤ A ≤ 1000
* 0 ≤ B ≤ 1000
* 1 ≤ C ≤ 1000000 (= 10 ^ 6)
Input Output
input
Input is given from standard input in the following format.
A B C
output
Output the minimum number of times JOI must log in to get at least C coins.
<!-
Subtask
1. (40 points) B = 0
2. (60 points) There are no additional restrictions.
->
Input / output example
Input example 1
3 0 10
Output example 1
Four
* I want to get 3 coins per login and collect 10 coins.
* JOI can get 12 coins by logging in for 4 consecutive days from Monday.
* Since you cannot get more than 10 coins by logging in 3 times or less, the minimum number of times JOI must log in is 4. Therefore, 4 is output.
Input example 2
1 2 10
Output example 2
8
* You can get 1 coin for each login. Apart from that, you can get 2 coins by logging in for a week in a row. I want to collect 10 coins.
* If you log in consecutively from Monday to Sunday, you will get 2 coins in addition to 7 daily coins, so you will get a total of 9 coins. Therefore, if you log in one more time, you will get 10 coins.
* Since you cannot get more than 10 coins by logging in 7 times or less, the minimum number of times JOI must log in is 8. Therefore, 8 is output.
Creative Commons License
Information Olympics Japan Committee work "18th Japan Information Olympics JOI 2018/2019 Qualifying Competition Tasks"
Example
Input
3 0 10
Output
4
Submitted Solution:
```
L=input().split( )
A,B,C=int(L[0]),int(L[1]),int(L[2])
w=C//(7*A+B)
m=C%(7*A+B)
d=m//A
if d>=7:
d=7
else:
if m%A!=0:
d+=1
print(7*w+d)
``` | instruction | 0 | 59,622 | 10 | 119,244 |
Yes | output | 1 | 59,622 | 10 | 119,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem statement
JOI decided to start a new social game from tomorrow.
In this social game, you can log in up to once a day, and you will get A coins each time you log in.
Also, if you log in for 7 consecutive days from Monday to Sunday, you will get an additional B coins each time.
No other coins will be given.
Tomorrow is Monday. Find the minimum number of times JOI must log in to get at least C coins.
Constraint
* 1 ≤ A ≤ 1000
* 0 ≤ B ≤ 1000
* 1 ≤ C ≤ 1000000 (= 10 ^ 6)
Input Output
input
Input is given from standard input in the following format.
A B C
output
Output the minimum number of times JOI must log in to get at least C coins.
<!-
Subtask
1. (40 points) B = 0
2. (60 points) There are no additional restrictions.
->
Input / output example
Input example 1
3 0 10
Output example 1
Four
* I want to get 3 coins per login and collect 10 coins.
* JOI can get 12 coins by logging in for 4 consecutive days from Monday.
* Since you cannot get more than 10 coins by logging in 3 times or less, the minimum number of times JOI must log in is 4. Therefore, 4 is output.
Input example 2
1 2 10
Output example 2
8
* You can get 1 coin for each login. Apart from that, you can get 2 coins by logging in for a week in a row. I want to collect 10 coins.
* If you log in consecutively from Monday to Sunday, you will get 2 coins in addition to 7 daily coins, so you will get a total of 9 coins. Therefore, if you log in one more time, you will get 10 coins.
* Since you cannot get more than 10 coins by logging in 7 times or less, the minimum number of times JOI must log in is 8. Therefore, 8 is output.
Creative Commons License
Information Olympics Japan Committee work "18th Japan Information Olympics JOI 2018/2019 Qualifying Competition Tasks"
Example
Input
3 0 10
Output
4
Submitted Solution:
```
inputs = inout().split(" ")
print(math.cell(inputs[2]/(inputs[0]+inputs[1]/7))
``` | instruction | 0 | 59,623 | 10 | 119,246 |
No | output | 1 | 59,623 | 10 | 119,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem statement
JOI decided to start a new social game from tomorrow.
In this social game, you can log in up to once a day, and you will get A coins each time you log in.
Also, if you log in for 7 consecutive days from Monday to Sunday, you will get an additional B coins each time.
No other coins will be given.
Tomorrow is Monday. Find the minimum number of times JOI must log in to get at least C coins.
Constraint
* 1 ≤ A ≤ 1000
* 0 ≤ B ≤ 1000
* 1 ≤ C ≤ 1000000 (= 10 ^ 6)
Input Output
input
Input is given from standard input in the following format.
A B C
output
Output the minimum number of times JOI must log in to get at least C coins.
<!-
Subtask
1. (40 points) B = 0
2. (60 points) There are no additional restrictions.
->
Input / output example
Input example 1
3 0 10
Output example 1
Four
* I want to get 3 coins per login and collect 10 coins.
* JOI can get 12 coins by logging in for 4 consecutive days from Monday.
* Since you cannot get more than 10 coins by logging in 3 times or less, the minimum number of times JOI must log in is 4. Therefore, 4 is output.
Input example 2
1 2 10
Output example 2
8
* You can get 1 coin for each login. Apart from that, you can get 2 coins by logging in for a week in a row. I want to collect 10 coins.
* If you log in consecutively from Monday to Sunday, you will get 2 coins in addition to 7 daily coins, so you will get a total of 9 coins. Therefore, if you log in one more time, you will get 10 coins.
* Since you cannot get more than 10 coins by logging in 7 times or less, the minimum number of times JOI must log in is 8. Therefore, 8 is output.
Creative Commons License
Information Olympics Japan Committee work "18th Japan Information Olympics JOI 2018/2019 Qualifying Competition Tasks"
Example
Input
3 0 10
Output
4
Submitted Solution:
```
import math
inputs = inout().split(" ")
print(math.cell(inputs[2]/(inputs[0]+inputs[1]/7))
``` | instruction | 0 | 59,624 | 10 | 119,248 |
No | output | 1 | 59,624 | 10 | 119,249 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem statement
JOI decided to start a new social game from tomorrow.
In this social game, you can log in up to once a day, and you will get A coins each time you log in.
Also, if you log in for 7 consecutive days from Monday to Sunday, you will get an additional B coins each time.
No other coins will be given.
Tomorrow is Monday. Find the minimum number of times JOI must log in to get at least C coins.
Constraint
* 1 ≤ A ≤ 1000
* 0 ≤ B ≤ 1000
* 1 ≤ C ≤ 1000000 (= 10 ^ 6)
Input Output
input
Input is given from standard input in the following format.
A B C
output
Output the minimum number of times JOI must log in to get at least C coins.
<!-
Subtask
1. (40 points) B = 0
2. (60 points) There are no additional restrictions.
->
Input / output example
Input example 1
3 0 10
Output example 1
Four
* I want to get 3 coins per login and collect 10 coins.
* JOI can get 12 coins by logging in for 4 consecutive days from Monday.
* Since you cannot get more than 10 coins by logging in 3 times or less, the minimum number of times JOI must log in is 4. Therefore, 4 is output.
Input example 2
1 2 10
Output example 2
8
* You can get 1 coin for each login. Apart from that, you can get 2 coins by logging in for a week in a row. I want to collect 10 coins.
* If you log in consecutively from Monday to Sunday, you will get 2 coins in addition to 7 daily coins, so you will get a total of 9 coins. Therefore, if you log in one more time, you will get 10 coins.
* Since you cannot get more than 10 coins by logging in 7 times or less, the minimum number of times JOI must log in is 8. Therefore, 8 is output.
Creative Commons License
Information Olympics Japan Committee work "18th Japan Information Olympics JOI 2018/2019 Qualifying Competition Tasks"
Example
Input
3 0 10
Output
4
Submitted Solution:
```
A,B,C = map(int, input().split())
days = C//(A+(B/7))
coins = A*days + B*days/7
days = int(days) if coins >= C else int(days+1)
print(days)
``` | instruction | 0 | 59,625 | 10 | 119,250 |
No | output | 1 | 59,625 | 10 | 119,251 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem statement
JOI decided to start a new social game from tomorrow.
In this social game, you can log in up to once a day, and you will get A coins each time you log in.
Also, if you log in for 7 consecutive days from Monday to Sunday, you will get an additional B coins each time.
No other coins will be given.
Tomorrow is Monday. Find the minimum number of times JOI must log in to get at least C coins.
Constraint
* 1 ≤ A ≤ 1000
* 0 ≤ B ≤ 1000
* 1 ≤ C ≤ 1000000 (= 10 ^ 6)
Input Output
input
Input is given from standard input in the following format.
A B C
output
Output the minimum number of times JOI must log in to get at least C coins.
<!-
Subtask
1. (40 points) B = 0
2. (60 points) There are no additional restrictions.
->
Input / output example
Input example 1
3 0 10
Output example 1
Four
* I want to get 3 coins per login and collect 10 coins.
* JOI can get 12 coins by logging in for 4 consecutive days from Monday.
* Since you cannot get more than 10 coins by logging in 3 times or less, the minimum number of times JOI must log in is 4. Therefore, 4 is output.
Input example 2
1 2 10
Output example 2
8
* You can get 1 coin for each login. Apart from that, you can get 2 coins by logging in for a week in a row. I want to collect 10 coins.
* If you log in consecutively from Monday to Sunday, you will get 2 coins in addition to 7 daily coins, so you will get a total of 9 coins. Therefore, if you log in one more time, you will get 10 coins.
* Since you cannot get more than 10 coins by logging in 7 times or less, the minimum number of times JOI must log in is 8. Therefore, 8 is output.
Creative Commons License
Information Olympics Japan Committee work "18th Japan Information Olympics JOI 2018/2019 Qualifying Competition Tasks"
Example
Input
3 0 10
Output
4
Submitted Solution:
```
A,B,C = map(int, input().split())
days = C / (A+(B/7))
coins = A*days + B*days/7
days = int(days) if coins >= C else int(days+1)
print(days)
``` | instruction | 0 | 59,626 | 10 | 119,252 |
No | output | 1 | 59,626 | 10 | 119,253 |
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