message stringlengths 2 11.9k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 137 108k | cluster float64 18 18 | __index_level_0__ int64 274 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Dwango Co., Ltd., there is a content distribution system named 'Dwango Media Cluster', and it is called 'DMC' for short.
The name 'DMC' sounds cool for Niwango-kun, so he starts to define DMC-ness of a string.
Given a string S of length N and an integer k (k \geq 3), he defines the k-DMC number of S as the number of triples (a, b, c) of integers that satisfy the following conditions:
* 0 \leq a < b < c \leq N - 1
* S[a] = `D`
* S[b] = `M`
* S[c] = `C`
* c-a < k
Here S[a] is the a-th character of the string S. Indexing is zero-based, that is, 0 \leq a \leq N - 1 holds.
For a string S and Q integers k_0, k_1, ..., k_{Q-1}, calculate the k_i-DMC number of S for each i (0 \leq i \leq Q-1).
Constraints
* 3 \leq N \leq 10^6
* S consists of uppercase English letters
* 1 \leq Q \leq 75
* 3 \leq k_i \leq N
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N
S
Q
k_{0} k_{1} ... k_{Q-1}
Output
Print Q lines. The i-th line should contain the k_i-DMC number of the string S.
Examples
Input
18
DWANGOMEDIACLUSTER
1
18
Output
1
Input
18
DDDDDDMMMMMCCCCCCC
1
18
Output
210
Input
54
DIALUPWIDEAREANETWORKGAMINGOPERATIONCORPORATIONLIMITED
3
20 30 40
Output
0
1
2
Input
30
DMCDMCDMCDMCDMCDMCDMCDMCDMCDMC
4
5 10 15 20
Output
10
52
110
140
Submitted Solution:
```
import sys
readline = sys.stdin.readline
MOD = 10 ** 9 + 7
INF = float('INF')
sys.setrecursionlimit(10 ** 5)
def main():
N = int(readline())
S = input()
Q = int(readline())
K = list(map(int, readline().split()))
for q in range(Q):
k = K[q]
res = 0
cnt = [0] * 3
for i in range(k):
cur = S[i]
if cur == "D":
cnt[0] += 1
elif cur == "M":
cnt[1] += 1
cnt[2] += cnt[0]
elif cur == "C":
res += cnt[2]
for i in range(k, N):
prev = S[i - k]
cur = S[i]
if prev == "D":
cnt[0] -= 1
cnt[2] -= cnt[1]
elif prev == "M":
cnt[1] -= 1
if cur == "D":
cnt[0] += 1
elif cur == "M":
cnt[1] += 1
cnt[2] += cnt[0]
elif cur == "C":
res += cnt[2]
print(res)
if __name__ == '__main__':
main()
``` | instruction | 0 | 14,838 | 18 | 29,676 |
Yes | output | 1 | 14,838 | 18 | 29,677 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Dwango Co., Ltd., there is a content distribution system named 'Dwango Media Cluster', and it is called 'DMC' for short.
The name 'DMC' sounds cool for Niwango-kun, so he starts to define DMC-ness of a string.
Given a string S of length N and an integer k (k \geq 3), he defines the k-DMC number of S as the number of triples (a, b, c) of integers that satisfy the following conditions:
* 0 \leq a < b < c \leq N - 1
* S[a] = `D`
* S[b] = `M`
* S[c] = `C`
* c-a < k
Here S[a] is the a-th character of the string S. Indexing is zero-based, that is, 0 \leq a \leq N - 1 holds.
For a string S and Q integers k_0, k_1, ..., k_{Q-1}, calculate the k_i-DMC number of S for each i (0 \leq i \leq Q-1).
Constraints
* 3 \leq N \leq 10^6
* S consists of uppercase English letters
* 1 \leq Q \leq 75
* 3 \leq k_i \leq N
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N
S
Q
k_{0} k_{1} ... k_{Q-1}
Output
Print Q lines. The i-th line should contain the k_i-DMC number of the string S.
Examples
Input
18
DWANGOMEDIACLUSTER
1
18
Output
1
Input
18
DDDDDDMMMMMCCCCCCC
1
18
Output
210
Input
54
DIALUPWIDEAREANETWORKGAMINGOPERATIONCORPORATIONLIMITED
3
20 30 40
Output
0
1
2
Input
30
DMCDMCDMCDMCDMCDMCDMCDMCDMCDMC
4
5 10 15 20
Output
10
52
110
140
Submitted Solution:
```
import sys
def main():
input = sys.stdin.readline
N=int(input())
S=input()
Q=int(input())
*K,=map(int, input().split())
for k in K:
ans = 0
d,m,dm=0,0,0
j=0
for i in range(N):
if i-j == k:
if S[j]=='D':
d -= 1
dm -= m
elif S[j]=='M':
m -= 1
j += 1
if S[i] == 'D':
d += 1
elif S[i] == 'M':
m += 1
dm += d
elif S[i] == 'C':
ans += dm
print(ans)
if __name__ == '__main__':
main()
``` | instruction | 0 | 14,839 | 18 | 29,678 |
Yes | output | 1 | 14,839 | 18 | 29,679 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Dwango Co., Ltd., there is a content distribution system named 'Dwango Media Cluster', and it is called 'DMC' for short.
The name 'DMC' sounds cool for Niwango-kun, so he starts to define DMC-ness of a string.
Given a string S of length N and an integer k (k \geq 3), he defines the k-DMC number of S as the number of triples (a, b, c) of integers that satisfy the following conditions:
* 0 \leq a < b < c \leq N - 1
* S[a] = `D`
* S[b] = `M`
* S[c] = `C`
* c-a < k
Here S[a] is the a-th character of the string S. Indexing is zero-based, that is, 0 \leq a \leq N - 1 holds.
For a string S and Q integers k_0, k_1, ..., k_{Q-1}, calculate the k_i-DMC number of S for each i (0 \leq i \leq Q-1).
Constraints
* 3 \leq N \leq 10^6
* S consists of uppercase English letters
* 1 \leq Q \leq 75
* 3 \leq k_i \leq N
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N
S
Q
k_{0} k_{1} ... k_{Q-1}
Output
Print Q lines. The i-th line should contain the k_i-DMC number of the string S.
Examples
Input
18
DWANGOMEDIACLUSTER
1
18
Output
1
Input
18
DDDDDDMMMMMCCCCCCC
1
18
Output
210
Input
54
DIALUPWIDEAREANETWORKGAMINGOPERATIONCORPORATIONLIMITED
3
20 30 40
Output
0
1
2
Input
30
DMCDMCDMCDMCDMCDMCDMCDMCDMCDMC
4
5 10 15 20
Output
10
52
110
140
Submitted Solution:
```
import numpy as np
N = int(input())
S = [i for i in list(input())]
Q = int(input())
k = [int(i) for i in input().split()]
S = np.array(S, dtype=None)
index_D = np.where(S == 'D')
index_M = np.where(S == 'M')
index_C = np.where(S == 'C')
count = 0
for k_i in k:
for a in index_D[0]:
for b in index_M[0]:
for c in index_C[0]:
if a < b < c:
if c-a < k_i:
count += 1
print(count)
``` | instruction | 0 | 14,840 | 18 | 29,680 |
No | output | 1 | 14,840 | 18 | 29,681 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Dwango Co., Ltd., there is a content distribution system named 'Dwango Media Cluster', and it is called 'DMC' for short.
The name 'DMC' sounds cool for Niwango-kun, so he starts to define DMC-ness of a string.
Given a string S of length N and an integer k (k \geq 3), he defines the k-DMC number of S as the number of triples (a, b, c) of integers that satisfy the following conditions:
* 0 \leq a < b < c \leq N - 1
* S[a] = `D`
* S[b] = `M`
* S[c] = `C`
* c-a < k
Here S[a] is the a-th character of the string S. Indexing is zero-based, that is, 0 \leq a \leq N - 1 holds.
For a string S and Q integers k_0, k_1, ..., k_{Q-1}, calculate the k_i-DMC number of S for each i (0 \leq i \leq Q-1).
Constraints
* 3 \leq N \leq 10^6
* S consists of uppercase English letters
* 1 \leq Q \leq 75
* 3 \leq k_i \leq N
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N
S
Q
k_{0} k_{1} ... k_{Q-1}
Output
Print Q lines. The i-th line should contain the k_i-DMC number of the string S.
Examples
Input
18
DWANGOMEDIACLUSTER
1
18
Output
1
Input
18
DDDDDDMMMMMCCCCCCC
1
18
Output
210
Input
54
DIALUPWIDEAREANETWORKGAMINGOPERATIONCORPORATIONLIMITED
3
20 30 40
Output
0
1
2
Input
30
DMCDMCDMCDMCDMCDMCDMCDMCDMCDMC
4
5 10 15 20
Output
10
52
110
140
Submitted Solution:
```
N=int(input())
D,M,C,A=map(ord,'DMCA')
S=[ord(c)-A for c in input()]
Q=int(input())
K=list(map(int,input().split()))
DMC=[D-A,M-A,C-A]
for k in K:
dmc=[0,0,0]
dm=0
a=0
for i,s in enumerate(S):
if i>=k:
if S[i-k] in DMC:
m=DMC.index(S[i-k])
dmc[m]-=1
if m==0:
dm-=dmc[1]
if s in DMC:
p=DMC.index(s)
dmc[p]+=1
if p==1:
dm+=dmc[0]
if p==2:
a+=dm
print(a)
``` | instruction | 0 | 14,841 | 18 | 29,682 |
No | output | 1 | 14,841 | 18 | 29,683 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Dwango Co., Ltd., there is a content distribution system named 'Dwango Media Cluster', and it is called 'DMC' for short.
The name 'DMC' sounds cool for Niwango-kun, so he starts to define DMC-ness of a string.
Given a string S of length N and an integer k (k \geq 3), he defines the k-DMC number of S as the number of triples (a, b, c) of integers that satisfy the following conditions:
* 0 \leq a < b < c \leq N - 1
* S[a] = `D`
* S[b] = `M`
* S[c] = `C`
* c-a < k
Here S[a] is the a-th character of the string S. Indexing is zero-based, that is, 0 \leq a \leq N - 1 holds.
For a string S and Q integers k_0, k_1, ..., k_{Q-1}, calculate the k_i-DMC number of S for each i (0 \leq i \leq Q-1).
Constraints
* 3 \leq N \leq 10^6
* S consists of uppercase English letters
* 1 \leq Q \leq 75
* 3 \leq k_i \leq N
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N
S
Q
k_{0} k_{1} ... k_{Q-1}
Output
Print Q lines. The i-th line should contain the k_i-DMC number of the string S.
Examples
Input
18
DWANGOMEDIACLUSTER
1
18
Output
1
Input
18
DDDDDDMMMMMCCCCCCC
1
18
Output
210
Input
54
DIALUPWIDEAREANETWORKGAMINGOPERATIONCORPORATIONLIMITED
3
20 30 40
Output
0
1
2
Input
30
DMCDMCDMCDMCDMCDMCDMCDMCDMCDMC
4
5 10 15 20
Output
10
52
110
140
Submitted Solution:
```
N=int(input())
D,M,C=map(ord,'DMC')
S=[ord(c) for c in input()]
input()
K=map(int,input().split())
for k in K:
d,m,c,dm=0,0,0,0
a=0
for i,s in enumerate(S):
if i>=k:
if S[i-k]==D:
d-=1
dm-=m
elif S[i-k]==M:
m-=1
elif S[i-k]==C:
c-=1
if s==D:
d+=1
elif s==M:
m+=1
dm+=d
elif s==C:
c+=1
a+=dm
print(a)
``` | instruction | 0 | 14,842 | 18 | 29,684 |
No | output | 1 | 14,842 | 18 | 29,685 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Dwango Co., Ltd., there is a content distribution system named 'Dwango Media Cluster', and it is called 'DMC' for short.
The name 'DMC' sounds cool for Niwango-kun, so he starts to define DMC-ness of a string.
Given a string S of length N and an integer k (k \geq 3), he defines the k-DMC number of S as the number of triples (a, b, c) of integers that satisfy the following conditions:
* 0 \leq a < b < c \leq N - 1
* S[a] = `D`
* S[b] = `M`
* S[c] = `C`
* c-a < k
Here S[a] is the a-th character of the string S. Indexing is zero-based, that is, 0 \leq a \leq N - 1 holds.
For a string S and Q integers k_0, k_1, ..., k_{Q-1}, calculate the k_i-DMC number of S for each i (0 \leq i \leq Q-1).
Constraints
* 3 \leq N \leq 10^6
* S consists of uppercase English letters
* 1 \leq Q \leq 75
* 3 \leq k_i \leq N
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
N
S
Q
k_{0} k_{1} ... k_{Q-1}
Output
Print Q lines. The i-th line should contain the k_i-DMC number of the string S.
Examples
Input
18
DWANGOMEDIACLUSTER
1
18
Output
1
Input
18
DDDDDDMMMMMCCCCCCC
1
18
Output
210
Input
54
DIALUPWIDEAREANETWORKGAMINGOPERATIONCORPORATIONLIMITED
3
20 30 40
Output
0
1
2
Input
30
DMCDMCDMCDMCDMCDMCDMCDMCDMCDMC
4
5 10 15 20
Output
10
52
110
140
Submitted Solution:
```
from itertools import accumulate
n = int(input())
s = input()
acc_cnt_a = [0] * (n + 1)
acc_cnt_b = [0] * (n + 1)
acc_cnt_ab = [0] * (n + 1)
c_loc = []
cnt_a = 0
for i, c in enumerate(s):
if c == 'D':
acc_cnt_a[i + 1] += 1
cnt_a += 1
elif c == 'M':
acc_cnt_b[i + 1] += 1
acc_cnt_ab[i + 1] += cnt_a
elif c == 'C':
c_loc.append(i + 1)
acc_cnt_a = list(accumulate(acc_cnt_a))
acc_cnt_b = list(accumulate(acc_cnt_b))
acc_cnt_ab = list(accumulate(acc_cnt_ab))
q = int(input())
buf = []
for k in map(int, input().split()):
ans = 0
for r in c_loc:
l = max(0, r - k)
rejected_a = acc_cnt_a[l]
cnt_internal_b = acc_cnt_b[r] - acc_cnt_b[l]
cnt_internal_ab = acc_cnt_ab[r] - acc_cnt_ab[l]
ans += cnt_internal_ab - rejected_a * cnt_internal_b
buf.append(ans)
print('\n'.join(map(str, buf)))
``` | instruction | 0 | 14,843 | 18 | 29,686 |
No | output | 1 | 14,843 | 18 | 29,687 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S.
Takahashi can insert the character `A` at any position in this string any number of times.
Can he change S into `AKIHABARA`?
Constraints
* 1 \leq |S| \leq 50
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If it is possible to change S into `AKIHABARA`, print `YES`; otherwise, print `NO`.
Examples
Input
KIHBR
Output
YES
Input
AKIBAHARA
Output
NO
Input
AAKIAHBAARA
Output
NO | instruction | 0 | 14,860 | 18 | 29,720 |
"Correct Solution:
```
import re
s = input()
if re.match('A?KIHA?BA?RA?$', s):
print('YES')
else:
print('NO')
``` | output | 1 | 14,860 | 18 | 29,721 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S.
Takahashi can insert the character `A` at any position in this string any number of times.
Can he change S into `AKIHABARA`?
Constraints
* 1 \leq |S| \leq 50
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If it is possible to change S into `AKIHABARA`, print `YES`; otherwise, print `NO`.
Examples
Input
KIHBR
Output
YES
Input
AKIBAHARA
Output
NO
Input
AAKIAHBAARA
Output
NO | instruction | 0 | 14,861 | 18 | 29,722 |
"Correct Solution:
```
import re
S = input()
if re.match('A?KIHA?BA?RA?$', S):
print('YES')
else:
print('NO')
``` | output | 1 | 14,861 | 18 | 29,723 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S.
Takahashi can insert the character `A` at any position in this string any number of times.
Can he change S into `AKIHABARA`?
Constraints
* 1 \leq |S| \leq 50
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If it is possible to change S into `AKIHABARA`, print `YES`; otherwise, print `NO`.
Examples
Input
KIHBR
Output
YES
Input
AKIBAHARA
Output
NO
Input
AAKIAHBAARA
Output
NO | instruction | 0 | 14,862 | 18 | 29,724 |
"Correct Solution:
```
S=input()
b=S.replace("A","")=="KIHBR"
if b:
L=list(map(len, "".join("*" if s!="A" else s for s in S).split("*")))
X=[1,0,0,1,1,1,1]
b&=all(L[i]<=X[i] for i in range(len(L)))
print("YNEOS"[not b::2])
``` | output | 1 | 14,862 | 18 | 29,725 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S.
Takahashi can insert the character `A` at any position in this string any number of times.
Can he change S into `AKIHABARA`?
Constraints
* 1 \leq |S| \leq 50
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If it is possible to change S into `AKIHABARA`, print `YES`; otherwise, print `NO`.
Examples
Input
KIHBR
Output
YES
Input
AKIBAHARA
Output
NO
Input
AAKIAHBAARA
Output
NO | instruction | 0 | 14,863 | 18 | 29,726 |
"Correct Solution:
```
s = input()
a = 'AKIHABARA'
a = list(a)
idxs = [0, 4, 6, 8]
for i in range(2**4):
for j in range(4):
a[idxs[j]] = 'A' if i>>j&1 else ''
if s==''.join(a):
print('YES')
exit(0)
print('NO')
``` | output | 1 | 14,863 | 18 | 29,727 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S.
Takahashi can insert the character `A` at any position in this string any number of times.
Can he change S into `AKIHABARA`?
Constraints
* 1 \leq |S| \leq 50
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If it is possible to change S into `AKIHABARA`, print `YES`; otherwise, print `NO`.
Examples
Input
KIHBR
Output
YES
Input
AKIBAHARA
Output
NO
Input
AAKIAHBAARA
Output
NO | instruction | 0 | 14,864 | 18 | 29,728 |
"Correct Solution:
```
import re
s = input()
if re.fullmatch(r'A?KIHA?BA?RA?', s) == None:
print('NO')
else:
print('YES')
``` | output | 1 | 14,864 | 18 | 29,729 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S.
Takahashi can insert the character `A` at any position in this string any number of times.
Can he change S into `AKIHABARA`?
Constraints
* 1 \leq |S| \leq 50
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If it is possible to change S into `AKIHABARA`, print `YES`; otherwise, print `NO`.
Examples
Input
KIHBR
Output
YES
Input
AKIBAHARA
Output
NO
Input
AAKIAHBAARA
Output
NO | instruction | 0 | 14,865 | 18 | 29,730 |
"Correct Solution:
```
import re
print("YES" if re.match(r"^A?KIHA?BA?RA?$", input()) else "NO")
``` | output | 1 | 14,865 | 18 | 29,731 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S.
Takahashi can insert the character `A` at any position in this string any number of times.
Can he change S into `AKIHABARA`?
Constraints
* 1 \leq |S| \leq 50
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If it is possible to change S into `AKIHABARA`, print `YES`; otherwise, print `NO`.
Examples
Input
KIHBR
Output
YES
Input
AKIBAHARA
Output
NO
Input
AAKIAHBAARA
Output
NO | instruction | 0 | 14,866 | 18 | 29,732 |
"Correct Solution:
```
S=input()
akh=list("AKIHABARA")
lis=[]
a=[0,4,6,8]
for i in range(16):
akh=list("AKIHABARA")
for j in range(4):
if (i>>j) & 1:
akh[a[j]]=""
lis.append("".join(akh))
if S in lis:print("YES")
else:print("NO")
``` | output | 1 | 14,866 | 18 | 29,733 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S.
Takahashi can insert the character `A` at any position in this string any number of times.
Can he change S into `AKIHABARA`?
Constraints
* 1 \leq |S| \leq 50
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If it is possible to change S into `AKIHABARA`, print `YES`; otherwise, print `NO`.
Examples
Input
KIHBR
Output
YES
Input
AKIBAHARA
Output
NO
Input
AAKIAHBAARA
Output
NO | instruction | 0 | 14,867 | 18 | 29,734 |
"Correct Solution:
```
S = input().strip()
if S.find('AA') != -1:
print('NO')
exit()
if S.find('KAI') != -1:
print('NO')
exit()
if S.find('IAH') != -1:
print('NO')
exit()
if S.replace('A', '') == 'KIHBR':
print('YES')
else:
print('NO')
``` | output | 1 | 14,867 | 18 | 29,735 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S.
Takahashi can insert the character `A` at any position in this string any number of times.
Can he change S into `AKIHABARA`?
Constraints
* 1 \leq |S| \leq 50
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If it is possible to change S into `AKIHABARA`, print `YES`; otherwise, print `NO`.
Examples
Input
KIHBR
Output
YES
Input
AKIBAHARA
Output
NO
Input
AAKIAHBAARA
Output
NO
Submitted Solution:
```
import re
pattern = '^A?KIHA?BA?RA?$'
print("YES" if re.match(pattern,input()) else "NO")
``` | instruction | 0 | 14,868 | 18 | 29,736 |
Yes | output | 1 | 14,868 | 18 | 29,737 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S.
Takahashi can insert the character `A` at any position in this string any number of times.
Can he change S into `AKIHABARA`?
Constraints
* 1 \leq |S| \leq 50
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If it is possible to change S into `AKIHABARA`, print `YES`; otherwise, print `NO`.
Examples
Input
KIHBR
Output
YES
Input
AKIBAHARA
Output
NO
Input
AAKIAHBAARA
Output
NO
Submitted Solution:
```
s = input()
flag = 1
if s.find("AA") != -1:
flag = 0
if s.find("KA") != -1:
flag = 0
if s.find("IA") != -1:
flag = 0
if s.replace("A","") != "KIHBR":
flag = 0
if flag == 1:
print("YES")
else:
print("NO")
``` | instruction | 0 | 14,869 | 18 | 29,738 |
Yes | output | 1 | 14,869 | 18 | 29,739 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S.
Takahashi can insert the character `A` at any position in this string any number of times.
Can he change S into `AKIHABARA`?
Constraints
* 1 \leq |S| \leq 50
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If it is possible to change S into `AKIHABARA`, print `YES`; otherwise, print `NO`.
Examples
Input
KIHBR
Output
YES
Input
AKIBAHARA
Output
NO
Input
AAKIAHBAARA
Output
NO
Submitted Solution:
```
S = input()
T = ["AKIHABARA","KIHABARA","AKIHBARA","AKIHABRA","AKIHABAR"\
,"KIHBARA","KIHABRA","KIHABAR","AKIHBRA","AKIHBAR","AKIHABR"\
,"KIHBRA","KIHBAR","KIHABR","AKIHBR","KIHBR"]
print("YES" if S in T else "NO")
``` | instruction | 0 | 14,870 | 18 | 29,740 |
Yes | output | 1 | 14,870 | 18 | 29,741 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S.
Takahashi can insert the character `A` at any position in this string any number of times.
Can he change S into `AKIHABARA`?
Constraints
* 1 \leq |S| \leq 50
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If it is possible to change S into `AKIHABARA`, print `YES`; otherwise, print `NO`.
Examples
Input
KIHBR
Output
YES
Input
AKIBAHARA
Output
NO
Input
AAKIAHBAARA
Output
NO
Submitted Solution:
```
import re;print('YNEOS'[re.match('A?KIHA?BA?RA?$',input())==None::2])
``` | instruction | 0 | 14,871 | 18 | 29,742 |
Yes | output | 1 | 14,871 | 18 | 29,743 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S.
Takahashi can insert the character `A` at any position in this string any number of times.
Can he change S into `AKIHABARA`?
Constraints
* 1 \leq |S| \leq 50
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If it is possible to change S into `AKIHABARA`, print `YES`; otherwise, print `NO`.
Examples
Input
KIHBR
Output
YES
Input
AKIBAHARA
Output
NO
Input
AAKIAHBAARA
Output
NO
Submitted Solution:
```
s = input()
if len(s) > 9:
print("NO")
exit()
for i in range(len(s)-1):
if s[i] + s[i+1] == "AA":
print("NO")
exit()
s = "".join([i for i in s if i != "A"])
c = "KIHBR"
print("YES") if s == c else print("NO")
``` | instruction | 0 | 14,872 | 18 | 29,744 |
No | output | 1 | 14,872 | 18 | 29,745 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S.
Takahashi can insert the character `A` at any position in this string any number of times.
Can he change S into `AKIHABARA`?
Constraints
* 1 \leq |S| \leq 50
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If it is possible to change S into `AKIHABARA`, print `YES`; otherwise, print `NO`.
Examples
Input
KIHBR
Output
YES
Input
AKIBAHARA
Output
NO
Input
AAKIAHBAARA
Output
NO
Submitted Solution:
```
S = input().strip()
if S.find('AA') >= 0:
print('NO')
exit()
if S.replace('A', '') == 'KIHBR':
print('YES')
else:
print('NO')
``` | instruction | 0 | 14,873 | 18 | 29,746 |
No | output | 1 | 14,873 | 18 | 29,747 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S.
Takahashi can insert the character `A` at any position in this string any number of times.
Can he change S into `AKIHABARA`?
Constraints
* 1 \leq |S| \leq 50
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If it is possible to change S into `AKIHABARA`, print `YES`; otherwise, print `NO`.
Examples
Input
KIHBR
Output
YES
Input
AKIBAHARA
Output
NO
Input
AAKIAHBAARA
Output
NO
Submitted Solution:
```
def main():
s = str(input())
t = 'AKIHABARA'
i = 0
j = 0
while i < len(t) and j < len(s):
if t[i] == s[j]:
i += 1
j += 1
else:
i += 1
print('YES' if j == len(s) - 1 else 'NO')
if __name__ == '__main__':
main()
``` | instruction | 0 | 14,874 | 18 | 29,748 |
No | output | 1 | 14,874 | 18 | 29,749 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S.
Takahashi can insert the character `A` at any position in this string any number of times.
Can he change S into `AKIHABARA`?
Constraints
* 1 \leq |S| \leq 50
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If it is possible to change S into `AKIHABARA`, print `YES`; otherwise, print `NO`.
Examples
Input
KIHBR
Output
YES
Input
AKIBAHARA
Output
NO
Input
AAKIAHBAARA
Output
NO
Submitted Solution:
```
s = input()
ans = "AKIHABARA"
now = 0
for i in s:
while now < len(ans) and ans[now] != i:
now += 1
if now <= 8:
print("YES")
else:
print("NO")
``` | instruction | 0 | 14,875 | 18 | 29,750 |
No | output | 1 | 14,875 | 18 | 29,751 |
Provide a correct Python 3 solution for this coding contest problem.
()
Problem Statement
There is a string S. Initially, S is an empty string.
Perform the following processing in order of n.
* Add x_i p_i (=" (" or") ") to the end of S.
After processing, determine if S is a well-balanced string.
"The string is balanced" is defined as follows.
* The empty string is well-balanced.
* For balanced strings a and b, a + b (+ represents a concatenation of strings) is balanced.
* For a balanced string a, "(" + a + ")" is balanced.
Constraints
* 1 ≤ n ≤ 1,000
* 1 ≤ x_i ≤ 10 ^ 6
* p_i is "(" or ")"
Input
Input follows the following format. All given numbers are integers.
n
p_1 x_1
.. ..
p_n x_n
Output
Output "YES" if balanced, otherwise output "NO" on one line.
Examples
Input
3
( 5
) 4
) 1
Output
YES
Input
5
( 2
) 2
( 3
) 1
) 2
Output
YES
Input
2
) 1
( 1
Output
NO | instruction | 0 | 15,020 | 18 | 30,040 |
"Correct Solution:
```
n = int(input())
pos = 0
flag = True
for _ in range(n):
k, num = input().split()
num = int(num)
if k == "(":
pos += num
else:
pos -= num
if pos < 0:
flag = False
if not flag or pos != 0:
print("NO")
else:
print("YES")
``` | output | 1 | 15,020 | 18 | 30,041 |
Provide a correct Python 3 solution for this coding contest problem.
()
Problem Statement
There is a string S. Initially, S is an empty string.
Perform the following processing in order of n.
* Add x_i p_i (=" (" or") ") to the end of S.
After processing, determine if S is a well-balanced string.
"The string is balanced" is defined as follows.
* The empty string is well-balanced.
* For balanced strings a and b, a + b (+ represents a concatenation of strings) is balanced.
* For a balanced string a, "(" + a + ")" is balanced.
Constraints
* 1 ≤ n ≤ 1,000
* 1 ≤ x_i ≤ 10 ^ 6
* p_i is "(" or ")"
Input
Input follows the following format. All given numbers are integers.
n
p_1 x_1
.. ..
p_n x_n
Output
Output "YES" if balanced, otherwise output "NO" on one line.
Examples
Input
3
( 5
) 4
) 1
Output
YES
Input
5
( 2
) 2
( 3
) 1
) 2
Output
YES
Input
2
) 1
( 1
Output
NO | instruction | 0 | 15,021 | 18 | 30,042 |
"Correct Solution:
```
a=0
for _ in range(int(input())):
b,c=input().split()
a+=int(c) if b=='(' else -int(c)
if a<0:break
print('NO' if a else 'YES')
``` | output | 1 | 15,021 | 18 | 30,043 |
Provide a correct Python 3 solution for this coding contest problem.
()
Problem Statement
There is a string S. Initially, S is an empty string.
Perform the following processing in order of n.
* Add x_i p_i (=" (" or") ") to the end of S.
After processing, determine if S is a well-balanced string.
"The string is balanced" is defined as follows.
* The empty string is well-balanced.
* For balanced strings a and b, a + b (+ represents a concatenation of strings) is balanced.
* For a balanced string a, "(" + a + ")" is balanced.
Constraints
* 1 ≤ n ≤ 1,000
* 1 ≤ x_i ≤ 10 ^ 6
* p_i is "(" or ")"
Input
Input follows the following format. All given numbers are integers.
n
p_1 x_1
.. ..
p_n x_n
Output
Output "YES" if balanced, otherwise output "NO" on one line.
Examples
Input
3
( 5
) 4
) 1
Output
YES
Input
5
( 2
) 2
( 3
) 1
) 2
Output
YES
Input
2
) 1
( 1
Output
NO | instruction | 0 | 15,022 | 18 | 30,044 |
"Correct Solution:
```
n=int(input())
a,b=0,0
flag=0
for i in range(n):
p,x=map(str,input().split())
if p=="(":a+=int(x)
else:b+=int(x)
if a<b:
flag=1
break
if flag==1 or a!=b:print("NO")
else:print("YES")
``` | output | 1 | 15,022 | 18 | 30,045 |
Provide a correct Python 3 solution for this coding contest problem.
()
Problem Statement
There is a string S. Initially, S is an empty string.
Perform the following processing in order of n.
* Add x_i p_i (=" (" or") ") to the end of S.
After processing, determine if S is a well-balanced string.
"The string is balanced" is defined as follows.
* The empty string is well-balanced.
* For balanced strings a and b, a + b (+ represents a concatenation of strings) is balanced.
* For a balanced string a, "(" + a + ")" is balanced.
Constraints
* 1 ≤ n ≤ 1,000
* 1 ≤ x_i ≤ 10 ^ 6
* p_i is "(" or ")"
Input
Input follows the following format. All given numbers are integers.
n
p_1 x_1
.. ..
p_n x_n
Output
Output "YES" if balanced, otherwise output "NO" on one line.
Examples
Input
3
( 5
) 4
) 1
Output
YES
Input
5
( 2
) 2
( 3
) 1
) 2
Output
YES
Input
2
) 1
( 1
Output
NO | instruction | 0 | 15,023 | 18 | 30,046 |
"Correct Solution:
```
n = int(input())
b = 0
ans = True
for _ in range(n):
p, x = input().split()
x = int(x)
if p == "(":
b += x
else:
b -= x
if b < 0:
ans = False
if ans:
if b == 0:
print("YES")
else:
print("NO")
else:
print("NO")
``` | output | 1 | 15,023 | 18 | 30,047 |
Provide a correct Python 3 solution for this coding contest problem.
()
Problem Statement
There is a string S. Initially, S is an empty string.
Perform the following processing in order of n.
* Add x_i p_i (=" (" or") ") to the end of S.
After processing, determine if S is a well-balanced string.
"The string is balanced" is defined as follows.
* The empty string is well-balanced.
* For balanced strings a and b, a + b (+ represents a concatenation of strings) is balanced.
* For a balanced string a, "(" + a + ")" is balanced.
Constraints
* 1 ≤ n ≤ 1,000
* 1 ≤ x_i ≤ 10 ^ 6
* p_i is "(" or ")"
Input
Input follows the following format. All given numbers are integers.
n
p_1 x_1
.. ..
p_n x_n
Output
Output "YES" if balanced, otherwise output "NO" on one line.
Examples
Input
3
( 5
) 4
) 1
Output
YES
Input
5
( 2
) 2
( 3
) 1
) 2
Output
YES
Input
2
) 1
( 1
Output
NO | instruction | 0 | 15,024 | 18 | 30,048 |
"Correct Solution:
```
a=0
for _ in range(int(input())):
b,c=input().split()
a+= int(c) if b=='(' else -int(c)
if a<0:print('NO');break
else: print('NO' if a else 'YES')
``` | output | 1 | 15,024 | 18 | 30,049 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Happy new year! The year 2020 is also known as Year Gyeongja (경자년, gyeongja-nyeon) in Korea. Where did the name come from? Let's briefly look at the Gapja system, which is traditionally used in Korea to name the years.
There are two sequences of n strings s_1, s_2, s_3, …, s_{n} and m strings t_1, t_2, t_3, …, t_{m}. These strings contain only lowercase letters. There might be duplicates among these strings.
Let's call a concatenation of strings x and y as the string that is obtained by writing down strings x and y one right after another without changing the order. For example, the concatenation of the strings "code" and "forces" is the string "codeforces".
The year 1 has a name which is the concatenation of the two strings s_1 and t_1. When the year increases by one, we concatenate the next two strings in order from each of the respective sequences. If the string that is currently being used is at the end of its sequence, we go back to the first string in that sequence.
For example, if n = 3, m = 4, s = {"a", "b", "c"}, t = {"d", "e", "f", "g"}, the following table denotes the resulting year names. Note that the names of the years may repeat.
<image>
You are given two sequences of strings of size n and m and also q queries. For each query, you will be given the current year. Could you find the name corresponding to the given year, according to the Gapja system?
Input
The first line contains two integers n, m (1 ≤ n, m ≤ 20).
The next line contains n strings s_1, s_2, …, s_{n}. Each string contains only lowercase letters, and they are separated by spaces. The length of each string is at least 1 and at most 10.
The next line contains m strings t_1, t_2, …, t_{m}. Each string contains only lowercase letters, and they are separated by spaces. The length of each string is at least 1 and at most 10.
Among the given n + m strings may be duplicates (that is, they are not necessarily all different).
The next line contains a single integer q (1 ≤ q ≤ 2 020).
In the next q lines, an integer y (1 ≤ y ≤ 10^9) is given, denoting the year we want to know the name for.
Output
Print q lines. For each line, print the name of the year as per the rule described above.
Example
Input
10 12
sin im gye gap eul byeong jeong mu gi gyeong
yu sul hae ja chuk in myo jin sa o mi sin
14
1
2
3
4
10
11
12
13
73
2016
2017
2018
2019
2020
Output
sinyu
imsul
gyehae
gapja
gyeongo
sinmi
imsin
gyeyu
gyeyu
byeongsin
jeongyu
musul
gihae
gyeongja
Note
The first example denotes the actual names used in the Gapja system. These strings usually are either a number or the name of some animal.
Submitted Solution:
```
n,m = map(int,input().split())
s = input().split()
t = input().split()
q = int(input())
mass = ""
for i in range(q):
y = int(input())
if y <= n:
mass += s[y-1]
else:
mass += s[y%n-1]
if y <= m:
mass += t[y-1] + ' '
else:
mass += t[y%m-1] + ' '
mass = mass.split()
for i in range(q):
print(mass[i])
``` | instruction | 0 | 15,189 | 18 | 30,378 |
Yes | output | 1 | 15,189 | 18 | 30,379 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Happy new year! The year 2020 is also known as Year Gyeongja (경자년, gyeongja-nyeon) in Korea. Where did the name come from? Let's briefly look at the Gapja system, which is traditionally used in Korea to name the years.
There are two sequences of n strings s_1, s_2, s_3, …, s_{n} and m strings t_1, t_2, t_3, …, t_{m}. These strings contain only lowercase letters. There might be duplicates among these strings.
Let's call a concatenation of strings x and y as the string that is obtained by writing down strings x and y one right after another without changing the order. For example, the concatenation of the strings "code" and "forces" is the string "codeforces".
The year 1 has a name which is the concatenation of the two strings s_1 and t_1. When the year increases by one, we concatenate the next two strings in order from each of the respective sequences. If the string that is currently being used is at the end of its sequence, we go back to the first string in that sequence.
For example, if n = 3, m = 4, s = {"a", "b", "c"}, t = {"d", "e", "f", "g"}, the following table denotes the resulting year names. Note that the names of the years may repeat.
<image>
You are given two sequences of strings of size n and m and also q queries. For each query, you will be given the current year. Could you find the name corresponding to the given year, according to the Gapja system?
Input
The first line contains two integers n, m (1 ≤ n, m ≤ 20).
The next line contains n strings s_1, s_2, …, s_{n}. Each string contains only lowercase letters, and they are separated by spaces. The length of each string is at least 1 and at most 10.
The next line contains m strings t_1, t_2, …, t_{m}. Each string contains only lowercase letters, and they are separated by spaces. The length of each string is at least 1 and at most 10.
Among the given n + m strings may be duplicates (that is, they are not necessarily all different).
The next line contains a single integer q (1 ≤ q ≤ 2 020).
In the next q lines, an integer y (1 ≤ y ≤ 10^9) is given, denoting the year we want to know the name for.
Output
Print q lines. For each line, print the name of the year as per the rule described above.
Example
Input
10 12
sin im gye gap eul byeong jeong mu gi gyeong
yu sul hae ja chuk in myo jin sa o mi sin
14
1
2
3
4
10
11
12
13
73
2016
2017
2018
2019
2020
Output
sinyu
imsul
gyehae
gapja
gyeongo
sinmi
imsin
gyeyu
gyeyu
byeongsin
jeongyu
musul
gihae
gyeongja
Note
The first example denotes the actual names used in the Gapja system. These strings usually are either a number or the name of some animal.
Submitted Solution:
```
n, m = map(int, input().split())
s = list(map(str, input().split()))
t = list(map(str, input().split()))
q=int(input())
y=[]
for i in range(q):
y.append(int(input()))
for i in range(0,q):
if y[i]>n:
ni=y[i]%n-1
else:
ni=y[i]-1
if y[i]>m:
mi=y[i]%m-1
else:
mi=y[i]-1
print(s[ni]+t[mi])
``` | instruction | 0 | 15,190 | 18 | 30,380 |
Yes | output | 1 | 15,190 | 18 | 30,381 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Happy new year! The year 2020 is also known as Year Gyeongja (경자년, gyeongja-nyeon) in Korea. Where did the name come from? Let's briefly look at the Gapja system, which is traditionally used in Korea to name the years.
There are two sequences of n strings s_1, s_2, s_3, …, s_{n} and m strings t_1, t_2, t_3, …, t_{m}. These strings contain only lowercase letters. There might be duplicates among these strings.
Let's call a concatenation of strings x and y as the string that is obtained by writing down strings x and y one right after another without changing the order. For example, the concatenation of the strings "code" and "forces" is the string "codeforces".
The year 1 has a name which is the concatenation of the two strings s_1 and t_1. When the year increases by one, we concatenate the next two strings in order from each of the respective sequences. If the string that is currently being used is at the end of its sequence, we go back to the first string in that sequence.
For example, if n = 3, m = 4, s = {"a", "b", "c"}, t = {"d", "e", "f", "g"}, the following table denotes the resulting year names. Note that the names of the years may repeat.
<image>
You are given two sequences of strings of size n and m and also q queries. For each query, you will be given the current year. Could you find the name corresponding to the given year, according to the Gapja system?
Input
The first line contains two integers n, m (1 ≤ n, m ≤ 20).
The next line contains n strings s_1, s_2, …, s_{n}. Each string contains only lowercase letters, and they are separated by spaces. The length of each string is at least 1 and at most 10.
The next line contains m strings t_1, t_2, …, t_{m}. Each string contains only lowercase letters, and they are separated by spaces. The length of each string is at least 1 and at most 10.
Among the given n + m strings may be duplicates (that is, they are not necessarily all different).
The next line contains a single integer q (1 ≤ q ≤ 2 020).
In the next q lines, an integer y (1 ≤ y ≤ 10^9) is given, denoting the year we want to know the name for.
Output
Print q lines. For each line, print the name of the year as per the rule described above.
Example
Input
10 12
sin im gye gap eul byeong jeong mu gi gyeong
yu sul hae ja chuk in myo jin sa o mi sin
14
1
2
3
4
10
11
12
13
73
2016
2017
2018
2019
2020
Output
sinyu
imsul
gyehae
gapja
gyeongo
sinmi
imsin
gyeyu
gyeyu
byeongsin
jeongyu
musul
gihae
gyeongja
Note
The first example denotes the actual names used in the Gapja system. These strings usually are either a number or the name of some animal.
Submitted Solution:
```
length1,length2=map(int,input().split())
s1=list(map(str,input().split()))
s2=list(map(str,input().split()))
nn=int(input())
for x in range(nn):
a=int(input()) - 1
x1=a % length1
x2=a % length2
print(s1[x1]+s2[x2])
``` | instruction | 0 | 15,191 | 18 | 30,382 |
Yes | output | 1 | 15,191 | 18 | 30,383 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Happy new year! The year 2020 is also known as Year Gyeongja (경자년, gyeongja-nyeon) in Korea. Where did the name come from? Let's briefly look at the Gapja system, which is traditionally used in Korea to name the years.
There are two sequences of n strings s_1, s_2, s_3, …, s_{n} and m strings t_1, t_2, t_3, …, t_{m}. These strings contain only lowercase letters. There might be duplicates among these strings.
Let's call a concatenation of strings x and y as the string that is obtained by writing down strings x and y one right after another without changing the order. For example, the concatenation of the strings "code" and "forces" is the string "codeforces".
The year 1 has a name which is the concatenation of the two strings s_1 and t_1. When the year increases by one, we concatenate the next two strings in order from each of the respective sequences. If the string that is currently being used is at the end of its sequence, we go back to the first string in that sequence.
For example, if n = 3, m = 4, s = {"a", "b", "c"}, t = {"d", "e", "f", "g"}, the following table denotes the resulting year names. Note that the names of the years may repeat.
<image>
You are given two sequences of strings of size n and m and also q queries. For each query, you will be given the current year. Could you find the name corresponding to the given year, according to the Gapja system?
Input
The first line contains two integers n, m (1 ≤ n, m ≤ 20).
The next line contains n strings s_1, s_2, …, s_{n}. Each string contains only lowercase letters, and they are separated by spaces. The length of each string is at least 1 and at most 10.
The next line contains m strings t_1, t_2, …, t_{m}. Each string contains only lowercase letters, and they are separated by spaces. The length of each string is at least 1 and at most 10.
Among the given n + m strings may be duplicates (that is, they are not necessarily all different).
The next line contains a single integer q (1 ≤ q ≤ 2 020).
In the next q lines, an integer y (1 ≤ y ≤ 10^9) is given, denoting the year we want to know the name for.
Output
Print q lines. For each line, print the name of the year as per the rule described above.
Example
Input
10 12
sin im gye gap eul byeong jeong mu gi gyeong
yu sul hae ja chuk in myo jin sa o mi sin
14
1
2
3
4
10
11
12
13
73
2016
2017
2018
2019
2020
Output
sinyu
imsul
gyehae
gapja
gyeongo
sinmi
imsin
gyeyu
gyeyu
byeongsin
jeongyu
musul
gihae
gyeongja
Note
The first example denotes the actual names used in the Gapja system. These strings usually are either a number or the name of some animal.
Submitted Solution:
```
nm = input().split()
n = int(nm[0])
m = int(nm[1])
nList = input().split()
mList = input().split()
q = int(input())
for x in range(q):
y = int(input())
ny = y % n
my = y % m
print(nList[ny-1]+mList[my-1])
``` | instruction | 0 | 15,192 | 18 | 30,384 |
Yes | output | 1 | 15,192 | 18 | 30,385 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Happy new year! The year 2020 is also known as Year Gyeongja (경자년, gyeongja-nyeon) in Korea. Where did the name come from? Let's briefly look at the Gapja system, which is traditionally used in Korea to name the years.
There are two sequences of n strings s_1, s_2, s_3, …, s_{n} and m strings t_1, t_2, t_3, …, t_{m}. These strings contain only lowercase letters. There might be duplicates among these strings.
Let's call a concatenation of strings x and y as the string that is obtained by writing down strings x and y one right after another without changing the order. For example, the concatenation of the strings "code" and "forces" is the string "codeforces".
The year 1 has a name which is the concatenation of the two strings s_1 and t_1. When the year increases by one, we concatenate the next two strings in order from each of the respective sequences. If the string that is currently being used is at the end of its sequence, we go back to the first string in that sequence.
For example, if n = 3, m = 4, s = {"a", "b", "c"}, t = {"d", "e", "f", "g"}, the following table denotes the resulting year names. Note that the names of the years may repeat.
<image>
You are given two sequences of strings of size n and m and also q queries. For each query, you will be given the current year. Could you find the name corresponding to the given year, according to the Gapja system?
Input
The first line contains two integers n, m (1 ≤ n, m ≤ 20).
The next line contains n strings s_1, s_2, …, s_{n}. Each string contains only lowercase letters, and they are separated by spaces. The length of each string is at least 1 and at most 10.
The next line contains m strings t_1, t_2, …, t_{m}. Each string contains only lowercase letters, and they are separated by spaces. The length of each string is at least 1 and at most 10.
Among the given n + m strings may be duplicates (that is, they are not necessarily all different).
The next line contains a single integer q (1 ≤ q ≤ 2 020).
In the next q lines, an integer y (1 ≤ y ≤ 10^9) is given, denoting the year we want to know the name for.
Output
Print q lines. For each line, print the name of the year as per the rule described above.
Example
Input
10 12
sin im gye gap eul byeong jeong mu gi gyeong
yu sul hae ja chuk in myo jin sa o mi sin
14
1
2
3
4
10
11
12
13
73
2016
2017
2018
2019
2020
Output
sinyu
imsul
gyehae
gapja
gyeongo
sinmi
imsin
gyeyu
gyeyu
byeongsin
jeongyu
musul
gihae
gyeongja
Note
The first example denotes the actual names used in the Gapja system. These strings usually are either a number or the name of some animal.
Submitted Solution:
```
l=list(map(int,input().split()))
s1=list(map(str,input().split()))
s2=list(map(str,input().split()))
q=int(input())
n=l[0]
m=l[1]
while(q):
yr=int(input())
if(yr>120):
yr=yr%120
y1=yr%n
y2=yr%m
if(y1==0):
y1=n
if(y2==0):
y2=m
ans=s1[y1-1]+s2[y2-1]
print(ans)
q-=1
``` | instruction | 0 | 15,193 | 18 | 30,386 |
No | output | 1 | 15,193 | 18 | 30,387 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Happy new year! The year 2020 is also known as Year Gyeongja (경자년, gyeongja-nyeon) in Korea. Where did the name come from? Let's briefly look at the Gapja system, which is traditionally used in Korea to name the years.
There are two sequences of n strings s_1, s_2, s_3, …, s_{n} and m strings t_1, t_2, t_3, …, t_{m}. These strings contain only lowercase letters. There might be duplicates among these strings.
Let's call a concatenation of strings x and y as the string that is obtained by writing down strings x and y one right after another without changing the order. For example, the concatenation of the strings "code" and "forces" is the string "codeforces".
The year 1 has a name which is the concatenation of the two strings s_1 and t_1. When the year increases by one, we concatenate the next two strings in order from each of the respective sequences. If the string that is currently being used is at the end of its sequence, we go back to the first string in that sequence.
For example, if n = 3, m = 4, s = {"a", "b", "c"}, t = {"d", "e", "f", "g"}, the following table denotes the resulting year names. Note that the names of the years may repeat.
<image>
You are given two sequences of strings of size n and m and also q queries. For each query, you will be given the current year. Could you find the name corresponding to the given year, according to the Gapja system?
Input
The first line contains two integers n, m (1 ≤ n, m ≤ 20).
The next line contains n strings s_1, s_2, …, s_{n}. Each string contains only lowercase letters, and they are separated by spaces. The length of each string is at least 1 and at most 10.
The next line contains m strings t_1, t_2, …, t_{m}. Each string contains only lowercase letters, and they are separated by spaces. The length of each string is at least 1 and at most 10.
Among the given n + m strings may be duplicates (that is, they are not necessarily all different).
The next line contains a single integer q (1 ≤ q ≤ 2 020).
In the next q lines, an integer y (1 ≤ y ≤ 10^9) is given, denoting the year we want to know the name for.
Output
Print q lines. For each line, print the name of the year as per the rule described above.
Example
Input
10 12
sin im gye gap eul byeong jeong mu gi gyeong
yu sul hae ja chuk in myo jin sa o mi sin
14
1
2
3
4
10
11
12
13
73
2016
2017
2018
2019
2020
Output
sinyu
imsul
gyehae
gapja
gyeongo
sinmi
imsin
gyeyu
gyeyu
byeongsin
jeongyu
musul
gihae
gyeongja
Note
The first example denotes the actual names used in the Gapja system. These strings usually are either a number or the name of some animal.
Submitted Solution:
```
n, m = map(int, input().split())
str1 = input().split()
str2 = input().split()
print(str1)
print(str2)
q = int(input())
while q:
q = q - 1
inp = int(input())
in1 = inp % len(str1)
in2 = inp % len(str2)
print(str1[in1-1] + str2[in2-1])
``` | instruction | 0 | 15,194 | 18 | 30,388 |
No | output | 1 | 15,194 | 18 | 30,389 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Happy new year! The year 2020 is also known as Year Gyeongja (경자년, gyeongja-nyeon) in Korea. Where did the name come from? Let's briefly look at the Gapja system, which is traditionally used in Korea to name the years.
There are two sequences of n strings s_1, s_2, s_3, …, s_{n} and m strings t_1, t_2, t_3, …, t_{m}. These strings contain only lowercase letters. There might be duplicates among these strings.
Let's call a concatenation of strings x and y as the string that is obtained by writing down strings x and y one right after another without changing the order. For example, the concatenation of the strings "code" and "forces" is the string "codeforces".
The year 1 has a name which is the concatenation of the two strings s_1 and t_1. When the year increases by one, we concatenate the next two strings in order from each of the respective sequences. If the string that is currently being used is at the end of its sequence, we go back to the first string in that sequence.
For example, if n = 3, m = 4, s = {"a", "b", "c"}, t = {"d", "e", "f", "g"}, the following table denotes the resulting year names. Note that the names of the years may repeat.
<image>
You are given two sequences of strings of size n and m and also q queries. For each query, you will be given the current year. Could you find the name corresponding to the given year, according to the Gapja system?
Input
The first line contains two integers n, m (1 ≤ n, m ≤ 20).
The next line contains n strings s_1, s_2, …, s_{n}. Each string contains only lowercase letters, and they are separated by spaces. The length of each string is at least 1 and at most 10.
The next line contains m strings t_1, t_2, …, t_{m}. Each string contains only lowercase letters, and they are separated by spaces. The length of each string is at least 1 and at most 10.
Among the given n + m strings may be duplicates (that is, they are not necessarily all different).
The next line contains a single integer q (1 ≤ q ≤ 2 020).
In the next q lines, an integer y (1 ≤ y ≤ 10^9) is given, denoting the year we want to know the name for.
Output
Print q lines. For each line, print the name of the year as per the rule described above.
Example
Input
10 12
sin im gye gap eul byeong jeong mu gi gyeong
yu sul hae ja chuk in myo jin sa o mi sin
14
1
2
3
4
10
11
12
13
73
2016
2017
2018
2019
2020
Output
sinyu
imsul
gyehae
gapja
gyeongo
sinmi
imsin
gyeyu
gyeyu
byeongsin
jeongyu
musul
gihae
gyeongja
Note
The first example denotes the actual names used in the Gapja system. These strings usually are either a number or the name of some animal.
Submitted Solution:
```
n,m = list(map(int,input().split()))
arr1 = list(map(str,input().split()))
arr2 = list(map(str,input().split()))
d = []
i = 0
j = 0
t = 0
while True:
i = i%n
j = j%m
x = arr1[i]+arr2[j]
if x not in d:
d.append(arr1[i]+arr2[j])
else:
break
i+=1
j+=1
t+=1
q = int(input())
for i in range(q):
x = int(input())
x = x%(len(d))
print(d[x-1])
``` | instruction | 0 | 15,195 | 18 | 30,390 |
No | output | 1 | 15,195 | 18 | 30,391 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Happy new year! The year 2020 is also known as Year Gyeongja (경자년, gyeongja-nyeon) in Korea. Where did the name come from? Let's briefly look at the Gapja system, which is traditionally used in Korea to name the years.
There are two sequences of n strings s_1, s_2, s_3, …, s_{n} and m strings t_1, t_2, t_3, …, t_{m}. These strings contain only lowercase letters. There might be duplicates among these strings.
Let's call a concatenation of strings x and y as the string that is obtained by writing down strings x and y one right after another without changing the order. For example, the concatenation of the strings "code" and "forces" is the string "codeforces".
The year 1 has a name which is the concatenation of the two strings s_1 and t_1. When the year increases by one, we concatenate the next two strings in order from each of the respective sequences. If the string that is currently being used is at the end of its sequence, we go back to the first string in that sequence.
For example, if n = 3, m = 4, s = {"a", "b", "c"}, t = {"d", "e", "f", "g"}, the following table denotes the resulting year names. Note that the names of the years may repeat.
<image>
You are given two sequences of strings of size n and m and also q queries. For each query, you will be given the current year. Could you find the name corresponding to the given year, according to the Gapja system?
Input
The first line contains two integers n, m (1 ≤ n, m ≤ 20).
The next line contains n strings s_1, s_2, …, s_{n}. Each string contains only lowercase letters, and they are separated by spaces. The length of each string is at least 1 and at most 10.
The next line contains m strings t_1, t_2, …, t_{m}. Each string contains only lowercase letters, and they are separated by spaces. The length of each string is at least 1 and at most 10.
Among the given n + m strings may be duplicates (that is, they are not necessarily all different).
The next line contains a single integer q (1 ≤ q ≤ 2 020).
In the next q lines, an integer y (1 ≤ y ≤ 10^9) is given, denoting the year we want to know the name for.
Output
Print q lines. For each line, print the name of the year as per the rule described above.
Example
Input
10 12
sin im gye gap eul byeong jeong mu gi gyeong
yu sul hae ja chuk in myo jin sa o mi sin
14
1
2
3
4
10
11
12
13
73
2016
2017
2018
2019
2020
Output
sinyu
imsul
gyehae
gapja
gyeongo
sinmi
imsin
gyeyu
gyeyu
byeongsin
jeongyu
musul
gihae
gyeongja
Note
The first example denotes the actual names used in the Gapja system. These strings usually are either a number or the name of some animal.
Submitted Solution:
```
s1 = []
s2 = []
s3 = []
n, m = map(int, input().split())
inp = input()
for i in inp.split():
s1.append(i)
inp2 = input()
for i in inp2.split():
s2.append(i)
q = int(input())
for i in range(q):
a = int(input())
s3.append(s1[(a-1)%n] + s2[(a-1)%m])
print(s3, end="\n")
``` | instruction | 0 | 15,196 | 18 | 30,392 |
No | output | 1 | 15,196 | 18 | 30,393 |
Provide a correct Python 3 solution for this coding contest problem.
Dr .: Peter, do you know "Yes, I have a number"?
Peter: I used to do it on TV the other day. You remember something by the number of characters in each word contained in a sentence. "Yes, I have a number", so it means "the number 3.14" and is a keyword for remembering pi.
Dr .: Peter, that's not the case. This should be interpreted as 3.1416. The circumference ratio is 3.14159 ... That's why.
Peter: Then, did you forcibly omit that program just because Japan teaches that the circumference ratio is 3.14? ??
Dr .: ... Let's just say that the pi taught in elementary school has finally returned from 3 to 3.14.
Peter: Do you really remember this in the first place?
Dr .: It may be difficult for Japanese people. It seems that English-speaking people use it, because it is difficult to make a ground ball in English.
Peter: Even so, it seems to be a hassle to check.
Dr .: Then, I want you to make a program that converts sentences into a sequence of the number of characters in a word.
Peter: I understand. Please tell me the detailed specifications.
Dr .: Let's enter one line of text. For simplicity, you can use sentences that contain only alphabets and blanks. For this sentence, output the length of the character string between the blank, the beginning of the sentence, and the end of the sentence in order. You can proceed if the number of characters in the string does not exceed 9.
For example, Yes in "Yes I have" has 3 characters because it is separated by the beginning and the first space, and I is separated by the first and second spaces and has 1 character.
Dr .: Then, don't forget the character string where the number of characters becomes 0 when there are consecutive blanks.
Input
Multiple datasets are given as input. For each dataset, a string containing alphabets and spaces is given on one line.
When the character string is "END OF INPUT", it is the end of input. Do not output to this input.
Output
For each dataset, output a sequence of the number of characters for the string on one line.
Example
Input
Yes I have a number
How I wish I could calculate an unused color for space
Thank you
END OF INPUT
Output
31416
31415926535
53 | instruction | 0 | 15,879 | 18 | 31,758 |
"Correct Solution:
```
# coding: utf-8
# Your code here!
while True:
n=input()
if n=='END OF INPUT':
break
X=list(n)
c=0
for i in range(len(X)):
if X[i]==' ':
print(c,end='')
c=0
else:
c+=1
print(c)
``` | output | 1 | 15,879 | 18 | 31,759 |
Provide a correct Python 3 solution for this coding contest problem.
Dr .: Peter, do you know "Yes, I have a number"?
Peter: I used to do it on TV the other day. You remember something by the number of characters in each word contained in a sentence. "Yes, I have a number", so it means "the number 3.14" and is a keyword for remembering pi.
Dr .: Peter, that's not the case. This should be interpreted as 3.1416. The circumference ratio is 3.14159 ... That's why.
Peter: Then, did you forcibly omit that program just because Japan teaches that the circumference ratio is 3.14? ??
Dr .: ... Let's just say that the pi taught in elementary school has finally returned from 3 to 3.14.
Peter: Do you really remember this in the first place?
Dr .: It may be difficult for Japanese people. It seems that English-speaking people use it, because it is difficult to make a ground ball in English.
Peter: Even so, it seems to be a hassle to check.
Dr .: Then, I want you to make a program that converts sentences into a sequence of the number of characters in a word.
Peter: I understand. Please tell me the detailed specifications.
Dr .: Let's enter one line of text. For simplicity, you can use sentences that contain only alphabets and blanks. For this sentence, output the length of the character string between the blank, the beginning of the sentence, and the end of the sentence in order. You can proceed if the number of characters in the string does not exceed 9.
For example, Yes in "Yes I have" has 3 characters because it is separated by the beginning and the first space, and I is separated by the first and second spaces and has 1 character.
Dr .: Then, don't forget the character string where the number of characters becomes 0 when there are consecutive blanks.
Input
Multiple datasets are given as input. For each dataset, a string containing alphabets and spaces is given on one line.
When the character string is "END OF INPUT", it is the end of input. Do not output to this input.
Output
For each dataset, output a sequence of the number of characters for the string on one line.
Example
Input
Yes I have a number
How I wish I could calculate an unused color for space
Thank you
END OF INPUT
Output
31416
31415926535
53 | instruction | 0 | 15,880 | 18 | 31,760 |
"Correct Solution:
```
while True:
s = input()
if s == "END OF INPUT":
break
l = 0
for i in s:
if i == ' ':
print(l,end='')
l = 0
else:
l += 1
print(l)
``` | output | 1 | 15,880 | 18 | 31,761 |
Provide a correct Python 3 solution for this coding contest problem.
Dr .: Peter, do you know "Yes, I have a number"?
Peter: I used to do it on TV the other day. You remember something by the number of characters in each word contained in a sentence. "Yes, I have a number", so it means "the number 3.14" and is a keyword for remembering pi.
Dr .: Peter, that's not the case. This should be interpreted as 3.1416. The circumference ratio is 3.14159 ... That's why.
Peter: Then, did you forcibly omit that program just because Japan teaches that the circumference ratio is 3.14? ??
Dr .: ... Let's just say that the pi taught in elementary school has finally returned from 3 to 3.14.
Peter: Do you really remember this in the first place?
Dr .: It may be difficult for Japanese people. It seems that English-speaking people use it, because it is difficult to make a ground ball in English.
Peter: Even so, it seems to be a hassle to check.
Dr .: Then, I want you to make a program that converts sentences into a sequence of the number of characters in a word.
Peter: I understand. Please tell me the detailed specifications.
Dr .: Let's enter one line of text. For simplicity, you can use sentences that contain only alphabets and blanks. For this sentence, output the length of the character string between the blank, the beginning of the sentence, and the end of the sentence in order. You can proceed if the number of characters in the string does not exceed 9.
For example, Yes in "Yes I have" has 3 characters because it is separated by the beginning and the first space, and I is separated by the first and second spaces and has 1 character.
Dr .: Then, don't forget the character string where the number of characters becomes 0 when there are consecutive blanks.
Input
Multiple datasets are given as input. For each dataset, a string containing alphabets and spaces is given on one line.
When the character string is "END OF INPUT", it is the end of input. Do not output to this input.
Output
For each dataset, output a sequence of the number of characters for the string on one line.
Example
Input
Yes I have a number
How I wish I could calculate an unused color for space
Thank you
END OF INPUT
Output
31416
31415926535
53 | instruction | 0 | 15,881 | 18 | 31,762 |
"Correct Solution:
```
while True:
text = input()
if text=='END OF INPUT':
break
TEXT = text.split(' ')
for i in range(len(TEXT)):
print(len(TEXT[i]),end='')
print('')
``` | output | 1 | 15,881 | 18 | 31,763 |
Provide a correct Python 3 solution for this coding contest problem.
Dr .: Peter, do you know "Yes, I have a number"?
Peter: I used to do it on TV the other day. You remember something by the number of characters in each word contained in a sentence. "Yes, I have a number", so it means "the number 3.14" and is a keyword for remembering pi.
Dr .: Peter, that's not the case. This should be interpreted as 3.1416. The circumference ratio is 3.14159 ... That's why.
Peter: Then, did you forcibly omit that program just because Japan teaches that the circumference ratio is 3.14? ??
Dr .: ... Let's just say that the pi taught in elementary school has finally returned from 3 to 3.14.
Peter: Do you really remember this in the first place?
Dr .: It may be difficult for Japanese people. It seems that English-speaking people use it, because it is difficult to make a ground ball in English.
Peter: Even so, it seems to be a hassle to check.
Dr .: Then, I want you to make a program that converts sentences into a sequence of the number of characters in a word.
Peter: I understand. Please tell me the detailed specifications.
Dr .: Let's enter one line of text. For simplicity, you can use sentences that contain only alphabets and blanks. For this sentence, output the length of the character string between the blank, the beginning of the sentence, and the end of the sentence in order. You can proceed if the number of characters in the string does not exceed 9.
For example, Yes in "Yes I have" has 3 characters because it is separated by the beginning and the first space, and I is separated by the first and second spaces and has 1 character.
Dr .: Then, don't forget the character string where the number of characters becomes 0 when there are consecutive blanks.
Input
Multiple datasets are given as input. For each dataset, a string containing alphabets and spaces is given on one line.
When the character string is "END OF INPUT", it is the end of input. Do not output to this input.
Output
For each dataset, output a sequence of the number of characters for the string on one line.
Example
Input
Yes I have a number
How I wish I could calculate an unused color for space
Thank you
END OF INPUT
Output
31416
31415926535
53 | instruction | 0 | 15,882 | 18 | 31,764 |
"Correct Solution:
```
while True:
string = input()
if string == 'END OF INPUT':
break
ans = []
temp = 0
space = 0
for i in range(len(string)):
if string[i] == ' ' and space == 0:
ans.append(temp)
temp = 0
space = 1
continue
elif space > 0 and string[i] == ' ':
ans.append(0)
continue
space = 0
temp += 1
ans.append(temp)
print(''.join([str(x) for x in ans]))
``` | output | 1 | 15,882 | 18 | 31,765 |
Provide a correct Python 3 solution for this coding contest problem.
Dr .: Peter, do you know "Yes, I have a number"?
Peter: I used to do it on TV the other day. You remember something by the number of characters in each word contained in a sentence. "Yes, I have a number", so it means "the number 3.14" and is a keyword for remembering pi.
Dr .: Peter, that's not the case. This should be interpreted as 3.1416. The circumference ratio is 3.14159 ... That's why.
Peter: Then, did you forcibly omit that program just because Japan teaches that the circumference ratio is 3.14? ??
Dr .: ... Let's just say that the pi taught in elementary school has finally returned from 3 to 3.14.
Peter: Do you really remember this in the first place?
Dr .: It may be difficult for Japanese people. It seems that English-speaking people use it, because it is difficult to make a ground ball in English.
Peter: Even so, it seems to be a hassle to check.
Dr .: Then, I want you to make a program that converts sentences into a sequence of the number of characters in a word.
Peter: I understand. Please tell me the detailed specifications.
Dr .: Let's enter one line of text. For simplicity, you can use sentences that contain only alphabets and blanks. For this sentence, output the length of the character string between the blank, the beginning of the sentence, and the end of the sentence in order. You can proceed if the number of characters in the string does not exceed 9.
For example, Yes in "Yes I have" has 3 characters because it is separated by the beginning and the first space, and I is separated by the first and second spaces and has 1 character.
Dr .: Then, don't forget the character string where the number of characters becomes 0 when there are consecutive blanks.
Input
Multiple datasets are given as input. For each dataset, a string containing alphabets and spaces is given on one line.
When the character string is "END OF INPUT", it is the end of input. Do not output to this input.
Output
For each dataset, output a sequence of the number of characters for the string on one line.
Example
Input
Yes I have a number
How I wish I could calculate an unused color for space
Thank you
END OF INPUT
Output
31416
31415926535
53 | instruction | 0 | 15,883 | 18 | 31,766 |
"Correct Solution:
```
import re
while 1:
n=input()
if n=='END OF INPUT':
break
s=re.split('[ ]', n)
a=len(s)
for i in range(a):
print(len(s[i]),end ='')
print("")
``` | output | 1 | 15,883 | 18 | 31,767 |
Provide a correct Python 3 solution for this coding contest problem.
Dr .: Peter, do you know "Yes, I have a number"?
Peter: I used to do it on TV the other day. You remember something by the number of characters in each word contained in a sentence. "Yes, I have a number", so it means "the number 3.14" and is a keyword for remembering pi.
Dr .: Peter, that's not the case. This should be interpreted as 3.1416. The circumference ratio is 3.14159 ... That's why.
Peter: Then, did you forcibly omit that program just because Japan teaches that the circumference ratio is 3.14? ??
Dr .: ... Let's just say that the pi taught in elementary school has finally returned from 3 to 3.14.
Peter: Do you really remember this in the first place?
Dr .: It may be difficult for Japanese people. It seems that English-speaking people use it, because it is difficult to make a ground ball in English.
Peter: Even so, it seems to be a hassle to check.
Dr .: Then, I want you to make a program that converts sentences into a sequence of the number of characters in a word.
Peter: I understand. Please tell me the detailed specifications.
Dr .: Let's enter one line of text. For simplicity, you can use sentences that contain only alphabets and blanks. For this sentence, output the length of the character string between the blank, the beginning of the sentence, and the end of the sentence in order. You can proceed if the number of characters in the string does not exceed 9.
For example, Yes in "Yes I have" has 3 characters because it is separated by the beginning and the first space, and I is separated by the first and second spaces and has 1 character.
Dr .: Then, don't forget the character string where the number of characters becomes 0 when there are consecutive blanks.
Input
Multiple datasets are given as input. For each dataset, a string containing alphabets and spaces is given on one line.
When the character string is "END OF INPUT", it is the end of input. Do not output to this input.
Output
For each dataset, output a sequence of the number of characters for the string on one line.
Example
Input
Yes I have a number
How I wish I could calculate an unused color for space
Thank you
END OF INPUT
Output
31416
31415926535
53 | instruction | 0 | 15,884 | 18 | 31,768 |
"Correct Solution:
```
while True:
a=input()
if a==('END OF INPUT') : break
s = list(a.split(' '))
for i in range(len(s)):
print(len(s[i]), end='')
print()
``` | output | 1 | 15,884 | 18 | 31,769 |
Provide a correct Python 3 solution for this coding contest problem.
Dr .: Peter, do you know "Yes, I have a number"?
Peter: I used to do it on TV the other day. You remember something by the number of characters in each word contained in a sentence. "Yes, I have a number", so it means "the number 3.14" and is a keyword for remembering pi.
Dr .: Peter, that's not the case. This should be interpreted as 3.1416. The circumference ratio is 3.14159 ... That's why.
Peter: Then, did you forcibly omit that program just because Japan teaches that the circumference ratio is 3.14? ??
Dr .: ... Let's just say that the pi taught in elementary school has finally returned from 3 to 3.14.
Peter: Do you really remember this in the first place?
Dr .: It may be difficult for Japanese people. It seems that English-speaking people use it, because it is difficult to make a ground ball in English.
Peter: Even so, it seems to be a hassle to check.
Dr .: Then, I want you to make a program that converts sentences into a sequence of the number of characters in a word.
Peter: I understand. Please tell me the detailed specifications.
Dr .: Let's enter one line of text. For simplicity, you can use sentences that contain only alphabets and blanks. For this sentence, output the length of the character string between the blank, the beginning of the sentence, and the end of the sentence in order. You can proceed if the number of characters in the string does not exceed 9.
For example, Yes in "Yes I have" has 3 characters because it is separated by the beginning and the first space, and I is separated by the first and second spaces and has 1 character.
Dr .: Then, don't forget the character string where the number of characters becomes 0 when there are consecutive blanks.
Input
Multiple datasets are given as input. For each dataset, a string containing alphabets and spaces is given on one line.
When the character string is "END OF INPUT", it is the end of input. Do not output to this input.
Output
For each dataset, output a sequence of the number of characters for the string on one line.
Example
Input
Yes I have a number
How I wish I could calculate an unused color for space
Thank you
END OF INPUT
Output
31416
31415926535
53 | instruction | 0 | 15,885 | 18 | 31,770 |
"Correct Solution:
```
while True:
a = input()
if a == "END OF INPUT":
break
a = a.split(' ')
for i in range(len(a)):
print(len(a[i]),end='')
print()
``` | output | 1 | 15,885 | 18 | 31,771 |
Provide a correct Python 3 solution for this coding contest problem.
Dr .: Peter, do you know "Yes, I have a number"?
Peter: I used to do it on TV the other day. You remember something by the number of characters in each word contained in a sentence. "Yes, I have a number", so it means "the number 3.14" and is a keyword for remembering pi.
Dr .: Peter, that's not the case. This should be interpreted as 3.1416. The circumference ratio is 3.14159 ... That's why.
Peter: Then, did you forcibly omit that program just because Japan teaches that the circumference ratio is 3.14? ??
Dr .: ... Let's just say that the pi taught in elementary school has finally returned from 3 to 3.14.
Peter: Do you really remember this in the first place?
Dr .: It may be difficult for Japanese people. It seems that English-speaking people use it, because it is difficult to make a ground ball in English.
Peter: Even so, it seems to be a hassle to check.
Dr .: Then, I want you to make a program that converts sentences into a sequence of the number of characters in a word.
Peter: I understand. Please tell me the detailed specifications.
Dr .: Let's enter one line of text. For simplicity, you can use sentences that contain only alphabets and blanks. For this sentence, output the length of the character string between the blank, the beginning of the sentence, and the end of the sentence in order. You can proceed if the number of characters in the string does not exceed 9.
For example, Yes in "Yes I have" has 3 characters because it is separated by the beginning and the first space, and I is separated by the first and second spaces and has 1 character.
Dr .: Then, don't forget the character string where the number of characters becomes 0 when there are consecutive blanks.
Input
Multiple datasets are given as input. For each dataset, a string containing alphabets and spaces is given on one line.
When the character string is "END OF INPUT", it is the end of input. Do not output to this input.
Output
For each dataset, output a sequence of the number of characters for the string on one line.
Example
Input
Yes I have a number
How I wish I could calculate an unused color for space
Thank you
END OF INPUT
Output
31416
31415926535
53 | instruction | 0 | 15,886 | 18 | 31,772 |
"Correct Solution:
```
while True:
i = input()
if i == "END OF INPUT":
break
print("".join(map(str, [0 if x == '' else len(x) for x in i.split(' ')])))
``` | output | 1 | 15,886 | 18 | 31,773 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr .: Peter, do you know "Yes, I have a number"?
Peter: I used to do it on TV the other day. You remember something by the number of characters in each word contained in a sentence. "Yes, I have a number", so it means "the number 3.14" and is a keyword for remembering pi.
Dr .: Peter, that's not the case. This should be interpreted as 3.1416. The circumference ratio is 3.14159 ... That's why.
Peter: Then, did you forcibly omit that program just because Japan teaches that the circumference ratio is 3.14? ??
Dr .: ... Let's just say that the pi taught in elementary school has finally returned from 3 to 3.14.
Peter: Do you really remember this in the first place?
Dr .: It may be difficult for Japanese people. It seems that English-speaking people use it, because it is difficult to make a ground ball in English.
Peter: Even so, it seems to be a hassle to check.
Dr .: Then, I want you to make a program that converts sentences into a sequence of the number of characters in a word.
Peter: I understand. Please tell me the detailed specifications.
Dr .: Let's enter one line of text. For simplicity, you can use sentences that contain only alphabets and blanks. For this sentence, output the length of the character string between the blank, the beginning of the sentence, and the end of the sentence in order. You can proceed if the number of characters in the string does not exceed 9.
For example, Yes in "Yes I have" has 3 characters because it is separated by the beginning and the first space, and I is separated by the first and second spaces and has 1 character.
Dr .: Then, don't forget the character string where the number of characters becomes 0 when there are consecutive blanks.
Input
Multiple datasets are given as input. For each dataset, a string containing alphabets and spaces is given on one line.
When the character string is "END OF INPUT", it is the end of input. Do not output to this input.
Output
For each dataset, output a sequence of the number of characters for the string on one line.
Example
Input
Yes I have a number
How I wish I could calculate an unused color for space
Thank you
END OF INPUT
Output
31416
31415926535
53
Submitted Solution:
```
while True:
s=input()
if s=="END OF INPUT":
break
c=0
for i in s:
if i!=" ":
c+=1
else:
print(c,end="")
c=0
print(c)
``` | instruction | 0 | 15,887 | 18 | 31,774 |
Yes | output | 1 | 15,887 | 18 | 31,775 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr .: Peter, do you know "Yes, I have a number"?
Peter: I used to do it on TV the other day. You remember something by the number of characters in each word contained in a sentence. "Yes, I have a number", so it means "the number 3.14" and is a keyword for remembering pi.
Dr .: Peter, that's not the case. This should be interpreted as 3.1416. The circumference ratio is 3.14159 ... That's why.
Peter: Then, did you forcibly omit that program just because Japan teaches that the circumference ratio is 3.14? ??
Dr .: ... Let's just say that the pi taught in elementary school has finally returned from 3 to 3.14.
Peter: Do you really remember this in the first place?
Dr .: It may be difficult for Japanese people. It seems that English-speaking people use it, because it is difficult to make a ground ball in English.
Peter: Even so, it seems to be a hassle to check.
Dr .: Then, I want you to make a program that converts sentences into a sequence of the number of characters in a word.
Peter: I understand. Please tell me the detailed specifications.
Dr .: Let's enter one line of text. For simplicity, you can use sentences that contain only alphabets and blanks. For this sentence, output the length of the character string between the blank, the beginning of the sentence, and the end of the sentence in order. You can proceed if the number of characters in the string does not exceed 9.
For example, Yes in "Yes I have" has 3 characters because it is separated by the beginning and the first space, and I is separated by the first and second spaces and has 1 character.
Dr .: Then, don't forget the character string where the number of characters becomes 0 when there are consecutive blanks.
Input
Multiple datasets are given as input. For each dataset, a string containing alphabets and spaces is given on one line.
When the character string is "END OF INPUT", it is the end of input. Do not output to this input.
Output
For each dataset, output a sequence of the number of characters for the string on one line.
Example
Input
Yes I have a number
How I wish I could calculate an unused color for space
Thank you
END OF INPUT
Output
31416
31415926535
53
Submitted Solution:
```
while True:
n=input()
if n=='END OF INPUT':
break
c=0
for i in n :
if i==' ':
print(c,end='')
c=0
else:
c+=1
print(c)
``` | instruction | 0 | 15,888 | 18 | 31,776 |
Yes | output | 1 | 15,888 | 18 | 31,777 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr .: Peter, do you know "Yes, I have a number"?
Peter: I used to do it on TV the other day. You remember something by the number of characters in each word contained in a sentence. "Yes, I have a number", so it means "the number 3.14" and is a keyword for remembering pi.
Dr .: Peter, that's not the case. This should be interpreted as 3.1416. The circumference ratio is 3.14159 ... That's why.
Peter: Then, did you forcibly omit that program just because Japan teaches that the circumference ratio is 3.14? ??
Dr .: ... Let's just say that the pi taught in elementary school has finally returned from 3 to 3.14.
Peter: Do you really remember this in the first place?
Dr .: It may be difficult for Japanese people. It seems that English-speaking people use it, because it is difficult to make a ground ball in English.
Peter: Even so, it seems to be a hassle to check.
Dr .: Then, I want you to make a program that converts sentences into a sequence of the number of characters in a word.
Peter: I understand. Please tell me the detailed specifications.
Dr .: Let's enter one line of text. For simplicity, you can use sentences that contain only alphabets and blanks. For this sentence, output the length of the character string between the blank, the beginning of the sentence, and the end of the sentence in order. You can proceed if the number of characters in the string does not exceed 9.
For example, Yes in "Yes I have" has 3 characters because it is separated by the beginning and the first space, and I is separated by the first and second spaces and has 1 character.
Dr .: Then, don't forget the character string where the number of characters becomes 0 when there are consecutive blanks.
Input
Multiple datasets are given as input. For each dataset, a string containing alphabets and spaces is given on one line.
When the character string is "END OF INPUT", it is the end of input. Do not output to this input.
Output
For each dataset, output a sequence of the number of characters for the string on one line.
Example
Input
Yes I have a number
How I wish I could calculate an unused color for space
Thank you
END OF INPUT
Output
31416
31415926535
53
Submitted Solution:
```
# AOJ 1042: Yes, I have a number
# Python3 2018.7.6 bal4u
while True:
s = input()
if s == "END OF INPUT": break
ans, i, ls = '', 0, len(s)
while i < ls:
w = 0
while i < ls and s[i] == ' ':
w, i = w+1, i+1
if w > 1: ans += '0'*(w-1)
w = 0
while i < ls and s[i].isalpha():
w, i = w+1, i+1
if w > 0: ans += chr(ord('0')+w)
print(ans)
``` | instruction | 0 | 15,889 | 18 | 31,778 |
Yes | output | 1 | 15,889 | 18 | 31,779 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr .: Peter, do you know "Yes, I have a number"?
Peter: I used to do it on TV the other day. You remember something by the number of characters in each word contained in a sentence. "Yes, I have a number", so it means "the number 3.14" and is a keyword for remembering pi.
Dr .: Peter, that's not the case. This should be interpreted as 3.1416. The circumference ratio is 3.14159 ... That's why.
Peter: Then, did you forcibly omit that program just because Japan teaches that the circumference ratio is 3.14? ??
Dr .: ... Let's just say that the pi taught in elementary school has finally returned from 3 to 3.14.
Peter: Do you really remember this in the first place?
Dr .: It may be difficult for Japanese people. It seems that English-speaking people use it, because it is difficult to make a ground ball in English.
Peter: Even so, it seems to be a hassle to check.
Dr .: Then, I want you to make a program that converts sentences into a sequence of the number of characters in a word.
Peter: I understand. Please tell me the detailed specifications.
Dr .: Let's enter one line of text. For simplicity, you can use sentences that contain only alphabets and blanks. For this sentence, output the length of the character string between the blank, the beginning of the sentence, and the end of the sentence in order. You can proceed if the number of characters in the string does not exceed 9.
For example, Yes in "Yes I have" has 3 characters because it is separated by the beginning and the first space, and I is separated by the first and second spaces and has 1 character.
Dr .: Then, don't forget the character string where the number of characters becomes 0 when there are consecutive blanks.
Input
Multiple datasets are given as input. For each dataset, a string containing alphabets and spaces is given on one line.
When the character string is "END OF INPUT", it is the end of input. Do not output to this input.
Output
For each dataset, output a sequence of the number of characters for the string on one line.
Example
Input
Yes I have a number
How I wish I could calculate an unused color for space
Thank you
END OF INPUT
Output
31416
31415926535
53
Submitted Solution:
```
while True:
s = input()
if s == 'END OF INPUT': break
while True:
if ' ' in s: s = s.replace(' ', ' 0 ')
else: break
s = s.split()
for i in s: print(len(i) if i != '0' else 0, end='')
print('')
``` | instruction | 0 | 15,890 | 18 | 31,780 |
Yes | output | 1 | 15,890 | 18 | 31,781 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr .: Peter, do you know "Yes, I have a number"?
Peter: I used to do it on TV the other day. You remember something by the number of characters in each word contained in a sentence. "Yes, I have a number", so it means "the number 3.14" and is a keyword for remembering pi.
Dr .: Peter, that's not the case. This should be interpreted as 3.1416. The circumference ratio is 3.14159 ... That's why.
Peter: Then, did you forcibly omit that program just because Japan teaches that the circumference ratio is 3.14? ??
Dr .: ... Let's just say that the pi taught in elementary school has finally returned from 3 to 3.14.
Peter: Do you really remember this in the first place?
Dr .: It may be difficult for Japanese people. It seems that English-speaking people use it, because it is difficult to make a ground ball in English.
Peter: Even so, it seems to be a hassle to check.
Dr .: Then, I want you to make a program that converts sentences into a sequence of the number of characters in a word.
Peter: I understand. Please tell me the detailed specifications.
Dr .: Let's enter one line of text. For simplicity, you can use sentences that contain only alphabets and blanks. For this sentence, output the length of the character string between the blank, the beginning of the sentence, and the end of the sentence in order. You can proceed if the number of characters in the string does not exceed 9.
For example, Yes in "Yes I have" has 3 characters because it is separated by the beginning and the first space, and I is separated by the first and second spaces and has 1 character.
Dr .: Then, don't forget the character string where the number of characters becomes 0 when there are consecutive blanks.
Input
Multiple datasets are given as input. For each dataset, a string containing alphabets and spaces is given on one line.
When the character string is "END OF INPUT", it is the end of input. Do not output to this input.
Output
For each dataset, output a sequence of the number of characters for the string on one line.
Example
Input
Yes I have a number
How I wish I could calculate an unused color for space
Thank you
END OF INPUT
Output
31416
31415926535
53
Submitted Solution:
```
while True:
dataset = input().split(' ')
if dataset == ["END", "OF", "INPUT"]: break
i = tmp = 0
ans = ''
while i < len(dataset):
if dataset[i] == '':
tmp += 1
else:
if tmp:
ans += str(tmp)
tmp = 0
ans += str(len(dataset[i]))
else:
ans += str(len(dataset[i]))
i += 1
print(ans)
``` | instruction | 0 | 15,891 | 18 | 31,782 |
No | output | 1 | 15,891 | 18 | 31,783 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr .: Peter, do you know "Yes, I have a number"?
Peter: I used to do it on TV the other day. You remember something by the number of characters in each word contained in a sentence. "Yes, I have a number", so it means "the number 3.14" and is a keyword for remembering pi.
Dr .: Peter, that's not the case. This should be interpreted as 3.1416. The circumference ratio is 3.14159 ... That's why.
Peter: Then, did you forcibly omit that program just because Japan teaches that the circumference ratio is 3.14? ??
Dr .: ... Let's just say that the pi taught in elementary school has finally returned from 3 to 3.14.
Peter: Do you really remember this in the first place?
Dr .: It may be difficult for Japanese people. It seems that English-speaking people use it, because it is difficult to make a ground ball in English.
Peter: Even so, it seems to be a hassle to check.
Dr .: Then, I want you to make a program that converts sentences into a sequence of the number of characters in a word.
Peter: I understand. Please tell me the detailed specifications.
Dr .: Let's enter one line of text. For simplicity, you can use sentences that contain only alphabets and blanks. For this sentence, output the length of the character string between the blank, the beginning of the sentence, and the end of the sentence in order. You can proceed if the number of characters in the string does not exceed 9.
For example, Yes in "Yes I have" has 3 characters because it is separated by the beginning and the first space, and I is separated by the first and second spaces and has 1 character.
Dr .: Then, don't forget the character string where the number of characters becomes 0 when there are consecutive blanks.
Input
Multiple datasets are given as input. For each dataset, a string containing alphabets and spaces is given on one line.
When the character string is "END OF INPUT", it is the end of input. Do not output to this input.
Output
For each dataset, output a sequence of the number of characters for the string on one line.
Example
Input
Yes I have a number
How I wish I could calculate an unused color for space
Thank you
END OF INPUT
Output
31416
31415926535
53
Submitted Solution:
```
while True:
s = input()
if s == 'END OF INPUT': break
if ' ' in s: s.replace(' ', ' 0 ')
s = s.split()
for i in s: print(len(i) if i != '0' else 0, end='')
print('')
``` | instruction | 0 | 15,892 | 18 | 31,784 |
No | output | 1 | 15,892 | 18 | 31,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr .: Peter, do you know "Yes, I have a number"?
Peter: I used to do it on TV the other day. You remember something by the number of characters in each word contained in a sentence. "Yes, I have a number", so it means "the number 3.14" and is a keyword for remembering pi.
Dr .: Peter, that's not the case. This should be interpreted as 3.1416. The circumference ratio is 3.14159 ... That's why.
Peter: Then, did you forcibly omit that program just because Japan teaches that the circumference ratio is 3.14? ??
Dr .: ... Let's just say that the pi taught in elementary school has finally returned from 3 to 3.14.
Peter: Do you really remember this in the first place?
Dr .: It may be difficult for Japanese people. It seems that English-speaking people use it, because it is difficult to make a ground ball in English.
Peter: Even so, it seems to be a hassle to check.
Dr .: Then, I want you to make a program that converts sentences into a sequence of the number of characters in a word.
Peter: I understand. Please tell me the detailed specifications.
Dr .: Let's enter one line of text. For simplicity, you can use sentences that contain only alphabets and blanks. For this sentence, output the length of the character string between the blank, the beginning of the sentence, and the end of the sentence in order. You can proceed if the number of characters in the string does not exceed 9.
For example, Yes in "Yes I have" has 3 characters because it is separated by the beginning and the first space, and I is separated by the first and second spaces and has 1 character.
Dr .: Then, don't forget the character string where the number of characters becomes 0 when there are consecutive blanks.
Input
Multiple datasets are given as input. For each dataset, a string containing alphabets and spaces is given on one line.
When the character string is "END OF INPUT", it is the end of input. Do not output to this input.
Output
For each dataset, output a sequence of the number of characters for the string on one line.
Example
Input
Yes I have a number
How I wish I could calculate an unused color for space
Thank you
END OF INPUT
Output
31416
31415926535
53
Submitted Solution:
```
while True:
s = input()
if s == 'END OF INPUT': break
if ' ' in s: s = s.replace(' ', ' 0 ')
s = s.split()
for i in s: print(len(i) if i != '0' else 0, end='')
print('')
``` | instruction | 0 | 15,893 | 18 | 31,786 |
No | output | 1 | 15,893 | 18 | 31,787 |
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