text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a tree with N vertices, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N - 1), the i-th edge connects Vertex x_i and y_i.
Taro has decided to paint each vertex in white or black, so that any black vertex can be reached from any other black vertex by passing through only black vertices.
You are given a positive integer M. For each v (1 \leq v \leq N), answer the following question:
* Assuming that Vertex v has to be black, find the number of ways in which the vertices can be painted, modulo M.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^5
* 2 \leq M \leq 10^9
* 1 \leq x_i, y_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N M
x_1 y_1
x_2 y_2
:
x_{N - 1} y_{N - 1}
Output
Print N lines. The v-th (1 \leq v \leq N) line should contain the answer to the following question:
* Assuming that Vertex v has to be black, find the number of ways in which the vertices can be painted, modulo M.
Examples
Input
3 100
1 2
2 3
Output
3
4
3
Input
4 100
1 2
1 3
1 4
Output
8
5
5
5
Input
1 100
Output
1
Input
10 2
8 5
10 8
6 5
1 5
4 8
2 10
3 6
9 2
1 7
Output
0
0
1
1
1
0
1
0
1
1
Submitted Solution:
```
import sys,os,io
input = sys.stdin.readline
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
N, M = map(int, input().split())
edge = [[] for _ in range(N)]
for i in range(N-1):
a,b = list(map(int, input().split()))
edge[a-1].append(b-1)
edge[b-1].append(a-1)
def dfs(start):
stack = [start]
parent = [N]*N
parent[start] = -1
while stack:
v = stack[-1]
marker = 0
for u in edge[v]:
if u==parent[v]:
continue
if parent[u]==N: #子へ降ろす
marker = 1
parent[u]=v
stack.append(u)
else: #子から吸い上げる
ans[v] *= ans[u]+1
if marker==0:
stack.pop()
ans[v]%=M
return
def dfs2(start):
stack = [start]
parent = [N]*N
parent[start] = -1
p_value = [0]*N
while stack:
v = stack.pop()
cum1 = [1]*(len(edge[v])+1)
cum2 = [1]*(len(edge[v])+1)
for i,u in enumerate(edge[v]):
if u==parent[v]:
cum1[i+1] = cum1[i]*(p_value[v]+1)
else:
cum1[i+1] = cum1[i]*(ans[u]+1)
cum1[i+1] %= M
for i,u in enumerate(edge[v][::-1]):
if u==parent[v]:
cum2[-i-2] = cum2[-i-1]*(p_value[v]+1)
else:
cum2[-i-2] = cum2[-i-1]*(ans[u]+1)
cum2[i+1]%=M
for i,u in enumerate(edge[v]):
if u==parent[v]:
continue
parent[u]=v
p_value[u] = cum1[i]*cum2[i+1]%M
ans[u] *= (p_value[u]+1)
ans[u] %= M
stack.append(u)
ans[v] %= M
return
ans = [1]*N
dfs(0)
dfs2(0)
print(*ans, sep='\n')
```
No
| 107,700 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a tree with N vertices, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N - 1), the i-th edge connects Vertex x_i and y_i.
Taro has decided to paint each vertex in white or black, so that any black vertex can be reached from any other black vertex by passing through only black vertices.
You are given a positive integer M. For each v (1 \leq v \leq N), answer the following question:
* Assuming that Vertex v has to be black, find the number of ways in which the vertices can be painted, modulo M.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^5
* 2 \leq M \leq 10^9
* 1 \leq x_i, y_i \leq N
* The given graph is a tree.
Input
Input is given from Standard Input in the following format:
N M
x_1 y_1
x_2 y_2
:
x_{N - 1} y_{N - 1}
Output
Print N lines. The v-th (1 \leq v \leq N) line should contain the answer to the following question:
* Assuming that Vertex v has to be black, find the number of ways in which the vertices can be painted, modulo M.
Examples
Input
3 100
1 2
2 3
Output
3
4
3
Input
4 100
1 2
1 3
1 4
Output
8
5
5
5
Input
1 100
Output
1
Input
10 2
8 5
10 8
6 5
1 5
4 8
2 10
3 6
9 2
1 7
Output
0
0
1
1
1
0
1
0
1
1
Submitted Solution:
```
import itertools
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10 ** 7)
N,MOD = map(int,input().split())
if N == 1:
print(1)
exit()
graph = [[] for _ in range(N+1)]
for _ in range(N-1):
x,y = map(int,input().split())
graph[x].append(y)
graph[y].append(x)
def dfs_1(v=1,parent=None):
# 子ごとに、そこを黒くして適切 or 完全白
# dp_1 + 1 の積
x = 1
for w in graph[v]:
if w == parent:
continue
x *= (dfs_1(w,v) + 1)
dp_1[v] = x
return x
dp_1 = [0] * (N+1) # 自分を黒く塗って、子側の部分木を適切に塗る方法
dfs_1()
def make_products(arr):
# 左右からの積を作る方法
f = lambda x,y: x*y%MOD
L = [1] + list(itertools.accumulate(arr[:-1], f))
R = list(itertools.accumulate(arr[::-1][:-1], f))[::-1] + [1]
return [f(x,y) for x,y in zip(L,R)]
def dfs_2(v=1,parent=None,x=1):
# それぞれの子以外の情報をマージして、子に渡さなければいけない
arr = [1 + dp_1[w] if w != parent else x for w in graph[v]]
products = make_products(arr)
dp[v] = arr[0] * products[0] % MOD
for w,p in zip(graph[v],products):
if w==parent:
continue
dfs_2(w,v,p+1)
dp = [0] * (N+1)
dfs_2()
print('\n'.join(map(str,dp[1:])))
```
No
| 107,701 |
Provide a correct Python 3 solution for this coding contest problem.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
"Correct Solution:
```
n = int(input())
res = n
for i in range(n + 1):
t = i
c = 0
while t > 0:
c += t % 9
t //= 9
t = n - i
while t > 0:
c += t % 6
t //= 6
if res > c:
res = c
print(res)
```
| 107,702 |
Provide a correct Python 3 solution for this coding contest problem.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
"Correct Solution:
```
n = int(input())
ans = n
for i in range(n+1):
c = 0
t = i
while t>0:
c+=t%6
t //= 6
t = n-i
while t>0:
c+=t%9
t //= 9
ans = min(ans,c)
print(ans)
```
| 107,703 |
Provide a correct Python 3 solution for this coding contest problem.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
"Correct Solution:
```
N=int(input())
dstb=[1]+[6**i for i in range(1,7)]+[9**j for j in range(1,6)]
DP=[10**5]*(10**5*2)
DP[0]=0
for i in range(N+1):
for j in dstb:
DP[i+j]=min(DP[i+j], DP[i]+1)
print(DP[N])
```
| 107,704 |
Provide a correct Python 3 solution for this coding contest problem.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
"Correct Solution:
```
n=int(input())
ans=float("INF")
for i in range(n+1):
d=0
temp=n-i
while (i>0):
d+=i%6
i//=6
while(temp>0):
d+=temp%9
temp//=9
ans=min(d,ans)
print(ans)
```
| 107,705 |
Provide a correct Python 3 solution for this coding contest problem.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
"Correct Solution:
```
n=int(input())
ans=float("INF")
for i in range(n+1):
d=0
temp=i
while (temp>0):
d+=temp%6
temp//=6
temp=n-i
while(temp>0):
d+=temp%9
temp//=9
ans=min(d,ans)
print(ans)
```
| 107,706 |
Provide a correct Python 3 solution for this coding contest problem.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
"Correct Solution:
```
N = int(input())
res = 10**10
for i in range(N+1):
count = 0
six = i
while six>0:
count += six%6
six = six//6
nine = N-i
while nine>0:
count += nine%9
nine = nine//9
res = min(res,count)
print(res)
```
| 107,707 |
Provide a correct Python 3 solution for this coding contest problem.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
"Correct Solution:
```
n=int(input())
ans=1000000000
for i in range(0,n+1,6):
j=n-i
a=0
while j:
a+=j%9
j//=9
while i:
a+=i%6
i//=6
ans=min(a,ans)
print(ans)
```
| 107,708 |
Provide a correct Python 3 solution for this coding contest problem.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
"Correct Solution:
```
N=int(input())
ans=N
for i in range(N+1):
cnt=0
t=i
while t>0: cnt+=t%6; t//=6
j=N-i
while j>0: cnt+=j%9; j//=9
ans = min(ans,cnt)
print(ans)
```
| 107,709 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
Submitted Solution:
```
n=int(input())
a=100000
for i in range(n+1):
j=n-i
count=0
while i>0:
count+=i%6
i//=6
while j>0:
count+=j%9
j//=9
a=min(a,count)
print(a)
```
Yes
| 107,710 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
Submitted Solution:
```
N=int(input())
M=[0]*(N+1)
M[1]=1
for i in range(2,N+1):
L=[M[i-1]+1]
j=0
while i-6**j>=0:
L.append(M[i-6**j]+1)
j+=1
j=0
while i-9**j>=0:
L.append(M[i-9**j]+1)
j+=1
M[i]=min(L)
#print(M)
print(M[N])
```
Yes
| 107,711 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
Submitted Solution:
```
N=int(input())
ans=N
for i in range(N+1):
cnt=0
t = i
while t>0:
cnt+=t%6
t//=6
t=N-i
while t>0:
cnt+=t%9
t//=9
ans=min(ans,cnt)
print(ans)
```
Yes
| 107,712 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
Submitted Solution:
```
n=int(input())
a=n
for i in range(n+1):
x=i;y=n-i;c=0
while x>=6:
c+=x%6
x=x//6
c+=x
while y>=9:
c+=y%9
y=y//9
c+=y
a=min(a,c)
print(a)
```
Yes
| 107,713 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
Submitted Solution:
```
N = int(input())
A = [1]
for i in range(1,7):
A.append(6 ** i)
for i in range(1, 6):
A.append(9 ** i)
A = sorted(A)
dp = [float("inf")] * (N+1)
dp[0] = 0
for i in range(1,N+1):
for k, v in enumerate(A):
if i >= v:
dp[i] = min(dp[i], dp[i-A[s]]+1)
print(dp[N])
```
No
| 107,714 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
Submitted Solution:
```
N = int(input())
for i in range(6):
a = 9**(i+1)
if N<a:
nine = i
N = N-(9**i)
break
for i in range(7):
a = 6**(i+1)
if N<a:
six = i
N = N-(6**i)
one = N
break
out = one + six + nine
print(out)
```
No
| 107,715 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
Submitted Solution:
```
# 引き出したいお金の額を入力する
N = int(input())
# 整数aの累乗のうち、整数N以下かつ最も大きいものを返す関数を定義
def func(N, a):
index = 0
while True:
if a ** (index + 1) <= N:
index += 1
else:
break
return a ** index
# 引き出し回数を再帰的に求める関数を定義
def search_times_re(N):
times = 0
if N < 6:
times = N
return times
elif N < 9:
times = 1 + (N - 6)
return times
else:
N1 = search_times_re(N - func(N, 6))
N2 = search_times_re(N - func(N, 9))
if N1 < N2:
return 1 + N1
else:
return 1 + N2
print(search_times_re(N))
```
No
| 107,716 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:
* 1 yen (the currency of Japan)
* 6 yen, 6^2(=36) yen, 6^3(=216) yen, ...
* 9 yen, 9^2(=81) yen, 9^3(=729) yen, ...
At least how many operations are required to withdraw exactly N yen in total?
It is not allowed to re-deposit the money you withdrew.
Constraints
* 1 \leq N \leq 100000
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If at least x operations are required to withdraw exactly N yen in total, print x.
Examples
Input
127
Output
4
Input
3
Output
3
Input
44852
Output
16
Submitted Solution:
```
#!/usr/bin/env python3
import math
# n進数の各位の和
def base_10_n(x,n):
a = int(math.log(x,n))
ans = 0
for i in range(a,-1,-1):
ans += x // n**i
x %= n**i
ans += x
return ans
def main():
N = int(input())
ans = N
for j in range(1,N//18 + 1):
a = base_10_n(N-18*j,6)
b = base_10_n(18*j,9)
ans = min(ans, a+b)
print(ans)
# print([i for i in range(2,-1,-1)])
if __name__ == "__main__":
main()
```
No
| 107,717 |
Provide a correct Python 3 solution for this coding contest problem.
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
Constraints
* The lengths of s and t are between 1 and 100 (inclusive).
* s and t consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If it is possible to satisfy s' < t', print `Yes`; if it is not, print `No`.
Examples
Input
yx
axy
Output
Yes
Input
ratcode
atlas
Output
Yes
Input
cd
abc
Output
No
Input
w
ww
Output
Yes
Input
zzz
zzz
Output
No
"Correct Solution:
```
print('Yes'*(sorted(input())<sorted(input())[::-1])or'No')
```
| 107,718 |
Provide a correct Python 3 solution for this coding contest problem.
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
Constraints
* The lengths of s and t are between 1 and 100 (inclusive).
* s and t consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If it is possible to satisfy s' < t', print `Yes`; if it is not, print `No`.
Examples
Input
yx
axy
Output
Yes
Input
ratcode
atlas
Output
Yes
Input
cd
abc
Output
No
Input
w
ww
Output
Yes
Input
zzz
zzz
Output
No
"Correct Solution:
```
s = sorted(input())
t = sorted(input())[::-1]
if s < t:
print("Yes")
else:
print("No")
```
| 107,719 |
Provide a correct Python 3 solution for this coding contest problem.
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
Constraints
* The lengths of s and t are between 1 and 100 (inclusive).
* s and t consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If it is possible to satisfy s' < t', print `Yes`; if it is not, print `No`.
Examples
Input
yx
axy
Output
Yes
Input
ratcode
atlas
Output
Yes
Input
cd
abc
Output
No
Input
w
ww
Output
Yes
Input
zzz
zzz
Output
No
"Correct Solution:
```
s=sorted(input())
t=sorted(input())
t.sort(reverse=True)
print("Yes" if s<t else "No")
```
| 107,720 |
Provide a correct Python 3 solution for this coding contest problem.
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
Constraints
* The lengths of s and t are between 1 and 100 (inclusive).
* s and t consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If it is possible to satisfy s' < t', print `Yes`; if it is not, print `No`.
Examples
Input
yx
axy
Output
Yes
Input
ratcode
atlas
Output
Yes
Input
cd
abc
Output
No
Input
w
ww
Output
Yes
Input
zzz
zzz
Output
No
"Correct Solution:
```
S = sorted(list(input()))
T = sorted(list(input()))[::-1]
print('Yes' if S<T else 'No')
```
| 107,721 |
Provide a correct Python 3 solution for this coding contest problem.
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
Constraints
* The lengths of s and t are between 1 and 100 (inclusive).
* s and t consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If it is possible to satisfy s' < t', print `Yes`; if it is not, print `No`.
Examples
Input
yx
axy
Output
Yes
Input
ratcode
atlas
Output
Yes
Input
cd
abc
Output
No
Input
w
ww
Output
Yes
Input
zzz
zzz
Output
No
"Correct Solution:
```
s=input()
t=input()
print(['No','Yes'][''.join(sorted(s))<''.join(sorted(t,reverse=True))])
```
| 107,722 |
Provide a correct Python 3 solution for this coding contest problem.
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
Constraints
* The lengths of s and t are between 1 and 100 (inclusive).
* s and t consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If it is possible to satisfy s' < t', print `Yes`; if it is not, print `No`.
Examples
Input
yx
axy
Output
Yes
Input
ratcode
atlas
Output
Yes
Input
cd
abc
Output
No
Input
w
ww
Output
Yes
Input
zzz
zzz
Output
No
"Correct Solution:
```
s = ''.join(sorted(input()))
t = ''.join(sorted(input())[::-1])
print('Yes' if s < t else 'No')
```
| 107,723 |
Provide a correct Python 3 solution for this coding contest problem.
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
Constraints
* The lengths of s and t are between 1 and 100 (inclusive).
* s and t consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If it is possible to satisfy s' < t', print `Yes`; if it is not, print `No`.
Examples
Input
yx
axy
Output
Yes
Input
ratcode
atlas
Output
Yes
Input
cd
abc
Output
No
Input
w
ww
Output
Yes
Input
zzz
zzz
Output
No
"Correct Solution:
```
s=input()
t=input()
s1=sorted(s)
t1=sorted(t)[::-1]
print('Yes' if s1<t1 else 'No' )
```
| 107,724 |
Provide a correct Python 3 solution for this coding contest problem.
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
Constraints
* The lengths of s and t are between 1 and 100 (inclusive).
* s and t consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If it is possible to satisfy s' < t', print `Yes`; if it is not, print `No`.
Examples
Input
yx
axy
Output
Yes
Input
ratcode
atlas
Output
Yes
Input
cd
abc
Output
No
Input
w
ww
Output
Yes
Input
zzz
zzz
Output
No
"Correct Solution:
```
s, t = [input() for _ in range(2)]
print('Yes' if sorted(s) < sorted(t)[::-1] else 'No')
```
| 107,725 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
Constraints
* The lengths of s and t are between 1 and 100 (inclusive).
* s and t consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If it is possible to satisfy s' < t', print `Yes`; if it is not, print `No`.
Examples
Input
yx
axy
Output
Yes
Input
ratcode
atlas
Output
Yes
Input
cd
abc
Output
No
Input
w
ww
Output
Yes
Input
zzz
zzz
Output
No
Submitted Solution:
```
s, t = sorted(input()), sorted(input())[::-1]
print("YNeos"[sorted([s,t])[0]==t::2])
```
Yes
| 107,726 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
Constraints
* The lengths of s and t are between 1 and 100 (inclusive).
* s and t consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If it is possible to satisfy s' < t', print `Yes`; if it is not, print `No`.
Examples
Input
yx
axy
Output
Yes
Input
ratcode
atlas
Output
Yes
Input
cd
abc
Output
No
Input
w
ww
Output
Yes
Input
zzz
zzz
Output
No
Submitted Solution:
```
s=sorted(input())
t=sorted(input(),reverse=True)
print("Yes" if s<t else "No")
```
Yes
| 107,727 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
Constraints
* The lengths of s and t are between 1 and 100 (inclusive).
* s and t consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If it is possible to satisfy s' < t', print `Yes`; if it is not, print `No`.
Examples
Input
yx
axy
Output
Yes
Input
ratcode
atlas
Output
Yes
Input
cd
abc
Output
No
Input
w
ww
Output
Yes
Input
zzz
zzz
Output
No
Submitted Solution:
```
s=list(input())
t=list(input())
s.sort()
t.sort(reverse=True)
if s<t:print('Yes')
else:print('No')
```
Yes
| 107,728 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
Constraints
* The lengths of s and t are between 1 and 100 (inclusive).
* s and t consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If it is possible to satisfy s' < t', print `Yes`; if it is not, print `No`.
Examples
Input
yx
axy
Output
Yes
Input
ratcode
atlas
Output
Yes
Input
cd
abc
Output
No
Input
w
ww
Output
Yes
Input
zzz
zzz
Output
No
Submitted Solution:
```
s = sorted(list(input()))
t = sorted(list(input()), reverse=True)
print("Yes" if s < t else "No")
```
Yes
| 107,729 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
Constraints
* The lengths of s and t are between 1 and 100 (inclusive).
* s and t consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If it is possible to satisfy s' < t', print `Yes`; if it is not, print `No`.
Examples
Input
yx
axy
Output
Yes
Input
ratcode
atlas
Output
Yes
Input
cd
abc
Output
No
Input
w
ww
Output
Yes
Input
zzz
zzz
Output
No
Submitted Solution:
```
input = [str(input()) for i in range(2)]
s_list = sorted(input[0])
t_list = sorted(input[1], reverse=True)
def main():
# tが小さければno
for s, t in zip(s_list, t_list):
if s < t:
return 'Yes'
elif s > t:
return 'No'
if len(s) >= len(t):
return 'No'
else:
return 'Yes'
print(main())
```
No
| 107,730 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
Constraints
* The lengths of s and t are between 1 and 100 (inclusive).
* s and t consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If it is possible to satisfy s' < t', print `Yes`; if it is not, print `No`.
Examples
Input
yx
axy
Output
Yes
Input
ratcode
atlas
Output
Yes
Input
cd
abc
Output
No
Input
w
ww
Output
Yes
Input
zzz
zzz
Output
No
Submitted Solution:
```
s = input()
t = input()
s = sorted(s)
t = sorted(t, reverse=True)
s_len = len(s)
t_len = len(t)
min_len = min(s_len, t_len)
for i in range(min_len):
if s[i] < t[i]:
print('Yes')
exit()
elif s[i] > t[i]:
print('No')
exit()
if s_len == t_len:
print('No')
else:
print('Yes')
```
No
| 107,731 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
Constraints
* The lengths of s and t are between 1 and 100 (inclusive).
* s and t consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If it is possible to satisfy s' < t', print `Yes`; if it is not, print `No`.
Examples
Input
yx
axy
Output
Yes
Input
ratcode
atlas
Output
Yes
Input
cd
abc
Output
No
Input
w
ww
Output
Yes
Input
zzz
zzz
Output
No
Submitted Solution:
```
print("Yes") if sorted(input()) < sorted(input()) else print("No")
```
No
| 107,732 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
Constraints
* The lengths of s and t are between 1 and 100 (inclusive).
* s and t consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If it is possible to satisfy s' < t', print `Yes`; if it is not, print `No`.
Examples
Input
yx
axy
Output
Yes
Input
ratcode
atlas
Output
Yes
Input
cd
abc
Output
No
Input
w
ww
Output
Yes
Input
zzz
zzz
Output
No
Submitted Solution:
```
import sys
import math
import itertools
import bisect
from copy import copy
from collections import deque,Counter
from decimal import Decimal
def s(): return input()
def i(): return int(input())
def S(): return input().split()
def I(): return map(int,input().split())
def X(): return list(input())
def L(): return list(input().split())
def l(): return list(map(int,input().split()))
def lcm(a,b): return a*b//math.gcd(a,b)
def gcd(*numbers): reduce(math.gcd, numbers)
sys.setrecursionlimit(10 ** 9)
mod = 10**9+7
count = 0
ans = 0
a = s()
t = s()
a.sort()
y = "".join(a)
t = sorted(t,reverse = True)
x = "".join(t)
if y < x:
print("Yes")
else:
print("No")
```
No
| 107,733 |
Provide a correct Python 3 solution for this coding contest problem.
There is a directed graph with N vertices and N edges. The vertices are numbered 1, 2, ..., N.
The graph has the following N edges: (p_1, 1), (p_2, 2), ..., (p_N, N), and the graph is weakly connected. Here, an edge from Vertex u to Vertex v is denoted by (u, v), and a weakly connected graph is a graph which would be connected if each edge was bidirectional.
We would like to assign a value to each of the vertices in this graph so that the following conditions are satisfied. Here, a_i is the value assigned to Vertex i.
* Each a_i is a non-negative integer.
* For each edge (i, j), a_i \neq a_j holds.
* For each i and each integer x(0 ≤ x < a_i), there exists a vertex j such that the edge (i, j) exists and x = a_j holds.
Determine whether there exists such an assignment.
Constraints
* 2 ≤ N ≤ 200 000
* 1 ≤ p_i ≤ N
* p_i \neq i
* The graph is weakly connected.
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 ... p_N
Output
If the assignment is possible, print `POSSIBLE`; otherwise, print `IMPOSSIBLE`.
Examples
Input
4
2 3 4 1
Output
POSSIBLE
Input
3
2 3 1
Output
IMPOSSIBLE
Input
4
2 3 1 1
Output
POSSIBLE
Input
6
4 5 6 5 6 4
Output
IMPOSSIBLE
"Correct Solution:
```
import sys
from itertools import chain
readline = sys.stdin.readline
#非再帰
def scc(Edge):
N = len(Edge)
Edgeinv = [[] for _ in range(N)]
for vn in range(N):
for vf in Edge[vn]:
Edgeinv[vf].append(vn)
used = [False]*N
dim = [len(Edge[i]) for i in range(N)]
order = []
for st in range(N):
if not used[st]:
stack = [st, 0]
while stack:
vn, i = stack[-2], stack[-1]
if not i and used[vn]:
stack.pop()
stack.pop()
else:
used[vn] = True
if i < dim[vn]:
stack[-1] += 1
stack.append(Edge[vn][i])
stack.append(0)
else:
stack.pop()
order.append(stack.pop())
res = [None]*N
used = [False]*N
cnt = -1
for st in order[::-1]:
if not used[st]:
cnt += 1
stack = [st]
res[st] = cnt
used[st] = True
while stack:
vn = stack.pop()
for vf in Edgeinv[vn]:
if not used[vf]:
used[vf] = True
res[vf] = cnt
stack.append(vf)
M = cnt+1
components = [[] for _ in range(M)]
for i in range(N):
components[res[i]].append(i)
tEdge = [[] for _ in range(M)]
teset = set()
for vn in range(N):
tn = res[vn]
for vf in Edge[vn]:
tf = res[vf]
if tn != tf and tn*M + tf not in teset:
teset.add(tn*M + tf)
tEdge[tn].append(tf)
return res, components, tEdge
N = int(readline())
P = list(map(lambda x: int(x)-1, readline().split()))
Edge = [[] for _ in range(N)]
for i in range(N):
Edge[P[i]].append(i)
R, Com, _ = scc(Edge)
Lord = list(chain(*Com[::-1]))
val = [None]*N
for vn in Lord:
if not R[vn]:
break
lvn = len(Edge[vn]) + 1
res = [0]*lvn
for vf in Edge[vn]:
if val[vf] < lvn:
res[val[vf]] += 1
for k in range(lvn):
if not res[k]:
val[vn] = k
break
st = Lord[-1]
lst = len(Edge[st]) + 2
res = [0]*lst
for vf in Edge[st]:
if val[vf] is None:
continue
if val[vf] < lst:
res[val[vf]] += 1
mc = []
for k in range(lst):
if not res[k]:
mc.append(k)
vn = st
Ls = []
while vn is not None:
for vf in Edge[vn]:
if R[vf]:
continue
if vf == st:
vn = None
else:
Ls.append(vf)
vn = vf
Ls.reverse()
ans = False
for idx in range(2):
vc = val[:]
vc[st] = mc[idx]
for vn in Ls:
lvn = len(Edge[vn])+1
res = [0]*lvn
for vf in Edge[vn]:
if vc[vf] < lvn:
res[vc[vf]] += 1
for k in range(lvn):
if not res[k]:
vc[vn] = k
break
for vn in range(N):
res = [False]*vc[vn]
for vf in Edge[vn]:
if vc[vn] == vc[vf]:
break
if vc[vf] < vc[vn]:
res[vc[vf]] = True
else:
if not all(res):
break
continue
break
else:
ans = True
if ans:
break
print('POSSIBLE' if ans else 'IMPOSSIBLE')
```
| 107,734 |
Provide a correct Python 3 solution for this coding contest problem.
There is a directed graph with N vertices and N edges. The vertices are numbered 1, 2, ..., N.
The graph has the following N edges: (p_1, 1), (p_2, 2), ..., (p_N, N), and the graph is weakly connected. Here, an edge from Vertex u to Vertex v is denoted by (u, v), and a weakly connected graph is a graph which would be connected if each edge was bidirectional.
We would like to assign a value to each of the vertices in this graph so that the following conditions are satisfied. Here, a_i is the value assigned to Vertex i.
* Each a_i is a non-negative integer.
* For each edge (i, j), a_i \neq a_j holds.
* For each i and each integer x(0 ≤ x < a_i), there exists a vertex j such that the edge (i, j) exists and x = a_j holds.
Determine whether there exists such an assignment.
Constraints
* 2 ≤ N ≤ 200 000
* 1 ≤ p_i ≤ N
* p_i \neq i
* The graph is weakly connected.
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 ... p_N
Output
If the assignment is possible, print `POSSIBLE`; otherwise, print `IMPOSSIBLE`.
Examples
Input
4
2 3 4 1
Output
POSSIBLE
Input
3
2 3 1
Output
IMPOSSIBLE
Input
4
2 3 1 1
Output
POSSIBLE
Input
6
4 5 6 5 6 4
Output
IMPOSSIBLE
"Correct Solution:
```
from collections import deque
N=int(input())
IN=[[] for i in range(N)]
OUT=[-1 for i in range(N)]
p=list(map(int,input().split()))
for i in range(N):
IN[p[i]-1].append(i)
OUT[i]=p[i]-1
deg=[len(IN[i]) for i in range(N)]
deq=deque([v for v in range(N) if deg[v]==0])
res=[]
while deq:
v=deq.popleft()
res.append(v)
deg[OUT[v]]-=1
if deg[OUT[v]]==0:
deq.append(OUT[v])
if len(res)==N:
print("POSSIBLE")
exit()
start=-1
for i in range(N):
if deg[i]>0:
start=i
break
cycle=[start]
while True:
nv=OUT[cycle[-1]]
if nv!=start:
cycle.append(nv)
else:
break
dp=[-1]*N
for v in res:
mex=[False]*(len(IN[v])+1)
for pv in IN[v]:
if dp[pv]<=len(IN[v]):
mex[dp[pv]]=True
for i in range(len(mex)):
if not mex[i]:
dp[v]=i
break
m0=-1
m1=-1
mex=[False]*(len(IN[start])+2)
for pv in IN[start]:
if dp[pv]!=-1 and dp[pv]<=len(IN[v])+1:
mex[dp[pv]]=True
for i in range(len(mex)):
if not mex[i]:
if m0==-1:
m0=i
else:
m1=i
break
#m0
dp[start]=m0
for i in range(1,len(cycle)):
v=cycle[i]
temp=-1
mex=[False]*(len(IN[v])+1)
for pv in IN[v]:
if dp[pv]<=len(IN[v]):
mex[dp[pv]]=True
for i in range(len(mex)):
if not mex[i]:
dp[v]=i
break
mex=[False]*(len(IN[start])+2)
for pv in IN[start]:
if dp[pv]!=-1 and dp[pv]<=len(IN[v])+1:
mex[dp[pv]]=True
check=-1
for i in range(len(mex)):
if not mex[i]:
check=i
break
if check==m0:
print("POSSIBLE")
exit()
for v in cycle:
dp[v]=-1
#m1
dp[start]=m1
for i in range(1,len(cycle)):
v=cycle[i]
temp=-1
mex=[False]*(len(IN[v])+1)
for pv in IN[v]:
if dp[pv]<=len(IN[v]):
mex[dp[pv]]=True
for i in range(len(mex)):
if not mex[i]:
dp[v]=i
break
mex=[False]*(len(IN[start])+2)
for pv in IN[start]:
if dp[pv]!=-1 and dp[pv]<=len(IN[start])+1:
mex[dp[pv]]=True
check=-1
for i in range(len(mex)):
if not mex[i]:
check=i
break
if check==m1:
print("POSSIBLE")
exit()
print("IMPOSSIBLE")
```
| 107,735 |
Provide a correct Python 3 solution for this coding contest problem.
There is a directed graph with N vertices and N edges. The vertices are numbered 1, 2, ..., N.
The graph has the following N edges: (p_1, 1), (p_2, 2), ..., (p_N, N), and the graph is weakly connected. Here, an edge from Vertex u to Vertex v is denoted by (u, v), and a weakly connected graph is a graph which would be connected if each edge was bidirectional.
We would like to assign a value to each of the vertices in this graph so that the following conditions are satisfied. Here, a_i is the value assigned to Vertex i.
* Each a_i is a non-negative integer.
* For each edge (i, j), a_i \neq a_j holds.
* For each i and each integer x(0 ≤ x < a_i), there exists a vertex j such that the edge (i, j) exists and x = a_j holds.
Determine whether there exists such an assignment.
Constraints
* 2 ≤ N ≤ 200 000
* 1 ≤ p_i ≤ N
* p_i \neq i
* The graph is weakly connected.
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 ... p_N
Output
If the assignment is possible, print `POSSIBLE`; otherwise, print `IMPOSSIBLE`.
Examples
Input
4
2 3 4 1
Output
POSSIBLE
Input
3
2 3 1
Output
IMPOSSIBLE
Input
4
2 3 1 1
Output
POSSIBLE
Input
6
4 5 6 5 6 4
Output
IMPOSSIBLE
"Correct Solution:
```
from collections import defaultdict
import heapq as hq
def helper(n):
q = child_L[n]
del child_L[n]
hq.heapify(q)
i = 0
while q:
t = hq.heappop(q)
if i < t:
break
else:
i += (i == t)
j = i + 1
while q:
t = hq.heappop(q)
if j < t:
break
else:
j += (j == t)
return (i, j)
if __name__ == '__main__':
N = int(input())
P = list(map(lambda x: int(x) - 1, input().split()))
# D:出している辺の本数
D = [0] * N
for p in P:
D[p] += 1
# print(D)
child_L = defaultdict(list)
# S:辺を出してないものリスト
S = [p for p in range(N) if D[p] == 0]
L = [None] * N
while S:
# print(child_L)
n = S.pop()
q = child_L[n]
# print(q)
del child_L[n]
# listのqをヒープキューに変換
hq.heapify(q)
i = 0
while q:
t = hq.heappop(q)
if i < t:
break
else:
i += (i == t)
L[n] = i
# print(L)
m = P[n]
# print("m:" + str(m))
child_L[m].append(i)
D[m] -= 1
if D[m] == 0:
S.append(m)
# print(D)
# cycle check
try:
start = D.index(1)
except ValueError:
print('POSSIBLE')
exit()
s1, s2 = helper(start)
G = []
n = P[start]
while n != start:
G.append(helper(n))
n = P[n]
# del N, P, D, child_L, S, L
# 可能な初期値をそれぞれシミュレート
# 1
n = s1
for g in G:
if g[0] == n:
n = g[1]
else:
n = g[0]
if n != s1:
print('POSSIBLE')
exit()
# 2
n = s2
for g in G:
if g[0] == n:
n = g[1]
else:
n = g[0]
if n == s1:
print('POSSIBLE')
exit()
print('IMPOSSIBLE')
```
| 107,736 |
Provide a correct Python 3 solution for this coding contest problem.
There is a directed graph with N vertices and N edges. The vertices are numbered 1, 2, ..., N.
The graph has the following N edges: (p_1, 1), (p_2, 2), ..., (p_N, N), and the graph is weakly connected. Here, an edge from Vertex u to Vertex v is denoted by (u, v), and a weakly connected graph is a graph which would be connected if each edge was bidirectional.
We would like to assign a value to each of the vertices in this graph so that the following conditions are satisfied. Here, a_i is the value assigned to Vertex i.
* Each a_i is a non-negative integer.
* For each edge (i, j), a_i \neq a_j holds.
* For each i and each integer x(0 ≤ x < a_i), there exists a vertex j such that the edge (i, j) exists and x = a_j holds.
Determine whether there exists such an assignment.
Constraints
* 2 ≤ N ≤ 200 000
* 1 ≤ p_i ≤ N
* p_i \neq i
* The graph is weakly connected.
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 ... p_N
Output
If the assignment is possible, print `POSSIBLE`; otherwise, print `IMPOSSIBLE`.
Examples
Input
4
2 3 4 1
Output
POSSIBLE
Input
3
2 3 1
Output
IMPOSSIBLE
Input
4
2 3 1 1
Output
POSSIBLE
Input
6
4 5 6 5 6 4
Output
IMPOSSIBLE
"Correct Solution:
```
# 結局は木+1辺
from collections import defaultdict
import heapq as hq
N = int(input())
P = list(map(lambda x:int(x)-1,input().split()))
D = [0]*N
for p in P:
D[p] += 1
child_L = defaultdict(list)
S = [p for p in range(N) if D[p] == 0]
L = [None]*N
while S:
n = S.pop()
q = child_L[n]
del child_L[n]
hq.heapify(q)
i = 0
while q:
t = hq.heappop(q)
if i < t:
break
else:
i += (i==t)
L[n] = i
m = P[n]
child_L[m].append(i)
D[m] -= 1
if D[m] == 0:
S.append(m)
# cycle check
try:
start = D.index(1)
except ValueError:
print('POSSIBLE')
exit()
def helper(n):
q = child_L[n]
del child_L[n]
hq.heapify(q)
i = 0
while q:
t = hq.heappop(q)
if i < t:
break
else:
i += (i==t)
j = i+1
while q:
t = hq.heappop(q)
if j < t:
break
else:
j += (j==t)
return (i,j)
s1,s2 = helper(start)
G = []
n = P[start]
while n != start:
G.append(helper(n))
n = P[n]
del N,P,D,child_L,S,L
# 可能な初期値をそれぞれシミュレート
# 1
n = s1
for g in G:
if g[0] == n:
n = g[1]
else:
n = g[0]
if n != s1:
print('POSSIBLE')
exit()
# 2
n = s2
for g in G:
if g[0] == n:
n = g[1]
else:
n = g[0]
if n == s1:
print('POSSIBLE')
exit()
print('IMPOSSIBLE')
```
| 107,737 |
Provide a correct Python 3 solution for this coding contest problem.
There is a directed graph with N vertices and N edges. The vertices are numbered 1, 2, ..., N.
The graph has the following N edges: (p_1, 1), (p_2, 2), ..., (p_N, N), and the graph is weakly connected. Here, an edge from Vertex u to Vertex v is denoted by (u, v), and a weakly connected graph is a graph which would be connected if each edge was bidirectional.
We would like to assign a value to each of the vertices in this graph so that the following conditions are satisfied. Here, a_i is the value assigned to Vertex i.
* Each a_i is a non-negative integer.
* For each edge (i, j), a_i \neq a_j holds.
* For each i and each integer x(0 ≤ x < a_i), there exists a vertex j such that the edge (i, j) exists and x = a_j holds.
Determine whether there exists such an assignment.
Constraints
* 2 ≤ N ≤ 200 000
* 1 ≤ p_i ≤ N
* p_i \neq i
* The graph is weakly connected.
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 ... p_N
Output
If the assignment is possible, print `POSSIBLE`; otherwise, print `IMPOSSIBLE`.
Examples
Input
4
2 3 4 1
Output
POSSIBLE
Input
3
2 3 1
Output
IMPOSSIBLE
Input
4
2 3 1 1
Output
POSSIBLE
Input
6
4 5 6 5 6 4
Output
IMPOSSIBLE
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
"""
・pseudo-tree graph
・in_deg = 1が保証されているので、向きの揃ったcycleがある
・1箇所の2択を決めれば全部決まる
"""
N = int(readline())
parent = [0] + list(map(int,read().split()))
child = [[] for _ in range(N+1)]
for i,x in enumerate(parent):
child[x].append(i)
out_deg = [len(x) for x in child]
out_deg
G = [-1] * (N+1)
# とりあえず唯一のサイクル以外を処理：out_deg=0の頂点を捨てていく
stack = [i for i,x in enumerate(out_deg) if not x]
while stack:
x = stack.pop()
se = set(G[c] for c in child[x])
g = 0
while g in se:
g += 1
G[x] = g
p = parent[x]
out_deg[p] -= 1
if not out_deg[p]:
stack.append(p)
"""
・grundy数の候補は2種。
・片方決め打って計算。2周grundy数を計算してstableになっていればよい。
"""
for i,x in enumerate(out_deg[1:],1):
if x==1:
root = i
break
cycle_len = sum(out_deg[1:])
is_stable = False
x = root
for _ in range(cycle_len*2+10):
se = set(G[c] for c in child[x])
g = 0
while g in se:
g += 1
if g == G[x]:
is_stable = True
break
G[x] = g
x = parent[x]
answer = 'POSSIBLE' if is_stable else 'IMPOSSIBLE'
print(answer)
```
| 107,738 |
Provide a correct Python 3 solution for this coding contest problem.
There is a directed graph with N vertices and N edges. The vertices are numbered 1, 2, ..., N.
The graph has the following N edges: (p_1, 1), (p_2, 2), ..., (p_N, N), and the graph is weakly connected. Here, an edge from Vertex u to Vertex v is denoted by (u, v), and a weakly connected graph is a graph which would be connected if each edge was bidirectional.
We would like to assign a value to each of the vertices in this graph so that the following conditions are satisfied. Here, a_i is the value assigned to Vertex i.
* Each a_i is a non-negative integer.
* For each edge (i, j), a_i \neq a_j holds.
* For each i and each integer x(0 ≤ x < a_i), there exists a vertex j such that the edge (i, j) exists and x = a_j holds.
Determine whether there exists such an assignment.
Constraints
* 2 ≤ N ≤ 200 000
* 1 ≤ p_i ≤ N
* p_i \neq i
* The graph is weakly connected.
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 ... p_N
Output
If the assignment is possible, print `POSSIBLE`; otherwise, print `IMPOSSIBLE`.
Examples
Input
4
2 3 4 1
Output
POSSIBLE
Input
3
2 3 1
Output
IMPOSSIBLE
Input
4
2 3 1 1
Output
POSSIBLE
Input
6
4 5 6 5 6 4
Output
IMPOSSIBLE
"Correct Solution:
```
import sys
sys.setrecursionlimit(10**6)
n = int(input())
p = list(map(int, input().split()))
c = [[] for _ in range(n)]
is_leaf = [True for _ in range(n)]
for i in range(n):
p[i] -= 1
c[p[i]].append(i)
is_leaf[p[i]] = False
if sum(is_leaf) == 0:
if n%2 == 0:
print("POSSIBLE")
else:
print("IMPOSSIBLE")
sys.exit()
for i in range(n):
if is_leaf[i]:
cur = i
break
visited_set = {cur}
visited_list = [cur]
while p[cur] not in visited_set:
visited_list.append(p[cur])
visited_set.add(p[cur])
cur = p[cur]
root = p[cur]
grundy = [-1 for _ in range(n)]
g_set = [set() for _ in range(n)]
def dfs(x):
res = 0
for v in c[x]:
dfs(v)
g_set[x].add(grundy[v])
while res in g_set[x]:
res += 1
grundy[x] = res
return res
loop = [False for _ in range(n)]
loop[root] = True
ind = len(visited_list)-1
while visited_list[ind] != root:
loop[visited_list[ind]] = True
ind -= 1
#print(loop)
for i in range(n):
if loop[i]:
for x in c[i]:
if not loop[x]:
dfs(x)
g_set[i].add(grundy[x])
cand = []
num = 0
while num in g_set[root]:
num += 1
cand.append(num)
num += 1
while num in g_set[root]:
num += 1
cand.append(num)
for x in cand:
cur = root
grundy[root] = x
while True:
num = 0
while num in g_set[p[cur]] or num == grundy[cur]:
num += 1
grundy[p[cur]] = num
if p[cur] == root:
break
cur = p[cur]
if grundy[root] == x:
#print(grundy)
print("POSSIBLE")
sys.exit()
print("IMPOSSIBLE")
```
| 107,739 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a directed graph with N vertices and N edges. The vertices are numbered 1, 2, ..., N.
The graph has the following N edges: (p_1, 1), (p_2, 2), ..., (p_N, N), and the graph is weakly connected. Here, an edge from Vertex u to Vertex v is denoted by (u, v), and a weakly connected graph is a graph which would be connected if each edge was bidirectional.
We would like to assign a value to each of the vertices in this graph so that the following conditions are satisfied. Here, a_i is the value assigned to Vertex i.
* Each a_i is a non-negative integer.
* For each edge (i, j), a_i \neq a_j holds.
* For each i and each integer x(0 ≤ x < a_i), there exists a vertex j such that the edge (i, j) exists and x = a_j holds.
Determine whether there exists such an assignment.
Constraints
* 2 ≤ N ≤ 200 000
* 1 ≤ p_i ≤ N
* p_i \neq i
* The graph is weakly connected.
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 ... p_N
Output
If the assignment is possible, print `POSSIBLE`; otherwise, print `IMPOSSIBLE`.
Examples
Input
4
2 3 4 1
Output
POSSIBLE
Input
3
2 3 1
Output
IMPOSSIBLE
Input
4
2 3 1 1
Output
POSSIBLE
Input
6
4 5 6 5 6 4
Output
IMPOSSIBLE
Submitted Solution:
```
from collections import deque
N=int(input())
IN=[[] for i in range(N)]
OUT=[-1 for i in range(N)]
p=list(map(int,input().split()))
for i in range(N):
IN[p[i]-1].append(i)
OUT[i]=p[i]-1
deg=[len(IN[i]) for i in range(N)]
deq=deque([v for v in range(N) if deg[v]==0])
res=[]
while deq:
v=deq.popleft()
res.append(v)
deg[OUT[v]]-=1
if deg[OUT[v]]==0:
deq.append(OUT[v])
if len(res)==N:
exit(print("POSSIBLE"))
start=-1
for i in range(N):
if deg[i]>0:
start=i
break
cycle=[start]
while True:
nv=OUT[cycle[-1]]
if nv!=start:
cycle.append(nv)
else:
break
dp=[-1]*N
for v in res:
mex=[False]*(len(IN[v])+1)
for pv in IN[v]:
if dp[pv]<=len(IN[v]):
mex[dp[pv]]=True
for i in range(len(mex)):
if not mex[i]:
dp[v]=i
break
m0=-1
m1=-1
mex=[False]*(len(IN[start])+2)
for pv in IN[start]:
if dp[pv]!=-1 and dp[pv]<=len(IN[v])+1:
mex[dp[pv]]=True
for i in range(len(mex)):
if not mex[i]:
if m0==-1:
m0=i
else:
m1=i
break
#m0
dp[start]=m0
for i in range(1,len(cycle)):
v=cycle[i]
temp=-1
mex=[False]*(len(IN[v])+1)
for pv in IN[v]:
if dp[pv]<=len(IN[v]):
mex[dp[pv]]=True
for i in range(len(mex)):
if not mex[i]:
dp[v]=i
break
mex=[False]*(len(IN[start])+2)
for pv in IN[start]:
if dp[pv]!=-1 and dp[pv]<=len(IN[v])+1:
mex[dp[pv]]=True
check=-1
for i in range(len(mex)):
if not mex[i]:
check=i
break
if i==m0:
exit(print("POSSIBLE"))
```
No
| 107,740 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a directed graph with N vertices and N edges. The vertices are numbered 1, 2, ..., N.
The graph has the following N edges: (p_1, 1), (p_2, 2), ..., (p_N, N), and the graph is weakly connected. Here, an edge from Vertex u to Vertex v is denoted by (u, v), and a weakly connected graph is a graph which would be connected if each edge was bidirectional.
We would like to assign a value to each of the vertices in this graph so that the following conditions are satisfied. Here, a_i is the value assigned to Vertex i.
* Each a_i is a non-negative integer.
* For each edge (i, j), a_i \neq a_j holds.
* For each i and each integer x(0 ≤ x < a_i), there exists a vertex j such that the edge (i, j) exists and x = a_j holds.
Determine whether there exists such an assignment.
Constraints
* 2 ≤ N ≤ 200 000
* 1 ≤ p_i ≤ N
* p_i \neq i
* The graph is weakly connected.
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 ... p_N
Output
If the assignment is possible, print `POSSIBLE`; otherwise, print `IMPOSSIBLE`.
Examples
Input
4
2 3 4 1
Output
POSSIBLE
Input
3
2 3 1
Output
IMPOSSIBLE
Input
4
2 3 1 1
Output
POSSIBLE
Input
6
4 5 6 5 6 4
Output
IMPOSSIBLE
Submitted Solution:
```
import sys
from itertools import chain
readline = sys.stdin.readline
#非再帰
def scc(Edge):
N = len(Edge)
Edgeinv = [[] for _ in range(N)]
for vn in range(N):
for vf in Edge[vn]:
Edgeinv[vf].append(vn)
used = [False]*N
dim = [len(Edge[i]) for i in range(N)]
order = []
for st in range(N):
if not used[st]:
stack = [st, 0]
while stack:
vn, i = stack[-2], stack[-1]
if not i and used[vn]:
stack.pop()
stack.pop()
else:
used[vn] = True
if i < dim[vn]:
stack[-1] += 1
stack.append(Edge[vn][i])
stack.append(0)
else:
stack.pop()
order.append(stack.pop())
res = [None]*N
used = [False]*N
cnt = -1
for st in order[::-1]:
if not used[st]:
cnt += 1
stack = [st]
res[st] = cnt
used[st] = True
while stack:
vn = stack.pop()
for vf in Edgeinv[vn]:
if not used[vf]:
used[vf] = True
res[vf] = cnt
stack.append(vf)
M = cnt+1
components = [[] for _ in range(M)]
for i in range(N):
components[res[i]].append(i)
tEdge = [[] for _ in range(M)]
teset = set()
for vn in range(N):
tn = res[vn]
for vf in Edge[vn]:
tf = res[vf]
if tn != tf and tn*M + tf not in teset:
teset.add(tn*M + tf)
tEdge[tn].append(tf)
return res, components, tEdge
N = int(readline())
P = list(map(lambda x: int(x)-1, readline().split()))
Edge = [[] for _ in range(N)]
for i in range(N):
Edge[P[i]].append(i)
R, Com, _ = scc(Edge)
Lord = list(chain(*Com[::-1]))
val = [None]*N
for vn in Lord:
if not R[vn]:
break
lvn = len(Edge[vn]) + 1
res = [0]*lvn
for vf in Edge[vn]:
if val[vf] < lvn:
res[val[vf]] += 1
for k in range(lvn):
if not res[k]:
val[vn] = k
break
st = Lord[-1]
lst = len(Edge[st]) + 2
res = [0]*lst
for vf in Edge[st]:
if val[vf] is None:
continue
if val[vf] < lst:
res[val[vf]] += 1
mc = []
for k in range(lst):
if not res[k]:
mc.append(k)
vn = st
Ls = []
while vn is not None:
for vf in Edge[vn]:
if R[vf]:
continue
if vf == st:
vn = None
else:
Ls.append(vf)
vn = vf
Ls.reverse()
ans = False
for idx in range(2):
vc = val[:]
vc[st] = mc[idx]
for vn in Ls:
lvn = len(Edge[vn])+1
res = [0]*lvn
for vf in Edge[vn]:
if vc[vf] < lvn:
res[vc[vf]] += 1
for k in range(lvn):
if not res[k]:
vc[vn] = k
break
for vn in range(N):
assert vc[vn] is not None, 'Er'
for vf in Edge[vn]:
if vc[vn] == vc[vf]:
break
else:
continue
break
else:
ans = True
break
print('POSSIBLE' if ans else 'IMPOSSIBLE')
```
No
| 107,741 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a directed graph with N vertices and N edges. The vertices are numbered 1, 2, ..., N.
The graph has the following N edges: (p_1, 1), (p_2, 2), ..., (p_N, N), and the graph is weakly connected. Here, an edge from Vertex u to Vertex v is denoted by (u, v), and a weakly connected graph is a graph which would be connected if each edge was bidirectional.
We would like to assign a value to each of the vertices in this graph so that the following conditions are satisfied. Here, a_i is the value assigned to Vertex i.
* Each a_i is a non-negative integer.
* For each edge (i, j), a_i \neq a_j holds.
* For each i and each integer x(0 ≤ x < a_i), there exists a vertex j such that the edge (i, j) exists and x = a_j holds.
Determine whether there exists such an assignment.
Constraints
* 2 ≤ N ≤ 200 000
* 1 ≤ p_i ≤ N
* p_i \neq i
* The graph is weakly connected.
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 ... p_N
Output
If the assignment is possible, print `POSSIBLE`; otherwise, print `IMPOSSIBLE`.
Examples
Input
4
2 3 4 1
Output
POSSIBLE
Input
3
2 3 1
Output
IMPOSSIBLE
Input
4
2 3 1 1
Output
POSSIBLE
Input
6
4 5 6 5 6 4
Output
IMPOSSIBLE
Submitted Solution:
```
# coding: utf-8
from collections import defaultdict
def II(): return int(input())
def ILI(): return list(map(int, input().split()))
def read():
N = II()
p = ILI()
return N, p
def solve(N, p):
ans = None
edges = defaultdict(list)
for i in range(N):
edges[p[i]].append(i + 1)
nodes_has_branch = []
for i in range(1, N + 1):
if len(edges[i]) == 2:
nodes_has_branch.append(i)
if len(nodes_has_branch) == 0:
if N % 2 == 0:
ans = "POSSIBLE"
else:
ans = "IMPOSSIBLE"
else:
first_node = nodes_has_branch[0]
edges_reverse = defaultdict(list)
for fro, to in edges.items():
for i in to:
edges_reverse[i].append(fro)
circles = []
next_node = edges_reverse[first_node]
while True:
if next_node[0] == first_node:
break
circles.append(next_node[0])
next_node = edges_reverse[next_node[0]]
circles = [first_node] + list(reversed(circles))
s_circles = set(circles)
s_branch = set(range(1, N + 1)) - s_circles
circles_depth = [0] * len(circles)
for i, node in enumerate(circles):
next_node = edges[node]
if len(next_node) == 1:
continue
else:
next_node = list((set(next_node) & s_branch))[0]
while True:
circles_depth[i] += 1
next_node = edges[next_node]
if len(next_node) == 0:
break
if not 0 in circles_depth:
ans = "IMPOSSIBLE"
else:
ans = "POSSIBLE"
return ans
def main():
params = read()
print(solve(*params))
if __name__ == "__main__":
main()
```
No
| 107,742 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a directed graph with N vertices and N edges. The vertices are numbered 1, 2, ..., N.
The graph has the following N edges: (p_1, 1), (p_2, 2), ..., (p_N, N), and the graph is weakly connected. Here, an edge from Vertex u to Vertex v is denoted by (u, v), and a weakly connected graph is a graph which would be connected if each edge was bidirectional.
We would like to assign a value to each of the vertices in this graph so that the following conditions are satisfied. Here, a_i is the value assigned to Vertex i.
* Each a_i is a non-negative integer.
* For each edge (i, j), a_i \neq a_j holds.
* For each i and each integer x(0 ≤ x < a_i), there exists a vertex j such that the edge (i, j) exists and x = a_j holds.
Determine whether there exists such an assignment.
Constraints
* 2 ≤ N ≤ 200 000
* 1 ≤ p_i ≤ N
* p_i \neq i
* The graph is weakly connected.
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 ... p_N
Output
If the assignment is possible, print `POSSIBLE`; otherwise, print `IMPOSSIBLE`.
Examples
Input
4
2 3 4 1
Output
POSSIBLE
Input
3
2 3 1
Output
IMPOSSIBLE
Input
4
2 3 1 1
Output
POSSIBLE
Input
6
4 5 6 5 6 4
Output
IMPOSSIBLE
Submitted Solution:
```
import sys
from itertools import chain
readline = sys.stdin.readline
#非再帰
def scc(Edge):
N = len(Edge)
Edgeinv = [[] for _ in range(N)]
for vn in range(N):
for vf in Edge[vn]:
Edgeinv[vf].append(vn)
used = [False]*N
dim = [len(Edge[i]) for i in range(N)]
order = []
for st in range(N):
if not used[st]:
stack = [st, 0]
while stack:
vn, i = stack[-2], stack[-1]
if not i and used[vn]:
stack.pop()
stack.pop()
else:
used[vn] = True
if i < dim[vn]:
stack[-1] += 1
stack.append(Edge[vn][i])
stack.append(0)
else:
stack.pop()
order.append(stack.pop())
res = [None]*N
used = [False]*N
cnt = -1
for st in order[::-1]:
if not used[st]:
cnt += 1
stack = [st]
res[st] = cnt
used[st] = True
while stack:
vn = stack.pop()
for vf in Edgeinv[vn]:
if not used[vf]:
used[vf] = True
res[vf] = cnt
stack.append(vf)
M = cnt+1
components = [[] for _ in range(M)]
for i in range(N):
components[res[i]].append(i)
tEdge = [[] for _ in range(M)]
teset = set()
for vn in range(N):
tn = res[vn]
for vf in Edge[vn]:
tf = res[vf]
if tn != tf and tn*M + tf not in teset:
teset.add(tn*M + tf)
tEdge[tn].append(tf)
return res, components, tEdge
N = int(readline())
P = list(map(lambda x: int(x)-1, readline().split()))
Edge = [[] for _ in range(N)]
for i in range(N):
Edge[P[i]].append(i)
R, Com, _ = scc(Edge)
assert len(Com[0]) > 1 ,''
Lord = list(chain(*Com[::-1]))
val = [None]*N
for vn in Lord:
if not R[vn]:
break
lvn = len(Edge[vn]) + 1
res = [0]*lvn
for vf in Edge[vn]:
if val[vf] < lvn:
res[val[vf]] += 1
for k in range(lvn):
if not res[k]:
val[vn] = k
break
st = Lord[-1]
lst = len(Edge[st]) + 2
res = [0]*lst
for vf in Edge[st]:
if val[vf] is None:
continue
if val[vf] < lst:
res[val[vf]] += 1
mc = []
for k in range(lst):
if not res[k]:
mc.append(k)
vn = st
Ls = []
while vn is not None:
for vf in Edge[vn]:
if R[vf]:
continue
if vf == st:
vn = None
else:
Ls.append(vf)
vn = vf
Ls.reverse()
ans = False
for idx in range(2):
vc = val[:]
vc[st] = mc[idx]
for vn in Ls:
lvn = len(Edge[vn])+1
res = [0]*lvn
for vf in Edge[vn]:
if vc[vf] < lvn:
res[vc[vf]] += 1
for k in range(lvn):
if not res[k]:
vc[vn] = k
break
for vn in range(N):
for vf in Edge[vn]:
if vc[vn] == vc[vf]:
break
else:
continue
break
else:
ans = True
break
print('POSSIBLE' if ans else 'IMPOSSIBLE')
```
No
| 107,743 |
Provide a correct Python 3 solution for this coding contest problem.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
"Correct Solution:
```
N,A,B=map(int,input().split())
inf=float("inf")
dp=[[inf]*401 for _ in range(401)]
dp[0][0]=0
for i in range(N):
a,b,c=map(int,input().split())
for j in range(-1,-402,-1):
for k in range(-1,-402,-1):
if dp[j][k]!=inf:
dp[j+a][k+b]=min(dp[j+a][k+b],dp[j][k]+c)
ans=inf
for i in range(1,401):
for j in range(1,401):
if i/j==A/B:
ans=min(ans,dp[i][j])
if ans==inf:
ans=-1
print(ans)
```
| 107,744 |
Provide a correct Python 3 solution for this coding contest problem.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
"Correct Solution:
```
# -*- coding: utf-8 -*-
def inpl(): return map(int, input().split())
N, Ma, Mb = inpl()
INF = 4001
DP = [[INF]*(411) for _ in range(411)]
DP[0][0] = 0
for _ in range(N):
DP2 = [[INF]*(411) for _ in range(411)]
a, b, c = inpl()
for i in range(400):
for j in range(400):
DP2[i][j] = min(DP[i][j], DP2[i][j])
DP2[i+a][j+b] = min(DP[i+a][j+b], DP[i][j] + c)
DP = DP2
x = 1
ans = INF
while Ma*x <= 400 and Mb*x <= 400:
ans = min(ans, DP[Ma*x][Mb*x])
x += 1
if ans < INF:
print(ans)
else:
print(-1)
```
| 107,745 |
Provide a correct Python 3 solution for this coding contest problem.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
"Correct Solution:
```
def inpl(): return [int(i) for i in input().split()]
N,Ma,Mb = inpl()
inf = float('inf')
H = {(0,0): 0}
for i in range(N):
a, b, c = inpl()
for (ia, ib), ic in H.copy().items():
H[(ia+a,ib+b)] = min(H.get((ia+a, ib+b), inf),ic+c)
ans = min(H.get((i*Ma, i*Mb), inf) for i in range(1,401))
print(-1 if ans == inf else ans)
```
| 107,746 |
Provide a correct Python 3 solution for this coding contest problem.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
"Correct Solution:
```
n,ma,mb=map(int,input().split())
dp=[[[10**30 for j in range(401)]for i in range(401)]for k in range(n+1)]
dp[0][0][0]=0
l=[]
for i in range(n):
a,b,c=map(int,input().split())
l.append([a,b,c])
for i in range(1,n+1):
a,b,c=l[i-1][0],l[i-1][1],l[i-1][2]
for x in range(401):
for y in range(401):
if x<a or y<b:
dp[i][x][y]=dp[i-1][x][y]
continue
dp[i][x][y]=min(dp[i-1][x][y],dp[i-1][x-a][y-b]+c)
mini=10**30
for i in range(1,40):
mini=min(mini,dp[n][ma*i][mb*i])
if mini>=10**30 :
print(-1)
else:
print(mini)
```
| 107,747 |
Provide a correct Python 3 solution for this coding contest problem.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
"Correct Solution:
```
inf = float('inf')
N,Ma,Mb = map(int,input().split())
abc = [list(map(int,input().split())) for _ in range(N)]
# dp[i][j][k]: i番目までみてj(g), k(g)となるような最小予算
dp = [[[inf]*401 for _ in range(401)] for _ in range(N+1)]
dp[0][0][0] = 0
for i in range(N):
a,b,c = abc[i]
for j in range(401):
for k in range(401):
if j - a >= 0 and k - b >= 0:
dp[i+1][j][k] = min(dp[i][j][k], dp[i][j-a][k-b] + c)
else:
dp[i+1][j][k] = dp[i][j][k]
ans = inf
for j in range(1,401):
for k in range(1,401):
if j * Mb == Ma * k:
ans = min(ans, dp[N][j][k])
if ans == inf:
ans = -1
print(ans)
```
| 107,748 |
Provide a correct Python 3 solution for this coding contest problem.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
"Correct Solution:
```
INFTY = 10**4
N,Ma,Mb = map(int,input().split())
A = [list(map(int,input().split())) for _ in range(N)]
dp = [[[INFTY for _ in range(401)] for _ in range(401)] for _ in range(N+1)]
dp[0][0][0] = 0
for i in range(1,N+1):
for j in range(401):
for k in range(401):
dp[i][j][k] = dp[i-1][j][k]
a = A[i-1][0]
b = A[i-1][1]
c = A[i-1][2]
if j>=a and k>=b:
dp[i][j][k] = min(dp[i][j][k],dp[i-1][j-a][k-b]+c)
cmin = INFTY
for j in range(1,401):
for k in range(1,401):
if j*Mb==k*Ma:
cmin = min(cmin,dp[N][j][k])
if cmin>=INFTY:
print(-1)
else:
print(cmin)
```
| 107,749 |
Provide a correct Python 3 solution for this coding contest problem.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
"Correct Solution:
```
N, Ma, Mb = map(int, input().split())
table = []
for i in range(N):
a,b,c = map(int, input().split())
table.append([a,b,c])
inf = 10**9
dp = [[[inf]*401 for i in range(401)] for i in range(N+1)]
dp[0][0][0] = 0
for i in range(1,N+1):
a,b,c = table[i-1]
for j in range(401):
for k in range(401):
dp[i][j][k] = dp[i-1][j][k]
if 400>=j-a>=0 and 400>=k-b>=0:
dp[i][j][k] = min(dp[i-1][j][k],dp[i-1][j-a][k-b]+c)
ans = 10**9
i = 1
while max(Ma,Mb)*i <= 400:
ans = min(ans,dp[N][Ma*i][Mb*i])
i += 1
if ans == 10**9:
ans = -1
print(ans)
```
| 107,750 |
Provide a correct Python 3 solution for this coding contest problem.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
"Correct Solution:
```
INF = float('inf')
MAXAB = 10
n, ma, mb = map(int, input().split())
t = [[INF] * (n * MAXAB + 1) for _ in range(n * MAXAB + 1)]
t[0][0] = 0
for _ in range(n):
a, b, c = map(int, input().split())
for aa in range(n * MAXAB, -1, -1):
for bb in range(n * MAXAB, -1, -1):
if t[aa][bb] == INF:
continue
if t[a + aa][b + bb] > t[aa][bb] + c:
t[a + aa][b + bb] = t[aa][bb] + c
result = INF
for a in range(1, n * MAXAB + 1):
for b in range(1, n * MAXAB + 1):
if a * mb == b * ma and result > t[a][b]:
result = t[a][b]
if result == INF:
result = -1
print(result)
```
| 107,751 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
Submitted Solution:
```
# DP
INF = float('inf')
N, Ma, Mb = map(int, input().split())
cmax = N * 10
t = [[INF] * (cmax + 1) for _ in range(cmax + 1)]
for _ in range(N):
a, b, c = map(int, input().split())
for aa in range(cmax, 0, -1):
for bb in range(cmax, 0, -1):
if t[aa][bb] == INF:
continue
t[aa + a][bb + b] = min(t[aa + a][bb + b], t[aa][bb] + c)
t[a][b] = min(t[a][b], c)
result = INF
for a in range(1, cmax):
for b in range(1, cmax):
if a * Mb == b * Ma:
result = min(result, t[a][b])
if result == INF:
result = -1
print(result)
```
Yes
| 107,752 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
Submitted Solution:
```
n, ma, mb = map(int, input().split())
abc = []
for _ in range(n):
a, b, c = map(int, input().split())
abc.append((a, b, c))
dp = {(0, 0): 0}
for a, b, c in abc:
newdp = dp.copy()
for (ka, kb), p in dp.items():
k = (a+ka, b+kb)
if k in newdp:
newdp[k] = min(newdp[k], p+c)
else:
newdp[k] = p+c
dp = newdp
INF = 10**10
ans = INF
for (a, b), c in dp.items():
if ma*b!=mb*a or a==b==0:
continue
ans = min(ans, c)
if ans==INF:
print(-1)
else:
print(ans)
```
Yes
| 107,753 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
Submitted Solution:
```
N,Ma,Mb=map(int,input().split())
a,b,c=map(list,zip(*[list(map(int,input().split())) for i in range(N)]))
ma,mb=sum(a),sum(b)
MAX=5000
dp=[[[MAX]*(mb+1) for i in range(ma+1)] for j in range(N+1)]
dp[0][0][0]=0
for i in range(N):
for j in range(ma):
for k in range(mb):
dp[i+1][j][k]=min(dp[i][j][k],dp[i+1][j][k])
if j+a[i]<=ma and k+b[i]<=mb:
dp[i+1][j+a[i]][k+b[i]]=min(dp[i+1][j+a[i]][k+b[i]],dp[i][j][k]+c[i])
ans=MAX
x=1
while x*Ma<=ma and x*Mb<=mb:
ans=min(ans,dp[N][x*Ma][x*Mb])
x+=1
print(ans if ans<MAX else -1)
```
Yes
| 107,754 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
Submitted Solution:
```
n, A, B = map(int, input().split())
z = [tuple(map(int, input().split())) for _ in range(n)]
inf = 10 ** 18
d = {(0, 0):0}
for a, b, cost in z:
newd = []
for (x, y), c in d.items():
new = (a+x, b+y)
if new not in d or d[new] > c + cost:
newd.append((new, c+cost))
for new, newc in newd:
d[new] = newc
ans = inf
for i in range(1, 10 * n):
t = (A * i, B * i)
if t in d:
ans = min(ans, d[t])
if ans == inf:
print(-1)
else:
print(ans)
```
Yes
| 107,755 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
Submitted Solution:
```
N, Ma, Mb = map(int, input().split())
A = [0]*N
B = [0]*N
C = [0]*N
for i in range(N):
A[i], B[i], C[i] = map(int, input().split())
# print(N, Ma, Mb, A, B, C)
A = [Mb * a for a in A]
B = [Ma * b for b in B]
D = []
for i in range(N):
D.append((A[i] - B[i], C[i]))
# print(N, Ma, Mb, A, B, C, D)
Dp = []
Dn = []
Dz = []
for d in D:
if d[0] == 0:
Dz.append(d)
elif d[0] > 0:
Dp.append(d)
else:
Dn.append(d)
Dp.sort(key=lambda x: x[0])
Dn.sort(key=lambda x: x[0])
Dz.sort(key=lambda x: x[1])
min_c = 100*41
if len(Dz) > 0:
min_c = Dz[0][1]
# print(Dp, Dn, Dz, min_c)
Ds = []
Dl = []
if len(Dp) < len(Dn):
Ds, Dl = Dp, Dn
else:
Ds, Dl = Dn, Dp
def calc(D, l):
s = 0
c = 0
for i in range(len(D)):
if (l >> i) & 1 != 0:
s += D[i][0]
c += D[i][1]
return s, c
Ls = len(Ds)
Ll = len(Dl)
for ls in range(Ls):
s, c = calc(Ds, ls + 1)
# print('ls', ls, s, c)
if c >= min_c:
continue
for ll in range(Ll):
sl, cl = calc(Dl, ll + 1)
# print('ll', ll, sl, cl)
if c >= min_c:
continue
if s + sl == 0:
min_c = c + cl
if min_c == 100*41:
print(-1)
else:
print(min_c)
```
No
| 107,756 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
Submitted Solution:
```
import math
import copy
from operator import mul
from functools import reduce
from collections import defaultdict
from collections import Counter
from collections import deque
# 直積 A={a, b, c}, B={d, e}:のとき，A×B={(a,d),(a,e),(b,d),(b,e),(c,d),(c,e)}: product(A, B)
from itertools import product
# 階乗 P!: permutations(seq), 順列 {}_len(seq) P_n: permutations(seq, n)
from itertools import permutations
# 組み合わせ {}_len(seq) C_n: combinations(seq, n)
from itertools import combinations
# 一次元累積和
from itertools import accumulate
from bisect import bisect_left, bisect_right
import re
# import numpy as np
# from scipy.misc import comb
# 再帰がやばいとき
# import sys
# sys.setrecursionlimit(10**9)
def inside(y, x, H, W):
return 0 <= y < H and 0 <= x < W
# 四方向: 右, 下, 左, 上
dy = [0, -1, 0, 1]
dx = [1, 0, -1, 0]
def i_inpl(): return int(input())
def l_inpl(): return list(map(int, input().split()))
def line_inpl(x): return [i_inpl() for _ in range(x)]
INF = int(1e50)
MOD = int(1e9)+7 # 10^9 + 7
# field[H][W]
def create_grid(H, W, value = 0):
return [[ value for _ in range(W)] for _ in range(H)]
########
def main():
N, Ma, Mb = l_inpl()
abc = []
for _ in range(N):
abci = l_inpl()
abc.append(abci)
max_ab = 40*10
# dp[i][j][k]: i番目までの薬品の組み合わせで，物質aがjグラム，物質bがkグラムのときのコスト
dp = [[[INF for _ in range(max_ab+1)] for _ in range(max_ab+1) ] for _ in range(N+1)]
dp[0][0][0] = 0
for i in range(N):
for j in range(max_ab+1):
for k in range(max_ab+1):
# ここでのINFはNoneと同じ意味なので
if dp[i][j][k] == INF:
continue
# i番目の薬品を加えない
dp[i+1][j][k] = min(dp[i+1][j][k], dp[i][j][k])
# i番目の薬品を加える
dp[i+1][j+abc[i][0]][k+abc[i][1]]
dp[i+1][j+abc[i][0]][k+abc[i][1]] = min(dp[i+1][j+abc[i][0]][k+abc[i][1]], dp[i+1][j][k] + abc[i][2])
ans = INF
for ca in range(1, max_ab+1):
for cb in range(1, max_ab+1):
if ca*Mb == cb*Ma:
ans = min(ans, dp[N][ca][cb])
if ans == INF:
print(-1)
else:
print(ans)
if __name__ == "__main__":
main()
```
No
| 107,757 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
Submitted Solution:
```
n,ma,mb=map(int,input().split())
dp=[[[9999]*401for _ in range(401)]for _ in range(n+1)];dp[0][0][0]=0
sa=sb=0
for i in range(1,n+1):
a,b,c=map(int,input().split())
sa+=a;sb+=b
for j in range(401):
for k in range(401):
if j>=a and k>=b:dp[i][j][k]=min(dp[i-1][j-a][k-b]+c,dp[i-1][j][k])
else:dp[i][j][k]=dp[i-1][j][k]
a=9999
for i in range(1,min(sa//ma,sb//mb)+1):
if dp[n][ma*i][mb*i]!=9999:a=min(a,dp[n][ma*i][mb*i])
print(-1if a==9999else a)
```
No
| 107,758 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dolphin is planning to generate a small amount of a certain chemical substance C.
In order to generate the substance C, he must prepare a solution which is a mixture of two substances A and B in the ratio of M_a:M_b.
He does not have any stock of chemicals, however, so he will purchase some chemicals at a local pharmacy.
The pharmacy sells N kinds of chemicals. For each kind of chemical, there is exactly one package of that chemical in stock.
The package of chemical i contains a_i grams of the substance A and b_i grams of the substance B, and is sold for c_i yen (the currency of Japan).
Dolphin will purchase some of these packages. For some reason, he must use all contents of the purchased packages to generate the substance C.
Find the minimum amount of money required to generate the substance C.
If it is not possible to generate the substance C by purchasing any combination of packages at the pharmacy, report that fact.
Constraints
* 1≦N≦40
* 1≦a_i,b_i≦10
* 1≦c_i≦100
* 1≦M_a,M_b≦10
* gcd(M_a,M_b)=1
* a_i, b_i, c_i, M_a and M_b are integers.
Input
The input is given from Standard Input in the following format:
N M_a M_b
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N
Output
Print the minimum amount of money required to generate the substance C. If it is not possible to generate the substance C, print `-1` instead.
Examples
Input
3 1 1
1 2 1
2 1 2
3 3 10
Output
3
Input
1 1 10
10 10 10
Output
-1
Submitted Solution:
```
N, Ma, Mb = map(int, input().split())
INF = 10 ** 8
lst = [[INF] * 401 for i in range(401)]
lst[0][0] = 0
for _ in range(N):
a, b, c = map(int, input().split())
for i in range(a, 401):
for j in range(b, 401):
lst[i][j] = min(lst[i][j], lst[i - a][j - b] + c)
ans = INF
n = 400 // max(Ma, Mb)
for i in range(1, n + 1):
ans = min(ans, lst[Ma * i][Mb * i])
if ans == INF:
print (-1)
else:
print (ans)
```
No
| 107,759 |
Provide a correct Python 3 solution for this coding contest problem.
There are N strings of lowercase alphabet only. The i-th string is S_i. Every string is unique.
Provide answers for the Q queries below. The i-th query has the following format:
Query: An integer k_i and a string p_{i,1}p_{i,2}...p_{i,26} that results from permuting {`a`,`b`,...,`z`} are given. Output the sequence of the string S_{k_i} among the N strings in lexicographical order when the literal sequence is p_{i,1}<p_{i,2}<...<p_{i,26}.
Constraints
* 1 ≦ N,Q ≦ 100000
* 1 ≦ |S_i| (1 ≦ i ≦ N)
* S_i (1 ≦ i ≦ N) is a string of lowercase alphabet.
* The sum of |S_i| is no more than 400000.
* Every S_i is unique.
* 1 ≦ k_i ≦ N (1 ≦ i ≦ Q)
* For all 1 ≦ i ≦ Q, p_{i,1}p_{i,2}...p_{i,26} is a permutation of `abcd...z`.
Input
Inputs are provided from standard inputs in the following form.
N
S_1
:
S_N
Q
k_1 p_{1,1}p_{1,2}...p_{1,26}
:
k_Q p_{Q,1}p_{Q,2}...p_{Q,26}
Output
Output Q lines.
On line i, for the i-th query, output an integer indicating the sequence of the string S_{k_i} among the N strings in lexicographical order.
Examples
Input
5
aa
abbaa
abbba
aaab
aaaaaba
5
1 abcdefghijklmnopqrstuvwxyz
2 bacdefghijklmnopqrstuvwxyz
3 abcdefghijklmnopqrstuvwxyz
4 bacdefghijklmnopqrstuvwxyz
5 abcdefghijklmnopqrstuvwxyz
Output
1
2
5
4
2
Input
8
abrakatabra
abadaba
abracadabra
atcoder
grand
contest
ababa
a
6
3 abcdefghijklmnopqrstuvwxyz
6 qwertyuiopasdfghjklzxcvbnm
8 poiuytrewqlkjhgfdsamnbvcxz
2 qazwsxedcrfvtgbyhnujmikolp
1 plokmijnuhbygvtfcrdxeszwaq
4 mnbvcxzasdfghjklpoiuytrewq
Output
4
8
2
3
4
7
"Correct Solution:
```
# 文字列の問題で、同じような文字列をまとめたいので、
# どう考えても Trie を使いたい気持ちになる
# あとは適当にやったらなんとかなった
# 1100 点の割には簡単っぽい
class TrieNode:
def __init__(self):
self.child = {}
self.count = 0 # 自身を含めた部分木の大きさ
class Trie:
def __init__(self):
self.root = TrieNode()
def add(self, s):
node = self.root
node.count += 1
for c in s:
c = ord(c) - 97
if c not in node.child:
node.child[c] = TrieNode()
node = node.child[c]
node.count += 1
node.child[26] = TrieNode()
node.child[26].count += 1
def query(self, s):
res = [[0]*27 for _ in range(26)]
node = self.root
for c in s:
c = ord(c) - 97
res_c = res[c]
for i, child in node.child.items():
res_c[i] += child.count
node = node.child[c]
return res
from itertools import groupby
import sys
input = sys.stdin.readline
N = int(input())
S = [input()[:-1] for _ in range(N)]
trie = Trie()
for s in S:
trie.add(s)
Q = int(input())
KP = [input().split() for _ in range(Q)]
Ans = [-1] * Q
Idx_Q = sorted(range(Q), key=lambda x: KP[x][0])
for v, g in groupby(Idx_Q, key=lambda x: KP[x][0]):
K = int(v) - 1
L_K = trie.query(S[K])
for idx in g:
_, P = KP[idx]
P = [ord(c) - 97 for c in P]
ans = 1
for i, pi in enumerate(P):
for pj in P[:i]:
ans += L_K[pi][pj]
ans += L_K[pi][26]
Ans[idx] = ans
print("\n".join(map(str, Ans)))
```
| 107,760 |
Provide a correct Python 3 solution for this coding contest problem.
There are N strings of lowercase alphabet only. The i-th string is S_i. Every string is unique.
Provide answers for the Q queries below. The i-th query has the following format:
Query: An integer k_i and a string p_{i,1}p_{i,2}...p_{i,26} that results from permuting {`a`,`b`,...,`z`} are given. Output the sequence of the string S_{k_i} among the N strings in lexicographical order when the literal sequence is p_{i,1}<p_{i,2}<...<p_{i,26}.
Constraints
* 1 ≦ N,Q ≦ 100000
* 1 ≦ |S_i| (1 ≦ i ≦ N)
* S_i (1 ≦ i ≦ N) is a string of lowercase alphabet.
* The sum of |S_i| is no more than 400000.
* Every S_i is unique.
* 1 ≦ k_i ≦ N (1 ≦ i ≦ Q)
* For all 1 ≦ i ≦ Q, p_{i,1}p_{i,2}...p_{i,26} is a permutation of `abcd...z`.
Input
Inputs are provided from standard inputs in the following form.
N
S_1
:
S_N
Q
k_1 p_{1,1}p_{1,2}...p_{1,26}
:
k_Q p_{Q,1}p_{Q,2}...p_{Q,26}
Output
Output Q lines.
On line i, for the i-th query, output an integer indicating the sequence of the string S_{k_i} among the N strings in lexicographical order.
Examples
Input
5
aa
abbaa
abbba
aaab
aaaaaba
5
1 abcdefghijklmnopqrstuvwxyz
2 bacdefghijklmnopqrstuvwxyz
3 abcdefghijklmnopqrstuvwxyz
4 bacdefghijklmnopqrstuvwxyz
5 abcdefghijklmnopqrstuvwxyz
Output
1
2
5
4
2
Input
8
abrakatabra
abadaba
abracadabra
atcoder
grand
contest
ababa
a
6
3 abcdefghijklmnopqrstuvwxyz
6 qwertyuiopasdfghjklzxcvbnm
8 poiuytrewqlkjhgfdsamnbvcxz
2 qazwsxedcrfvtgbyhnujmikolp
1 plokmijnuhbygvtfcrdxeszwaq
4 mnbvcxzasdfghjklpoiuytrewq
Output
4
8
2
3
4
7
"Correct Solution:
```
# 文字列の問題で、同じような文字列をまとめたいので、
# どう考えても Trie を使いたい気持ちになる
# あとは適当にやったらなんとかなった
# 1100 点の割には簡単っぽい
class TrieNode:
def __init__(self):
self.child = {}
self.count = 0 # 自身を含めた部分木の大きさ
class Trie:
def __init__(self):
self.root = TrieNode()
def add(self, s):
node = self.root
node.count += 1
for c in s:
c = ord(c) - 97
if c not in node.child:
node.child[c] = TrieNode()
node = node.child[c]
node.count += 1
node.child[26] = TrieNode()
node.child[26].count += 1
def query(self, s):
res = [[0]*27 for _ in range(26)]
node = self.root
for c in s:
c = ord(c) - 97
res_c = res[c]
for i, child in node.child.items():
res_c[i] += child.count
node = node.child[c]
return res
from itertools import groupby
import sys
def main():
input = sys.stdin.readline
N = int(input())
S = [input()[:-1] for _ in range(N)]
trie = Trie()
for s in S:
trie.add(s)
Q = int(input())
KP = [input().split() for _ in range(Q)]
Ans = [-1] * Q
Idx_Q = sorted(range(Q), key=lambda x: KP[x][0])
for v, g in groupby(Idx_Q, key=lambda x: KP[x][0]):
K = int(v) - 1
L_K = trie.query(S[K])
for idx in g:
_, P = KP[idx]
P = [ord(c) - 97 for c in P]
ans = 1
for i, pi in enumerate(P):
for pj in P[:i]:
ans += L_K[pi][pj]
ans += L_K[pi][26]
Ans[idx] = ans
print("\n".join(map(str, Ans)))
main()
```
| 107,761 |
Provide a correct Python 3 solution for this coding contest problem.
There are N strings of lowercase alphabet only. The i-th string is S_i. Every string is unique.
Provide answers for the Q queries below. The i-th query has the following format:
Query: An integer k_i and a string p_{i,1}p_{i,2}...p_{i,26} that results from permuting {`a`,`b`,...,`z`} are given. Output the sequence of the string S_{k_i} among the N strings in lexicographical order when the literal sequence is p_{i,1}<p_{i,2}<...<p_{i,26}.
Constraints
* 1 ≦ N,Q ≦ 100000
* 1 ≦ |S_i| (1 ≦ i ≦ N)
* S_i (1 ≦ i ≦ N) is a string of lowercase alphabet.
* The sum of |S_i| is no more than 400000.
* Every S_i is unique.
* 1 ≦ k_i ≦ N (1 ≦ i ≦ Q)
* For all 1 ≦ i ≦ Q, p_{i,1}p_{i,2}...p_{i,26} is a permutation of `abcd...z`.
Input
Inputs are provided from standard inputs in the following form.
N
S_1
:
S_N
Q
k_1 p_{1,1}p_{1,2}...p_{1,26}
:
k_Q p_{Q,1}p_{Q,2}...p_{Q,26}
Output
Output Q lines.
On line i, for the i-th query, output an integer indicating the sequence of the string S_{k_i} among the N strings in lexicographical order.
Examples
Input
5
aa
abbaa
abbba
aaab
aaaaaba
5
1 abcdefghijklmnopqrstuvwxyz
2 bacdefghijklmnopqrstuvwxyz
3 abcdefghijklmnopqrstuvwxyz
4 bacdefghijklmnopqrstuvwxyz
5 abcdefghijklmnopqrstuvwxyz
Output
1
2
5
4
2
Input
8
abrakatabra
abadaba
abracadabra
atcoder
grand
contest
ababa
a
6
3 abcdefghijklmnopqrstuvwxyz
6 qwertyuiopasdfghjklzxcvbnm
8 poiuytrewqlkjhgfdsamnbvcxz
2 qazwsxedcrfvtgbyhnujmikolp
1 plokmijnuhbygvtfcrdxeszwaq
4 mnbvcxzasdfghjklpoiuytrewq
Output
4
8
2
3
4
7
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
from collections import defaultdict
"""
・trie木
・各ノードから、アルファベットごとの子ノードの番号
・親ノードの番号
"""
N = int(readline())
data = tuple(read().split())
S = data[:N]
Q = int(data[N])
query_target = list(map(int,data[N+1::2]))
query_order = data[N+2::2]
class Node():
def __init__(self,parent):
self.parent=parent
self.child = defaultdict(int)
self.is_word_node = False
self.Nword = 0 # 部分木にあるword_nodeの個数
def __repr__(self):
return 'parent:{}\nchild:{}\nNword:{}'.format(self.parent,self.child,self.Nword)
def add(word):
n = 0
next_idx = len(nodes)
for x in word:
nodes[n].Nword += 1
c = nodes[n].child[x]
if c == 0:
c = next_idx
nodes[n].child[x] = c
nodes.append(Node(n))
next_idx += 1
n = c
nodes[n].is_word_node = True
nodes[n].Nword += 1
def solve(word):
"""
・wordに沿って進む。
・「char1<char2ならば自身より弱い文字列1個」という情報を調べる・
・自身のprefixの個数も調べる
"""
data = defaultdict(int)
prefix = 0
n = 0
for x in word:
# 進む前に、他のルートの文字列数も確認する
for k,v in nodes[n].child.items():
if k == x:
continue
data[1000*k+x] += nodes[v].Nword # k<xのときの加算量
# prefixを見つけたら加算
if nodes[n].is_word_node:
prefix += 1
# 進む
n = nodes[n].child[x]
return data, prefix
root = Node(0)
nodes = [root]
for word in S:
add(word)
# 同じ文字列に対するクエリを一括処理
word_to_query = [[] for _ in range(N)]
for i,x in enumerate(query_target):
word_to_query[x-1].append(i)
answer = [0]*Q
for n,word in enumerate(S):
if not word_to_query[n]:
continue
data,pref = solve(word)
for q in word_to_query[n]:
order = query_order[q]
alphabet_rank = {x:i for i,x in enumerate(order)}
rank = pref+1
for x,v in data.items():
a,b = divmod(x,1000)
if alphabet_rank[a] < alphabet_rank[b]:
rank += v
answer[q] = rank
print('\n'.join(map(str,answer)))
```
| 107,762 |
Provide a correct Python 3 solution for this coding contest problem.
There are N strings of lowercase alphabet only. The i-th string is S_i. Every string is unique.
Provide answers for the Q queries below. The i-th query has the following format:
Query: An integer k_i and a string p_{i,1}p_{i,2}...p_{i,26} that results from permuting {`a`,`b`,...,`z`} are given. Output the sequence of the string S_{k_i} among the N strings in lexicographical order when the literal sequence is p_{i,1}<p_{i,2}<...<p_{i,26}.
Constraints
* 1 ≦ N,Q ≦ 100000
* 1 ≦ |S_i| (1 ≦ i ≦ N)
* S_i (1 ≦ i ≦ N) is a string of lowercase alphabet.
* The sum of |S_i| is no more than 400000.
* Every S_i is unique.
* 1 ≦ k_i ≦ N (1 ≦ i ≦ Q)
* For all 1 ≦ i ≦ Q, p_{i,1}p_{i,2}...p_{i,26} is a permutation of `abcd...z`.
Input
Inputs are provided from standard inputs in the following form.
N
S_1
:
S_N
Q
k_1 p_{1,1}p_{1,2}...p_{1,26}
:
k_Q p_{Q,1}p_{Q,2}...p_{Q,26}
Output
Output Q lines.
On line i, for the i-th query, output an integer indicating the sequence of the string S_{k_i} among the N strings in lexicographical order.
Examples
Input
5
aa
abbaa
abbba
aaab
aaaaaba
5
1 abcdefghijklmnopqrstuvwxyz
2 bacdefghijklmnopqrstuvwxyz
3 abcdefghijklmnopqrstuvwxyz
4 bacdefghijklmnopqrstuvwxyz
5 abcdefghijklmnopqrstuvwxyz
Output
1
2
5
4
2
Input
8
abrakatabra
abadaba
abracadabra
atcoder
grand
contest
ababa
a
6
3 abcdefghijklmnopqrstuvwxyz
6 qwertyuiopasdfghjklzxcvbnm
8 poiuytrewqlkjhgfdsamnbvcxz
2 qazwsxedcrfvtgbyhnujmikolp
1 plokmijnuhbygvtfcrdxeszwaq
4 mnbvcxzasdfghjklpoiuytrewq
Output
4
8
2
3
4
7
"Correct Solution:
```
import sys
alp = "abcdefghijklmnopqrstuvwxyz"
D = {}
for i in range(26):
D[alp[i]] = i
N = int(input())
c = [{}]
num = [N]
dep = [0]
Slis = []
lon = 0
for loop in range(N):
S = input()
S += " "
Slis.append(S)
#Slen += len(S)
nind = 0
for i in S:
if i not in c[nind]:
lon += 1
if len(c[nind]) == 1:
lon -= 1
c[nind][i] = len(c)
c.append({})
num.append(0)
dep.append(dep[nind]+1)
nind = c[nind][i]
num[nind] += 1
Q = int(input())
#print (lon)
if lon < 70000:
always = [0] * N
table = []
#print (c)
#print (num)
#table構築
for loop in range(N):
nt = [[0] * 26 for i in range(26)]
nind = 0
for i in Slis[loop]:
if i == " ":
continue
#print ("now char",i)
for j in c[nind]:
#print ("nowj",j)
if j == " ":
always[loop] += 1
#print ("always pl+1")
elif j != i:
#print ("if",i,">",j,"plus:",num[c[nind][j]])
nt[D[i]][D[j]] += num[c[nind][j]]
nind = c[nind][i]
table.append(nt)
#print (always)
#print (table[0])
#Q = int(input())
for loop in range(Q):
k,p = input().split()
k = int(k)
k-=1
ans = always[k] + 1
for i in range(26):
for j in range(i):
ind1 = D[p[i]]
ind2 = D[p[j]]
ans += table[k][ind1][ind2]
print (ans)
sys.exit()
#print (c)
#print (num)
lpc = 0
for loop in range(Q):
k,p = input().split()
k = int(k)
p = " " + p
nind = 0
ans = 0
ncr = 0
while ncr < len(Slis[k-1]):
i = Slis[k-1][ncr]
for j in p:
if j == i:
break
elif j in c[nind]:
ans += num[c[nind][j]]
while len( c[c[nind][i]] ) == 1:
for x in c[c[nind][i]]:
c[nind][i] = c[c[nind][i]][x]
nind = c[nind][i]
ncr = dep[nind]
print (ans+1)
```
| 107,763 |
Provide a correct Python 3 solution for this coding contest problem.
There are N strings of lowercase alphabet only. The i-th string is S_i. Every string is unique.
Provide answers for the Q queries below. The i-th query has the following format:
Query: An integer k_i and a string p_{i,1}p_{i,2}...p_{i,26} that results from permuting {`a`,`b`,...,`z`} are given. Output the sequence of the string S_{k_i} among the N strings in lexicographical order when the literal sequence is p_{i,1}<p_{i,2}<...<p_{i,26}.
Constraints
* 1 ≦ N,Q ≦ 100000
* 1 ≦ |S_i| (1 ≦ i ≦ N)
* S_i (1 ≦ i ≦ N) is a string of lowercase alphabet.
* The sum of |S_i| is no more than 400000.
* Every S_i is unique.
* 1 ≦ k_i ≦ N (1 ≦ i ≦ Q)
* For all 1 ≦ i ≦ Q, p_{i,1}p_{i,2}...p_{i,26} is a permutation of `abcd...z`.
Input
Inputs are provided from standard inputs in the following form.
N
S_1
:
S_N
Q
k_1 p_{1,1}p_{1,2}...p_{1,26}
:
k_Q p_{Q,1}p_{Q,2}...p_{Q,26}
Output
Output Q lines.
On line i, for the i-th query, output an integer indicating the sequence of the string S_{k_i} among the N strings in lexicographical order.
Examples
Input
5
aa
abbaa
abbba
aaab
aaaaaba
5
1 abcdefghijklmnopqrstuvwxyz
2 bacdefghijklmnopqrstuvwxyz
3 abcdefghijklmnopqrstuvwxyz
4 bacdefghijklmnopqrstuvwxyz
5 abcdefghijklmnopqrstuvwxyz
Output
1
2
5
4
2
Input
8
abrakatabra
abadaba
abracadabra
atcoder
grand
contest
ababa
a
6
3 abcdefghijklmnopqrstuvwxyz
6 qwertyuiopasdfghjklzxcvbnm
8 poiuytrewqlkjhgfdsamnbvcxz
2 qazwsxedcrfvtgbyhnujmikolp
1 plokmijnuhbygvtfcrdxeszwaq
4 mnbvcxzasdfghjklpoiuytrewq
Output
4
8
2
3
4
7
"Correct Solution:
```
from array import*;R=range;T=[0]*28;N=int(input());S=[[ord(C)-95 for C in input()]for _ in[0]*N];U=N*26**2*array('i',[0]);V=N*[0];Q=int(input())
for X in S:
P=0
for C in X:
if T[P+C]==0:T[P+C]=len(T);T+=[0]*28
T[P]+=1;P=T[P+C]
T[P]+=1;T[P+1]=1
for I in R(N):
P=0
for C in S[I]:
for A in R(26):
X=T[P+A+2]
if(A!=C-2)*X:U[676*I+26*A+C-2]+=T[X]
V[I]+=T[P+1]
P=T[P+C]
while Q:
K,P=input().split();K=int(K)-1;P=[ord(C)-97 for C in P];X=1+V[K]
for A in R(26):
for B in R(A+1,26):
X+=U[676*K+26*P[A]+P[B]]
print(X);Q-=1
```
| 107,764 |
Provide a correct Python 3 solution for this coding contest problem.
There are N strings of lowercase alphabet only. The i-th string is S_i. Every string is unique.
Provide answers for the Q queries below. The i-th query has the following format:
Query: An integer k_i and a string p_{i,1}p_{i,2}...p_{i,26} that results from permuting {`a`,`b`,...,`z`} are given. Output the sequence of the string S_{k_i} among the N strings in lexicographical order when the literal sequence is p_{i,1}<p_{i,2}<...<p_{i,26}.
Constraints
* 1 ≦ N,Q ≦ 100000
* 1 ≦ |S_i| (1 ≦ i ≦ N)
* S_i (1 ≦ i ≦ N) is a string of lowercase alphabet.
* The sum of |S_i| is no more than 400000.
* Every S_i is unique.
* 1 ≦ k_i ≦ N (1 ≦ i ≦ Q)
* For all 1 ≦ i ≦ Q, p_{i,1}p_{i,2}...p_{i,26} is a permutation of `abcd...z`.
Input
Inputs are provided from standard inputs in the following form.
N
S_1
:
S_N
Q
k_1 p_{1,1}p_{1,2}...p_{1,26}
:
k_Q p_{Q,1}p_{Q,2}...p_{Q,26}
Output
Output Q lines.
On line i, for the i-th query, output an integer indicating the sequence of the string S_{k_i} among the N strings in lexicographical order.
Examples
Input
5
aa
abbaa
abbba
aaab
aaaaaba
5
1 abcdefghijklmnopqrstuvwxyz
2 bacdefghijklmnopqrstuvwxyz
3 abcdefghijklmnopqrstuvwxyz
4 bacdefghijklmnopqrstuvwxyz
5 abcdefghijklmnopqrstuvwxyz
Output
1
2
5
4
2
Input
8
abrakatabra
abadaba
abracadabra
atcoder
grand
contest
ababa
a
6
3 abcdefghijklmnopqrstuvwxyz
6 qwertyuiopasdfghjklzxcvbnm
8 poiuytrewqlkjhgfdsamnbvcxz
2 qazwsxedcrfvtgbyhnujmikolp
1 plokmijnuhbygvtfcrdxeszwaq
4 mnbvcxzasdfghjklpoiuytrewq
Output
4
8
2
3
4
7
"Correct Solution:
```
import sys
readline = sys.stdin.readline
class Node:
def __init__(self, sigma, depth):
self.end = False
self.child = [None] * sigma
self.depth = depth
self.count = 0
def __setitem__(self, i, x):
self.child[i] = x
def __getitem__(self, i):
return self.child[i]
class Trie():
def __init__(self, sigma):
self.sigma = sigma
self.root = Node(sigma, 0)
def add(self, S):
vn = self.root
vn.count += 1
for cs in S:
if vn[cs] is None:
vn[cs] = Node(self.sigma, vn.depth + 1)
vn = vn[cs]
vn.count += 1
vn.end = True
T = Trie(26)
N = int(readline())
Di = [None]*N
for i in range(N):
S = list(map(lambda x: ord(x)-97, readline().strip()))
Di[i] = S
T.add(S)
Q = int(readline())
Ans = [None]*Q
Query = [[] for _ in range(N)]
for qu in range(Q):
k, p = readline().strip().split()
k = int(k)-1
p = list(map(lambda x:ord(x)-97, p))
pinv = [None]*26
for i in range(26):
pinv[p[i]] = i
Query[k].append((qu, pinv))
for k in range(N):
if not Query[k]:
continue
cost = [[0]*26 for _ in range(26)]
geta = 0
vn = T.root
for si in Di[k]:
for ti in range(26):
if si != ti and vn[ti] is not None:
cost[si][ti] += vn[ti].count
vn = vn[si]
if vn.end:
geta += 1
L, P = map(list, zip(*Query[k]))
M = len(P)
for idx in range(M):
res = 0
invp = P[idx]
for i in range(26):
for j in range(26):
if invp[i] > invp[j]:
res += cost[i][j]
Ans[L[idx]] = res+geta
print('\n'.join(map(str, Ans)))
```
| 107,765 |
Provide a correct Python 3 solution for this coding contest problem.
There are N strings of lowercase alphabet only. The i-th string is S_i. Every string is unique.
Provide answers for the Q queries below. The i-th query has the following format:
Query: An integer k_i and a string p_{i,1}p_{i,2}...p_{i,26} that results from permuting {`a`,`b`,...,`z`} are given. Output the sequence of the string S_{k_i} among the N strings in lexicographical order when the literal sequence is p_{i,1}<p_{i,2}<...<p_{i,26}.
Constraints
* 1 ≦ N,Q ≦ 100000
* 1 ≦ |S_i| (1 ≦ i ≦ N)
* S_i (1 ≦ i ≦ N) is a string of lowercase alphabet.
* The sum of |S_i| is no more than 400000.
* Every S_i is unique.
* 1 ≦ k_i ≦ N (1 ≦ i ≦ Q)
* For all 1 ≦ i ≦ Q, p_{i,1}p_{i,2}...p_{i,26} is a permutation of `abcd...z`.
Input
Inputs are provided from standard inputs in the following form.
N
S_1
:
S_N
Q
k_1 p_{1,1}p_{1,2}...p_{1,26}
:
k_Q p_{Q,1}p_{Q,2}...p_{Q,26}
Output
Output Q lines.
On line i, for the i-th query, output an integer indicating the sequence of the string S_{k_i} among the N strings in lexicographical order.
Examples
Input
5
aa
abbaa
abbba
aaab
aaaaaba
5
1 abcdefghijklmnopqrstuvwxyz
2 bacdefghijklmnopqrstuvwxyz
3 abcdefghijklmnopqrstuvwxyz
4 bacdefghijklmnopqrstuvwxyz
5 abcdefghijklmnopqrstuvwxyz
Output
1
2
5
4
2
Input
8
abrakatabra
abadaba
abracadabra
atcoder
grand
contest
ababa
a
6
3 abcdefghijklmnopqrstuvwxyz
6 qwertyuiopasdfghjklzxcvbnm
8 poiuytrewqlkjhgfdsamnbvcxz
2 qazwsxedcrfvtgbyhnujmikolp
1 plokmijnuhbygvtfcrdxeszwaq
4 mnbvcxzasdfghjklpoiuytrewq
Output
4
8
2
3
4
7
"Correct Solution:
```
import array;R=range;L=input;T=[0]*28;N=int(L());S=[[ord(C)-95 for C in L()]for _ in[0]*N];U=N*26**2*array.array('i',[0]);V=N*[0];Q=int(L())
for X in S:
P=0
for C in X:
if T[P+C]==0:T[P+C]=len(T);T+=[0]*28
T[P]+=1;P=T[P+C]
T[P]+=1;T[P+1]=1
while N:
N-=1;P=0
for C in S[N]:
for A in R(26):X=T[P+A+2];U[676*N+26*A+C-2]+=T[X]*(X and A!=C-2)
V[N]+=T[P+1];P=T[P+C]
while Q:K,P=L().split();K=int(K)-1;P=[ord(C)-97 for C in P];X=1+V[K]+sum(U[676*K+26*P[A]+P[B]]for A in R(26)for B in R(A+1,26));print(X);Q-=1
```
| 107,766 |
Provide a correct Python 3 solution for this coding contest problem.
There are N strings of lowercase alphabet only. The i-th string is S_i. Every string is unique.
Provide answers for the Q queries below. The i-th query has the following format:
Query: An integer k_i and a string p_{i,1}p_{i,2}...p_{i,26} that results from permuting {`a`,`b`,...,`z`} are given. Output the sequence of the string S_{k_i} among the N strings in lexicographical order when the literal sequence is p_{i,1}<p_{i,2}<...<p_{i,26}.
Constraints
* 1 ≦ N,Q ≦ 100000
* 1 ≦ |S_i| (1 ≦ i ≦ N)
* S_i (1 ≦ i ≦ N) is a string of lowercase alphabet.
* The sum of |S_i| is no more than 400000.
* Every S_i is unique.
* 1 ≦ k_i ≦ N (1 ≦ i ≦ Q)
* For all 1 ≦ i ≦ Q, p_{i,1}p_{i,2}...p_{i,26} is a permutation of `abcd...z`.
Input
Inputs are provided from standard inputs in the following form.
N
S_1
:
S_N
Q
k_1 p_{1,1}p_{1,2}...p_{1,26}
:
k_Q p_{Q,1}p_{Q,2}...p_{Q,26}
Output
Output Q lines.
On line i, for the i-th query, output an integer indicating the sequence of the string S_{k_i} among the N strings in lexicographical order.
Examples
Input
5
aa
abbaa
abbba
aaab
aaaaaba
5
1 abcdefghijklmnopqrstuvwxyz
2 bacdefghijklmnopqrstuvwxyz
3 abcdefghijklmnopqrstuvwxyz
4 bacdefghijklmnopqrstuvwxyz
5 abcdefghijklmnopqrstuvwxyz
Output
1
2
5
4
2
Input
8
abrakatabra
abadaba
abracadabra
atcoder
grand
contest
ababa
a
6
3 abcdefghijklmnopqrstuvwxyz
6 qwertyuiopasdfghjklzxcvbnm
8 poiuytrewqlkjhgfdsamnbvcxz
2 qazwsxedcrfvtgbyhnujmikolp
1 plokmijnuhbygvtfcrdxeszwaq
4 mnbvcxzasdfghjklpoiuytrewq
Output
4
8
2
3
4
7
"Correct Solution:
```
import array;R=range;L=input;T=[0]*28;N=int(L());S=[[ord(C)-95 for C in L()]for _ in[0]*N];U=N*26**2*array.array('i',[0]);V=N*[0];Q=int(L())
for X in S:
P=0
for C in X:
if T[P+C]==0:T[P+C]=len(T);T+=[0]*28
T[P]+=1;P=T[P+C]
T[P]+=1;T[P+1]=1
while N:
N-=1;P=0
for C in S[N]:
for A in R(26):X=T[P+A+2];U[676*N+26*A+C-2]+=T[X]*(X and A!=C-2)
V[N]+=T[P+1];P=T[P+C]
while Q:
K,P=L().split();K=int(K)-1;P=[ord(C)-97 for C in P];X=1+V[K]
for A in R(26):
for B in R(A+1,26):X+=U[676*K+26*P[A]+P[B]]
print(X);Q-=1
```
| 107,767 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N strings of lowercase alphabet only. The i-th string is S_i. Every string is unique.
Provide answers for the Q queries below. The i-th query has the following format:
Query: An integer k_i and a string p_{i,1}p_{i,2}...p_{i,26} that results from permuting {`a`,`b`,...,`z`} are given. Output the sequence of the string S_{k_i} among the N strings in lexicographical order when the literal sequence is p_{i,1}<p_{i,2}<...<p_{i,26}.
Constraints
* 1 ≦ N,Q ≦ 100000
* 1 ≦ |S_i| (1 ≦ i ≦ N)
* S_i (1 ≦ i ≦ N) is a string of lowercase alphabet.
* The sum of |S_i| is no more than 400000.
* Every S_i is unique.
* 1 ≦ k_i ≦ N (1 ≦ i ≦ Q)
* For all 1 ≦ i ≦ Q, p_{i,1}p_{i,2}...p_{i,26} is a permutation of `abcd...z`.
Input
Inputs are provided from standard inputs in the following form.
N
S_1
:
S_N
Q
k_1 p_{1,1}p_{1,2}...p_{1,26}
:
k_Q p_{Q,1}p_{Q,2}...p_{Q,26}
Output
Output Q lines.
On line i, for the i-th query, output an integer indicating the sequence of the string S_{k_i} among the N strings in lexicographical order.
Examples
Input
5
aa
abbaa
abbba
aaab
aaaaaba
5
1 abcdefghijklmnopqrstuvwxyz
2 bacdefghijklmnopqrstuvwxyz
3 abcdefghijklmnopqrstuvwxyz
4 bacdefghijklmnopqrstuvwxyz
5 abcdefghijklmnopqrstuvwxyz
Output
1
2
5
4
2
Input
8
abrakatabra
abadaba
abracadabra
atcoder
grand
contest
ababa
a
6
3 abcdefghijklmnopqrstuvwxyz
6 qwertyuiopasdfghjklzxcvbnm
8 poiuytrewqlkjhgfdsamnbvcxz
2 qazwsxedcrfvtgbyhnujmikolp
1 plokmijnuhbygvtfcrdxeszwaq
4 mnbvcxzasdfghjklpoiuytrewq
Output
4
8
2
3
4
7
Submitted Solution:
```
import array;R=range;L=input;T=[0]*28;N=int(L());S=[[ord(C)-95 for C in L()]for _ in[0]*N];U=N*26**2*array.array('i',[0]);V=N*[0];Q=int(L())
for X in S:
P=0
for C in X:
if T[P+C]==0:T[P+C]=len(T);T+=[0]*28
T[P]+=1;P=T[P+C]
T[P]+=1;T[P+1]=1
while N:
N-=1;P=0
for C in S[N]:
for A in R(26):
X=T[P+A+2]
if(A!=C-2)*X:U[676*N+26*A+C-2]+=T[X]
V[N]+=T[P+1]
P=T[P+C]
while Q:
K,P=L().split();K=int(K)-1;P=[ord(C)-97 for C in P];X=1+V[K]
for A in R(26):
for B in R(A+1,26):
X+=U[676*K+26*P[A]+P[B]]
print(X);Q-=1
```
Yes
| 107,768 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N strings of lowercase alphabet only. The i-th string is S_i. Every string is unique.
Provide answers for the Q queries below. The i-th query has the following format:
Query: An integer k_i and a string p_{i,1}p_{i,2}...p_{i,26} that results from permuting {`a`,`b`,...,`z`} are given. Output the sequence of the string S_{k_i} among the N strings in lexicographical order when the literal sequence is p_{i,1}<p_{i,2}<...<p_{i,26}.
Constraints
* 1 ≦ N,Q ≦ 100000
* 1 ≦ |S_i| (1 ≦ i ≦ N)
* S_i (1 ≦ i ≦ N) is a string of lowercase alphabet.
* The sum of |S_i| is no more than 400000.
* Every S_i is unique.
* 1 ≦ k_i ≦ N (1 ≦ i ≦ Q)
* For all 1 ≦ i ≦ Q, p_{i,1}p_{i,2}...p_{i,26} is a permutation of `abcd...z`.
Input
Inputs are provided from standard inputs in the following form.
N
S_1
:
S_N
Q
k_1 p_{1,1}p_{1,2}...p_{1,26}
:
k_Q p_{Q,1}p_{Q,2}...p_{Q,26}
Output
Output Q lines.
On line i, for the i-th query, output an integer indicating the sequence of the string S_{k_i} among the N strings in lexicographical order.
Examples
Input
5
aa
abbaa
abbba
aaab
aaaaaba
5
1 abcdefghijklmnopqrstuvwxyz
2 bacdefghijklmnopqrstuvwxyz
3 abcdefghijklmnopqrstuvwxyz
4 bacdefghijklmnopqrstuvwxyz
5 abcdefghijklmnopqrstuvwxyz
Output
1
2
5
4
2
Input
8
abrakatabra
abadaba
abracadabra
atcoder
grand
contest
ababa
a
6
3 abcdefghijklmnopqrstuvwxyz
6 qwertyuiopasdfghjklzxcvbnm
8 poiuytrewqlkjhgfdsamnbvcxz
2 qazwsxedcrfvtgbyhnujmikolp
1 plokmijnuhbygvtfcrdxeszwaq
4 mnbvcxzasdfghjklpoiuytrewq
Output
4
8
2
3
4
7
Submitted Solution:
```
import sys
readline = sys.stdin.readline
class Node:
def __init__(self, sigma, depth):
self.end = False
self.child = [None] * sigma
self.depth = depth
self.count = 0
def __setitem__(self, i, x):
self.child[i] = x
def __getitem__(self, i):
return self.child[i]
class Trie():
def __init__(self, sigma):
self.sigma = sigma
self.root = Node(sigma, 0)
def add(self, S):
vn = self.root
vn.count += 1
for cs in S:
if vn[cs] is None:
vn[cs] = Node(self.sigma, vn.depth + 1)
vn = vn[cs]
vn.count += 1
vn.end = True
T = Trie(26)
N = int(readline())
Di = [None]*N
for i in range(N):
S = list(map(lambda x: ord(x)-97, readline().strip()))
Di[i] = S
T.add(S)
Q = int(readline())
Ans = [None]*Q
for qu in range(Q):
k, p = readline().strip().split()
k = int(k)-1
p = list(map(lambda x:ord(x)-97, p))
vn = T.root
res = 0
for si in Di[k]:
for al in p:
if al == si:
break
if vn[al] is not None:
res += vn[al].count
vn = vn[si]
if vn.end:
res += 1
Ans[qu] = res
print('\n'.join(map(str, Ans)))
```
No
| 107,769 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N strings of lowercase alphabet only. The i-th string is S_i. Every string is unique.
Provide answers for the Q queries below. The i-th query has the following format:
Query: An integer k_i and a string p_{i,1}p_{i,2}...p_{i,26} that results from permuting {`a`,`b`,...,`z`} are given. Output the sequence of the string S_{k_i} among the N strings in lexicographical order when the literal sequence is p_{i,1}<p_{i,2}<...<p_{i,26}.
Constraints
* 1 ≦ N,Q ≦ 100000
* 1 ≦ |S_i| (1 ≦ i ≦ N)
* S_i (1 ≦ i ≦ N) is a string of lowercase alphabet.
* The sum of |S_i| is no more than 400000.
* Every S_i is unique.
* 1 ≦ k_i ≦ N (1 ≦ i ≦ Q)
* For all 1 ≦ i ≦ Q, p_{i,1}p_{i,2}...p_{i,26} is a permutation of `abcd...z`.
Input
Inputs are provided from standard inputs in the following form.
N
S_1
:
S_N
Q
k_1 p_{1,1}p_{1,2}...p_{1,26}
:
k_Q p_{Q,1}p_{Q,2}...p_{Q,26}
Output
Output Q lines.
On line i, for the i-th query, output an integer indicating the sequence of the string S_{k_i} among the N strings in lexicographical order.
Examples
Input
5
aa
abbaa
abbba
aaab
aaaaaba
5
1 abcdefghijklmnopqrstuvwxyz
2 bacdefghijklmnopqrstuvwxyz
3 abcdefghijklmnopqrstuvwxyz
4 bacdefghijklmnopqrstuvwxyz
5 abcdefghijklmnopqrstuvwxyz
Output
1
2
5
4
2
Input
8
abrakatabra
abadaba
abracadabra
atcoder
grand
contest
ababa
a
6
3 abcdefghijklmnopqrstuvwxyz
6 qwertyuiopasdfghjklzxcvbnm
8 poiuytrewqlkjhgfdsamnbvcxz
2 qazwsxedcrfvtgbyhnujmikolp
1 plokmijnuhbygvtfcrdxeszwaq
4 mnbvcxzasdfghjklpoiuytrewq
Output
4
8
2
3
4
7
Submitted Solution:
```
N = int(input())
S = [input() for _ in range(N)]
Q = int(input())
S.sort()
k, p = zip(*map(lambda x: (int(x[0]), x[1]), [input().split() for _ in range(Q)]))
for q in range(Q):
print(q)
```
No
| 107,770 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N strings of lowercase alphabet only. The i-th string is S_i. Every string is unique.
Provide answers for the Q queries below. The i-th query has the following format:
Query: An integer k_i and a string p_{i,1}p_{i,2}...p_{i,26} that results from permuting {`a`,`b`,...,`z`} are given. Output the sequence of the string S_{k_i} among the N strings in lexicographical order when the literal sequence is p_{i,1}<p_{i,2}<...<p_{i,26}.
Constraints
* 1 ≦ N,Q ≦ 100000
* 1 ≦ |S_i| (1 ≦ i ≦ N)
* S_i (1 ≦ i ≦ N) is a string of lowercase alphabet.
* The sum of |S_i| is no more than 400000.
* Every S_i is unique.
* 1 ≦ k_i ≦ N (1 ≦ i ≦ Q)
* For all 1 ≦ i ≦ Q, p_{i,1}p_{i,2}...p_{i,26} is a permutation of `abcd...z`.
Input
Inputs are provided from standard inputs in the following form.
N
S_1
:
S_N
Q
k_1 p_{1,1}p_{1,2}...p_{1,26}
:
k_Q p_{Q,1}p_{Q,2}...p_{Q,26}
Output
Output Q lines.
On line i, for the i-th query, output an integer indicating the sequence of the string S_{k_i} among the N strings in lexicographical order.
Examples
Input
5
aa
abbaa
abbba
aaab
aaaaaba
5
1 abcdefghijklmnopqrstuvwxyz
2 bacdefghijklmnopqrstuvwxyz
3 abcdefghijklmnopqrstuvwxyz
4 bacdefghijklmnopqrstuvwxyz
5 abcdefghijklmnopqrstuvwxyz
Output
1
2
5
4
2
Input
8
abrakatabra
abadaba
abracadabra
atcoder
grand
contest
ababa
a
6
3 abcdefghijklmnopqrstuvwxyz
6 qwertyuiopasdfghjklzxcvbnm
8 poiuytrewqlkjhgfdsamnbvcxz
2 qazwsxedcrfvtgbyhnujmikolp
1 plokmijnuhbygvtfcrdxeszwaq
4 mnbvcxzasdfghjklpoiuytrewq
Output
4
8
2
3
4
7
Submitted Solution:
```
N = int(input())
S = [input() for _ in range(N)]
Q = int(input())
k, p = zip(*map(lambda x: (int(x[0]), x[1]), [input().split() for _ in range(Q)]))
for q in range(Q):
print(q)
```
No
| 107,771 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N strings of lowercase alphabet only. The i-th string is S_i. Every string is unique.
Provide answers for the Q queries below. The i-th query has the following format:
Query: An integer k_i and a string p_{i,1}p_{i,2}...p_{i,26} that results from permuting {`a`,`b`,...,`z`} are given. Output the sequence of the string S_{k_i} among the N strings in lexicographical order when the literal sequence is p_{i,1}<p_{i,2}<...<p_{i,26}.
Constraints
* 1 ≦ N,Q ≦ 100000
* 1 ≦ |S_i| (1 ≦ i ≦ N)
* S_i (1 ≦ i ≦ N) is a string of lowercase alphabet.
* The sum of |S_i| is no more than 400000.
* Every S_i is unique.
* 1 ≦ k_i ≦ N (1 ≦ i ≦ Q)
* For all 1 ≦ i ≦ Q, p_{i,1}p_{i,2}...p_{i,26} is a permutation of `abcd...z`.
Input
Inputs are provided from standard inputs in the following form.
N
S_1
:
S_N
Q
k_1 p_{1,1}p_{1,2}...p_{1,26}
:
k_Q p_{Q,1}p_{Q,2}...p_{Q,26}
Output
Output Q lines.
On line i, for the i-th query, output an integer indicating the sequence of the string S_{k_i} among the N strings in lexicographical order.
Examples
Input
5
aa
abbaa
abbba
aaab
aaaaaba
5
1 abcdefghijklmnopqrstuvwxyz
2 bacdefghijklmnopqrstuvwxyz
3 abcdefghijklmnopqrstuvwxyz
4 bacdefghijklmnopqrstuvwxyz
5 abcdefghijklmnopqrstuvwxyz
Output
1
2
5
4
2
Input
8
abrakatabra
abadaba
abracadabra
atcoder
grand
contest
ababa
a
6
3 abcdefghijklmnopqrstuvwxyz
6 qwertyuiopasdfghjklzxcvbnm
8 poiuytrewqlkjhgfdsamnbvcxz
2 qazwsxedcrfvtgbyhnujmikolp
1 plokmijnuhbygvtfcrdxeszwaq
4 mnbvcxzasdfghjklpoiuytrewq
Output
4
8
2
3
4
7
Submitted Solution:
```
import sys
readline = sys.stdin.readline
class Node:
def __init__(self, sigma, depth):
self.end = False
self.child = [None] * sigma
self.depth = depth
self.count = 0
def __setitem__(self, i, x):
self.child[i] = x
def __getitem__(self, i):
return self.child[i]
class Trie():
def __init__(self, sigma):
self.sigma = sigma
self.root = Node(sigma, 0)
def add(self, S):
vn = self.root
vn.count += 1
for cs in S:
if vn[cs] is None:
vn[cs] = Node(self.sigma, vn.depth + 1)
vn = vn[cs]
vn.count += 1
vn.end = True
T = Trie(26)
N = int(readline())
Di = [None]*N
for i in range(N):
S = list(map(lambda x: ord(x)-97, readline().strip()))
Di[i] = S
T.add(S)
Q = int(readline())
Ans = [None]*Q
Query = [[] for _ in range(N)]
for qu in range(Q):
k, p = readline().strip().split()
k = int(k)-1
p = list(map(lambda x:ord(x)-97, p))
pinv = [None]*26
for i in range(26):
pinv[p[i]] = i
Query[k].append((qu, pinv))
for i in range(N):
if not Query[i]:
continue
L = []
P = []
for qu, p in Query[i]:
L.append(qu)
P.append(p)
M = len(P)
vn = T.root
ans = [0]*M
for si in Di[i]:
for al in range(26):
for idx in range(M):
pinv = P[idx]
if pinv[al] < pinv[si]:
if vn[al] is not None:
ans[idx] += vn[al].count
vn = vn[si]
if vn.end:
ans = [1+a for a in ans]
for j in range(M):
Ans[L[j]] = ans[j]
print('\n'.join(map(str, Ans)))
```
No
| 107,772 |
Provide a correct Python 3 solution for this coding contest problem.
There is data of up to 100 characters per line, consisting of half-width alphabetic character strings. Some lines are symmetric (same whether read from the left edge or the right edge). Create a program that reads this data and outputs the number of symmetric strings in it. Note that lines consisting of only one character are symmetrical.
Input
Multiple strings are given over multiple lines. One string is given for each line. The number of strings does not exceed 50.
Output
Outputs the number of symmetric strings on one line.
Example
Input
abcba
sx
abcddcba
rttrd
Output
2
"Correct Solution:
```
import sys
readlines = sys.stdin.readlines
write = sys.stdout.write
def solve():
ans = 0
for line in readlines():
s = line.strip()
if s == s[::-1]:
ans += 1
write("%d\n" % ans)
solve()
```
| 107,773 |
Provide a correct Python 3 solution for this coding contest problem.
There is data of up to 100 characters per line, consisting of half-width alphabetic character strings. Some lines are symmetric (same whether read from the left edge or the right edge). Create a program that reads this data and outputs the number of symmetric strings in it. Note that lines consisting of only one character are symmetrical.
Input
Multiple strings are given over multiple lines. One string is given for each line. The number of strings does not exceed 50.
Output
Outputs the number of symmetric strings on one line.
Example
Input
abcba
sx
abcddcba
rttrd
Output
2
"Correct Solution:
```
cnt = 0
while True :
try :
n = list(input())
N = list(reversed(n))
if n == N :
cnt += 1
except :
print(cnt)
break
```
| 107,774 |
Provide a correct Python 3 solution for this coding contest problem.
There is data of up to 100 characters per line, consisting of half-width alphabetic character strings. Some lines are symmetric (same whether read from the left edge or the right edge). Create a program that reads this data and outputs the number of symmetric strings in it. Note that lines consisting of only one character are symmetrical.
Input
Multiple strings are given over multiple lines. One string is given for each line. The number of strings does not exceed 50.
Output
Outputs the number of symmetric strings on one line.
Example
Input
abcba
sx
abcddcba
rttrd
Output
2
"Correct Solution:
```
q=0
while True:
try:
x = input()
c = 0
if len(x)%2==0:
for i in range(len(x)//2):
if x[i] == x[len(x)-i-1]:
c += 1
else:
pass
if c == len(x)//2:
q += 1
else:
for j in range((len(x)-1)//2):
if x[j] == x[len(x)-j-1]:
c += 1
else:
pass
if c == (len(x)-1)//2:
q += 1
except EOFError:
break
print(q)
```
| 107,775 |
Provide a correct Python 3 solution for this coding contest problem.
There is data of up to 100 characters per line, consisting of half-width alphabetic character strings. Some lines are symmetric (same whether read from the left edge or the right edge). Create a program that reads this data and outputs the number of symmetric strings in it. Note that lines consisting of only one character are symmetrical.
Input
Multiple strings are given over multiple lines. One string is given for each line. The number of strings does not exceed 50.
Output
Outputs the number of symmetric strings on one line.
Example
Input
abcba
sx
abcddcba
rttrd
Output
2
"Correct Solution:
```
def get_input():
while True:
try:
yield ''.join(input())
except EOFError:
break
N = list(get_input())
ans = 0
for ll in range(len(N)):
S = N[ll]
flag = True
for i in range(len(S) // 2):
if S[i] != S[len(S)-i-1]:
flag = False
break
if flag:
ans += 1
print(ans)
```
| 107,776 |
Provide a correct Python 3 solution for this coding contest problem.
There is data of up to 100 characters per line, consisting of half-width alphabetic character strings. Some lines are symmetric (same whether read from the left edge or the right edge). Create a program that reads this data and outputs the number of symmetric strings in it. Note that lines consisting of only one character are symmetrical.
Input
Multiple strings are given over multiple lines. One string is given for each line. The number of strings does not exceed 50.
Output
Outputs the number of symmetric strings on one line.
Example
Input
abcba
sx
abcddcba
rttrd
Output
2
"Correct Solution:
```
c = 0
while True:
try:
s = input()
t = s[::-1]
print
if s == t:
c += 1
except EOFError:
break
print(c)
```
| 107,777 |
Provide a correct Python 3 solution for this coding contest problem.
There is data of up to 100 characters per line, consisting of half-width alphabetic character strings. Some lines are symmetric (same whether read from the left edge or the right edge). Create a program that reads this data and outputs the number of symmetric strings in it. Note that lines consisting of only one character are symmetrical.
Input
Multiple strings are given over multiple lines. One string is given for each line. The number of strings does not exceed 50.
Output
Outputs the number of symmetric strings on one line.
Example
Input
abcba
sx
abcddcba
rttrd
Output
2
"Correct Solution:
```
import sys
f = sys.stdin
def is_symmetry(s):
return s == s[::-1]
print(sum(1 for line in f if is_symmetry(line.strip())))
```
| 107,778 |
Provide a correct Python 3 solution for this coding contest problem.
There is data of up to 100 characters per line, consisting of half-width alphabetic character strings. Some lines are symmetric (same whether read from the left edge or the right edge). Create a program that reads this data and outputs the number of symmetric strings in it. Note that lines consisting of only one character are symmetrical.
Input
Multiple strings are given over multiple lines. One string is given for each line. The number of strings does not exceed 50.
Output
Outputs the number of symmetric strings on one line.
Example
Input
abcba
sx
abcddcba
rttrd
Output
2
"Correct Solution:
```
ans=0
while True:
try :
n=input()
if n==n[::-1]:
ans+=1
except :
break
print(ans)
```
| 107,779 |
Provide a correct Python 3 solution for this coding contest problem.
There is data of up to 100 characters per line, consisting of half-width alphabetic character strings. Some lines are symmetric (same whether read from the left edge or the right edge). Create a program that reads this data and outputs the number of symmetric strings in it. Note that lines consisting of only one character are symmetrical.
Input
Multiple strings are given over multiple lines. One string is given for each line. The number of strings does not exceed 50.
Output
Outputs the number of symmetric strings on one line.
Example
Input
abcba
sx
abcddcba
rttrd
Output
2
"Correct Solution:
```
# coding: utf-8
# Your code here!
s=0
while True:
try:
x=input()
X=x[::-1]
if x==X:
s+=1
else:
pass
except EOFError:
break
print(s)
```
| 107,780 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is data of up to 100 characters per line, consisting of half-width alphabetic character strings. Some lines are symmetric (same whether read from the left edge or the right edge). Create a program that reads this data and outputs the number of symmetric strings in it. Note that lines consisting of only one character are symmetrical.
Input
Multiple strings are given over multiple lines. One string is given for each line. The number of strings does not exceed 50.
Output
Outputs the number of symmetric strings on one line.
Example
Input
abcba
sx
abcddcba
rttrd
Output
2
Submitted Solution:
```
ans=0
while True:
try:
b = input()
a = list(b)
c=len(a)
if c==0:
break
c=int(c/2)
e=0
for i in range(c):
d=i+1
if a[i]==a[-d]:
e+=1
else:
break
if e==c:
ans+=1
except:
break
print(ans)
```
Yes
| 107,781 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is data of up to 100 characters per line, consisting of half-width alphabetic character strings. Some lines are symmetric (same whether read from the left edge or the right edge). Create a program that reads this data and outputs the number of symmetric strings in it. Note that lines consisting of only one character are symmetrical.
Input
Multiple strings are given over multiple lines. One string is given for each line. The number of strings does not exceed 50.
Output
Outputs the number of symmetric strings on one line.
Example
Input
abcba
sx
abcddcba
rttrd
Output
2
Submitted Solution:
```
s=0
while True:
try:
x=input()
X=x[::-1]
if x==X:
s+=1
else:
pass
except EOFError:
break
print(s)
```
Yes
| 107,782 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is data of up to 100 characters per line, consisting of half-width alphabetic character strings. Some lines are symmetric (same whether read from the left edge or the right edge). Create a program that reads this data and outputs the number of symmetric strings in it. Note that lines consisting of only one character are symmetrical.
Input
Multiple strings are given over multiple lines. One string is given for each line. The number of strings does not exceed 50.
Output
Outputs the number of symmetric strings on one line.
Example
Input
abcba
sx
abcddcba
rttrd
Output
2
Submitted Solution:
```
import sys
print(sum([1 for i in sys.stdin if i.strip()==i.strip()[::-1]]))
```
Yes
| 107,783 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is data of up to 100 characters per line, consisting of half-width alphabetic character strings. Some lines are symmetric (same whether read from the left edge or the right edge). Create a program that reads this data and outputs the number of symmetric strings in it. Note that lines consisting of only one character are symmetrical.
Input
Multiple strings are given over multiple lines. One string is given for each line. The number of strings does not exceed 50.
Output
Outputs the number of symmetric strings on one line.
Example
Input
abcba
sx
abcddcba
rttrd
Output
2
Submitted Solution:
```
count = 0
while True:
try:
text = list(input())
a = len(text)
b = a//2
_te = []
te_ = []
for i in range(0,b):
_te.append(text[i])
te_.append(text[a-1-i])
if _te==te_:
count += 1
except:
break
print(count)
```
Yes
| 107,784 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is data of up to 100 characters per line, consisting of half-width alphabetic character strings. Some lines are symmetric (same whether read from the left edge or the right edge). Create a program that reads this data and outputs the number of symmetric strings in it. Note that lines consisting of only one character are symmetrical.
Input
Multiple strings are given over multiple lines. One string is given for each line. The number of strings does not exceed 50.
Output
Outputs the number of symmetric strings on one line.
Example
Input
abcba
sx
abcddcba
rttrd
Output
2
Submitted Solution:
```
res = 0
while True:
try:
cc = list(input())
except:
break
l = len(cc)
if l % 2 == 0:
if cc[:l // 2] == cc[:l // 2 - 1:-1]:
res += 1
else:
if cc[:(l + 1) // 2] == cc[:(l - 1) // 2 - 1:-1]:
res += 1
print(res)
```
No
| 107,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is data of up to 100 characters per line, consisting of half-width alphabetic character strings. Some lines are symmetric (same whether read from the left edge or the right edge). Create a program that reads this data and outputs the number of symmetric strings in it. Note that lines consisting of only one character are symmetrical.
Input
Multiple strings are given over multiple lines. One string is given for each line. The number of strings does not exceed 50.
Output
Outputs the number of symmetric strings on one line.
Example
Input
abcba
sx
abcddcba
rttrd
Output
2
Submitted Solution:
```
res = 0
while True:
try:
cc = list(input())
if not len(cc):
break
except:
break
l = len(cc)
if l % 2 == 0:
if cc[:l // 2] == cc[:l // 2 - 1:-1]:
res += 1
else:
if cc[:(l + 1) // 2] == cc[:(l - 1) // 2 - 1:-1]:
res += 1
print(res)
```
No
| 107,786 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is data of up to 100 characters per line, consisting of half-width alphabetic character strings. Some lines are symmetric (same whether read from the left edge or the right edge). Create a program that reads this data and outputs the number of symmetric strings in it. Note that lines consisting of only one character are symmetrical.
Input
Multiple strings are given over multiple lines. One string is given for each line. The number of strings does not exceed 50.
Output
Outputs the number of symmetric strings on one line.
Example
Input
abcba
sx
abcddcba
rttrd
Output
2
Submitted Solution:
```
import sys
print(e[:-1]==e[-2::-1]for e in sys.stdin)
```
No
| 107,787 |
Provide a correct Python 3 solution for this coding contest problem.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247
"Correct Solution:
```
if __name__ == '__main__':
A = ["A","B","C","D","E"]
while True:
try:
n,m = map(int,input().split())
max_num = n + m
number = 0
if n == 0 and m == 0:
break
for i in range(4):
n,m = map(int,input().split())
if max_num < n + m:
max_num = n + m
number = i + 1
print(A[number],max_num)
except EOFError:
break
```
| 107,788 |
Provide a correct Python 3 solution for this coding contest problem.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247
"Correct Solution:
```
while True :
a, p = map(int, input().split())
if(a == 0 and p== 0) :
break
else :
m = ["A", "B", "C", "D", "E"]
shop = list()
shop.append(a + p)
for i in range(4) :
a, p = map(int, input().split())
shop.append(a + p)
top = max(shop)
Shop = shop.index(top)
print(m[Shop], top)
```
| 107,789 |
Provide a correct Python 3 solution for this coding contest problem.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247
"Correct Solution:
```
# Aizu Problem 0195: What is the Most Popular Shop in Tokaichi?
#
import sys, math, os
# read input:
PYDEV = os.environ.get('PYDEV')
if PYDEV=="True":
sys.stdin = open("sample-input.txt", "rt")
while True:
most = 0
shop = 'A'
a, b = [int(_) for _ in input().split()]
if a == b == 0:
break
most = a + b
for s in range(4):
a = sum([int(_) for _ in input().split()])
if a > most:
most = a
shop = chr(66 + s)
print(shop, most)
```
| 107,790 |
Provide a correct Python 3 solution for this coding contest problem.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0195
"""
import sys
from sys import stdin
input = stdin.readline
def main(args):
while True:
am, pm = map(int, input().split())
if am == 0 and pm == 0:
break
totals = [0] * 5
totals[0] = am + pm
for i in range(1, 5):
am, pm = map(int, input().split())
totals[i] = am + pm
top = max(totals)
shop = totals.index(top)
print('{} {}'.format(chr(ord('A')+shop), top))
if __name__ == '__main__':
main(sys.argv[1:])
```
| 107,791 |
Provide a correct Python 3 solution for this coding contest problem.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247
"Correct Solution:
```
S = ["A","B","C","D","E"]
while True:
L = []
a,p = map(int,input().split())
if a == 0:
break
L.append(a+p)
for i in range(4):
a,p = map(int,input().split())
L.append(a+p)
m = max(L)
s = L.index(m)
print(S[s],m)
```
| 107,792 |
Provide a correct Python 3 solution for this coding contest problem.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247
"Correct Solution:
```
while True:
try:
A=[]
for i in range(5):
s1,s2=map(int,input().split())
if s1==0 and s2==0:
break
s=s1+s2
A.append(s)
if s==0:
break
for i in range(len(A)):
if A[i]==max(A):
if i==0:
print("A",A[i])
if i==1:
print("B",A[i])
if i==2:
print("C",A[i])
if i==3:
print("D",A[i])
if i==4:
print("E",A[i])
except EOFError:
break
```
| 107,793 |
Provide a correct Python 3 solution for this coding contest problem.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247
"Correct Solution:
```
try:
while True:
num_list = []
num_di = {}
#リストに数値を格納して１番大きい数を出力する
for i in range(0,5):
line = list(map(int,input().split()))
num_list.append(sum(line))
num_di[sum(line)] = 65 + i
num_list.sort()
print(chr(num_di[num_list[4]]) + " " + str(num_list[4]))
#EOFErrorを検知する
except EOFError:
pass
```
| 107,794 |
Provide a correct Python 3 solution for this coding contest problem.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247
"Correct Solution:
```
# AOJ 0195 What is the Most Popular Shop in Tokaich
# Python3 2018.6.20 bal4u
import sys
while True:
tbl = {}
for i in range(5):
s = sum(list(map(int, input().split())))
if i == 0 and s == 0: sys.exit()
tbl[chr(ord('A')+i)] = s
ans = sorted(tbl.items(), key=lambda x: x[1], reverse = True)
print(*ans[0])
```
| 107,795 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247
Submitted Solution:
```
end = 0
while end == 0 :
shop = -1
max_s = 0
for i in range(5) :
try :
s1, s2 = map(int, input().split())
except EOFError :
end = 1
break
if s1 + s2 > max_s :
max_s = s1 + s2
shop = i
if shop == 0 :
shop_name = "A"
elif shop == 1 :
shop_name = "B"
elif shop == 2 :
shop_name = "C"
elif shop == 3 :
shop_name = "D"
elif shop == 4 :
shop_name = "E"
if end == 0 :
print(shop_name, max_s)
```
Yes
| 107,796 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247
Submitted Solution:
```
import sys
while True:
sells=[]
for i in range(5):
s1,s2=[int(j) for j in input().split(" ")]
if s1==0 and s2==0:
sys.exit()
sells.append(s1+s2)
print(chr(ord('A')+sells.index(max(sells))),max(sells))
```
Yes
| 107,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247
Submitted Solution:
```
def solve():
from sys import stdin
f_i = stdin
while True:
s1A, s2A = map(int, f_i.readline().split())
if s1A == 0 and s2A == 0:
break
sales = [(s1A + s2A, 'A')]
for shop in 'BCDE':
cnt = sum(map(int, f_i.readline().split()))
sales.append((cnt, shop))
cnt, shop = max(sales)
print(f"{shop} {cnt}")
solve()
```
Yes
| 107,798 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Aizuwakamatsu City, there is a first city called "Tokaichi" on January 10th every year. This Tokaichi has a history of about 600 years and is the largest first city in the Aizu region. It is also well known that Okiagari-koboshi, a familiar lucky charm, is sold in the Aizu region. Okiagari-koboshi is a papier-mâché with a center of gravity of about 3 cm in size, and it got up immediately after rolling, so it got its name. At each household, be sure to buy one more than your family and offer it to the Kamidana. This one has the meaning of "to increase the number of families" and "to carry troubles".
| <image>
--- | ---
The Tokaichi Executive Committee has decided to investigate the stores that have the highest number of Okiagari-koboshi sold for the next Tokaichi. The number of stores opened this year is 5 (A, B, C, D, E: half-width alphabetic characters), and the number of units sold is reported to the Tokaichi Executive Committee in the morning and afternoon.
Enter the information of each store and create a program that outputs the name of the store with the highest number of units sold per day and the number of stores.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
s1A s2A
s1B s2B
s1C s2C
s1D s2D
s1E s2E
Line i is given the morning sales quantity s1i and the afternoon sales quantity s2i (1 ≤ s1i, s2i ≤ 10000) for A, B, C, D, and E, respectively. However, it is assumed that no store has the same number of units sold per day.
The number of datasets does not exceed 100.
Output
For each dataset, the name of the store with the highest sales volume per day and the number of stores are output on one line.
Example
Input
1593 4311
4321 2155
1256 6421
5310 1455
2152 5421
1549 3386
4528 3719
1234 4321
3330 3109
2739 2199
0 0
Output
C 7677
B 8247
Submitted Solution:
```
while True :
a, b = map(int, input().split())
if(a == b and b == 0) :
break
else :
S = ["A", "B", "C", "D", "E"]
shop = list()
shop.append(a + b)
for i in range(4) :
a, b = map(int, input().split())
shop.append(a + b)
win = max(shop)
Shop = shop.index(win)
print(S[Shop], win)
```
Yes
| 107,799 |
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