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.
Kenkoooo is planning a trip in Republic of Snuke. In this country, there are n cities and m trains running. The cities are numbered 1 through n, and the i-th train connects City u_i and v_i bidirectionally. Any city can be reached from any city by changing trains.
Two currencies are used in the country: yen and snuuk. Any train fare can be paid by both yen and snuuk. The fare of the i-th train is a_i yen if paid in yen, and b_i snuuk if paid in snuuk.
In a city with a money exchange office, you can change 1 yen into 1 snuuk. However, when you do a money exchange, you have to change all your yen into snuuk. That is, if Kenkoooo does a money exchange when he has X yen, he will then have X snuuk. Currently, there is a money exchange office in every city, but the office in City i will shut down in i years and can never be used in and after that year.
Kenkoooo is planning to depart City s with 10^{15} yen in his pocket and head for City t, and change his yen into snuuk in some city while traveling. It is acceptable to do the exchange in City s or City t.
Kenkoooo would like to have as much snuuk as possible when he reaches City t by making the optimal choices for the route to travel and the city to do the exchange. For each i=0,...,n-1, find the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i years. You can assume that the trip finishes within the year.
Constraints
* 2 \leq n \leq 10^5
* 1 \leq m \leq 10^5
* 1 \leq s,t \leq n
* s \neq t
* 1 \leq u_i < v_i \leq n
* 1 \leq a_i,b_i \leq 10^9
* If i\neq j, then u_i \neq u_j or v_i \neq v_j.
* Any city can be reached from any city by changing trains.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
n m s t
u_1 v_1 a_1 b_1
:
u_m v_m a_m b_m
Output
Print n lines. In the i-th line, print the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i-1 years.
Examples
Input
4 3 2 3
1 4 1 100
1 2 1 10
1 3 20 1
Output
999999999999998
999999999999989
999999999999979
999999999999897
Input
8 12 3 8
2 8 685087149 857180777
6 7 298270585 209942236
2 4 346080035 234079976
2 5 131857300 22507157
4 8 30723332 173476334
2 6 480845267 448565596
1 4 181424400 548830121
4 5 57429995 195056405
7 8 160277628 479932440
1 6 475692952 203530153
3 5 336869679 160714712
2 7 389775999 199123879
Output
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
Submitted Solution:
```
import sys
input=sys.stdin.readline
from collections import defaultdict
from heapq import heappush,heappop
N,m,s,t=map(int,input().split())
en=defaultdict(set)
sneek=defaultdict(set)
for i in range(m):
u,v,a,b=map(int,input().split())
en[u]|={(a,v)}
en[v]|={(a,u)}
sneek[u]|={(b,v)}
sneek[v]|={(b,u)}
inf=float('inf')
def makedist(first,branch):
dist=[inf]*(N+1)
dist[first]=0
check=[(0,first)]
while len(check)>0:
d,now=heappop(check)
if dist[now]<d:
continue
for lenth,nex in branch[now]:
if d+lenth<dist[nex]:
dist[nex]=d+lenth
heappush(check,(d+lenth,nex))
return dist
disten=makedist(s,en)
distsneek=makedist(t,sneek)
ans=[0]*(N+1)
ans[N]=disten[N]+distsneek[N]
for i in range(N-1,0,-1):
ans[i]=min(ans[i+1],disten[i]+distsneek[i])
big=10**15
for i in range(1,N+1):
print(big-ans[i])
``` | instruction | 0 | 41,903 | 10 | 83,806 |
Yes | output | 1 | 41,903 | 10 | 83,807 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kenkoooo is planning a trip in Republic of Snuke. In this country, there are n cities and m trains running. The cities are numbered 1 through n, and the i-th train connects City u_i and v_i bidirectionally. Any city can be reached from any city by changing trains.
Two currencies are used in the country: yen and snuuk. Any train fare can be paid by both yen and snuuk. The fare of the i-th train is a_i yen if paid in yen, and b_i snuuk if paid in snuuk.
In a city with a money exchange office, you can change 1 yen into 1 snuuk. However, when you do a money exchange, you have to change all your yen into snuuk. That is, if Kenkoooo does a money exchange when he has X yen, he will then have X snuuk. Currently, there is a money exchange office in every city, but the office in City i will shut down in i years and can never be used in and after that year.
Kenkoooo is planning to depart City s with 10^{15} yen in his pocket and head for City t, and change his yen into snuuk in some city while traveling. It is acceptable to do the exchange in City s or City t.
Kenkoooo would like to have as much snuuk as possible when he reaches City t by making the optimal choices for the route to travel and the city to do the exchange. For each i=0,...,n-1, find the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i years. You can assume that the trip finishes within the year.
Constraints
* 2 \leq n \leq 10^5
* 1 \leq m \leq 10^5
* 1 \leq s,t \leq n
* s \neq t
* 1 \leq u_i < v_i \leq n
* 1 \leq a_i,b_i \leq 10^9
* If i\neq j, then u_i \neq u_j or v_i \neq v_j.
* Any city can be reached from any city by changing trains.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
n m s t
u_1 v_1 a_1 b_1
:
u_m v_m a_m b_m
Output
Print n lines. In the i-th line, print the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i-1 years.
Examples
Input
4 3 2 3
1 4 1 100
1 2 1 10
1 3 20 1
Output
999999999999998
999999999999989
999999999999979
999999999999897
Input
8 12 3 8
2 8 685087149 857180777
6 7 298270585 209942236
2 4 346080035 234079976
2 5 131857300 22507157
4 8 30723332 173476334
2 6 480845267 448565596
1 4 181424400 548830121
4 5 57429995 195056405
7 8 160277628 479932440
1 6 475692952 203530153
3 5 336869679 160714712
2 7 389775999 199123879
Output
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
Submitted Solution:
```
import heapq
def dijkstra(start: int, graph: list) -> list:
"""dijkstra法: 始点startから各頂点への最短距離を求める
計算量: O((E+V)logV)
"""
INF = 10 ** 18
n = len(graph)
dist = [INF] * n
dist[start] = 0
q = [(0, start)] # q = [(startからの距離, 現在地)]
while q:
d, v = heapq.heappop(q)
if dist[v] < d:
continue
for nxt_v, cost in graph[v]:
if dist[v] + cost < dist[nxt_v]:
dist[nxt_v] = dist[v] + cost
heapq.heappush(q, (dist[nxt_v], nxt_v))
return dist
n, m, s, t = map(int, input().split())
info = [list(map(int, input().split())) for i in range(m)]
graph_y = [[] for i in range(n)]
graph_s = [[] for i in range(n)]
for u, v, a, b in info:
u -= 1
v -= 1
graph_y[u].append((v, a))
graph_y[v].append((u, a))
graph_s[u].append((v, b))
graph_s[v].append((u, b))
s -= 1
t -= 1
dist_y = dijkstra(s, graph_y)
dist_s = dijkstra(t, graph_s)
ans = 10 ** 18
ret = [10 ** 15] * n
for i in range(n)[::-1]:
tmp = dist_y[i] + dist_s[i]
ans = min(ans, tmp)
ret[i] -= ans
for i in ret:
print(i)
``` | instruction | 0 | 41,904 | 10 | 83,808 |
Yes | output | 1 | 41,904 | 10 | 83,809 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kenkoooo is planning a trip in Republic of Snuke. In this country, there are n cities and m trains running. The cities are numbered 1 through n, and the i-th train connects City u_i and v_i bidirectionally. Any city can be reached from any city by changing trains.
Two currencies are used in the country: yen and snuuk. Any train fare can be paid by both yen and snuuk. The fare of the i-th train is a_i yen if paid in yen, and b_i snuuk if paid in snuuk.
In a city with a money exchange office, you can change 1 yen into 1 snuuk. However, when you do a money exchange, you have to change all your yen into snuuk. That is, if Kenkoooo does a money exchange when he has X yen, he will then have X snuuk. Currently, there is a money exchange office in every city, but the office in City i will shut down in i years and can never be used in and after that year.
Kenkoooo is planning to depart City s with 10^{15} yen in his pocket and head for City t, and change his yen into snuuk in some city while traveling. It is acceptable to do the exchange in City s or City t.
Kenkoooo would like to have as much snuuk as possible when he reaches City t by making the optimal choices for the route to travel and the city to do the exchange. For each i=0,...,n-1, find the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i years. You can assume that the trip finishes within the year.
Constraints
* 2 \leq n \leq 10^5
* 1 \leq m \leq 10^5
* 1 \leq s,t \leq n
* s \neq t
* 1 \leq u_i < v_i \leq n
* 1 \leq a_i,b_i \leq 10^9
* If i\neq j, then u_i \neq u_j or v_i \neq v_j.
* Any city can be reached from any city by changing trains.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
n m s t
u_1 v_1 a_1 b_1
:
u_m v_m a_m b_m
Output
Print n lines. In the i-th line, print the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i-1 years.
Examples
Input
4 3 2 3
1 4 1 100
1 2 1 10
1 3 20 1
Output
999999999999998
999999999999989
999999999999979
999999999999897
Input
8 12 3 8
2 8 685087149 857180777
6 7 298270585 209942236
2 4 346080035 234079976
2 5 131857300 22507157
4 8 30723332 173476334
2 6 480845267 448565596
1 4 181424400 548830121
4 5 57429995 195056405
7 8 160277628 479932440
1 6 475692952 203530153
3 5 336869679 160714712
2 7 389775999 199123879
Output
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
Submitted Solution:
```
from heapq import heapify, heappop, heappush, heappushpop
class PriorityQueue:
def __init__(self, heap):
self.heap = heap
heapify(self.heap)
def push(self, item):
heappush(self.heap, item)
def pop(self):
return heappop(self.heap)
def pushpop(self, item):
return heappushpop(self.heap, item)
def __call__(self):
return self.heap
def __len__(self):
return len(self.heap)
def main():
N, M, s, t = map(int, input().split())
s, t = s - 1, t - 1
G = [{} for _ in range(N)]
for _ in range(M):
u, v, a, b = map(int, input().split())
u, v = u - 1, v - 1
G[u][v] = (a, b) # (yen, snu)
G[v][u] = (a, b)
# Dijkstra
ans = [0] * N
for i in range(N):
# exchange at ith station
# s to i
q = PriorityQueue([(0, s)]) # (cost, id)
vis = [False] * N
while len(q):
c, v = q.pop()
if v == i:
ans[i] = c
break
vis[v] = True
for n in G[v].keys():
if vis[n] is True:
continue
q.push((c + G[v][n][0], n)) # pay by yen
# i to t
q = PriorityQueue([(0, i)]) # (cost, id)
vis = [False] * N
while len(q):
c, v = q.pop()
if v == t:
ans[i] += c
break
vis[v] = True
for n in G[v].keys():
if vis[n] is True:
continue
q.push((c + G[v][n][1], n)) # pay by snu
# calc ans
MONEY = pow(10, 15)
for i in range(1, N):
if ans[N - i] < ans[N - i - 1]:
ans[N - i - 1] = ans[N - 1]
print('\n'.join(map(lambda x: str(MONEY - x), ans)))
if __name__ == '__main__':
main()
``` | instruction | 0 | 41,905 | 10 | 83,810 |
No | output | 1 | 41,905 | 10 | 83,811 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kenkoooo is planning a trip in Republic of Snuke. In this country, there are n cities and m trains running. The cities are numbered 1 through n, and the i-th train connects City u_i and v_i bidirectionally. Any city can be reached from any city by changing trains.
Two currencies are used in the country: yen and snuuk. Any train fare can be paid by both yen and snuuk. The fare of the i-th train is a_i yen if paid in yen, and b_i snuuk if paid in snuuk.
In a city with a money exchange office, you can change 1 yen into 1 snuuk. However, when you do a money exchange, you have to change all your yen into snuuk. That is, if Kenkoooo does a money exchange when he has X yen, he will then have X snuuk. Currently, there is a money exchange office in every city, but the office in City i will shut down in i years and can never be used in and after that year.
Kenkoooo is planning to depart City s with 10^{15} yen in his pocket and head for City t, and change his yen into snuuk in some city while traveling. It is acceptable to do the exchange in City s or City t.
Kenkoooo would like to have as much snuuk as possible when he reaches City t by making the optimal choices for the route to travel and the city to do the exchange. For each i=0,...,n-1, find the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i years. You can assume that the trip finishes within the year.
Constraints
* 2 \leq n \leq 10^5
* 1 \leq m \leq 10^5
* 1 \leq s,t \leq n
* s \neq t
* 1 \leq u_i < v_i \leq n
* 1 \leq a_i,b_i \leq 10^9
* If i\neq j, then u_i \neq u_j or v_i \neq v_j.
* Any city can be reached from any city by changing trains.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
n m s t
u_1 v_1 a_1 b_1
:
u_m v_m a_m b_m
Output
Print n lines. In the i-th line, print the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i-1 years.
Examples
Input
4 3 2 3
1 4 1 100
1 2 1 10
1 3 20 1
Output
999999999999998
999999999999989
999999999999979
999999999999897
Input
8 12 3 8
2 8 685087149 857180777
6 7 298270585 209942236
2 4 346080035 234079976
2 5 131857300 22507157
4 8 30723332 173476334
2 6 480845267 448565596
1 4 181424400 548830121
4 5 57429995 195056405
7 8 160277628 479932440
1 6 475692952 203530153
3 5 336869679 160714712
2 7 389775999 199123879
Output
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
Submitted Solution:
```
n,m,s,t=map(int,input().split())
TRAIN=[None]*m
for i in range(m):
TRAIN[i]=list(map(int,input().split()))
YEN=[10**16]*(n+1)#sから国iまで行くときの最小円
SNOOK=[10**16]*(n+1)#国iからtまで行くときの最小スヌーク
YEN[s]=0
SNOOK[t]=0
TRAINMAP=[[] for i in range(n+1)]
for i in range(m):
TRAINMAP[TRAIN[i][0]]+=[(TRAIN[i][1],TRAIN[i][2],TRAIN[i][3])]
TRAINMAP[TRAIN[i][1]]+=[(TRAIN[i][0],TRAIN[i][2],TRAIN[i][3])]
checktown=s
checked=set()
alltown=set(list(range(1,n+1)))
checking=set()
while checked != alltown:
for i in TRAINMAP[checktown]:
if YEN[i[0]]>YEN[checktown]+i[1]:
YEN[i[0]]=YEN[checktown]+i[1]
checking=checking|{i[0]}
checked=checked|{checktown}
YENcheck=10**16
for i in (checking - checked):
if YENcheck>YEN[i]:
YENcheck=YEN[i]
checktown=i
checktown=t
checked=set()
alltown=set(list(range(1,n+1)))
checking=set()
while checked != alltown:
for i in TRAINMAP[checktown]:
if SNOOK[i[0]]>SNOOK[checktown]+i[2]:
SNOOK[i[0]]=SNOOK[checktown]+i[2]
checking=checking|{i[0]}
checked=checked|{checktown}
SNOOKcheck=10**16
for i in (checking - checked):
if SNOOKcheck>SNOOK[i]:
SNOOKcheck=SNOOK[i]
checktown=i
MONEY=[None]*(n+1)
for i in range(1,n+1):
MONEY[i]=YEN[i]+SNOOK[i]
for i in range(1,n+1):
print(10**15-min(MONEY[i:]))
``` | instruction | 0 | 41,906 | 10 | 83,812 |
No | output | 1 | 41,906 | 10 | 83,813 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kenkoooo is planning a trip in Republic of Snuke. In this country, there are n cities and m trains running. The cities are numbered 1 through n, and the i-th train connects City u_i and v_i bidirectionally. Any city can be reached from any city by changing trains.
Two currencies are used in the country: yen and snuuk. Any train fare can be paid by both yen and snuuk. The fare of the i-th train is a_i yen if paid in yen, and b_i snuuk if paid in snuuk.
In a city with a money exchange office, you can change 1 yen into 1 snuuk. However, when you do a money exchange, you have to change all your yen into snuuk. That is, if Kenkoooo does a money exchange when he has X yen, he will then have X snuuk. Currently, there is a money exchange office in every city, but the office in City i will shut down in i years and can never be used in and after that year.
Kenkoooo is planning to depart City s with 10^{15} yen in his pocket and head for City t, and change his yen into snuuk in some city while traveling. It is acceptable to do the exchange in City s or City t.
Kenkoooo would like to have as much snuuk as possible when he reaches City t by making the optimal choices for the route to travel and the city to do the exchange. For each i=0,...,n-1, find the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i years. You can assume that the trip finishes within the year.
Constraints
* 2 \leq n \leq 10^5
* 1 \leq m \leq 10^5
* 1 \leq s,t \leq n
* s \neq t
* 1 \leq u_i < v_i \leq n
* 1 \leq a_i,b_i \leq 10^9
* If i\neq j, then u_i \neq u_j or v_i \neq v_j.
* Any city can be reached from any city by changing trains.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
n m s t
u_1 v_1 a_1 b_1
:
u_m v_m a_m b_m
Output
Print n lines. In the i-th line, print the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i-1 years.
Examples
Input
4 3 2 3
1 4 1 100
1 2 1 10
1 3 20 1
Output
999999999999998
999999999999989
999999999999979
999999999999897
Input
8 12 3 8
2 8 685087149 857180777
6 7 298270585 209942236
2 4 346080035 234079976
2 5 131857300 22507157
4 8 30723332 173476334
2 6 480845267 448565596
1 4 181424400 548830121
4 5 57429995 195056405
7 8 160277628 479932440
1 6 475692952 203530153
3 5 336869679 160714712
2 7 389775999 199123879
Output
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
Submitted Solution:
```
from heapq import *
class PriorityQueue(object):
def __init__(self):
self.queue = []
def push(self, value):
heappush(self.queue, value)
def pop(self):
return heappop(self.queue)
def empty(self):
return len(self.queue) == 0
def size(self):
return len(self)
from numpy import array,inf
def djikstra(s):
que = PriorityQueue()
d = [inf]*V
d[s] = 0
que.push((0,s))
while(not que.empty()):
p = que.pop()
v = p[1]
if d[v] < p[0]:
continue
for i in range(len(G[v])):
e = G[v][i]
if d[e[0]] > d[v] + e[1]:
d[e[0]] = d[v] + e[1]
que.push((d[e[0]],e[0]))
return d
n,m,s,t = map(int,input().split())
V = n
Ga = [[]]*n
Gb = [[]]*n
for i in range(n):
Ga[i]=[]
Gb[i]=[]
for i in range(m):
u,v,a,b = map(int,input().split())
Ga[u-1].append((v-1,a))
Ga[v-1].append((u-1,a))
Gb[u-1].append((v-1,b))
Gb[v-1].append((u-1,b))
G = Ga
S = djikstra(s-1)
G = Gb
T = djikstra(t-1)
M = array(S)+array(T)
for i in range(n):
print(10**15-min(M[i:]))
``` | instruction | 0 | 41,907 | 10 | 83,814 |
No | output | 1 | 41,907 | 10 | 83,815 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kenkoooo is planning a trip in Republic of Snuke. In this country, there are n cities and m trains running. The cities are numbered 1 through n, and the i-th train connects City u_i and v_i bidirectionally. Any city can be reached from any city by changing trains.
Two currencies are used in the country: yen and snuuk. Any train fare can be paid by both yen and snuuk. The fare of the i-th train is a_i yen if paid in yen, and b_i snuuk if paid in snuuk.
In a city with a money exchange office, you can change 1 yen into 1 snuuk. However, when you do a money exchange, you have to change all your yen into snuuk. That is, if Kenkoooo does a money exchange when he has X yen, he will then have X snuuk. Currently, there is a money exchange office in every city, but the office in City i will shut down in i years and can never be used in and after that year.
Kenkoooo is planning to depart City s with 10^{15} yen in his pocket and head for City t, and change his yen into snuuk in some city while traveling. It is acceptable to do the exchange in City s or City t.
Kenkoooo would like to have as much snuuk as possible when he reaches City t by making the optimal choices for the route to travel and the city to do the exchange. For each i=0,...,n-1, find the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i years. You can assume that the trip finishes within the year.
Constraints
* 2 \leq n \leq 10^5
* 1 \leq m \leq 10^5
* 1 \leq s,t \leq n
* s \neq t
* 1 \leq u_i < v_i \leq n
* 1 \leq a_i,b_i \leq 10^9
* If i\neq j, then u_i \neq u_j or v_i \neq v_j.
* Any city can be reached from any city by changing trains.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
n m s t
u_1 v_1 a_1 b_1
:
u_m v_m a_m b_m
Output
Print n lines. In the i-th line, print the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i-1 years.
Examples
Input
4 3 2 3
1 4 1 100
1 2 1 10
1 3 20 1
Output
999999999999998
999999999999989
999999999999979
999999999999897
Input
8 12 3 8
2 8 685087149 857180777
6 7 298270585 209942236
2 4 346080035 234079976
2 5 131857300 22507157
4 8 30723332 173476334
2 6 480845267 448565596
1 4 181424400 548830121
4 5 57429995 195056405
7 8 160277628 479932440
1 6 475692952 203530153
3 5 336869679 160714712
2 7 389775999 199123879
Output
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
Submitted Solution:
```
import sys
import numpy as np
from scipy.sparse.csgraph import dijkstra
from scipy.sparse import csr_matrix
read = sys.stdin.read
readline = sys.stdin.readline
n, m, s, t, *uvab = map(int, read().split())
u = np.array(uvab[::4], np.int64)
v = np.array(uvab[1::4], np.int64)
a = np.array(uvab[2::4], np.int64)
b = np.array(uvab[3::4], np.int64)
yen_matrix = csr_matrix((a, (u, v)), shape=(n + 1, n + 1))
snuuk_matrix = csr_matrix((b, (u, v)), shape=(n + 1, n + 1))
yen = dijkstra(yen_matrix, directed=False, indices=s)
snuuk = dijkstra(snuuk_matrix, directed=False, indices=t)
costs = (yen[1:] + snuuk[1:]).astype(int)
answer = 10 ** 15 - np.minimum.accumulate(costs[::-1])[::-1]
print('\n'.join(map(str, answer.tolist())))
``` | instruction | 0 | 41,908 | 10 | 83,816 |
No | output | 1 | 41,908 | 10 | 83,817 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 ≤ A ≤ 105), the desired number of ways.
Output
In the first line print integers N and M (1 ≤ N ≤ 106, 1 ≤ M ≤ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 ≤ Di ≤ 106), the denominations of the coins. All denominations must be distinct: for any i ≠ j we must have Di ≠ Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139 | instruction | 0 | 42,578 | 10 | 85,156 |
Tags: constructive algorithms
Correct Solution:
```
a = int(input())
if a==1:
print(a, 1)
print(1)
else:
print((a-1)*2, 2)
print(1, 2)
``` | output | 1 | 42,578 | 10 | 85,157 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 ≤ A ≤ 105), the desired number of ways.
Output
In the first line print integers N and M (1 ≤ N ≤ 106, 1 ≤ M ≤ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 ≤ Di ≤ 106), the denominations of the coins. All denominations must be distinct: for any i ≠ j we must have Di ≠ Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139 | instruction | 0 | 42,579 | 10 | 85,158 |
Tags: constructive algorithms
Correct Solution:
```
A=int(input());print([str((A-1)*2)+" 2\n1 2","1 1\n1"][A==1])
``` | output | 1 | 42,579 | 10 | 85,159 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 ≤ A ≤ 105), the desired number of ways.
Output
In the first line print integers N and M (1 ≤ N ≤ 106, 1 ≤ M ≤ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 ≤ Di ≤ 106), the denominations of the coins. All denominations must be distinct: for any i ≠ j we must have Di ≠ Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139 | instruction | 0 | 42,580 | 10 | 85,160 |
Tags: constructive algorithms
Correct Solution:
```
n=int(input())
if n==1:
print(1,1)
print(1)
else:
print((n-1)*2,2)
print(1,2)
``` | output | 1 | 42,580 | 10 | 85,161 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 ≤ A ≤ 105), the desired number of ways.
Output
In the first line print integers N and M (1 ≤ N ≤ 106, 1 ≤ M ≤ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 ≤ Di ≤ 106), the denominations of the coins. All denominations must be distinct: for any i ≠ j we must have Di ≠ Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139 | instruction | 0 | 42,581 | 10 | 85,162 |
Tags: constructive algorithms
Correct Solution:
```
n = int(input())
print('%d 2\n1 2' % (2 * n - 1))
``` | output | 1 | 42,581 | 10 | 85,163 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 ≤ A ≤ 105), the desired number of ways.
Output
In the first line print integers N and M (1 ≤ N ≤ 106, 1 ≤ M ≤ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 ≤ Di ≤ 106), the denominations of the coins. All denominations must be distinct: for any i ≠ j we must have Di ≠ Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139 | instruction | 0 | 42,582 | 10 | 85,164 |
Tags: constructive algorithms
Correct Solution:
```
print(2*int(input())-1,"2 1 2")
``` | output | 1 | 42,582 | 10 | 85,165 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 ≤ A ≤ 105), the desired number of ways.
Output
In the first line print integers N and M (1 ≤ N ≤ 106, 1 ≤ M ≤ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 ≤ Di ≤ 106), the denominations of the coins. All denominations must be distinct: for any i ≠ j we must have Di ≠ Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139 | instruction | 0 | 42,583 | 10 | 85,166 |
Tags: constructive algorithms
Correct Solution:
```
n = int(input())
print(n * 2 - 1, 2)
print("1 2")
``` | output | 1 | 42,583 | 10 | 85,167 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 ≤ A ≤ 105), the desired number of ways.
Output
In the first line print integers N and M (1 ≤ N ≤ 106, 1 ≤ M ≤ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 ≤ Di ≤ 106), the denominations of the coins. All denominations must be distinct: for any i ≠ j we must have Di ≠ Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139 | instruction | 0 | 42,584 | 10 | 85,168 |
Tags: constructive algorithms
Correct Solution:
```
a = int(input())
if(a == 1):
print(1, 1)
print(1)
exit()
n = (a-1)*2
print(n, 2)
print(1, 2)
``` | output | 1 | 42,584 | 10 | 85,169 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 ≤ A ≤ 105), the desired number of ways.
Output
In the first line print integers N and M (1 ≤ N ≤ 106, 1 ≤ M ≤ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 ≤ Di ≤ 106), the denominations of the coins. All denominations must be distinct: for any i ≠ j we must have Di ≠ Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139 | instruction | 0 | 42,585 | 10 | 85,170 |
Tags: constructive algorithms
Correct Solution:
```
a = int(input())
n = 2*(a-1)+1
m = 2
print(n, m)
print(1, 2)
``` | output | 1 | 42,585 | 10 | 85,171 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 ≤ A ≤ 105), the desired number of ways.
Output
In the first line print integers N and M (1 ≤ N ≤ 106, 1 ≤ M ≤ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 ≤ Di ≤ 106), the denominations of the coins. All denominations must be distinct: for any i ≠ j we must have Di ≠ Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139
Submitted Solution:
```
ways = int(input())
if ways==1:
print(1,1)
print(1)
else:
print(2*(ways-1),2)
print(1,2)
``` | instruction | 0 | 42,586 | 10 | 85,172 |
Yes | output | 1 | 42,586 | 10 | 85,173 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 ≤ A ≤ 105), the desired number of ways.
Output
In the first line print integers N and M (1 ≤ N ≤ 106, 1 ≤ M ≤ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 ≤ Di ≤ 106), the denominations of the coins. All denominations must be distinct: for any i ≠ j we must have Di ≠ Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139
Submitted Solution:
```
from sys import stdin, stdout
n = int(stdin.readline())
if n == 1:
stdout.write("2 1\n2")
else:
stdout.write(str((n - 1) * 2)+" 2\n")
stdout.write("1 2")
``` | instruction | 0 | 42,587 | 10 | 85,174 |
Yes | output | 1 | 42,587 | 10 | 85,175 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 ≤ A ≤ 105), the desired number of ways.
Output
In the first line print integers N and M (1 ≤ N ≤ 106, 1 ≤ M ≤ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 ≤ Di ≤ 106), the denominations of the coins. All denominations must be distinct: for any i ≠ j we must have Di ≠ Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139
Submitted Solution:
```
n=int(input())
print(2*n-1,'2')
print('1 2')
``` | instruction | 0 | 42,588 | 10 | 85,176 |
Yes | output | 1 | 42,588 | 10 | 85,177 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 ≤ A ≤ 105), the desired number of ways.
Output
In the first line print integers N and M (1 ≤ N ≤ 106, 1 ≤ M ≤ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 ≤ Di ≤ 106), the denominations of the coins. All denominations must be distinct: for any i ≠ j we must have Di ≠ Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139
Submitted Solution:
```
A = int(input())
if A == 1:
print("1 1")
print("1")
else:
print(str(2*(A - 1)) + " 2")
print("1 2")
``` | instruction | 0 | 42,589 | 10 | 85,178 |
Yes | output | 1 | 42,589 | 10 | 85,179 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 ≤ A ≤ 105), the desired number of ways.
Output
In the first line print integers N and M (1 ≤ N ≤ 106, 1 ≤ M ≤ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 ≤ Di ≤ 106), the denominations of the coins. All denominations must be distinct: for any i ≠ j we must have Di ≠ Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139
Submitted Solution:
```
a = int(input())
print((a-1)*10, end=' ')
print(2)
print(1, 10)
``` | instruction | 0 | 42,590 | 10 | 85,180 |
No | output | 1 | 42,590 | 10 | 85,181 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 ≤ A ≤ 105), the desired number of ways.
Output
In the first line print integers N and M (1 ≤ N ≤ 106, 1 ≤ M ≤ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 ≤ Di ≤ 106), the denominations of the coins. All denominations must be distinct: for any i ≠ j we must have Di ≠ Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139
Submitted Solution:
```
n = int(input())
number = 2*n -1
m = 2
denom = [1,2]
print(n,m)
print(*denom)
``` | instruction | 0 | 42,591 | 10 | 85,182 |
No | output | 1 | 42,591 | 10 | 85,183 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 ≤ A ≤ 105), the desired number of ways.
Output
In the first line print integers N and M (1 ≤ N ≤ 106, 1 ≤ M ≤ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 ≤ Di ≤ 106), the denominations of the coins. All denominations must be distinct: for any i ≠ j we must have Di ≠ Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139
Submitted Solution:
```
n=int(input())
print(2*(n-1),2)
print('1 2')
``` | instruction | 0 | 42,592 | 10 | 85,184 |
No | output | 1 | 42,592 | 10 | 85,185 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input
Input will consist of a single integer A (1 ≤ A ≤ 105), the desired number of ways.
Output
In the first line print integers N and M (1 ≤ N ≤ 106, 1 ≤ M ≤ 10), the amount of change to be made, and the number of denominations, respectively.
Then print M integers D1, D2, ..., DM (1 ≤ Di ≤ 106), the denominations of the coins. All denominations must be distinct: for any i ≠ j we must have Di ≠ Dj.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Examples
Input
18
Output
30 4
1 5 10 25
Input
3
Output
20 2
5 2
Input
314
Output
183 4
6 5 2 139
Submitted Solution:
```
n=int(input())
print((n-1)*2,2)
print(1,2)
``` | instruction | 0 | 42,593 | 10 | 85,186 |
No | output | 1 | 42,593 | 10 | 85,187 |
Provide a correct Python 3 solution for this coding contest problem.
The squirrel Chokudai has N acorns. One day, he decides to do some trades in multiple precious metal exchanges to make more acorns.
His plan is as follows:
1. Get out of the nest with N acorns in his hands.
2. Go to Exchange A and do some trades.
3. Go to Exchange B and do some trades.
4. Go to Exchange A and do some trades.
5. Go back to the nest.
In Exchange X (X = A, B), he can perform the following operations any integer number of times (possibly zero) in any order:
* Lose g_{X} acorns and gain 1 gram of gold.
* Gain g_{X} acorns and lose 1 gram of gold.
* Lose s_{X} acorns and gain 1 gram of silver.
* Gain s_{X} acorns and lose 1 gram of silver.
* Lose b_{X} acorns and gain 1 gram of bronze.
* Gain b_{X} acorns and lose 1 gram of bronze.
Naturally, he cannot perform an operation that would leave him with a negative amount of acorns, gold, silver, or bronze.
What is the maximum number of acorns that he can bring to the nest? Note that gold, silver, or bronze brought to the nest would be worthless because he is just a squirrel.
Constraints
* 1 \leq N \leq 5000
* 1 \leq g_{X} \leq 5000
* 1 \leq s_{X} \leq 5000
* 1 \leq b_{X} \leq 5000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
g_A s_A b_A
g_B s_B b_B
Output
Print the maximum number of acorns that Chokudai can bring to the nest.
Example
Input
23
1 1 1
2 1 1
Output
46 | instruction | 0 | 42,688 | 10 | 85,376 |
"Correct Solution:
```
N = int(input())
A = [int(a) for a in input().split()]
B = [int(a) for a in input().split()]
def calc(n, X, Y):
Z = [(X[i], Y[i]) for i in range(3) if Y[i] > X[i]]
k = len(Z)
if k == 0:
return n
if k == 1:
a = n//Z[0][0]
return n + a*(Z[0][1]-Z[0][0])
if k == 2:
ma = 0
for a in range(n+1):
b = (n-a*Z[0][0])//Z[1][0]
if b >= 0:
ma = max(ma, n + a*(Z[0][1]-Z[0][0]) + b*(Z[1][1]-Z[1][0]))
return ma
ma = 0
for a in range(n+1):
for b in range(n+1-a):
c = (n-a*Z[0][0]-b*Z[1][0])//Z[2][0]
if c >= 0:
ma = max(ma, n + a*(Z[0][1]-Z[0][0]) + b*(Z[1][1]-Z[1][0]) + c*(Z[2][1]-Z[2][0]))
return ma
print(calc(calc(N, A, B), B, A))
``` | output | 1 | 42,688 | 10 | 85,377 |
Provide a correct Python 3 solution for this coding contest problem.
The squirrel Chokudai has N acorns. One day, he decides to do some trades in multiple precious metal exchanges to make more acorns.
His plan is as follows:
1. Get out of the nest with N acorns in his hands.
2. Go to Exchange A and do some trades.
3. Go to Exchange B and do some trades.
4. Go to Exchange A and do some trades.
5. Go back to the nest.
In Exchange X (X = A, B), he can perform the following operations any integer number of times (possibly zero) in any order:
* Lose g_{X} acorns and gain 1 gram of gold.
* Gain g_{X} acorns and lose 1 gram of gold.
* Lose s_{X} acorns and gain 1 gram of silver.
* Gain s_{X} acorns and lose 1 gram of silver.
* Lose b_{X} acorns and gain 1 gram of bronze.
* Gain b_{X} acorns and lose 1 gram of bronze.
Naturally, he cannot perform an operation that would leave him with a negative amount of acorns, gold, silver, or bronze.
What is the maximum number of acorns that he can bring to the nest? Note that gold, silver, or bronze brought to the nest would be worthless because he is just a squirrel.
Constraints
* 1 \leq N \leq 5000
* 1 \leq g_{X} \leq 5000
* 1 \leq s_{X} \leq 5000
* 1 \leq b_{X} \leq 5000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
g_A s_A b_A
g_B s_B b_B
Output
Print the maximum number of acorns that Chokudai can bring to the nest.
Example
Input
23
1 1 1
2 1 1
Output
46 | instruction | 0 | 42,689 | 10 | 85,378 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
def small_dp(N,A,B):
*X, = filter(lambda c: int.__sub__(*c)<0, zip(A,B))
k = len(X)
if k == 0:
return N
if k == 1:
return X[0][1]*(N//X[0][0]) + N%X[0][0]
if k == 2:
c = 0
for i in range(N//X[0][0]+1):
j = (N-i*X[0][0])//X[1][0]
c = max(X[0][1]*i + X[1][1]*j + (N-i*X[0][0])%X[1][0], c)
return c
if k == 3:
if (N//X[1][0]+1)*(N//X[0][0]+1) <= 10**6 :
return max(X[0][1]*i + max(X[1][1]*j + X[2][1]*((N-i*X[0][0]-j*X[1][0])//X[2][0]) + ((N-i*X[0][0]-j*X[1][0])%X[2][0]) for j in range((N-i*X[0][0])//X[1][0]+1)) for i in range(N//X[0][0]+1))
else:
dp = [0]*(N+1)
for w in range(1,N+1):
dp[w] = max(dp[w-X[i][0]]+X[i][1] if w>=X[i][0] else w for i in range(k))
N = dp[-1]
return N
def solve():
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
res = small_dp(small_dp(N, A, B), B, A)
return str(res)
if __name__ == '__main__':
print(solve())
``` | output | 1 | 42,689 | 10 | 85,379 |
Provide a correct Python 3 solution for this coding contest problem.
The squirrel Chokudai has N acorns. One day, he decides to do some trades in multiple precious metal exchanges to make more acorns.
His plan is as follows:
1. Get out of the nest with N acorns in his hands.
2. Go to Exchange A and do some trades.
3. Go to Exchange B and do some trades.
4. Go to Exchange A and do some trades.
5. Go back to the nest.
In Exchange X (X = A, B), he can perform the following operations any integer number of times (possibly zero) in any order:
* Lose g_{X} acorns and gain 1 gram of gold.
* Gain g_{X} acorns and lose 1 gram of gold.
* Lose s_{X} acorns and gain 1 gram of silver.
* Gain s_{X} acorns and lose 1 gram of silver.
* Lose b_{X} acorns and gain 1 gram of bronze.
* Gain b_{X} acorns and lose 1 gram of bronze.
Naturally, he cannot perform an operation that would leave him with a negative amount of acorns, gold, silver, or bronze.
What is the maximum number of acorns that he can bring to the nest? Note that gold, silver, or bronze brought to the nest would be worthless because he is just a squirrel.
Constraints
* 1 \leq N \leq 5000
* 1 \leq g_{X} \leq 5000
* 1 \leq s_{X} \leq 5000
* 1 \leq b_{X} \leq 5000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
g_A s_A b_A
g_B s_B b_B
Output
Print the maximum number of acorns that Chokudai can bring to the nest.
Example
Input
23
1 1 1
2 1 1
Output
46 | instruction | 0 | 42,690 | 10 | 85,380 |
"Correct Solution:
```
n = int(input())
aa = list(map(int, input().split()))
bb = list(map(int, input().split()))
rates_list = [[],[]]
for i in range(3):
if aa[i] > bb[i]:
rates_list[1].append([bb[i], aa[i]])
else:
rates_list[0].append([aa[i], bb[i]])
max_d = n
for j in range(2):
rates = rates_list[j]
if len(rates) == 1:
max_d = (max_d // rates[0][0]) * rates[0][1] + max_d % rates[0][0]
elif len(rates) == 2:
val = [rates[0][1] * i for i in range(max_d // rates[0][0] + 1)]
rest = [max_d - rates[0][0] * i for i in range(max_d // rates[0][0] + 1)]
val = [val[i] + rates[1][1] * (rest[i] // rates[1][0]) + (rest[i] % rates[1][0]) if rest[i] > 0 else val[i] for i in range(len(val))]
max_d = max(val)
elif len(rates) == 3:
dp = [(d // rates[0][0]) * rates[0][1] + d % rates[0][0] for d in range(max_d + 1)]
for d in range(1, max_d + 1):
dp[d] = max([dp[d]] + [dp[d - rates[i][0]] + rates[i][1] if d - rates[i][0] >= 0 else d for i in range(1, len(rates))])
max_d = dp[-1]
print (max_d)
``` | output | 1 | 42,690 | 10 | 85,381 |
Provide a correct Python 3 solution for this coding contest problem.
The squirrel Chokudai has N acorns. One day, he decides to do some trades in multiple precious metal exchanges to make more acorns.
His plan is as follows:
1. Get out of the nest with N acorns in his hands.
2. Go to Exchange A and do some trades.
3. Go to Exchange B and do some trades.
4. Go to Exchange A and do some trades.
5. Go back to the nest.
In Exchange X (X = A, B), he can perform the following operations any integer number of times (possibly zero) in any order:
* Lose g_{X} acorns and gain 1 gram of gold.
* Gain g_{X} acorns and lose 1 gram of gold.
* Lose s_{X} acorns and gain 1 gram of silver.
* Gain s_{X} acorns and lose 1 gram of silver.
* Lose b_{X} acorns and gain 1 gram of bronze.
* Gain b_{X} acorns and lose 1 gram of bronze.
Naturally, he cannot perform an operation that would leave him with a negative amount of acorns, gold, silver, or bronze.
What is the maximum number of acorns that he can bring to the nest? Note that gold, silver, or bronze brought to the nest would be worthless because he is just a squirrel.
Constraints
* 1 \leq N \leq 5000
* 1 \leq g_{X} \leq 5000
* 1 \leq s_{X} \leq 5000
* 1 \leq b_{X} \leq 5000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
g_A s_A b_A
g_B s_B b_B
Output
Print the maximum number of acorns that Chokudai can bring to the nest.
Example
Input
23
1 1 1
2 1 1
Output
46 | instruction | 0 | 42,691 | 10 | 85,382 |
"Correct Solution:
```
N=int(input())
A=[int(i) for i in input().split()]
B=[int(i) for i in input().split()]
C=[(B[i]-A[i],A[i],B[i]) for i in range(3)]
C.sort()
ans=0
#print(C)
if C[2][0]<=0:
for i in range(3):
C[i]=(A[i]-B[i],B[i],A[i])
C.sort()
A,B=B,A
#print(A,B)
if C[0][0]>=0:
for p in range(N+1):
for q in range(N+1):
if p*A[0]+q*A[1]<=N:
r=(N-p*A[0]-q*A[1])//A[2]
s=N-(p*A[0]+q*A[1]+r*A[2])
ans=max(ans,p*B[0]+q*B[1]+r*B[2]+s)
elif C[0][0]<0 and C[1][0]>=0:
for p in range(N+1):
if p*C[1][1]<=N:
q=(N-C[1][1]*p)//C[2][1]
s=N-(p*C[1][1]+q*C[2][1])
n=p*C[1][2]+q*C[2][2]+s
r,m=n//C[0][2],n-C[0][2]*(n//C[0][2])
n=m+C[0][1]*r
ans=max(ans,n)
else:
r,m=N//C[2][1],N-C[2][1]*(N//C[2][1])
N_=m+C[2][2]*r
for p in range(N_+1):
if p*C[0][2]<=N_:
q=(N_-C[0][2]*p)//C[1][2]
s=N_-(p*C[0][2]+q*C[1][2])
n=p*C[0][1]+q*C[1][1]+s
#print(p,q,s,n)
ans=max(ans,n)
print(ans)
``` | output | 1 | 42,691 | 10 | 85,383 |
Provide a correct Python 3 solution for this coding contest problem.
The squirrel Chokudai has N acorns. One day, he decides to do some trades in multiple precious metal exchanges to make more acorns.
His plan is as follows:
1. Get out of the nest with N acorns in his hands.
2. Go to Exchange A and do some trades.
3. Go to Exchange B and do some trades.
4. Go to Exchange A and do some trades.
5. Go back to the nest.
In Exchange X (X = A, B), he can perform the following operations any integer number of times (possibly zero) in any order:
* Lose g_{X} acorns and gain 1 gram of gold.
* Gain g_{X} acorns and lose 1 gram of gold.
* Lose s_{X} acorns and gain 1 gram of silver.
* Gain s_{X} acorns and lose 1 gram of silver.
* Lose b_{X} acorns and gain 1 gram of bronze.
* Gain b_{X} acorns and lose 1 gram of bronze.
Naturally, he cannot perform an operation that would leave him with a negative amount of acorns, gold, silver, or bronze.
What is the maximum number of acorns that he can bring to the nest? Note that gold, silver, or bronze brought to the nest would be worthless because he is just a squirrel.
Constraints
* 1 \leq N \leq 5000
* 1 \leq g_{X} \leq 5000
* 1 \leq s_{X} \leq 5000
* 1 \leq b_{X} \leq 5000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
g_A s_A b_A
g_B s_B b_B
Output
Print the maximum number of acorns that Chokudai can bring to the nest.
Example
Input
23
1 1 1
2 1 1
Output
46 | instruction | 0 | 42,692 | 10 | 85,384 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
N=int(input())
ga,sa,ba=map(int,input().split())
gb,sb,bb=map(int,input().split())
ANS=N
def acorns(N,ga,sa,ba,gb,sb,bb):
X=[]
if ga<gb:
X.append((ga,gb))
if sa<sb:
X.append((sa,sb))
if ba<bb:
X.append((ba,bb))
if len(X)==0:
return N
if len(X)==1:
ga,gb=X[0][0],X[0][1]
t=N//ga
return N-t*ga+t*gb
if len(X)==2:
ga,gb=X[0][0],X[0][1]
sa,sb=X[1][0],X[1][1]
ANS=N
t=N//ga
for i in range(t+1):
M=i*gb
R=N-i*ga
u=R//sa
ANS=max(ANS,M+R+u*(sb-sa))
return ANS
if len(X)==3:
ga,gb=X[0][0],X[0][1]
sa,sb=X[1][0],X[1][1]
ba,bb=X[2][0],X[2][1]
ANS=N
t=N//ga
for i in range(t+1):
R=N-i*ga
u=R//sa
for j in range(u+1):
R2=R-j*sa
ANS=max(ANS,i*gb+j*sb+R2+R2//ba*(bb-ba))
return ANS
N2=acorns(N,ga,sa,ba,gb,sb,bb)
print(acorns(N2,gb,sb,bb,ga,sa,ba))
``` | output | 1 | 42,692 | 10 | 85,385 |
Provide a correct Python 3 solution for this coding contest problem.
The squirrel Chokudai has N acorns. One day, he decides to do some trades in multiple precious metal exchanges to make more acorns.
His plan is as follows:
1. Get out of the nest with N acorns in his hands.
2. Go to Exchange A and do some trades.
3. Go to Exchange B and do some trades.
4. Go to Exchange A and do some trades.
5. Go back to the nest.
In Exchange X (X = A, B), he can perform the following operations any integer number of times (possibly zero) in any order:
* Lose g_{X} acorns and gain 1 gram of gold.
* Gain g_{X} acorns and lose 1 gram of gold.
* Lose s_{X} acorns and gain 1 gram of silver.
* Gain s_{X} acorns and lose 1 gram of silver.
* Lose b_{X} acorns and gain 1 gram of bronze.
* Gain b_{X} acorns and lose 1 gram of bronze.
Naturally, he cannot perform an operation that would leave him with a negative amount of acorns, gold, silver, or bronze.
What is the maximum number of acorns that he can bring to the nest? Note that gold, silver, or bronze brought to the nest would be worthless because he is just a squirrel.
Constraints
* 1 \leq N \leq 5000
* 1 \leq g_{X} \leq 5000
* 1 \leq s_{X} \leq 5000
* 1 \leq b_{X} \leq 5000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
g_A s_A b_A
g_B s_B b_B
Output
Print the maximum number of acorns that Chokudai can bring to the nest.
Example
Input
23
1 1 1
2 1 1
Output
46 | instruction | 0 | 42,693 | 10 | 85,386 |
"Correct Solution:
```
N=int(input())
A=list(map(int,input().split()))
B=list(map(int,input().split()))
dp=[N]*(N+1)
#dp[i];=i個目まで選んで交換したときの最大
C=[B[i]-A[i] for i in range(3)]
for i in range(N):
dp[i+1]=max(dp[i+1],dp[i])
for j in range(3):
if i+1-A[j]>=0:
dp[i+1]=max(dp[i+1],dp[i+1-A[j]]+C[j])
M=dp[N]
dp=[0]*(M+1)
C=[A[i]-B[i] for i in range(3) if A[i]-B[i]>=0]
C=[]
b=[]
a=[]
for i in range(3):
if A[i]-B[i]>=0:
C.append(A[i]-B[i])
b.append(B[i])
a.append(A[i])
B=b
A=a
for i in range(M):
dp[i+1]=max(dp[i+1],dp[i])
for j in range(len(C)):
if i+1-B[j]>=0:
dp[i+1]=max(dp[i+1],dp[i+1-B[j]]+C[j])
print(M+dp[M])
``` | output | 1 | 42,693 | 10 | 85,387 |
Provide a correct Python 3 solution for this coding contest problem.
The squirrel Chokudai has N acorns. One day, he decides to do some trades in multiple precious metal exchanges to make more acorns.
His plan is as follows:
1. Get out of the nest with N acorns in his hands.
2. Go to Exchange A and do some trades.
3. Go to Exchange B and do some trades.
4. Go to Exchange A and do some trades.
5. Go back to the nest.
In Exchange X (X = A, B), he can perform the following operations any integer number of times (possibly zero) in any order:
* Lose g_{X} acorns and gain 1 gram of gold.
* Gain g_{X} acorns and lose 1 gram of gold.
* Lose s_{X} acorns and gain 1 gram of silver.
* Gain s_{X} acorns and lose 1 gram of silver.
* Lose b_{X} acorns and gain 1 gram of bronze.
* Gain b_{X} acorns and lose 1 gram of bronze.
Naturally, he cannot perform an operation that would leave him with a negative amount of acorns, gold, silver, or bronze.
What is the maximum number of acorns that he can bring to the nest? Note that gold, silver, or bronze brought to the nest would be worthless because he is just a squirrel.
Constraints
* 1 \leq N \leq 5000
* 1 \leq g_{X} \leq 5000
* 1 \leq s_{X} \leq 5000
* 1 \leq b_{X} \leq 5000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
g_A s_A b_A
g_B s_B b_B
Output
Print the maximum number of acorns that Chokudai can bring to the nest.
Example
Input
23
1 1 1
2 1 1
Output
46 | instruction | 0 | 42,694 | 10 | 85,388 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
N = int(input())
ga,sa,ba = map(int,input().split())
gb,sb,bb = map(int,input().split())
t1 = []
t2 = []
if ga < gb:
t1.append((ga,gb))
else:
t2.append((gb,ga))
if sa < sb:
t1.append((sa,sb))
else:
t2.append((sb,sa))
if ba < bb:
t1.append((ba,bb))
else:
t2.append((bb,ba))
dp = [0] * (N+1)
N1 = N
for i in range(1,N+1):
for c,r in t1:
if c <= i:
dp[i] = max(dp[i], dp[i-c] + r)
N1 = max(N1, N-i+dp[i])
ans = N1
dp = [0] * (N1+1)
for i in range(1,N1+1):
for c,r in t2:
if c <= i:
dp[i] = max(dp[i], dp[i-c] + r)
ans = max(ans, N1-i+dp[i])
print(ans)
``` | output | 1 | 42,694 | 10 | 85,389 |
Provide a correct Python 3 solution for this coding contest problem.
The squirrel Chokudai has N acorns. One day, he decides to do some trades in multiple precious metal exchanges to make more acorns.
His plan is as follows:
1. Get out of the nest with N acorns in his hands.
2. Go to Exchange A and do some trades.
3. Go to Exchange B and do some trades.
4. Go to Exchange A and do some trades.
5. Go back to the nest.
In Exchange X (X = A, B), he can perform the following operations any integer number of times (possibly zero) in any order:
* Lose g_{X} acorns and gain 1 gram of gold.
* Gain g_{X} acorns and lose 1 gram of gold.
* Lose s_{X} acorns and gain 1 gram of silver.
* Gain s_{X} acorns and lose 1 gram of silver.
* Lose b_{X} acorns and gain 1 gram of bronze.
* Gain b_{X} acorns and lose 1 gram of bronze.
Naturally, he cannot perform an operation that would leave him with a negative amount of acorns, gold, silver, or bronze.
What is the maximum number of acorns that he can bring to the nest? Note that gold, silver, or bronze brought to the nest would be worthless because he is just a squirrel.
Constraints
* 1 \leq N \leq 5000
* 1 \leq g_{X} \leq 5000
* 1 \leq s_{X} \leq 5000
* 1 \leq b_{X} \leq 5000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
g_A s_A b_A
g_B s_B b_B
Output
Print the maximum number of acorns that Chokudai can bring to the nest.
Example
Input
23
1 1 1
2 1 1
Output
46 | instruction | 0 | 42,695 | 10 | 85,390 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
import bisect
import heapq
import math
import random
import sys
from collections import Counter, defaultdict, deque
from decimal import ROUND_CEILING, ROUND_HALF_UP, Decimal
from functools import lru_cache, reduce
from itertools import combinations, combinations_with_replacement, product, permutations
from operator import add, mul, sub
sys.setrecursionlimit(100000)
input = sys.stdin.readline
def read_int():
return int(input())
def read_int_n():
return list(map(int, input().split()))
def read_float():
return float(input())
def read_float_n():
return list(map(float, input().split()))
def read_str():
return input().strip()
def read_str_n():
return list(map(str, input().split()))
def error_print(*args):
print(*args, file=sys.stderr)
def mt(f):
import time
def wrap(*args, **kwargs):
s = time.time()
ret = f(*args, **kwargs)
e = time.time()
error_print(e - s, 'sec')
return ret
return wrap
@mt
def slv(N, GSB):
def f(N, gsb):
dp = [[0] * (N+1) for _ in range(4)]
for i in range(3):
v = gsb[1][i] - gsb[0][i]
a = gsb[0][i]
if v > 0:
for w in range(N+1):
dp[i+1][w] = dp[i][w]
if w >= a:
dp[i+1][w] = max(dp[i][w], dp[i+1][w-a] + v)
else:
dp[i+1] = dp[i]
return dp[3][N] + N
tmp = f(N, GSB)
# print(tmp)
GSB[0], GSB[1] = GSB[1], GSB[0]
return f(tmp, GSB)
def main():
N = read_int()
GSB = [read_int_n() for _ in range(2)]
print(slv(N, GSB))
if __name__ == '__main__':
main()
``` | output | 1 | 42,695 | 10 | 85,391 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The squirrel Chokudai has N acorns. One day, he decides to do some trades in multiple precious metal exchanges to make more acorns.
His plan is as follows:
1. Get out of the nest with N acorns in his hands.
2. Go to Exchange A and do some trades.
3. Go to Exchange B and do some trades.
4. Go to Exchange A and do some trades.
5. Go back to the nest.
In Exchange X (X = A, B), he can perform the following operations any integer number of times (possibly zero) in any order:
* Lose g_{X} acorns and gain 1 gram of gold.
* Gain g_{X} acorns and lose 1 gram of gold.
* Lose s_{X} acorns and gain 1 gram of silver.
* Gain s_{X} acorns and lose 1 gram of silver.
* Lose b_{X} acorns and gain 1 gram of bronze.
* Gain b_{X} acorns and lose 1 gram of bronze.
Naturally, he cannot perform an operation that would leave him with a negative amount of acorns, gold, silver, or bronze.
What is the maximum number of acorns that he can bring to the nest? Note that gold, silver, or bronze brought to the nest would be worthless because he is just a squirrel.
Constraints
* 1 \leq N \leq 5000
* 1 \leq g_{X} \leq 5000
* 1 \leq s_{X} \leq 5000
* 1 \leq b_{X} \leq 5000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
g_A s_A b_A
g_B s_B b_B
Output
Print the maximum number of acorns that Chokudai can bring to the nest.
Example
Input
23
1 1 1
2 1 1
Output
46
Submitted Solution:
```
def solve(n, ga, sa, ba, gb, sb, bb):
gd, sd, bd = gb - ga, sb - sa, bb - ba
if bd <= 0:
if sd <= 0:
if gd <= 0:
return n
return n + n // ga * gd
if gd <= 0:
return n + n // sa * sd
return max(n + g * gd + (n - g * ga) // sa * sd for g in range(n // ga + 1))
if sd <= 0:
if gd <= 0:
return n + n // ba * bd
return max(n + g * gd + (n - g * ga) // ba * bd for g in range(n // ga + 1))
if gd <= 0:
return max(n + s * sd + (n - s * sa) // ba * bd for s in range(n // sa + 1))
(_, ga, gd), (_, sa, sd), (_, ba, bd) = \
sorted([(gb / ga, -ga, gd), (sb / sa, -sa, sd), (bb / ba, -ba, bd)], reverse=True)
ga, sa, ba = -ga, -sa, -ba
sga = 0 if n // ga < 1000 else n // ga // 2
return max(n + g * gd + s * sd + (n - g * ga - s * sa) // ba * bd
for g in range(sga, n // ga + 1)
for s in range((n - g * ga) // sa + 1))
n = int(input())
ga, sa, ba = list(map(int, input().split()))
gb, sb, bb = list(map(int, input().split()))
n2 = solve(n, ga, sa, ba, gb, sb, bb)
n3 = solve(n2, gb, sb, bb, ga, sa, ba)
print(n3)
``` | instruction | 0 | 42,696 | 10 | 85,392 |
Yes | output | 1 | 42,696 | 10 | 85,393 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The squirrel Chokudai has N acorns. One day, he decides to do some trades in multiple precious metal exchanges to make more acorns.
His plan is as follows:
1. Get out of the nest with N acorns in his hands.
2. Go to Exchange A and do some trades.
3. Go to Exchange B and do some trades.
4. Go to Exchange A and do some trades.
5. Go back to the nest.
In Exchange X (X = A, B), he can perform the following operations any integer number of times (possibly zero) in any order:
* Lose g_{X} acorns and gain 1 gram of gold.
* Gain g_{X} acorns and lose 1 gram of gold.
* Lose s_{X} acorns and gain 1 gram of silver.
* Gain s_{X} acorns and lose 1 gram of silver.
* Lose b_{X} acorns and gain 1 gram of bronze.
* Gain b_{X} acorns and lose 1 gram of bronze.
Naturally, he cannot perform an operation that would leave him with a negative amount of acorns, gold, silver, or bronze.
What is the maximum number of acorns that he can bring to the nest? Note that gold, silver, or bronze brought to the nest would be worthless because he is just a squirrel.
Constraints
* 1 \leq N \leq 5000
* 1 \leq g_{X} \leq 5000
* 1 \leq s_{X} \leq 5000
* 1 \leq b_{X} \leq 5000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
g_A s_A b_A
g_B s_B b_B
Output
Print the maximum number of acorns that Chokudai can bring to the nest.
Example
Input
23
1 1 1
2 1 1
Output
46
Submitted Solution:
```
import sys
#input = sys.stdin.readline
from sys import setrecursionlimit
setrecursionlimit(10**9)
def inpl(): return list(map(int, input().split()))
N = int(input())
GA = inpl()
GB = inpl()
typ = [GA[i] <= GB[i] for i in range(3)]
if sum(typ) == 3:
ans = 0
for g in range(N//GA[0] + 1):
D = N - GA[0]*g
for s in range(D//GA[1] + 1):
b, D2 = divmod(D - GA[1]*s, GA[2])
ans = max(ans, D2 + g*GB[0] + s*GB[1] + b*GB[2])
elif sum(typ) == 2:
ans = 0
x, y = [i for i in range(3) if typ[i]]
z = [i for i in range(3) if not typ[i]][0]
for a in range(N//GA[x] + 1):
b, D = divmod(N - GA[x]*a, GA[y])
c, D2 = divmod(D + GB[x]*a + GB[y]*b, GB[z])
ans = max(ans, D2 + c*GA[z])
elif sum(typ) == 1:
ans = 0
x, y = [i for i in range(3) if not typ[i]]
z = [i for i in range(3) if typ[i]][0]
c, D = divmod(N, GA[z])
D += GB[z]*c
for a in range(D//GB[x] + 1):
b, D2 = divmod(D-GB[x]*a, GB[y])
ans = max(ans, D2 + GA[x]*a + GA[y]*b)
for b in range(D//GB[y] + 1):
a, D2 = divmod(D-GB[y]*b, GB[x])
ans = max(ans, D2 + GA[x]*a + GA[y]*b)
else:
ans = 0
for g in range(N//GB[0] + 1):
D = N - GB[0]*g
for s in range(D//GB[1] + 1):
b, D2 = divmod(D - GB[1]*s, GB[2])
ans = max(ans, D2 + g*GA[0] + s*GA[1] + b*GA[2])
print(ans)
``` | instruction | 0 | 42,697 | 10 | 85,394 |
Yes | output | 1 | 42,697 | 10 | 85,395 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The squirrel Chokudai has N acorns. One day, he decides to do some trades in multiple precious metal exchanges to make more acorns.
His plan is as follows:
1. Get out of the nest with N acorns in his hands.
2. Go to Exchange A and do some trades.
3. Go to Exchange B and do some trades.
4. Go to Exchange A and do some trades.
5. Go back to the nest.
In Exchange X (X = A, B), he can perform the following operations any integer number of times (possibly zero) in any order:
* Lose g_{X} acorns and gain 1 gram of gold.
* Gain g_{X} acorns and lose 1 gram of gold.
* Lose s_{X} acorns and gain 1 gram of silver.
* Gain s_{X} acorns and lose 1 gram of silver.
* Lose b_{X} acorns and gain 1 gram of bronze.
* Gain b_{X} acorns and lose 1 gram of bronze.
Naturally, he cannot perform an operation that would leave him with a negative amount of acorns, gold, silver, or bronze.
What is the maximum number of acorns that he can bring to the nest? Note that gold, silver, or bronze brought to the nest would be worthless because he is just a squirrel.
Constraints
* 1 \leq N \leq 5000
* 1 \leq g_{X} \leq 5000
* 1 \leq s_{X} \leq 5000
* 1 \leq b_{X} \leq 5000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
g_A s_A b_A
g_B s_B b_B
Output
Print the maximum number of acorns that Chokudai can bring to the nest.
Example
Input
23
1 1 1
2 1 1
Output
46
Submitted Solution:
```
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
Achange = []
Bchange = []
for i in range(3):
if A[i] > B[i]:
Bchange.append(i)
elif A[i] < B[i]:
Achange.append(i)
#print(Achange, Bchange)
# 最初のA交換所→B交換所
Dmax1 = N
if Achange:
if len(Achange)==3:
i = Achange[0]
j = Achange[1]
k = Achange[2]
for x in range(N+1):
for y in range(N+1):
if A[i]*x+A[j]*y <= N:
Dmax1 = max(Dmax1, B[i]*x + B[j]*y + (N-A[i]*x-A[j]*y)%A[k] + (N-A[i]*x-A[j]*y)//A[k] * B[k])
if len(Achange)==2:
i = Achange[0]
j = Achange[1]
for x in range(N+1):
if A[i]*x <= N:
Dmax1 = max(Dmax1, B[i]*x + (N-A[i]*x)%A[j] + (N-A[i]*x)//A[j] * B[j])
if len(Achange)==1:
i = Achange[0]
Dmax1 = max(Dmax1, N%A[i] + (N//A[i] * B[i]))
#print(Dmax1)
#print(Achange, Bchange)
#次のB交換所→A交換所
Achange = Bchange
A,B=B,A
N = Dmax1
#print(Achange, Bchange)
if Achange:
if len(Achange)==3:
i = Achange[0]
j = Achange[1]
k = Achange[2]
for x in range(N+1):
for y in range(N+1):
if A[i]*x+A[j]*y <= N:
Dmax1 = max(Dmax1, B[i]*x + B[j]*y + (N-A[i]*x-A[j]*y)%A[k] + (N-A[i]*x-A[j]*y)//A[k] * B[k])
if len(Achange)==2:
i = Achange[0]
j = Achange[1]
for x in range(N+1):
if A[i]*x <= N:
Dmax1 = max(Dmax1, B[i]*x + (N-A[i]*x)%A[j] + (N-A[i]*x)//A[j] * B[j])
if len(Achange)==1:
i = Achange[0]
Dmax1 = max(Dmax1, N%A[i] + (N//A[i] * B[i]))
print(Dmax1)
``` | instruction | 0 | 42,698 | 10 | 85,396 |
Yes | output | 1 | 42,698 | 10 | 85,397 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The squirrel Chokudai has N acorns. One day, he decides to do some trades in multiple precious metal exchanges to make more acorns.
His plan is as follows:
1. Get out of the nest with N acorns in his hands.
2. Go to Exchange A and do some trades.
3. Go to Exchange B and do some trades.
4. Go to Exchange A and do some trades.
5. Go back to the nest.
In Exchange X (X = A, B), he can perform the following operations any integer number of times (possibly zero) in any order:
* Lose g_{X} acorns and gain 1 gram of gold.
* Gain g_{X} acorns and lose 1 gram of gold.
* Lose s_{X} acorns and gain 1 gram of silver.
* Gain s_{X} acorns and lose 1 gram of silver.
* Lose b_{X} acorns and gain 1 gram of bronze.
* Gain b_{X} acorns and lose 1 gram of bronze.
Naturally, he cannot perform an operation that would leave him with a negative amount of acorns, gold, silver, or bronze.
What is the maximum number of acorns that he can bring to the nest? Note that gold, silver, or bronze brought to the nest would be worthless because he is just a squirrel.
Constraints
* 1 \leq N \leq 5000
* 1 \leq g_{X} \leq 5000
* 1 \leq s_{X} \leq 5000
* 1 \leq b_{X} \leq 5000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
g_A s_A b_A
g_B s_B b_B
Output
Print the maximum number of acorns that Chokudai can bring to the nest.
Example
Input
23
1 1 1
2 1 1
Output
46
Submitted Solution:
```
###############################################################################
import sys
from sys import stdout
from bisect import bisect_left as binl
from copy import copy, deepcopy
from collections import defaultdict
import math
mod = 1
def intin():
input_tuple = input().split()
if len(input_tuple) <= 1:
return int(input_tuple[0])
return tuple(map(int, input_tuple))
def intina():
return [int(i) for i in input().split()]
def intinl(count):
return [intin() for _ in range(count)]
def modadd(x, y):
global mod
return (x + y) % mod
def modmlt(x, y):
global mod
return (x * y) % mod
def mod_inv(x):
# available only when mod is prime
global mod
return pow(x, mod - 2, mod)
def gcm(x, y):
while y != 0:
z = x % y
x = y
y = z
return x
def combination(x, y):
assert(x >= y)
ret = math.factorial(x)
ret = ret // (math.factorial(x - y) * math.factorial(y))
return ret
def get_divisors(x):
retlist = []
for i in range(1, int(x**0.5) + 3):
if x % i == 0:
retlist.append(i)
retlist.append(x // i)
return retlist
def get_factors(x):
retlist = []
for i in range(2, int(x**0.5) + 3):
while x % i == 0:
retlist.append(i)
x = x // i
retlist.append(x)
return retlist
def make_linklist(xylist):
linklist = {}
for a, b in xylist:
linklist.setdefault(a, [])
linklist.setdefault(b, [])
linklist[a].append(b)
linklist[b].append(a)
return linklist
def calc_longest_distance(linklist, v=1):
distance_list = {}
distance_count = 0
distance = 0
vlist_previous = []
vlist = [v]
nodecount = len(linklist)
while distance_count < nodecount:
vlist_next = []
for v in vlist:
distance_list[v] = distance
distance_count += 1
vlist_next.extend(linklist[v])
distance += 1
vlist_to_del = vlist_previous
vlist_previous = vlist
vlist = list(set(vlist_next) - set(vlist_to_del))
max_distance = -1
max_v = None
for v, distance in distance_list.items():
if distance > max_distance:
max_distance = distance
max_v = v
return (max_distance, max_v)
def calc_tree_diameter(linklist, v=1):
_, u = calc_longest_distance(linklist, v)
distance, _ = calc_longest_distance(linklist, u)
return distance
class UnionFind:
def __init__(self, n):
self.parent = [i for i in range(n)]
def root(self, i):
if self.parent[i] == i:
return i
self.parent[i] = self.root(self.parent[i])
return self.parent[i]
def unite(self, i, j):
rooti = self.root(i)
rootj = self.root(j)
if rooti == rootj:
return
if rooti < rootj:
self.parent[rootj] = rooti
else:
self.parent[rooti] = rootj
def same(self, i, j):
return self.root(i) == self.root(j)
###############################################################################
wvalue = []
cache = {}
def f(n, wvaluelen):
global cache, wvalue
if (n, wvaluelen) in cache:
return cache[(n, wvaluelen)]
if wvaluelen == 0:
return 0
if wvaluelen == 1:
w, value = wvalue[0]
ret = value * (n // w)
cache[(n, wvaluelen)] = ret
return ret
ret = -1
w, value = wvalue[wvaluelen-1]
for i in range(n // w + 1):
ret = max(ret, i * value + f(n - i * w, wvaluelen - 1))
cache[(n, wvaluelen)] = ret
return ret
def main():
global cache, wvalue
sys.setrecursionlimit(10000)
n = intin()
gsb_a = intina()
gsb_b = intina()
ga = gsb_a[0]
sa = gsb_a[1]
ba = gsb_a[2]
gb = gsb_b[0]
sb = gsb_b[1]
bb = gsb_b[2]
# a: buy
if gb - ga > 0:
wvalue.append((ga, gb - ga))
if sb - sa > 0:
wvalue.append((sa, sb - sa))
if bb - ba > 0:
wvalue.append((ba, bb - ba))
value = f(n, len(wvalue))
cache = {}
wvalue = []
# b: sell
n += value
# b: buy
if ga - gb > 0:
wvalue.append((gb, ga - gb))
if sa - sb > 0:
wvalue.append((sb, sa - sb))
if ba - bb > 0:
wvalue.append((bb, ba - bb))
value = f(n, len(wvalue))
cache = {}
wvalue = []
# a: sell
n += value
print(n)
if __name__ == '__main__':
main()
``` | instruction | 0 | 42,699 | 10 | 85,398 |
Yes | output | 1 | 42,699 | 10 | 85,399 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The squirrel Chokudai has N acorns. One day, he decides to do some trades in multiple precious metal exchanges to make more acorns.
His plan is as follows:
1. Get out of the nest with N acorns in his hands.
2. Go to Exchange A and do some trades.
3. Go to Exchange B and do some trades.
4. Go to Exchange A and do some trades.
5. Go back to the nest.
In Exchange X (X = A, B), he can perform the following operations any integer number of times (possibly zero) in any order:
* Lose g_{X} acorns and gain 1 gram of gold.
* Gain g_{X} acorns and lose 1 gram of gold.
* Lose s_{X} acorns and gain 1 gram of silver.
* Gain s_{X} acorns and lose 1 gram of silver.
* Lose b_{X} acorns and gain 1 gram of bronze.
* Gain b_{X} acorns and lose 1 gram of bronze.
Naturally, he cannot perform an operation that would leave him with a negative amount of acorns, gold, silver, or bronze.
What is the maximum number of acorns that he can bring to the nest? Note that gold, silver, or bronze brought to the nest would be worthless because he is just a squirrel.
Constraints
* 1 \leq N \leq 5000
* 1 \leq g_{X} \leq 5000
* 1 \leq s_{X} \leq 5000
* 1 \leq b_{X} \leq 5000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
g_A s_A b_A
g_B s_B b_B
Output
Print the maximum number of acorns that Chokudai can bring to the nest.
Example
Input
23
1 1 1
2 1 1
Output
46
Submitted Solution:
```
n = int(input())
ra = [ int(v) for v in input().split() ]
rb = [ int(v) for v in input().split() ]
ab_rate = sorted([ (rb[i] / ra[i],i,ra[i],rb[i]) for i in range(3)],reverse = True)
ba_rate = sorted([ (ra[i] / rb[i],i,rb[i],ra[i]) for i in range(3)],reverse = True)
def excange(r,x):
if r[0] <= 1:
return x, 0
else:
y = (x // r[2]) * r[3]
x = x % r[2]
return x, y
prof,amari = 0, n
for i in range(3):
res = excange(ab_rate[i],amari)
amari = res[0]
prof += res[1]
prof,amari = 0, prof+amari
for i in range(3):
res = excange(ba_rate[i],amari)
amari = res[0]
prof += res[1]
print(prof+amari)
``` | instruction | 0 | 42,700 | 10 | 85,400 |
No | output | 1 | 42,700 | 10 | 85,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The squirrel Chokudai has N acorns. One day, he decides to do some trades in multiple precious metal exchanges to make more acorns.
His plan is as follows:
1. Get out of the nest with N acorns in his hands.
2. Go to Exchange A and do some trades.
3. Go to Exchange B and do some trades.
4. Go to Exchange A and do some trades.
5. Go back to the nest.
In Exchange X (X = A, B), he can perform the following operations any integer number of times (possibly zero) in any order:
* Lose g_{X} acorns and gain 1 gram of gold.
* Gain g_{X} acorns and lose 1 gram of gold.
* Lose s_{X} acorns and gain 1 gram of silver.
* Gain s_{X} acorns and lose 1 gram of silver.
* Lose b_{X} acorns and gain 1 gram of bronze.
* Gain b_{X} acorns and lose 1 gram of bronze.
Naturally, he cannot perform an operation that would leave him with a negative amount of acorns, gold, silver, or bronze.
What is the maximum number of acorns that he can bring to the nest? Note that gold, silver, or bronze brought to the nest would be worthless because he is just a squirrel.
Constraints
* 1 \leq N \leq 5000
* 1 \leq g_{X} \leq 5000
* 1 \leq s_{X} \leq 5000
* 1 \leq b_{X} \leq 5000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
g_A s_A b_A
g_B s_B b_B
Output
Print the maximum number of acorns that Chokudai can bring to the nest.
Example
Input
23
1 1 1
2 1 1
Output
46
Submitted Solution:
```
N = int(input())
A = tuple(map(int, input().split()))
B = tuple(map(int, input().split()))
if len([0 for a, b in zip(A, B) if a != b]) == 0:
print(N)
exit()
if len([0 for a, b in zip(A, B) if a > b]) == 1:
M = N
i = [i for i, (a, b) in zip(A, B) if a > b][0]
for k in range(N // A[i] + 1):
D = N - k * A[i]
D += k * B[i]
M = max(M, D)
i, j = [i for i, (a, b) in zip(A, B) if not a > b]
ans = M
for k in range(N // B[i] + 1):
D = M - k * B[i]
l, D = divmod(D, B[j])
D += k * A[i]
D += l * A[j]
ans = max(ans, D)
print(ans)
exit()
if len([0 for a, b in zip(A, B) if a > b]) == 2:
M = N
i, j = [i for i, (a, b) in zip(A, B) if not a > b]
for k in range(N // B[i] + 1):
D = N - k * B[i]
l, D = divmod(D, B[j])
D += k * A[i]
D += l * A[j]
M = max(M, D)
i = [i for i, (a, b) in zip(A, B) if a > b][0]
ans = M
for k in range(N // A[i] + 1):
D = M - k * A[i]
D += k * B[i]
ans = max(ans, D)
print(ans)
exit()
def calc(fr, to):
ret = N
for g in range(N // fr[0] + 1):
for s in range(N // fr[1] + 1):
D = N - g * fr[0] - s * fr[1]
if D < 0:
break
q, r = divmod(D, fr[2])
D += g * to[0]
D += s * to[1]
r += g * to[0]
r += s * to[1]
r += q * to[2]
ret = max(ret, D, r)
return ret
print(calc(A, B))
``` | instruction | 0 | 42,701 | 10 | 85,402 |
No | output | 1 | 42,701 | 10 | 85,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The squirrel Chokudai has N acorns. One day, he decides to do some trades in multiple precious metal exchanges to make more acorns.
His plan is as follows:
1. Get out of the nest with N acorns in his hands.
2. Go to Exchange A and do some trades.
3. Go to Exchange B and do some trades.
4. Go to Exchange A and do some trades.
5. Go back to the nest.
In Exchange X (X = A, B), he can perform the following operations any integer number of times (possibly zero) in any order:
* Lose g_{X} acorns and gain 1 gram of gold.
* Gain g_{X} acorns and lose 1 gram of gold.
* Lose s_{X} acorns and gain 1 gram of silver.
* Gain s_{X} acorns and lose 1 gram of silver.
* Lose b_{X} acorns and gain 1 gram of bronze.
* Gain b_{X} acorns and lose 1 gram of bronze.
Naturally, he cannot perform an operation that would leave him with a negative amount of acorns, gold, silver, or bronze.
What is the maximum number of acorns that he can bring to the nest? Note that gold, silver, or bronze brought to the nest would be worthless because he is just a squirrel.
Constraints
* 1 \leq N \leq 5000
* 1 \leq g_{X} \leq 5000
* 1 \leq s_{X} \leq 5000
* 1 \leq b_{X} \leq 5000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
g_A s_A b_A
g_B s_B b_B
Output
Print the maximum number of acorns that Chokudai can bring to the nest.
Example
Input
23
1 1 1
2 1 1
Output
46
Submitted Solution:
```
#D問題
N = int(input())
GSB1 = list(map(int,input().split()))
GSB2 = list(map(int,input().split()))
gper = float(GSB2[0]/GSB1[0])
sper = float(GSB2[1]/GSB1[1])
bper = float(GSB2[2]/GSB1[2])
if gper >= sper and gper >= bper:
n1 = N//GSB1[0]
N-=n1
if sper >= bper:
n2 = N//GSB1[1]
n3 = N//GSB1[2]
N+=n3*GSB2[2]
N+=n2*GSB2[1]
else:
n2 = N//GSB1[2]
n3 = N//GSB1[1]
N+=n3*GSB2[1]
N+=n2*GSB2[2]
N+=n1*GSB2[0]
elif sper >= gper and sper >= bper:
n1 = N//GSB1[1]
N-=n1
if gper >= bper:
n2 = N//GSB1[0]
n3 = N//GSB1[2]
N+=n3*GSB2[2]
N+=n2*GSB2[0]
else:
n2 = N//GSB1[2]
n3 = N//GSB1[0]
N+=n3*GSB2[0]
N+=n2*GSB2[2]
N+=n1*GSB2[1]
elif bper >= sper and bper >= gper:
n1 = N//GSB1[2]
N-=n1
if gper >= sper:
n2 = N//GSB1[0]
n3 = N//GSB1[1]
N+=n3*GSB2[1]
N+=n2*GSB2[0]
else:
n2 = N//GSB1[1]
n3 = N//GSB1[0]
N+=n3*GSB2[0]
N+=n2*GSB2[1]
N+=n1*GSB2[2]
gper = float(GSB1[0]/GSB2[0])
sper = float(GSB1[1]/GSB2[1])
bper = float(GSB1[2]/GSB2[2])
if gper >= sper and gper >= bper:
n1 = N//GSB2[0]
N-=n1
if sper >= bper:
n2 = N//GSB2[1]
n3 = N//GSB2[2]
N+=n3*GSB1[2]
N+=n2*GSB1[1]
else:
n2 = N//GSB2[2]
n3 = N//GSB2[1]
N+=n3*GSB1[1]
N+=n2*GSB1[2]
N+=n1*GSB1[0]
elif sper >= gper and sper >= bper:
n1 = N//GSB2[1]
N-=n1
if gper >= bper:
n2 = N//GSB2[0]
n3 = N//GSB2[2]
N+=n3*GSB1[2]
N+=n2*GSB1[0]
else:
n2 = N//GSB2[2]
n3 = N//GSB2[0]
N+=n3*GSB1[0]
N+=n2*GSB1[2]
N+=n1*GSB1[1]
elif bper >= sper and bper >= gper:
n1 = N//GSB2[2]
N-=n1
if gper >= sper:
n2 = N//GSB2[0]
n3 = N//GSB2[1]
N+=n3*GSB1[1]
N+=n2*GSB1[0]
else:
n2 = N//GSB2[1]
n3 = N//GSB2[0]
N+=n3*GSB1[0]
N+=n2*GSB1[1]
N+=n1*GSB1[2]
print(N)
``` | instruction | 0 | 42,702 | 10 | 85,404 |
No | output | 1 | 42,702 | 10 | 85,405 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The squirrel Chokudai has N acorns. One day, he decides to do some trades in multiple precious metal exchanges to make more acorns.
His plan is as follows:
1. Get out of the nest with N acorns in his hands.
2. Go to Exchange A and do some trades.
3. Go to Exchange B and do some trades.
4. Go to Exchange A and do some trades.
5. Go back to the nest.
In Exchange X (X = A, B), he can perform the following operations any integer number of times (possibly zero) in any order:
* Lose g_{X} acorns and gain 1 gram of gold.
* Gain g_{X} acorns and lose 1 gram of gold.
* Lose s_{X} acorns and gain 1 gram of silver.
* Gain s_{X} acorns and lose 1 gram of silver.
* Lose b_{X} acorns and gain 1 gram of bronze.
* Gain b_{X} acorns and lose 1 gram of bronze.
Naturally, he cannot perform an operation that would leave him with a negative amount of acorns, gold, silver, or bronze.
What is the maximum number of acorns that he can bring to the nest? Note that gold, silver, or bronze brought to the nest would be worthless because he is just a squirrel.
Constraints
* 1 \leq N \leq 5000
* 1 \leq g_{X} \leq 5000
* 1 \leq s_{X} \leq 5000
* 1 \leq b_{X} \leq 5000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
g_A s_A b_A
g_B s_B b_B
Output
Print the maximum number of acorns that Chokudai can bring to the nest.
Example
Input
23
1 1 1
2 1 1
Output
46
Submitted Solution:
```
def gcd(a,b):
if b>a:
a,b=b,a
while b:
a,b=b, a%b
return a
def f(a,b,N):
lcm=a[0]*a[1]//gcd(a[0],a[1])
lcm=lcm*a[2]//gcd(a[2],lcm)
F=[[a[i],b[i]] for i in range(3)]
F.sort()
a=sorted([F[i][0] for i in range(3)],reverse=True)
b=sorted([F[i][1] for i in range(3)],reverse=True)
n1,n2=N-N%lcm,N%lcm
ans1=max(n1*b[i]//a[i] for i in range(3))
ans2=n2
i=0
while i*a[0]<=n2:
j=0
while i*a[0]+j*a[1]<=n2:
k=(n2-i*a[0]-j*a[1])//a[2]
ans2=max(ans2, i*b[0]+j*b[1]+k*b[2] + (n2-i*a[0]-j*a[1]-k*a[2]) )
j+=1
i+=1
return ans1+ans2
import sys
N=int(input())
A=list(map(int,input().split()))
B=list(map(int,input().split()))
N=f(A,B,N)
N=f(B,A,N)
print(int(N))
``` | instruction | 0 | 42,703 | 10 | 85,406 |
No | output | 1 | 42,703 | 10 | 85,407 |
Provide a correct Python 3 solution for this coding contest problem.
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total?
Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
Constraints
* 0 \leq A, B, C \leq 50
* A + B + C \geq 1
* 50 \leq X \leq 20 000
* A, B and C are integers.
* X is a multiple of 50.
Input
Input is given from Standard Input in the following format:
A
B
C
X
Output
Print the number of ways to select coins.
Examples
Input
2
2
2
100
Output
2
Input
5
1
0
150
Output
0
Input
30
40
50
6000
Output
213 | instruction | 0 | 42,736 | 10 | 85,472 |
"Correct Solution:
```
a=int(input())
b=int(input())
c=int(input())
x=int(input())
print([500*i+100*j+50*k for i in range(a+1) for j in range(b+1) for k in range(c+1)].count(x))
``` | output | 1 | 42,736 | 10 | 85,473 |
Provide a correct Python 3 solution for this coding contest problem.
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total?
Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
Constraints
* 0 \leq A, B, C \leq 50
* A + B + C \geq 1
* 50 \leq X \leq 20 000
* A, B and C are integers.
* X is a multiple of 50.
Input
Input is given from Standard Input in the following format:
A
B
C
X
Output
Print the number of ways to select coins.
Examples
Input
2
2
2
100
Output
2
Input
5
1
0
150
Output
0
Input
30
40
50
6000
Output
213 | instruction | 0 | 42,737 | 10 | 85,474 |
"Correct Solution:
```
a,b,c,x=[int(input()) for i in range(4)]
print(sum(0<=(x-500*A-100*B)//50<=c for A in range(a+1) for B in range(b+1)))
``` | output | 1 | 42,737 | 10 | 85,475 |
Provide a correct Python 3 solution for this coding contest problem.
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total?
Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
Constraints
* 0 \leq A, B, C \leq 50
* A + B + C \geq 1
* 50 \leq X \leq 20 000
* A, B and C are integers.
* X is a multiple of 50.
Input
Input is given from Standard Input in the following format:
A
B
C
X
Output
Print the number of ways to select coins.
Examples
Input
2
2
2
100
Output
2
Input
5
1
0
150
Output
0
Input
30
40
50
6000
Output
213 | instruction | 0 | 42,738 | 10 | 85,476 |
"Correct Solution:
```
a,b,c,x=[int(input())+1 for _ in range(4)]
print(sum([1 for i in range(a) for j in range(b) for k in range(c) if 500*i+100*j+50*k==x-1]))
``` | output | 1 | 42,738 | 10 | 85,477 |
Provide a correct Python 3 solution for this coding contest problem.
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total?
Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
Constraints
* 0 \leq A, B, C \leq 50
* A + B + C \geq 1
* 50 \leq X \leq 20 000
* A, B and C are integers.
* X is a multiple of 50.
Input
Input is given from Standard Input in the following format:
A
B
C
X
Output
Print the number of ways to select coins.
Examples
Input
2
2
2
100
Output
2
Input
5
1
0
150
Output
0
Input
30
40
50
6000
Output
213 | instruction | 0 | 42,739 | 10 | 85,478 |
"Correct Solution:
```
A=int(input())
B=int(input())
C=int(input())
X=int(input())
a=0
for i in range(A+1):
for j in range(B+1):
if 0<=X-500*i-100*j<=50*C:
a+=1
print(a)
``` | output | 1 | 42,739 | 10 | 85,479 |
Provide a correct Python 3 solution for this coding contest problem.
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total?
Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
Constraints
* 0 \leq A, B, C \leq 50
* A + B + C \geq 1
* 50 \leq X \leq 20 000
* A, B and C are integers.
* X is a multiple of 50.
Input
Input is given from Standard Input in the following format:
A
B
C
X
Output
Print the number of ways to select coins.
Examples
Input
2
2
2
100
Output
2
Input
5
1
0
150
Output
0
Input
30
40
50
6000
Output
213 | instruction | 0 | 42,740 | 10 | 85,480 |
"Correct Solution:
```
A, B, C, X = [int(input()) for i in range(4)]
print(sum([500*a+100*b+50*c == X for a in range(A+1) for b in range(B+1) for c in range(C+1)]))
``` | output | 1 | 42,740 | 10 | 85,481 |
Provide a correct Python 3 solution for this coding contest problem.
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total?
Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
Constraints
* 0 \leq A, B, C \leq 50
* A + B + C \geq 1
* 50 \leq X \leq 20 000
* A, B and C are integers.
* X is a multiple of 50.
Input
Input is given from Standard Input in the following format:
A
B
C
X
Output
Print the number of ways to select coins.
Examples
Input
2
2
2
100
Output
2
Input
5
1
0
150
Output
0
Input
30
40
50
6000
Output
213 | instruction | 0 | 42,741 | 10 | 85,482 |
"Correct Solution:
```
a,b,c,x=[1+int(input())for i in[0]*4];print(sum(1if~-x==500*i+100*j+50*k else 0for k in range(c)for j in range(b)for i in range(a)))
``` | output | 1 | 42,741 | 10 | 85,483 |
Provide a correct Python 3 solution for this coding contest problem.
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total?
Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
Constraints
* 0 \leq A, B, C \leq 50
* A + B + C \geq 1
* 50 \leq X \leq 20 000
* A, B and C are integers.
* X is a multiple of 50.
Input
Input is given from Standard Input in the following format:
A
B
C
X
Output
Print the number of ways to select coins.
Examples
Input
2
2
2
100
Output
2
Input
5
1
0
150
Output
0
Input
30
40
50
6000
Output
213 | instruction | 0 | 42,742 | 10 | 85,484 |
"Correct Solution:
```
A=int(input())
B=int(input())
C=int(input())
X=int(input())
K=0
for a in range(A+1):
for b in range(B+1):
for c in range(C+1):
K+=(500*a+100*b+50*c==X)
print(K)
``` | output | 1 | 42,742 | 10 | 85,485 |
Provide a correct Python 3 solution for this coding contest problem.
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total?
Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
Constraints
* 0 \leq A, B, C \leq 50
* A + B + C \geq 1
* 50 \leq X \leq 20 000
* A, B and C are integers.
* X is a multiple of 50.
Input
Input is given from Standard Input in the following format:
A
B
C
X
Output
Print the number of ways to select coins.
Examples
Input
2
2
2
100
Output
2
Input
5
1
0
150
Output
0
Input
30
40
50
6000
Output
213 | instruction | 0 | 42,743 | 10 | 85,486 |
"Correct Solution:
```
a,b,c,x=[int(input()) for _ in range(4)]
count=0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
if 500*i+100*j+50*k == x:
count+=1
print(count)
``` | output | 1 | 42,743 | 10 | 85,487 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total?
Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
Constraints
* 0 \leq A, B, C \leq 50
* A + B + C \geq 1
* 50 \leq X \leq 20 000
* A, B and C are integers.
* X is a multiple of 50.
Input
Input is given from Standard Input in the following format:
A
B
C
X
Output
Print the number of ways to select coins.
Examples
Input
2
2
2
100
Output
2
Input
5
1
0
150
Output
0
Input
30
40
50
6000
Output
213
Submitted Solution:
```
list=[int(input()) for i in range(4)]
m=0
for i in range(list[0]+1):
for j in range(list[1]+1):
for k in range(list[2]+1):
if list[3]==50*k+100*j+500*i:
m += 1
print(m)
``` | instruction | 0 | 42,744 | 10 | 85,488 |
Yes | output | 1 | 42,744 | 10 | 85,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total?
Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
Constraints
* 0 \leq A, B, C \leq 50
* A + B + C \geq 1
* 50 \leq X \leq 20 000
* A, B and C are integers.
* X is a multiple of 50.
Input
Input is given from Standard Input in the following format:
A
B
C
X
Output
Print the number of ways to select coins.
Examples
Input
2
2
2
100
Output
2
Input
5
1
0
150
Output
0
Input
30
40
50
6000
Output
213
Submitted Solution:
```
a,b,c,X = [int(input()) for i in range(4)]
l = []
for x in range(a + 1):
for y in range(b + 1):
for z in range(c + 1):
s = 500*x + 100*y + 50*z
l.append(s)
print(l.count(X))
``` | instruction | 0 | 42,745 | 10 | 85,490 |
Yes | output | 1 | 42,745 | 10 | 85,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total?
Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
Constraints
* 0 \leq A, B, C \leq 50
* A + B + C \geq 1
* 50 \leq X \leq 20 000
* A, B and C are integers.
* X is a multiple of 50.
Input
Input is given from Standard Input in the following format:
A
B
C
X
Output
Print the number of ways to select coins.
Examples
Input
2
2
2
100
Output
2
Input
5
1
0
150
Output
0
Input
30
40
50
6000
Output
213
Submitted Solution:
```
[print(sum(i * 500 + j * 100 + k * 50 == sum_ for k in range(c + 1) for j in range(b + 1) for i in range(a + 1))) for a, b, c, sum_ in [[int(input()) for _ in range(4)]]]
``` | instruction | 0 | 42,746 | 10 | 85,492 |
Yes | output | 1 | 42,746 | 10 | 85,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total?
Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
Constraints
* 0 \leq A, B, C \leq 50
* A + B + C \geq 1
* 50 \leq X \leq 20 000
* A, B and C are integers.
* X is a multiple of 50.
Input
Input is given from Standard Input in the following format:
A
B
C
X
Output
Print the number of ways to select coins.
Examples
Input
2
2
2
100
Output
2
Input
5
1
0
150
Output
0
Input
30
40
50
6000
Output
213
Submitted Solution:
```
A, B, C, X = (int(input()) for i in range(4))
print(sum(500 * a + 100 * b + 50 * c == X for a in range(A + 1) for b in range(B + 1) for c in range(C + 1)))
``` | instruction | 0 | 42,747 | 10 | 85,494 |
Yes | output | 1 | 42,747 | 10 | 85,495 |
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