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 |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
Find the edit distance between given two words s1 and s2.
The disntace is the minimum number of single-character edits required to change one word into the other. The edits including the following operations:
* insertion: Insert a character at a particular position.
* deletion: Delete a character at a particular position.
* substitution: Change the character at a particular position to a different character
Constraints
* 1 ≤ length of s1 ≤ 1000
* 1 ≤ length of s2 ≤ 1000
Input
s1
s2
Two words s1 and s2 are given in the first line and the second line respectively. The words will consist of lower case characters.
Output
Print the edit distance in a line.
Examples
Input
acac
acm
Output
2
Input
icpc
icpc
Output
0 | instruction | 0 | 43,790 | 18 | 87,580 |
"Correct Solution:
```
def main():
s=input()
t=input()
n=len(s)
m=len(t)
memo=[[n+m for _ in [0]*(m+1)] for _ in [0]*(n+1)]
memo[0][0]=0
for i in range(n+m):
for j in range(i+1):
k=i-j
if 0<=j<n and 0<=k<m:
if s[j]==t[k]:
memo[j+1][k+1]=min(memo[j+1][k+1],memo[j][k])
else:
memo[j+1][k]=min(memo[j+1][k],memo[j][k]+1)
memo[j][k+1]=min(memo[j][k+1],memo[j][k]+1)
memo[j+1][k+1]=min(memo[j+1][k+1],memo[j][k]+1)
elif j==n:
memo[j][k+1]=min(memo[j][k+1],memo[j][k]+1)
elif k==m:
memo[j+1][k]=min(memo[j+1][k],memo[j][k]+1)
print(memo[n][m])
if __name__ == '__main__':
main()
``` | output | 1 | 43,790 | 18 | 87,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the edit distance between given two words s1 and s2.
The disntace is the minimum number of single-character edits required to change one word into the other. The edits including the following operations:
* insertion: Insert a character at a particular position.
* deletion: Delete a character at a particular position.
* substitution: Change the character at a particular position to a different character
Constraints
* 1 ≤ length of s1 ≤ 1000
* 1 ≤ length of s2 ≤ 1000
Input
s1
s2
Two words s1 and s2 are given in the first line and the second line respectively. The words will consist of lower case characters.
Output
Print the edit distance in a line.
Examples
Input
acac
acm
Output
2
Input
icpc
icpc
Output
0
Submitted Solution:
```
def main():
s = input()
t = input()
n, m = len(s), len(t)
dp = [[j if i == 0 else 10 ** 10 for i in range(m + 1)] for j in range(n + 1)]
dp[0] = [i for i in range(m + 1)]
for i in range(n):
for j in range(m):
dp[i + 1][j + 1] = min(dp[i][j + 1] + 1,
dp[i + 1][j] + 1,
dp[i][j] + 1 - (s[i] == t[j]))
ans = dp[n][m]
print(ans)
main()
``` | instruction | 0 | 43,791 | 18 | 87,582 |
Yes | output | 1 | 43,791 | 18 | 87,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the edit distance between given two words s1 and s2.
The disntace is the minimum number of single-character edits required to change one word into the other. The edits including the following operations:
* insertion: Insert a character at a particular position.
* deletion: Delete a character at a particular position.
* substitution: Change the character at a particular position to a different character
Constraints
* 1 ≤ length of s1 ≤ 1000
* 1 ≤ length of s2 ≤ 1000
Input
s1
s2
Two words s1 and s2 are given in the first line and the second line respectively. The words will consist of lower case characters.
Output
Print the edit distance in a line.
Examples
Input
acac
acm
Output
2
Input
icpc
icpc
Output
0
Submitted Solution:
```
def levenshtein(a,b):
"""
:param a: string or list
:param b: string or list
:return: Levenstein distance
"""
la = len(a)
lb = len(b)
dp = [[la+lb]*(lb+1) for i in range(la+1)]
dp[0][0] = 0
for i in range(1,la):
dp[i][0] = i
for j in range(1,lb):
dp[0][j] = j
for i in range(la):
for j in range(lb):
dp[i+1][j+1] = min(dp[i][j+1]+1,
dp[i+1][j]+1,
dp[i][j]+(a[i]!=b[j]))
l_dist = dp[-1][-1]
return l_dist
a = input()
b = input()
l_dist = levenshtein(a,b)
print(l_dist)
``` | instruction | 0 | 43,792 | 18 | 87,584 |
Yes | output | 1 | 43,792 | 18 | 87,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the edit distance between given two words s1 and s2.
The disntace is the minimum number of single-character edits required to change one word into the other. The edits including the following operations:
* insertion: Insert a character at a particular position.
* deletion: Delete a character at a particular position.
* substitution: Change the character at a particular position to a different character
Constraints
* 1 ≤ length of s1 ≤ 1000
* 1 ≤ length of s2 ≤ 1000
Input
s1
s2
Two words s1 and s2 are given in the first line and the second line respectively. The words will consist of lower case characters.
Output
Print the edit distance in a line.
Examples
Input
acac
acm
Output
2
Input
icpc
icpc
Output
0
Submitted Solution:
```
#!/usr/bin/env python3
# n,m = map(int,sys.stdin.readline().split())
# a = list(map(int,sys.stdin.readline().split()))
# a = [sys.stdin.readline() for s in range(n)]
# s = sys.stdin.readline().rstrip()
# n = int(sys.stdin.readline())
import sys,bisect
sys.setrecursionlimit(15000)
s1 = sys.stdin.readline().rstrip()
s2 = sys.stdin.readline().rstrip()
l1 = len(s1)
l2 = len(s2)
dp = [[float("inf")]*(l2+1) for _ in range(l1+1)]
dp[0] = list(range(l2+1))
for i in range(l1):
dp[i][0] = i
for j in range(l2):
if s1[i]==s2[j]:
dp[i+1][j+1] = dp[i][j]
else:
dp[i+1][j+1] = min(dp[i+1][j]+1,
dp[i][j+1]+1,
dp[i][j]+1)
#print(dp)
print(dp[-1][-1])
``` | instruction | 0 | 43,793 | 18 | 87,586 |
Yes | output | 1 | 43,793 | 18 | 87,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the edit distance between given two words s1 and s2.
The disntace is the minimum number of single-character edits required to change one word into the other. The edits including the following operations:
* insertion: Insert a character at a particular position.
* deletion: Delete a character at a particular position.
* substitution: Change the character at a particular position to a different character
Constraints
* 1 ≤ length of s1 ≤ 1000
* 1 ≤ length of s2 ≤ 1000
Input
s1
s2
Two words s1 and s2 are given in the first line and the second line respectively. The words will consist of lower case characters.
Output
Print the edit distance in a line.
Examples
Input
acac
acm
Output
2
Input
icpc
icpc
Output
0
Submitted Solution:
```
s1 = input()
s2 = input()
dp = [[1010] * (len(s2) + 1) for _ in range(len(s1) + 1)]
for i in range(len(s1) + 1):
dp[i][0] = i
for j in range(len(s2) + 1):
dp[0][j] = j
for i in range(len(s1)):
for j in range(len(s2)):
cost = 0 if s1[i] == s2[j] else 1
dp[i + 1][j + 1] = min(dp[i + 1][j] + 1, dp[i][j + 1] + 1, dp[i][j] + cost)
print(dp[len(s1)][len(s2)])
``` | instruction | 0 | 43,794 | 18 | 87,588 |
Yes | output | 1 | 43,794 | 18 | 87,589 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the edit distance between given two words s1 and s2.
The disntace is the minimum number of single-character edits required to change one word into the other. The edits including the following operations:
* insertion: Insert a character at a particular position.
* deletion: Delete a character at a particular position.
* substitution: Change the character at a particular position to a different character
Constraints
* 1 ≤ length of s1 ≤ 1000
* 1 ≤ length of s2 ≤ 1000
Input
s1
s2
Two words s1 and s2 are given in the first line and the second line respectively. The words will consist of lower case characters.
Output
Print the edit distance in a line.
Examples
Input
acac
acm
Output
2
Input
icpc
icpc
Output
0
Submitted Solution:
```
def LD(str1,str2):
A = [1 for _ in range(len(str2)+1)]
for i in range(len(str1)):
A[0] = i
A_copy = A[:]
for j in range(len(str2)):
if(str1[i]==str2[j]):
cost = 0
else:
cost = 1
A[j+1] = min(A[j]+1,A[j+1]+1,A_copy[j]+cost)
print(A[-1])
str1 = input()
str2 = input()
LD(str1,str2)
``` | instruction | 0 | 43,795 | 18 | 87,590 |
No | output | 1 | 43,795 | 18 | 87,591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the edit distance between given two words s1 and s2.
The disntace is the minimum number of single-character edits required to change one word into the other. The edits including the following operations:
* insertion: Insert a character at a particular position.
* deletion: Delete a character at a particular position.
* substitution: Change the character at a particular position to a different character
Constraints
* 1 ≤ length of s1 ≤ 1000
* 1 ≤ length of s2 ≤ 1000
Input
s1
s2
Two words s1 and s2 are given in the first line and the second line respectively. The words will consist of lower case characters.
Output
Print the edit distance in a line.
Examples
Input
acac
acm
Output
2
Input
icpc
icpc
Output
0
Submitted Solution:
```
#dj 16D8101012J ito jun
import sys
def search(a,b):
na=len(a)
nb=len(b)
array = [[0]*(nb+1) for _ in range(na+1)]
for i in range(na):
array[i][0]=i
for j in range(nb):
array[0][j]=j
for i,x in enumerate(a,1):
prerow=array[i-1]
row=array[i]
for j,y in enumerate(b,1):
if x == y:
row[j]=prerow[j-1]
else:
row[j]=min(prerow[j], prerow[j-1], row[j-1])+1
return array[-1][-1]
a=input()
b=input()
print(search(a,b))
``` | instruction | 0 | 43,796 | 18 | 87,592 |
No | output | 1 | 43,796 | 18 | 87,593 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the edit distance between given two words s1 and s2.
The disntace is the minimum number of single-character edits required to change one word into the other. The edits including the following operations:
* insertion: Insert a character at a particular position.
* deletion: Delete a character at a particular position.
* substitution: Change the character at a particular position to a different character
Constraints
* 1 ≤ length of s1 ≤ 1000
* 1 ≤ length of s2 ≤ 1000
Input
s1
s2
Two words s1 and s2 are given in the first line and the second line respectively. The words will consist of lower case characters.
Output
Print the edit distance in a line.
Examples
Input
acac
acm
Output
2
Input
icpc
icpc
Output
0
Submitted Solution:
```
def main():
X=input()
Y=input()
dp=[]
for y in range(len(Y)):
now=0
for i in range(len(dp)):
d=dp[i]
if Y[y] in X[now:d]:
dp[i]=now+X[now:d].index(Y[y])
now=d+1
if Y[y] in X[now:]:
dp.append(now+X[now:].index(Y[y]))
return max(len(X),len(Y))-len(dp)
print(main())
``` | instruction | 0 | 43,797 | 18 | 87,594 |
No | output | 1 | 43,797 | 18 | 87,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the edit distance between given two words s1 and s2.
The disntace is the minimum number of single-character edits required to change one word into the other. The edits including the following operations:
* insertion: Insert a character at a particular position.
* deletion: Delete a character at a particular position.
* substitution: Change the character at a particular position to a different character
Constraints
* 1 ≤ length of s1 ≤ 1000
* 1 ≤ length of s2 ≤ 1000
Input
s1
s2
Two words s1 and s2 are given in the first line and the second line respectively. The words will consist of lower case characters.
Output
Print the edit distance in a line.
Examples
Input
acac
acm
Output
2
Input
icpc
icpc
Output
0
Submitted Solution:
```
list1=list(input())
list2=list(input())
cost=0
list_d=[[0 for i in range(len(list2)+1)] for j in range(list1)+1]
for i in range(1,len(list1)+1):
list_d[i][0]=i
for j in range(1,len(list2)+1):
list_d[0][j]=j
for i in range(1,len(list1)+1):
for j in range(1,len(list2)+1):
if list1[i-1]!=list2[j-1]:
cost=1
else:
cost=0
list_d[i][j]=min(list_d[i-1][j-1]+cost,list_d[i-1][j]+1,list_d[i][j-1]+1)
print(list_d[len(list1)][len(list2)])
``` | instruction | 0 | 43,798 | 18 | 87,596 |
No | output | 1 | 43,798 | 18 | 87,597 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder Inc. holds a contest every Saturday.
There are two types of contests called ABC and ARC, and just one of them is held at a time.
The company holds these two types of contests alternately: an ARC follows an ABC and vice versa.
Given a string S representing the type of the contest held last week, print the string representing the type of the contest held this week.
Constraints
* S is `ABC` or `ARC`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the string representing the type of the contest held this week.
Example
Input
ABC
Output
ARC | instruction | 0 | 44,421 | 18 | 88,842 |
"Correct Solution:
```
N = input()
print('ABC' if N == 'ARC' else 'ARC')
``` | output | 1 | 44,421 | 18 | 88,843 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder Inc. holds a contest every Saturday.
There are two types of contests called ABC and ARC, and just one of them is held at a time.
The company holds these two types of contests alternately: an ARC follows an ABC and vice versa.
Given a string S representing the type of the contest held last week, print the string representing the type of the contest held this week.
Constraints
* S is `ABC` or `ARC`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the string representing the type of the contest held this week.
Example
Input
ABC
Output
ARC | instruction | 0 | 44,422 | 18 | 88,844 |
"Correct Solution:
```
s = input()
se = {'ABC','ARC'}
print(list(se-{s})[0])
``` | output | 1 | 44,422 | 18 | 88,845 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder Inc. holds a contest every Saturday.
There are two types of contests called ABC and ARC, and just one of them is held at a time.
The company holds these two types of contests alternately: an ARC follows an ABC and vice versa.
Given a string S representing the type of the contest held last week, print the string representing the type of the contest held this week.
Constraints
* S is `ABC` or `ARC`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the string representing the type of the contest held this week.
Example
Input
ABC
Output
ARC | instruction | 0 | 44,423 | 18 | 88,846 |
"Correct Solution:
```
a = input()[1]
print(f'A{"R" if a=="B" else "B"}C')
``` | output | 1 | 44,423 | 18 | 88,847 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder Inc. holds a contest every Saturday.
There are two types of contests called ABC and ARC, and just one of them is held at a time.
The company holds these two types of contests alternately: an ARC follows an ABC and vice versa.
Given a string S representing the type of the contest held last week, print the string representing the type of the contest held this week.
Constraints
* S is `ABC` or `ARC`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the string representing the type of the contest held this week.
Example
Input
ABC
Output
ARC | instruction | 0 | 44,424 | 18 | 88,848 |
"Correct Solution:
```
s = input()
print('ARC') if s == 'ABC' else print('ABC')
``` | output | 1 | 44,424 | 18 | 88,849 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder Inc. holds a contest every Saturday.
There are two types of contests called ABC and ARC, and just one of them is held at a time.
The company holds these two types of contests alternately: an ARC follows an ABC and vice versa.
Given a string S representing the type of the contest held last week, print the string representing the type of the contest held this week.
Constraints
* S is `ABC` or `ARC`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the string representing the type of the contest held this week.
Example
Input
ABC
Output
ARC | instruction | 0 | 44,425 | 18 | 88,850 |
"Correct Solution:
```
print(["ABC","ARC"]["ABC"==input()])
``` | output | 1 | 44,425 | 18 | 88,851 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder Inc. holds a contest every Saturday.
There are two types of contests called ABC and ARC, and just one of them is held at a time.
The company holds these two types of contests alternately: an ARC follows an ABC and vice versa.
Given a string S representing the type of the contest held last week, print the string representing the type of the contest held this week.
Constraints
* S is `ABC` or `ARC`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the string representing the type of the contest held this week.
Example
Input
ABC
Output
ARC | instruction | 0 | 44,426 | 18 | 88,852 |
"Correct Solution:
```
print("ARC" if input() == "ABC" else "ABC")
``` | output | 1 | 44,426 | 18 | 88,853 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder Inc. holds a contest every Saturday.
There are two types of contests called ABC and ARC, and just one of them is held at a time.
The company holds these two types of contests alternately: an ARC follows an ABC and vice versa.
Given a string S representing the type of the contest held last week, print the string representing the type of the contest held this week.
Constraints
* S is `ABC` or `ARC`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the string representing the type of the contest held this week.
Example
Input
ABC
Output
ARC | instruction | 0 | 44,427 | 18 | 88,854 |
"Correct Solution:
```
print('AABRCC'[input()=='ABC'::2])
``` | output | 1 | 44,427 | 18 | 88,855 |
Provide a correct Python 3 solution for this coding contest problem.
AtCoder Inc. holds a contest every Saturday.
There are two types of contests called ABC and ARC, and just one of them is held at a time.
The company holds these two types of contests alternately: an ARC follows an ABC and vice versa.
Given a string S representing the type of the contest held last week, print the string representing the type of the contest held this week.
Constraints
* S is `ABC` or `ARC`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the string representing the type of the contest held this week.
Example
Input
ABC
Output
ARC | instruction | 0 | 44,428 | 18 | 88,856 |
"Correct Solution:
```
print(['ABC', 'ARC'][input() == 'ABC'])
``` | output | 1 | 44,428 | 18 | 88,857 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder Inc. holds a contest every Saturday.
There are two types of contests called ABC and ARC, and just one of them is held at a time.
The company holds these two types of contests alternately: an ARC follows an ABC and vice versa.
Given a string S representing the type of the contest held last week, print the string representing the type of the contest held this week.
Constraints
* S is `ABC` or `ARC`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the string representing the type of the contest held this week.
Example
Input
ABC
Output
ARC
Submitted Solution:
```
S = input('ABC、ARCのどちらかを入力してください：')
if S == ABC:
print('ARC')
else:
print('ABC')
``` | instruction | 0 | 44,435 | 18 | 88,870 |
No | output | 1 | 44,435 | 18 | 88,871 |
Provide a correct Python 3 solution for this coding contest problem.
Mobile phones are equipped with a function that displays input candidates in order to efficiently input texts such as emails. This records frequently used words and presents words that have the entered letters as initial letters as input candidates. For example, if you usually enter the word "computer" a lot, just enter "c" and "computer" will be presented as a candidate. Let's create the basic part of such a feature.
Create a program that inputs sentences and letters and outputs words that have the letters as the first letter in the order of the number of occurrences. However, if there are multiple words with the same number of occurrences, they are output in lexicographic order. Output up to 5 words. If the corresponding word does not exist, output "NA".
input
Given multiple datasets. The end of the input is represented by a single zero. Each dataset is given in the following format:
n
line1
line2
::
linen
k
The first line is given the number of lines of text n (1 ≤ n ≤ 10). The text is given on the next n lines. The text consists of half-width lowercase letters and half-width spaces, and the number of characters in one line is 1 to 1024 characters. Words separated by spaces are between 1 and 20 characters.
The first letter k is given in the following line as a half-width alphabetic character.
The number of datasets does not exceed 100.
output
For each dataset, it prints a word or "NA" with the specified letter as the first letter.
Example
Input
1
ben likes bananas the best among many fruits because bananas are sweet and cheap
b
1
winners get to visit aizu and the university of aizu and make many friends as well
a
3
ask alex about
the answer for the
assignment on android apps
a
2
programming is both
a sport and an intellectual puzzle
c
0
Output
bananas because ben best
aizu and as
about alex android answer apps
NA | instruction | 0 | 44,567 | 18 | 89,134 |
"Correct Solution:
```
# Aizu Problem 0242: Input Candidates
import sys, math, os, bisect
# read input:
PYDEV = os.environ.get('PYDEV')
if PYDEV=="True":
sys.stdin = open("sample-input.txt", "rt")
while True:
n = int(input())
if n == 0:
break
text = []
count = {}
for _ in range(n):
for word in input().split():
count[word] = count.get(word, 0) + 1
char = input().strip()
out = []
for key, value in sorted(sorted(count.items()), key=lambda x: -x[1]):
if key[0] == char:
out.append(key)
if len(out) == 5:
break
if len(out) == 0:
print("NA")
else:
print(' '.join(out))
``` | output | 1 | 44,567 | 18 | 89,135 |
Provide a correct Python 3 solution for this coding contest problem.
Mobile phones are equipped with a function that displays input candidates in order to efficiently input texts such as emails. This records frequently used words and presents words that have the entered letters as initial letters as input candidates. For example, if you usually enter the word "computer" a lot, just enter "c" and "computer" will be presented as a candidate. Let's create the basic part of such a feature.
Create a program that inputs sentences and letters and outputs words that have the letters as the first letter in the order of the number of occurrences. However, if there are multiple words with the same number of occurrences, they are output in lexicographic order. Output up to 5 words. If the corresponding word does not exist, output "NA".
input
Given multiple datasets. The end of the input is represented by a single zero. Each dataset is given in the following format:
n
line1
line2
::
linen
k
The first line is given the number of lines of text n (1 ≤ n ≤ 10). The text is given on the next n lines. The text consists of half-width lowercase letters and half-width spaces, and the number of characters in one line is 1 to 1024 characters. Words separated by spaces are between 1 and 20 characters.
The first letter k is given in the following line as a half-width alphabetic character.
The number of datasets does not exceed 100.
output
For each dataset, it prints a word or "NA" with the specified letter as the first letter.
Example
Input
1
ben likes bananas the best among many fruits because bananas are sweet and cheap
b
1
winners get to visit aizu and the university of aizu and make many friends as well
a
3
ask alex about
the answer for the
assignment on android apps
a
2
programming is both
a sport and an intellectual puzzle
c
0
Output
bananas because ben best
aizu and as
about alex android answer apps
NA | instruction | 0 | 44,568 | 18 | 89,136 |
"Correct Solution:
```
while 1:
n = int(input())
if n == 0:break
c = {}
for _ in range(n):
s = input().split()
for e in s:
if e in c:c[e] += 1
else:c[e] = 1
c = [(e, c[e]) for e in c.keys()]
c.sort(key = lambda x:x[0])
c.sort(key = lambda x:x[1], reverse = True)
s = input()
a = []
for i in range(len(c)):
if c[i][0][0] == s:a.append(c[i][0])
if a == []:print("NA")
else:print(*a[:5])
``` | output | 1 | 44,568 | 18 | 89,137 |
Provide a correct Python 3 solution for this coding contest problem.
Mobile phones are equipped with a function that displays input candidates in order to efficiently input texts such as emails. This records frequently used words and presents words that have the entered letters as initial letters as input candidates. For example, if you usually enter the word "computer" a lot, just enter "c" and "computer" will be presented as a candidate. Let's create the basic part of such a feature.
Create a program that inputs sentences and letters and outputs words that have the letters as the first letter in the order of the number of occurrences. However, if there are multiple words with the same number of occurrences, they are output in lexicographic order. Output up to 5 words. If the corresponding word does not exist, output "NA".
input
Given multiple datasets. The end of the input is represented by a single zero. Each dataset is given in the following format:
n
line1
line2
::
linen
k
The first line is given the number of lines of text n (1 ≤ n ≤ 10). The text is given on the next n lines. The text consists of half-width lowercase letters and half-width spaces, and the number of characters in one line is 1 to 1024 characters. Words separated by spaces are between 1 and 20 characters.
The first letter k is given in the following line as a half-width alphabetic character.
The number of datasets does not exceed 100.
output
For each dataset, it prints a word or "NA" with the specified letter as the first letter.
Example
Input
1
ben likes bananas the best among many fruits because bananas are sweet and cheap
b
1
winners get to visit aizu and the university of aizu and make many friends as well
a
3
ask alex about
the answer for the
assignment on android apps
a
2
programming is both
a sport and an intellectual puzzle
c
0
Output
bananas because ben best
aizu and as
about alex android answer apps
NA | instruction | 0 | 44,569 | 18 | 89,138 |
"Correct Solution:
```
import sys
import string
import operator
from collections import Counter
f = sys.stdin
while True:
n = int(f.readline())
if n == 0:
break
counters = {c:Counter() for c in string.ascii_lowercase}
for i in range(n):
for word in f.readline().strip().split():
counters[word[0]].update([word])
c = f.readline().strip()
candidate = counters[c].most_common()
candidate.sort(key=operator.itemgetter(0))
candidate.sort(key=operator.itemgetter(1), reverse=True)
candidate = [word for word, count in candidate[:5]]
if len(candidate) == 0:
print('NA')
else:
print(*candidate)
``` | output | 1 | 44,569 | 18 | 89,139 |
Provide a correct Python 3 solution for this coding contest problem.
Mobile phones are equipped with a function that displays input candidates in order to efficiently input texts such as emails. This records frequently used words and presents words that have the entered letters as initial letters as input candidates. For example, if you usually enter the word "computer" a lot, just enter "c" and "computer" will be presented as a candidate. Let's create the basic part of such a feature.
Create a program that inputs sentences and letters and outputs words that have the letters as the first letter in the order of the number of occurrences. However, if there are multiple words with the same number of occurrences, they are output in lexicographic order. Output up to 5 words. If the corresponding word does not exist, output "NA".
input
Given multiple datasets. The end of the input is represented by a single zero. Each dataset is given in the following format:
n
line1
line2
::
linen
k
The first line is given the number of lines of text n (1 ≤ n ≤ 10). The text is given on the next n lines. The text consists of half-width lowercase letters and half-width spaces, and the number of characters in one line is 1 to 1024 characters. Words separated by spaces are between 1 and 20 characters.
The first letter k is given in the following line as a half-width alphabetic character.
The number of datasets does not exceed 100.
output
For each dataset, it prints a word or "NA" with the specified letter as the first letter.
Example
Input
1
ben likes bananas the best among many fruits because bananas are sweet and cheap
b
1
winners get to visit aizu and the university of aizu and make many friends as well
a
3
ask alex about
the answer for the
assignment on android apps
a
2
programming is both
a sport and an intellectual puzzle
c
0
Output
bananas because ben best
aizu and as
about alex android answer apps
NA | instruction | 0 | 44,570 | 18 | 89,140 |
"Correct Solution:
```
from collections import defaultdict, Counter
while 1:
N = int(input())
if N == 0:
break
mp = defaultdict(list)
for i in range(N):
for word in input().split():
mp[word[0]].append(word)
k = input()
c = Counter(mp[k])
if len(c) == 0:
print("NA")
else:
print(*sorted(c, key=lambda x: (-c[x], x))[:5])
``` | output | 1 | 44,570 | 18 | 89,141 |
Provide a correct Python 3 solution for this coding contest problem.
Mobile phones are equipped with a function that displays input candidates in order to efficiently input texts such as emails. This records frequently used words and presents words that have the entered letters as initial letters as input candidates. For example, if you usually enter the word "computer" a lot, just enter "c" and "computer" will be presented as a candidate. Let's create the basic part of such a feature.
Create a program that inputs sentences and letters and outputs words that have the letters as the first letter in the order of the number of occurrences. However, if there are multiple words with the same number of occurrences, they are output in lexicographic order. Output up to 5 words. If the corresponding word does not exist, output "NA".
input
Given multiple datasets. The end of the input is represented by a single zero. Each dataset is given in the following format:
n
line1
line2
::
linen
k
The first line is given the number of lines of text n (1 ≤ n ≤ 10). The text is given on the next n lines. The text consists of half-width lowercase letters and half-width spaces, and the number of characters in one line is 1 to 1024 characters. Words separated by spaces are between 1 and 20 characters.
The first letter k is given in the following line as a half-width alphabetic character.
The number of datasets does not exceed 100.
output
For each dataset, it prints a word or "NA" with the specified letter as the first letter.
Example
Input
1
ben likes bananas the best among many fruits because bananas are sweet and cheap
b
1
winners get to visit aizu and the university of aizu and make many friends as well
a
3
ask alex about
the answer for the
assignment on android apps
a
2
programming is both
a sport and an intellectual puzzle
c
0
Output
bananas because ben best
aizu and as
about alex android answer apps
NA | instruction | 0 | 44,571 | 18 | 89,142 |
"Correct Solution:
```
import heapq
from collections import deque
from enum import Enum
import sys
import math
from _heapq import heappush, heappop
import copy
BIG_NUM = 2000000000
HUGE_NUM = 99999999999999999
MOD = 1000000007
EPS = 0.000000001
sys.setrecursionlimit(100000)
class Info:
def __init__(self,arg_word,arg_count):
self.word= arg_word
self.count = arg_count
def __lt__(self,another):
if self.count != another.count:
return self.count > another.count
else:
return self.word < another.word
while True:
num_row = int(input())
if num_row == 0:
break
DICT = {} #単語→ID
rev_DICT = {} #ID→単語
COUNT = {} #IDをキーとして、出現数を計上
num_words = 0
for _ in range(num_row):
words = input().split()
for i in range(len(words)):
if words[i] in DICT:
COUNT[DICT[words[i]]] += 1
else:
DICT[words[i]] = num_words
rev_DICT[num_words] = words[i]
num_words += 1
COUNT[DICT[words[i]]] = 1
head_word = str(input())
table = []
for i in range(num_words):
word = rev_DICT[i]
if word[0] != head_word:
continue
table.append(Info(word,COUNT[DICT[word]]))
if len(table) == 0:
print("NA")
continue
table.sort()
print("%s"%(table[0].word),end = "")
for i in range(1,min(5,len(table))):
print(" %s"%(table[i].word),end="")
print()
``` | output | 1 | 44,571 | 18 | 89,143 |
Provide a correct Python 3 solution for this coding contest problem.
Mobile phones are equipped with a function that displays input candidates in order to efficiently input texts such as emails. This records frequently used words and presents words that have the entered letters as initial letters as input candidates. For example, if you usually enter the word "computer" a lot, just enter "c" and "computer" will be presented as a candidate. Let's create the basic part of such a feature.
Create a program that inputs sentences and letters and outputs words that have the letters as the first letter in the order of the number of occurrences. However, if there are multiple words with the same number of occurrences, they are output in lexicographic order. Output up to 5 words. If the corresponding word does not exist, output "NA".
input
Given multiple datasets. The end of the input is represented by a single zero. Each dataset is given in the following format:
n
line1
line2
::
linen
k
The first line is given the number of lines of text n (1 ≤ n ≤ 10). The text is given on the next n lines. The text consists of half-width lowercase letters and half-width spaces, and the number of characters in one line is 1 to 1024 characters. Words separated by spaces are between 1 and 20 characters.
The first letter k is given in the following line as a half-width alphabetic character.
The number of datasets does not exceed 100.
output
For each dataset, it prints a word or "NA" with the specified letter as the first letter.
Example
Input
1
ben likes bananas the best among many fruits because bananas are sweet and cheap
b
1
winners get to visit aizu and the university of aizu and make many friends as well
a
3
ask alex about
the answer for the
assignment on android apps
a
2
programming is both
a sport and an intellectual puzzle
c
0
Output
bananas because ben best
aizu and as
about alex android answer apps
NA | instruction | 0 | 44,572 | 18 | 89,144 |
"Correct Solution:
```
# 参考:http://judge.u-aizu.ac.jp/onlinejudge/review.jsp?rid=3286382#1
while 1:
n = int(input())
if n == 0:
break
words,dic = [],[]
for i in range(n):
data = list(input().split())
for d in data:
words.append(d)
k = input()
setWords = list(set(words))
for w in setWords:
if w[0] == k:
dic.append([w, words.count(w)])
dic = sorted(dic, key=lambda x: (-x[1], x[0]))
if dic == []:
print("NA")
else:
print(' '.join(d[0] for d in dic[:5]))
``` | output | 1 | 44,572 | 18 | 89,145 |
Provide a correct Python 3 solution for this coding contest problem.
Mobile phones are equipped with a function that displays input candidates in order to efficiently input texts such as emails. This records frequently used words and presents words that have the entered letters as initial letters as input candidates. For example, if you usually enter the word "computer" a lot, just enter "c" and "computer" will be presented as a candidate. Let's create the basic part of such a feature.
Create a program that inputs sentences and letters and outputs words that have the letters as the first letter in the order of the number of occurrences. However, if there are multiple words with the same number of occurrences, they are output in lexicographic order. Output up to 5 words. If the corresponding word does not exist, output "NA".
input
Given multiple datasets. The end of the input is represented by a single zero. Each dataset is given in the following format:
n
line1
line2
::
linen
k
The first line is given the number of lines of text n (1 ≤ n ≤ 10). The text is given on the next n lines. The text consists of half-width lowercase letters and half-width spaces, and the number of characters in one line is 1 to 1024 characters. Words separated by spaces are between 1 and 20 characters.
The first letter k is given in the following line as a half-width alphabetic character.
The number of datasets does not exceed 100.
output
For each dataset, it prints a word or "NA" with the specified letter as the first letter.
Example
Input
1
ben likes bananas the best among many fruits because bananas are sweet and cheap
b
1
winners get to visit aizu and the university of aizu and make many friends as well
a
3
ask alex about
the answer for the
assignment on android apps
a
2
programming is both
a sport and an intellectual puzzle
c
0
Output
bananas because ben best
aizu and as
about alex android answer apps
NA | instruction | 0 | 44,573 | 18 | 89,146 |
"Correct Solution:
```
from collections import Counter
dic = {}
def add(x):
if x[0] not in dic:
dic[x[0]] = Counter()
dic[x[0]][x] += 1
while 1:
dic = {}
n = int(input())
if n == 0:break
for l in [input() for i in range(n)]:
for w in l.split():
add(w)
k = input()
if k not in dic:
print('NA')
else:
print(' '.join(map(lambda x:x[0],sorted(dic[k].items(), key=lambda x: (-x[1],x[0]))[:5])))
``` | output | 1 | 44,573 | 18 | 89,147 |
Provide a correct Python 3 solution for this coding contest problem.
Mobile phones are equipped with a function that displays input candidates in order to efficiently input texts such as emails. This records frequently used words and presents words that have the entered letters as initial letters as input candidates. For example, if you usually enter the word "computer" a lot, just enter "c" and "computer" will be presented as a candidate. Let's create the basic part of such a feature.
Create a program that inputs sentences and letters and outputs words that have the letters as the first letter in the order of the number of occurrences. However, if there are multiple words with the same number of occurrences, they are output in lexicographic order. Output up to 5 words. If the corresponding word does not exist, output "NA".
input
Given multiple datasets. The end of the input is represented by a single zero. Each dataset is given in the following format:
n
line1
line2
::
linen
k
The first line is given the number of lines of text n (1 ≤ n ≤ 10). The text is given on the next n lines. The text consists of half-width lowercase letters and half-width spaces, and the number of characters in one line is 1 to 1024 characters. Words separated by spaces are between 1 and 20 characters.
The first letter k is given in the following line as a half-width alphabetic character.
The number of datasets does not exceed 100.
output
For each dataset, it prints a word or "NA" with the specified letter as the first letter.
Example
Input
1
ben likes bananas the best among many fruits because bananas are sweet and cheap
b
1
winners get to visit aizu and the university of aizu and make many friends as well
a
3
ask alex about
the answer for the
assignment on android apps
a
2
programming is both
a sport and an intellectual puzzle
c
0
Output
bananas because ben best
aizu and as
about alex android answer apps
NA | instruction | 0 | 44,574 | 18 | 89,148 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Input Candidates
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0242
"""
import sys
from string import ascii_lowercase
from collections import defaultdict
def solve(n):
d = {ch: defaultdict(int) for ch in ascii_lowercase}
for _ in range(n):
for word in input().split():
d[word[0]][word] += 1
k = input()
return [a for _, a in sorted([(-v, k) for k, v in d[k].items()])[:5]] \
if d[k] else ['NA']
def main(args):
while True:
n = int(input())
if n == 0:
break
ans = solve(n)
print(*ans)
if __name__ == '__main__':
main(sys.argv[1:])
``` | output | 1 | 44,574 | 18 | 89,149 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mobile phones are equipped with a function that displays input candidates in order to efficiently input texts such as emails. This records frequently used words and presents words that have the entered letters as initial letters as input candidates. For example, if you usually enter the word "computer" a lot, just enter "c" and "computer" will be presented as a candidate. Let's create the basic part of such a feature.
Create a program that inputs sentences and letters and outputs words that have the letters as the first letter in the order of the number of occurrences. However, if there are multiple words with the same number of occurrences, they are output in lexicographic order. Output up to 5 words. If the corresponding word does not exist, output "NA".
input
Given multiple datasets. The end of the input is represented by a single zero. Each dataset is given in the following format:
n
line1
line2
::
linen
k
The first line is given the number of lines of text n (1 ≤ n ≤ 10). The text is given on the next n lines. The text consists of half-width lowercase letters and half-width spaces, and the number of characters in one line is 1 to 1024 characters. Words separated by spaces are between 1 and 20 characters.
The first letter k is given in the following line as a half-width alphabetic character.
The number of datasets does not exceed 100.
output
For each dataset, it prints a word or "NA" with the specified letter as the first letter.
Example
Input
1
ben likes bananas the best among many fruits because bananas are sweet and cheap
b
1
winners get to visit aizu and the university of aizu and make many friends as well
a
3
ask alex about
the answer for the
assignment on android apps
a
2
programming is both
a sport and an intellectual puzzle
c
0
Output
bananas because ben best
aizu and as
about alex android answer apps
NA
Submitted Solution:
```
# AOJ 0242: Input Candidates
# Python3 2018.6.25 bal4u
while 1:
n = int(input())
if n == 0: break
idx = {}
dict = {}
for i in range(n):
for j in list(input().split()):
if j in dict: dict[j] += 1
else:
dict[j] = 1
if j[0] not in idx:
idx[j[0]] = []
idx[j[0]].append(j)
k = input()
if k not in idx: print("NA")
else:
a = []
for j in idx[k]: a.append((j, dict[j]))
a = sorted(a, key=lambda x:(-x[1],x[0]))
w = len(a)
if w > 5: w = 5
ans = []
for j in range(w): ans.append(a[j][0])
print(*ans)
``` | instruction | 0 | 44,575 | 18 | 89,150 |
Yes | output | 1 | 44,575 | 18 | 89,151 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mobile phones are equipped with a function that displays input candidates in order to efficiently input texts such as emails. This records frequently used words and presents words that have the entered letters as initial letters as input candidates. For example, if you usually enter the word "computer" a lot, just enter "c" and "computer" will be presented as a candidate. Let's create the basic part of such a feature.
Create a program that inputs sentences and letters and outputs words that have the letters as the first letter in the order of the number of occurrences. However, if there are multiple words with the same number of occurrences, they are output in lexicographic order. Output up to 5 words. If the corresponding word does not exist, output "NA".
input
Given multiple datasets. The end of the input is represented by a single zero. Each dataset is given in the following format:
n
line1
line2
::
linen
k
The first line is given the number of lines of text n (1 ≤ n ≤ 10). The text is given on the next n lines. The text consists of half-width lowercase letters and half-width spaces, and the number of characters in one line is 1 to 1024 characters. Words separated by spaces are between 1 and 20 characters.
The first letter k is given in the following line as a half-width alphabetic character.
The number of datasets does not exceed 100.
output
For each dataset, it prints a word or "NA" with the specified letter as the first letter.
Example
Input
1
ben likes bananas the best among many fruits because bananas are sweet and cheap
b
1
winners get to visit aizu and the university of aizu and make many friends as well
a
3
ask alex about
the answer for the
assignment on android apps
a
2
programming is both
a sport and an intellectual puzzle
c
0
Output
bananas because ben best
aizu and as
about alex android answer apps
NA
Submitted Solution:
```
# AOJ 0242: Input Candidates
# Python3 2018.6.25 bal4u
while 1:
n = int(input())
if n == 0: break
idx, dict = {}, {}
for i in range(n):
for j in list(input().split()):
if j in dict: dict[j] += 1
else:
dict[j] = 1
if j[0] not in idx: idx[j[0]] = []
idx[j[0]].append(j)
k = input()
if k not in idx: print("NA")
else:
a, ans = [], []
for j in idx[k]: a.append((j, dict[j]))
a.sort(key=lambda x:(-x[1],x[0]))
w = min(len(a), 5)
for j in range(w): ans.append(a[j][0])
print(*ans)
``` | instruction | 0 | 44,576 | 18 | 89,152 |
Yes | output | 1 | 44,576 | 18 | 89,153 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mobile phones are equipped with a function that displays input candidates in order to efficiently input texts such as emails. This records frequently used words and presents words that have the entered letters as initial letters as input candidates. For example, if you usually enter the word "computer" a lot, just enter "c" and "computer" will be presented as a candidate. Let's create the basic part of such a feature.
Create a program that inputs sentences and letters and outputs words that have the letters as the first letter in the order of the number of occurrences. However, if there are multiple words with the same number of occurrences, they are output in lexicographic order. Output up to 5 words. If the corresponding word does not exist, output "NA".
input
Given multiple datasets. The end of the input is represented by a single zero. Each dataset is given in the following format:
n
line1
line2
::
linen
k
The first line is given the number of lines of text n (1 ≤ n ≤ 10). The text is given on the next n lines. The text consists of half-width lowercase letters and half-width spaces, and the number of characters in one line is 1 to 1024 characters. Words separated by spaces are between 1 and 20 characters.
The first letter k is given in the following line as a half-width alphabetic character.
The number of datasets does not exceed 100.
output
For each dataset, it prints a word or "NA" with the specified letter as the first letter.
Example
Input
1
ben likes bananas the best among many fruits because bananas are sweet and cheap
b
1
winners get to visit aizu and the university of aizu and make many friends as well
a
3
ask alex about
the answer for the
assignment on android apps
a
2
programming is both
a sport and an intellectual puzzle
c
0
Output
bananas because ben best
aizu and as
about alex android answer apps
NA
Submitted Solution:
```
import collections
while 1:
n=int(input())
if n==0:break
words=[]
words_match=[]
count=0
ans=[]
for i in range(n):
words+=list(map(str,input().split()))
keyword=input()
words.sort()
for i in words:
if keyword==i[0]:words_match.append(i)
count_dict = collections.Counter(words_match)
ans=count_dict.most_common(5)
if ans==[]:
print("NA")
else:
for i in range(len(ans)):
if i>0:print(" ",end="")
print(ans[i][0],end="")
print()
``` | instruction | 0 | 44,577 | 18 | 89,154 |
Yes | output | 1 | 44,577 | 18 | 89,155 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mobile phones are equipped with a function that displays input candidates in order to efficiently input texts such as emails. This records frequently used words and presents words that have the entered letters as initial letters as input candidates. For example, if you usually enter the word "computer" a lot, just enter "c" and "computer" will be presented as a candidate. Let's create the basic part of such a feature.
Create a program that inputs sentences and letters and outputs words that have the letters as the first letter in the order of the number of occurrences. However, if there are multiple words with the same number of occurrences, they are output in lexicographic order. Output up to 5 words. If the corresponding word does not exist, output "NA".
input
Given multiple datasets. The end of the input is represented by a single zero. Each dataset is given in the following format:
n
line1
line2
::
linen
k
The first line is given the number of lines of text n (1 ≤ n ≤ 10). The text is given on the next n lines. The text consists of half-width lowercase letters and half-width spaces, and the number of characters in one line is 1 to 1024 characters. Words separated by spaces are between 1 and 20 characters.
The first letter k is given in the following line as a half-width alphabetic character.
The number of datasets does not exceed 100.
output
For each dataset, it prints a word or "NA" with the specified letter as the first letter.
Example
Input
1
ben likes bananas the best among many fruits because bananas are sweet and cheap
b
1
winners get to visit aizu and the university of aizu and make many friends as well
a
3
ask alex about
the answer for the
assignment on android apps
a
2
programming is both
a sport and an intellectual puzzle
c
0
Output
bananas because ben best
aizu and as
about alex android answer apps
NA
Submitted Solution:
```
# coding: utf-8
# Your code here!
class Word:
def __init__(self, word, cnt):
self.word = word
self.cnt = cnt
def __lt__(self, other):
if self.cnt == other.cnt:
return self.word < other.word
else:
return self.cnt > other.cnt
while True:
N = int(input())
if N == 0:
break
str = []
for l in range(N):
str.extend(input().split())
str.sort()
k = input()
words = []
for i in range(len(str)):
if str[i][0] == k:
if len(words) > 0 and words[-1].word == str[i]:
words[-1].cnt = words[-1].cnt + 1
else:
t = Word(str[i],1)
words.append(t)
if len(words) == 0:
print("NA")
elif len(words) < 5:
words.sort()
for i in range(len(words)-1):
print(words[i].word, end = " ")
print(words[-1].word)
else:
words.sort()
for i in range(4):
print(words[i].word, end = " ")
print(words[4].word)
``` | instruction | 0 | 44,578 | 18 | 89,156 |
Yes | output | 1 | 44,578 | 18 | 89,157 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mobile phones are equipped with a function that displays input candidates in order to efficiently input texts such as emails. This records frequently used words and presents words that have the entered letters as initial letters as input candidates. For example, if you usually enter the word "computer" a lot, just enter "c" and "computer" will be presented as a candidate. Let's create the basic part of such a feature.
Create a program that inputs sentences and letters and outputs words that have the letters as the first letter in the order of the number of occurrences. However, if there are multiple words with the same number of occurrences, they are output in lexicographic order. Output up to 5 words. If the corresponding word does not exist, output "NA".
input
Given multiple datasets. The end of the input is represented by a single zero. Each dataset is given in the following format:
n
line1
line2
::
linen
k
The first line is given the number of lines of text n (1 ≤ n ≤ 10). The text is given on the next n lines. The text consists of half-width lowercase letters and half-width spaces, and the number of characters in one line is 1 to 1024 characters. Words separated by spaces are between 1 and 20 characters.
The first letter k is given in the following line as a half-width alphabetic character.
The number of datasets does not exceed 100.
output
For each dataset, it prints a word or "NA" with the specified letter as the first letter.
Example
Input
1
ben likes bananas the best among many fruits because bananas are sweet and cheap
b
1
winners get to visit aizu and the university of aizu and make many friends as well
a
3
ask alex about
the answer for the
assignment on android apps
a
2
programming is both
a sport and an intellectual puzzle
c
0
Output
bananas because ben best
aizu and as
about alex android answer apps
NA
Submitted Solution:
```
import collections as C
while True:
n = int(input())
if n == 0:
break
d = {}
for _ in range(n):
tmp = input().split()
for t in tmp:
k = t[0]
d[k] = [t] if k not in d else d[k]+[t]
k = input()
words = d.get(k, 0)
if words:
res = sorted(C.Counter(words).most_common(), key=lambda x: (x[0], x[1]))
print(*[k for k, v in res])
else:
print("NA")
``` | instruction | 0 | 44,579 | 18 | 89,158 |
No | output | 1 | 44,579 | 18 | 89,159 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mobile phones are equipped with a function that displays input candidates in order to efficiently input texts such as emails. This records frequently used words and presents words that have the entered letters as initial letters as input candidates. For example, if you usually enter the word "computer" a lot, just enter "c" and "computer" will be presented as a candidate. Let's create the basic part of such a feature.
Create a program that inputs sentences and letters and outputs words that have the letters as the first letter in the order of the number of occurrences. However, if there are multiple words with the same number of occurrences, they are output in lexicographic order. Output up to 5 words. If the corresponding word does not exist, output "NA".
input
Given multiple datasets. The end of the input is represented by a single zero. Each dataset is given in the following format:
n
line1
line2
::
linen
k
The first line is given the number of lines of text n (1 ≤ n ≤ 10). The text is given on the next n lines. The text consists of half-width lowercase letters and half-width spaces, and the number of characters in one line is 1 to 1024 characters. Words separated by spaces are between 1 and 20 characters.
The first letter k is given in the following line as a half-width alphabetic character.
The number of datasets does not exceed 100.
output
For each dataset, it prints a word or "NA" with the specified letter as the first letter.
Example
Input
1
ben likes bananas the best among many fruits because bananas are sweet and cheap
b
1
winners get to visit aizu and the university of aizu and make many friends as well
a
3
ask alex about
the answer for the
assignment on android apps
a
2
programming is both
a sport and an intellectual puzzle
c
0
Output
bananas because ben best
aizu and as
about alex android answer apps
NA
Submitted Solution:
```
import collections as C
while True:
n = int(input())
if n == 0:
break
d = {}
for _ in range(n):
tmp = input().split()
for t in tmp:
k = t[0]
d[k] = [t] if k not in d else d[k]+[t]
k = input()
words = d.get(k, 0)
if words:
res = sorted(C.Counter(words).most_common(5), key=lambda x: (x[0], x[1]))
print(*[k for k, v in res])
else:
print("NA")
``` | instruction | 0 | 44,580 | 18 | 89,160 |
No | output | 1 | 44,580 | 18 | 89,161 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mobile phones are equipped with a function that displays input candidates in order to efficiently input texts such as emails. This records frequently used words and presents words that have the entered letters as initial letters as input candidates. For example, if you usually enter the word "computer" a lot, just enter "c" and "computer" will be presented as a candidate. Let's create the basic part of such a feature.
Create a program that inputs sentences and letters and outputs words that have the letters as the first letter in the order of the number of occurrences. However, if there are multiple words with the same number of occurrences, they are output in lexicographic order. Output up to 5 words. If the corresponding word does not exist, output "NA".
input
Given multiple datasets. The end of the input is represented by a single zero. Each dataset is given in the following format:
n
line1
line2
::
linen
k
The first line is given the number of lines of text n (1 ≤ n ≤ 10). The text is given on the next n lines. The text consists of half-width lowercase letters and half-width spaces, and the number of characters in one line is 1 to 1024 characters. Words separated by spaces are between 1 and 20 characters.
The first letter k is given in the following line as a half-width alphabetic character.
The number of datasets does not exceed 100.
output
For each dataset, it prints a word or "NA" with the specified letter as the first letter.
Example
Input
1
ben likes bananas the best among many fruits because bananas are sweet and cheap
b
1
winners get to visit aizu and the university of aizu and make many friends as well
a
3
ask alex about
the answer for the
assignment on android apps
a
2
programming is both
a sport and an intellectual puzzle
c
0
Output
bananas because ben best
aizu and as
about alex android answer apps
NA
Submitted Solution:
```
while 1:
n = int(input())
if n == 0:break
c = {}
for _ in range(n):
s = input().split()
for e in s:
if e in c:c[e] += 1
else:c[e] = 1
c = [(e, c[e]) for e in c.keys()]
c.sort(key = lambda x:x[1], reverse = True)
c.sort(key = lambda x:x[0])
s = input()
a = []
for i in range(len(c)):
if c[i][0][0] == s:a.append(c[i][0])
if a == []:print("NA")
else:print(*a[:5])
``` | instruction | 0 | 44,581 | 18 | 89,162 |
No | output | 1 | 44,581 | 18 | 89,163 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mobile phones are equipped with a function that displays input candidates in order to efficiently input texts such as emails. This records frequently used words and presents words that have the entered letters as initial letters as input candidates. For example, if you usually enter the word "computer" a lot, just enter "c" and "computer" will be presented as a candidate. Let's create the basic part of such a feature.
Create a program that inputs sentences and letters and outputs words that have the letters as the first letter in the order of the number of occurrences. However, if there are multiple words with the same number of occurrences, they are output in lexicographic order. Output up to 5 words. If the corresponding word does not exist, output "NA".
input
Given multiple datasets. The end of the input is represented by a single zero. Each dataset is given in the following format:
n
line1
line2
::
linen
k
The first line is given the number of lines of text n (1 ≤ n ≤ 10). The text is given on the next n lines. The text consists of half-width lowercase letters and half-width spaces, and the number of characters in one line is 1 to 1024 characters. Words separated by spaces are between 1 and 20 characters.
The first letter k is given in the following line as a half-width alphabetic character.
The number of datasets does not exceed 100.
output
For each dataset, it prints a word or "NA" with the specified letter as the first letter.
Example
Input
1
ben likes bananas the best among many fruits because bananas are sweet and cheap
b
1
winners get to visit aizu and the university of aizu and make many friends as well
a
3
ask alex about
the answer for the
assignment on android apps
a
2
programming is both
a sport and an intellectual puzzle
c
0
Output
bananas because ben best
aizu and as
about alex android answer apps
NA
Submitted Solution:
```
from collections import Counter
while True:
n = int(input())
if n == 0:
break
dataset = [input().split() for _ in range(n)]
counter = Counter()
k = input()
for line in dataset:
for word in line:
if word[0] == k:
counter[word] += 1
candidates = sorted(counter.items(), key=lambda x: (x[0]))[:5]
print(" ".join(candidate[0] for candidate in candidates) if candidates else "NA")
``` | instruction | 0 | 44,582 | 18 | 89,164 |
No | output | 1 | 44,582 | 18 | 89,165 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is different with hard version only by constraints on total answers length
It is an interactive problem
Venya joined a tour to the madhouse, in which orderlies play with patients the following game. Orderlies pick a string s of length n, consisting only of lowercase English letters. The player can ask two types of queries:
* ? l r – ask to list all substrings of s[l..r]. Substrings will be returned in random order, and in every substring, all characters will be randomly shuffled.
* ! s – guess the string picked by the orderlies. This query can be asked exactly once, after that the game will finish. If the string is guessed correctly, the player wins, otherwise he loses.
The player can ask no more than 3 queries of the first type.
To make it easier for the orderlies, there is an additional limitation: the total number of returned substrings in all queries of the first type must not exceed (n+1)^2.
Venya asked you to write a program, which will guess the string by interacting with the orderlies' program and acting by the game's rules.
Your program should immediately terminate after guessing the string using a query of the second type. In case your program guessed the string incorrectly, or it violated the game rules, it will receive verdict Wrong answer.
Note that in every test case the string is fixed beforehand and will not change during the game, which means that the interactor is not adaptive.
Input
First line contains number n (1 ≤ n ≤ 100) — the length of the picked string.
Interaction
You start the interaction by reading the number n.
To ask a query about a substring from l to r inclusively (1 ≤ l ≤ r ≤ n), you should output
? l r
on a separate line. After this, all substrings of s[l..r] will be returned in random order, each substring exactly once. In every returned substring all characters will be randomly shuffled.
In the case, if you ask an incorrect query, ask more than 3 queries of the first type or there will be more than (n+1)^2 substrings returned in total, you will receive verdict Wrong answer.
To guess the string s, you should output
! s
on a separate line.
After printing each query, do not forget to flush the output. Otherwise, you will get Idleness limit exceeded. To flush the output, you can use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
If you received - (dash) as an answer to any query, you need to terminate your program with exit code 0 (for example, by calling exit(0)). This means that there was an error in the interaction protocol. If you don't terminate with exit code 0, you can receive any unsuccessful verdict.
Hack format
To hack a solution, use the following format:
The first line should contain one integer n (1 ≤ n ≤ 100) — the length of the string, and the following line should contain the string s.
Example
Input
4
a
aa
a
cb
b
c
c
Output
? 1 2
? 3 4
? 4 4
! aabc
Submitted Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
# M = mod = 998244353
def factors(n):return sorted(list(set(reduce(list.__add__,([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))))
# def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split(' ')]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n').split(' ')]
def li3():return [int(i) for i in input().rstrip('\n')]
n = val()
cnt1 = Counter()
cnt2 = Counter()
print('? 1 ' + str(n),flush = True)
le = 0
for i in range(n):
for j in range(i+1,n+1):le += 1
for j in range(le):
cnt1[st()] += 1
if n == 1:
for i in cnt1.keys():
print('! ' + str(i),flush = True)
exit()
print('? 2 ' + str(n),flush = True)
le = 0
for i in range(1,n):
for j in range(i+1,n+1):le += 1
# print(le)
for i in range(le):
cnt2[st()] += 1
cnt1 -= cnt2
cnt1 = sorted(list(cnt1),key = lambda x:len(x))
s = ''
currcount = Counter()
for i in cnt1:
currcount = Counter(s)
for j in i:
if not currcount[j]:
s += j
break
currcount[j] -= 1
print('! ' + s,flush = True)
``` | instruction | 0 | 44,726 | 18 | 89,452 |
No | output | 1 | 44,726 | 18 | 89,453 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is different with hard version only by constraints on total answers length
It is an interactive problem
Venya joined a tour to the madhouse, in which orderlies play with patients the following game. Orderlies pick a string s of length n, consisting only of lowercase English letters. The player can ask two types of queries:
* ? l r – ask to list all substrings of s[l..r]. Substrings will be returned in random order, and in every substring, all characters will be randomly shuffled.
* ! s – guess the string picked by the orderlies. This query can be asked exactly once, after that the game will finish. If the string is guessed correctly, the player wins, otherwise he loses.
The player can ask no more than 3 queries of the first type.
To make it easier for the orderlies, there is an additional limitation: the total number of returned substrings in all queries of the first type must not exceed (n+1)^2.
Venya asked you to write a program, which will guess the string by interacting with the orderlies' program and acting by the game's rules.
Your program should immediately terminate after guessing the string using a query of the second type. In case your program guessed the string incorrectly, or it violated the game rules, it will receive verdict Wrong answer.
Note that in every test case the string is fixed beforehand and will not change during the game, which means that the interactor is not adaptive.
Input
First line contains number n (1 ≤ n ≤ 100) — the length of the picked string.
Interaction
You start the interaction by reading the number n.
To ask a query about a substring from l to r inclusively (1 ≤ l ≤ r ≤ n), you should output
? l r
on a separate line. After this, all substrings of s[l..r] will be returned in random order, each substring exactly once. In every returned substring all characters will be randomly shuffled.
In the case, if you ask an incorrect query, ask more than 3 queries of the first type or there will be more than (n+1)^2 substrings returned in total, you will receive verdict Wrong answer.
To guess the string s, you should output
! s
on a separate line.
After printing each query, do not forget to flush the output. Otherwise, you will get Idleness limit exceeded. To flush the output, you can use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
If you received - (dash) as an answer to any query, you need to terminate your program with exit code 0 (for example, by calling exit(0)). This means that there was an error in the interaction protocol. If you don't terminate with exit code 0, you can receive any unsuccessful verdict.
Hack format
To hack a solution, use the following format:
The first line should contain one integer n (1 ≤ n ≤ 100) — the length of the string, and the following line should contain the string s.
Example
Input
4
a
aa
a
cb
b
c
c
Output
? 1 2
? 3 4
? 4 4
! aabc
Submitted Solution:
```
import sys
input = sys.stdin.readline
n=int(input())
ANS=[0]*n
print("?",1,n,flush=True)
S=["".join(sorted(input().strip())) for i in range(n*(n+1)//2)]
if n==1:
print(S[0])
sys.exit()
print("?",2,n,flush=True)
T=["".join(sorted(input().strip())) for i in range(n*(n-1)//2)]
from collections import Counter
U=Counter(S)-Counter(T)
K=sorted(U.keys(),key=lambda x:len(x))
K=[Counter(k) for k in K]
ANS=[]
for i in range(n):
x=list(K[i].keys())[0]
ANS.append(x)
for j in range(i+1,n):
K[j][x]-=1
if K[j][x]==0:
del(K[j][x])
print("!","".join(ANS),flush=True)
``` | instruction | 0 | 44,727 | 18 | 89,454 |
No | output | 1 | 44,727 | 18 | 89,455 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After you had helped Fedor to find friends in the «Call of Soldiers 3» game, he stopped studying completely. Today, the English teacher told him to prepare an essay. Fedor didn't want to prepare the essay, so he asked Alex for help. Alex came to help and wrote the essay for Fedor. But Fedor didn't like the essay at all. Now Fedor is going to change the essay using the synonym dictionary of the English language.
Fedor does not want to change the meaning of the essay. So the only change he would do: change a word from essay to one of its synonyms, basing on a replacement rule from the dictionary. Fedor may perform this operation any number of times.
As a result, Fedor wants to get an essay which contains as little letters «R» (the case doesn't matter) as possible. If there are multiple essays with minimum number of «R»s he wants to get the one with minimum length (length of essay is the sum of the lengths of all the words in it). Help Fedor get the required essay.
Please note that in this problem the case of letters doesn't matter. For example, if the synonym dictionary says that word cat can be replaced with word DOG, then it is allowed to replace the word Cat with the word doG.
Input
The first line contains a single integer m (1 ≤ m ≤ 105) — the number of words in the initial essay. The second line contains words of the essay. The words are separated by a single space. It is guaranteed that the total length of the words won't exceed 105 characters.
The next line contains a single integer n (0 ≤ n ≤ 105) — the number of pairs of words in synonym dictionary. The i-th of the next n lines contains two space-separated non-empty words xi and yi. They mean that word xi can be replaced with word yi (but not vise versa). It is guaranteed that the total length of all pairs of synonyms doesn't exceed 5·105 characters.
All the words at input can only consist of uppercase and lowercase letters of the English alphabet.
Output
Print two integers — the minimum number of letters «R» in an optimal essay and the minimum length of an optimal essay.
Examples
Input
3
AbRb r Zz
4
xR abRb
aA xr
zz Z
xr y
Output
2 6
Input
2
RuruRu fedya
1
ruruRU fedor
Output
1 10
Submitted Solution:
```
m=int(input())
a=input()
lv=len(a)-(m-1)
v=list(x.lower() for x in a.split())
vd={}
for i in v:
if i not in vd:
vd[i]=v.count(i)
vi=vd.copy()
n=int(input())
#print(lv)
#print(vd)
#print()
s={}
for i in range(n):
t=list(x.lower() for x in input().split())
# print(t)
if t[0] in vi.keys():
if t[0] not in vd.keys():
argh=t[0]
t[0]=s[t[0]]
c=vd[t[0]]
l0=len(t[0])
l1=len(t[1])
if t[1].count('r')<t[0].count('r'):
if t[0] in vi.keys():
s[t[0]]=t[1]
else:
s[argh]=t[1]
vd.pop(t[0])
vd[t[1]]=c
lv+=(l1-l0)*c
elif t[0].count('r')==t[1].count('r'):
if l1<l0:
if t[0] in vi.keys():
s[t[0]]=t[1]
else:
s[argh]=t[1]
vd.pop(t[0])
vd[t[1]]=c
lv+=(l1-l0)*c
# print(vd)
# print()
r=0
for i in vd.keys():
r+=i.count('r')*vd[i]
#r=(''.join(vd.keys())).count('r')
print (r,lv)
``` | instruction | 0 | 44,947 | 18 | 89,894 |
No | output | 1 | 44,947 | 18 | 89,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After you had helped Fedor to find friends in the «Call of Soldiers 3» game, he stopped studying completely. Today, the English teacher told him to prepare an essay. Fedor didn't want to prepare the essay, so he asked Alex for help. Alex came to help and wrote the essay for Fedor. But Fedor didn't like the essay at all. Now Fedor is going to change the essay using the synonym dictionary of the English language.
Fedor does not want to change the meaning of the essay. So the only change he would do: change a word from essay to one of its synonyms, basing on a replacement rule from the dictionary. Fedor may perform this operation any number of times.
As a result, Fedor wants to get an essay which contains as little letters «R» (the case doesn't matter) as possible. If there are multiple essays with minimum number of «R»s he wants to get the one with minimum length (length of essay is the sum of the lengths of all the words in it). Help Fedor get the required essay.
Please note that in this problem the case of letters doesn't matter. For example, if the synonym dictionary says that word cat can be replaced with word DOG, then it is allowed to replace the word Cat with the word doG.
Input
The first line contains a single integer m (1 ≤ m ≤ 105) — the number of words in the initial essay. The second line contains words of the essay. The words are separated by a single space. It is guaranteed that the total length of the words won't exceed 105 characters.
The next line contains a single integer n (0 ≤ n ≤ 105) — the number of pairs of words in synonym dictionary. The i-th of the next n lines contains two space-separated non-empty words xi and yi. They mean that word xi can be replaced with word yi (but not vise versa). It is guaranteed that the total length of all pairs of synonyms doesn't exceed 5·105 characters.
All the words at input can only consist of uppercase and lowercase letters of the English alphabet.
Output
Print two integers — the minimum number of letters «R» in an optimal essay and the minimum length of an optimal essay.
Examples
Input
3
AbRb r Zz
4
xR abRb
aA xr
zz Z
xr y
Output
2 6
Input
2
RuruRu fedya
1
ruruRU fedor
Output
1 10
Submitted Solution:
```
n = int(input())
ref = list(map(str, input().split(" ")))
rrr = [0 for i in range(n)]
for i in range(n):
ref[i] = ref[i].upper()
for j in range(len(ref[i])):
if ref[i][j] == 'R':
rrr[i] += 1
m = int(input())
for i in range(m):
s1, s2 = map(str, input().split(" "))
r1 = 0
r2 = 0
s1 = s1.upper()
s2 = s2.upper()
for j in range(len(s1)):
if s1[j] == 'R':
r1 += 1
for j in range(len(s2)):
if s2[j] == 'R':
r2 += 1
j = 0
while j < n and s1 in ref:
if rrr[ref.index(s1)] >= r2 and len(s1) >= len(s2):
ind = ref.index(s1)
ref.insert(ind, s2)
ref.__delitem__(ind + 1)
rrr[ind] = r2
elif rrr[ref.index(s1)] >= r2:
ind = ref.index(s1)
ref.insert(ind, s2)
ref.__delitem__(ind + 1)
rrr[ind] = r2
j += 1
ans1 = 0
ans2 = 0
for i in range(n):
ans1 += rrr[i]
ans2 += len(ref[i])
print(ans1, ans2)
``` | instruction | 0 | 44,948 | 18 | 89,896 |
No | output | 1 | 44,948 | 18 | 89,897 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After you had helped Fedor to find friends in the «Call of Soldiers 3» game, he stopped studying completely. Today, the English teacher told him to prepare an essay. Fedor didn't want to prepare the essay, so he asked Alex for help. Alex came to help and wrote the essay for Fedor. But Fedor didn't like the essay at all. Now Fedor is going to change the essay using the synonym dictionary of the English language.
Fedor does not want to change the meaning of the essay. So the only change he would do: change a word from essay to one of its synonyms, basing on a replacement rule from the dictionary. Fedor may perform this operation any number of times.
As a result, Fedor wants to get an essay which contains as little letters «R» (the case doesn't matter) as possible. If there are multiple essays with minimum number of «R»s he wants to get the one with minimum length (length of essay is the sum of the lengths of all the words in it). Help Fedor get the required essay.
Please note that in this problem the case of letters doesn't matter. For example, if the synonym dictionary says that word cat can be replaced with word DOG, then it is allowed to replace the word Cat with the word doG.
Input
The first line contains a single integer m (1 ≤ m ≤ 105) — the number of words in the initial essay. The second line contains words of the essay. The words are separated by a single space. It is guaranteed that the total length of the words won't exceed 105 characters.
The next line contains a single integer n (0 ≤ n ≤ 105) — the number of pairs of words in synonym dictionary. The i-th of the next n lines contains two space-separated non-empty words xi and yi. They mean that word xi can be replaced with word yi (but not vise versa). It is guaranteed that the total length of all pairs of synonyms doesn't exceed 5·105 characters.
All the words at input can only consist of uppercase and lowercase letters of the English alphabet.
Output
Print two integers — the minimum number of letters «R» in an optimal essay and the minimum length of an optimal essay.
Examples
Input
3
AbRb r Zz
4
xR abRb
aA xr
zz Z
xr y
Output
2 6
Input
2
RuruRu fedya
1
ruruRU fedor
Output
1 10
Submitted Solution:
```
m=int(input())
a=input()
lv=len(a)-(m-1)
v=list(x.lower() for x in a.split())
vd={}
for i in v:
if i not in vd:
vd[i]=v.count(i)
vi=vd.copy()
n=int(input())
#print(lv)
#print(vd)
#print()
s={}
for i in range(n):
t=list(x.lower() for x in input().split())
# print(t)
if t[0] in vi.keys():
if t[0] not in vd.keys():
argh=t[0]
t[0]=s[t[0]]
c=vd[t[0]]
l0=len(t[0])
l1=len(t[1])
if t[1].count('r')<t[0].count('r'):
s[t[0]]=t[1]
vd.pop(t[0])
vd[t[1]]=c
lv+=(l1-l0)*c
elif t[0].count('r')==t[1].count('r'):
if l1<l0:
if t[0] in vi.keys():
s[t[0]]=t[1]
else:
s[argh]=t[1]
vd.pop(t[0])
vd[t[1]]=c
lv+=(l1-l0)*c
# print(vd)
# print()
r=0
for i in vd.keys():
r+=i.count('r')*vd[i]
#r=(''.join(vd.keys())).count('r')
print (r,lv)
``` | instruction | 0 | 44,949 | 18 | 89,898 |
No | output | 1 | 44,949 | 18 | 89,899 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After you had helped Fedor to find friends in the «Call of Soldiers 3» game, he stopped studying completely. Today, the English teacher told him to prepare an essay. Fedor didn't want to prepare the essay, so he asked Alex for help. Alex came to help and wrote the essay for Fedor. But Fedor didn't like the essay at all. Now Fedor is going to change the essay using the synonym dictionary of the English language.
Fedor does not want to change the meaning of the essay. So the only change he would do: change a word from essay to one of its synonyms, basing on a replacement rule from the dictionary. Fedor may perform this operation any number of times.
As a result, Fedor wants to get an essay which contains as little letters «R» (the case doesn't matter) as possible. If there are multiple essays with minimum number of «R»s he wants to get the one with minimum length (length of essay is the sum of the lengths of all the words in it). Help Fedor get the required essay.
Please note that in this problem the case of letters doesn't matter. For example, if the synonym dictionary says that word cat can be replaced with word DOG, then it is allowed to replace the word Cat with the word doG.
Input
The first line contains a single integer m (1 ≤ m ≤ 105) — the number of words in the initial essay. The second line contains words of the essay. The words are separated by a single space. It is guaranteed that the total length of the words won't exceed 105 characters.
The next line contains a single integer n (0 ≤ n ≤ 105) — the number of pairs of words in synonym dictionary. The i-th of the next n lines contains two space-separated non-empty words xi and yi. They mean that word xi can be replaced with word yi (but not vise versa). It is guaranteed that the total length of all pairs of synonyms doesn't exceed 5·105 characters.
All the words at input can only consist of uppercase and lowercase letters of the English alphabet.
Output
Print two integers — the minimum number of letters «R» in an optimal essay and the minimum length of an optimal essay.
Examples
Input
3
AbRb r Zz
4
xR abRb
aA xr
zz Z
xr y
Output
2 6
Input
2
RuruRu fedya
1
ruruRU fedor
Output
1 10
Submitted Solution:
```
import collections
def solve():
m = int(input())
essay = [s for s in input().lower().split()]
n = int(input())
sti = dict()
pack = lambda word: (word.count('r'), len(word), sti.setdefault(word, len(sti)))
edge = collections.defaultdict(list)
nodes = list()
for _ in range(n):
word, synon = map(pack, input().lower().split())
edge[synon[-1]].append(word[-1])
nodes.append(synon)
nodes.sort()
best = dict()
for node in nodes:
if node[2] not in best:
stack = [node[2]]
while stack:
top = stack.pop()
if top not in best:
best[top] = node[:2]
for n in edge[top]:
if n is not best:
stack.append(n)
tr = 0
tl = 0
for word in essay:
if word in sti:
wid = sti[word]
tr += best[wid][0]
tl += best[wid][1]
else:
tr += word.count('r')
tl += len(word)
print(tr, ' ', tl)
if __name__ == '__main__':
solve()
``` | instruction | 0 | 44,950 | 18 | 89,900 |
No | output | 1 | 44,950 | 18 | 89,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string s consisting of n lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking!
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the length of the interview.
The second line contains the string s of length n, consisting of lowercase English letters.
Output
Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences.
Examples
Input
7
aogogob
Output
a***b
Input
13
ogogmgogogogo
Output
***gmg***
Input
9
ogoogoogo
Output
*********
Note
The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***".
Submitted Solution:
```
#ROUNIAAUDI
number=int(input())
stringBeutiful=input()
rangeoff=range(number,0,-1)
for i in rangeoff:
w="o"+"go"*i
stringBeutiful=stringBeutiful.replace(w,"***")
print(stringBeutiful)
``` | instruction | 0 | 45,023 | 18 | 90,046 |
Yes | output | 1 | 45,023 | 18 | 90,047 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string s consisting of n lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking!
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the length of the interview.
The second line contains the string s of length n, consisting of lowercase English letters.
Output
Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences.
Examples
Input
7
aogogob
Output
a***b
Input
13
ogogmgogogogo
Output
***gmg***
Input
9
ogoogoogo
Output
*********
Note
The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***".
Submitted Solution:
```
a = int(input())
string = input()
first = []
last = []
sas = False
i = 1
while i < a:
i += 1
if not sas:
if string[i-2:i+1] == "ogo":
first += [i-2]
sas = True
i += 1
else:
if string[i-1:i+1] == "go":
i += 1
else:
last += [i - 2]
sas = False
if sas:
last += [a - 1]
s = ""
counter = 0
n = 0
while counter < len(string):
if counter not in first:
s += string[counter]
counter += 1
else:
s += "***"
counter = last[n] + 1
n += 1
print(s)
``` | instruction | 0 | 45,024 | 18 | 90,048 |
Yes | output | 1 | 45,024 | 18 | 90,049 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string s consisting of n lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking!
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the length of the interview.
The second line contains the string s of length n, consisting of lowercase English letters.
Output
Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences.
Examples
Input
7
aogogob
Output
a***b
Input
13
ogogmgogogogo
Output
***gmg***
Input
9
ogoogoogo
Output
*********
Note
The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***".
Submitted Solution:
```
from math import ceil
n = int(input())
x = input()
p = ['ogo']+['ogo'+'go'*i for i in range(1,50)]
j = ceil((n-3)/2)
if n<3:
print(x)
else:
while j>=0:
if p[j] in x:
x = x.replace(p[j],'***')
else:
j = j-1
print(x)
``` | instruction | 0 | 45,025 | 18 | 90,050 |
Yes | output | 1 | 45,025 | 18 | 90,051 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string s consisting of n lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking!
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the length of the interview.
The second line contains the string s of length n, consisting of lowercase English letters.
Output
Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences.
Examples
Input
7
aogogob
Output
a***b
Input
13
ogogmgogogogo
Output
***gmg***
Input
9
ogoogoogo
Output
*********
Note
The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***".
Submitted Solution:
```
l = input()
s = input()
for i in range(len(s) // 2, -1, -1):
s = s.replace('ogo' + 'go' * i, '***')
print(s)
``` | instruction | 0 | 45,026 | 18 | 90,052 |
Yes | output | 1 | 45,026 | 18 | 90,053 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string s consisting of n lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking!
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the length of the interview.
The second line contains the string s of length n, consisting of lowercase English letters.
Output
Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences.
Examples
Input
7
aogogob
Output
a***b
Input
13
ogogmgogogogo
Output
***gmg***
Input
9
ogoogoogo
Output
*********
Note
The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***".
Submitted Solution:
```
n = int(input())
s = input()
res = ""
ogo = False
i = 0
while i < n:
if i < n - 2 and s[i: i + 3] == "ogo":
ogo = True
i += 3
elif i < n - 1 and s[i: i + 2] == "go" and ogo:
i += 2
else:
if ogo:
res += "***"
ogo = False
res += s[i]
i += 1
if ogo:
res += "***"
print(res)
``` | instruction | 0 | 45,027 | 18 | 90,054 |
No | output | 1 | 45,027 | 18 | 90,055 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string s consisting of n lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking!
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the length of the interview.
The second line contains the string s of length n, consisting of lowercase English letters.
Output
Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences.
Examples
Input
7
aogogob
Output
a***b
Input
13
ogogmgogogogo
Output
***gmg***
Input
9
ogoogoogo
Output
*********
Note
The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***".
Submitted Solution:
```
from sys import stdin, stdout
def ogo(string):
i = 0
string += ' '
for j in range(len(string)):
if string[j] == 'o' and i == j:
continue
elif string[j] == 'g' and j - i == 1:
continue
elif string[j] == 'o' and j - i == 2:
continue
elif string[j] == 'g' and j - i > 2 and (j - i) % 2 == 1:
continue
elif string[j] == 'o' and j - i > 3 and (j - i) % 2 == 0:
continue
else:
if j - i > 1:
if (j - i - 1) % 2 == 0:
pass
elif (j - i - 2) % 2 == 0:
j -= 1
string = string[:i] + "*"*(j-i) + string[j:]
if string[j] == 'o':
i = j-1
else:
i = j
i += 1
return string[:-1]
k = int(stdin.readline().strip())
s = stdin.readline().strip()
stdout.write(ogo(s))
``` | instruction | 0 | 45,028 | 18 | 90,056 |
No | output | 1 | 45,028 | 18 | 90,057 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string s consisting of n lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking!
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the length of the interview.
The second line contains the string s of length n, consisting of lowercase English letters.
Output
Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences.
Examples
Input
7
aogogob
Output
a***b
Input
13
ogogmgogogogo
Output
***gmg***
Input
9
ogoogoogo
Output
*********
Note
The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***".
Submitted Solution:
```
n=int(input())
s=input()
for i in range(n):
if 'ogo' in s:
s=s.replace('ogo','***')
j=s.split('go')
k=''.join(j)
l=k.split('*g**')
print(''.join(l))
``` | instruction | 0 | 45,029 | 18 | 90,058 |
No | output | 1 | 45,029 | 18 | 90,059 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string s consisting of n lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking!
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the length of the interview.
The second line contains the string s of length n, consisting of lowercase English letters.
Output
Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences.
Examples
Input
7
aogogob
Output
a***b
Input
13
ogogmgogogogo
Output
***gmg***
Input
9
ogoogoogo
Output
*********
Note
The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***".
Submitted Solution:
```
n = int(input())
t = input()
l = list(t)
j = 0
flag = True
for i in l:
if i=="o" or i=="g" or i=="o":
l.remove(i)
l.insert(j, "*")
print("a***b")
``` | instruction | 0 | 45,030 | 18 | 90,060 |
No | output | 1 | 45,030 | 18 | 90,061 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak can repeatedly remove one of the first two characters of a string, for example abcxyx \rightarrow acxyx \rightarrow cxyx \rightarrow cyx.
You are given N different strings S_1, S_2, \ldots, S_N. Among N \cdot (N-1) / 2 pairs (S_i, S_j), in how many pairs could Limak obtain one string from the other?
Constraints
* 2 \leq N \leq 200\,000
* S_i consists of lowercase English letters `a`-`z`.
* S_i \neq S_j
* 1 \leq |S_i|
* |S_1| + |S_2| + \ldots + |S_N| \leq 10^6
Input
Input is given from Standard Input in the following format.
N
S_1
S_2
\vdots
S_N
Output
Print the number of unordered pairs (S_i, S_j) where i \neq j and Limak can obtain one string from the other.
Examples
Input
3
abcxyx
cyx
abc
Output
1
Input
6
b
a
abc
c
d
ab
Output
5
Submitted Solution:
```
class RollingHash:
def __init__(self,S,base,M):
self.N=len(S)
self.M=M
self.H=[0]*(self.N+1)
self.B=[0]*(self.N+1)
self.B[0]=1
for i in range(self.N):
self.B[i+1]=self.B[i]*base%M
self.H[i+1]=(self.H[i]*base+(ord(S[i])-96))%M
def get(self,l,r):
return (self.H[r]-self.H[l]*self.B[r-l]+self.M)%self.M
def main():
import random
N=int(input())
M=2**61-1
base=random.randrange(100,M)
RH,S=[],[]
RH,S=[None for i in range(N)],[None]*N
for i in range(N):
s=input()
S[i]=s
RH[i]=RollingHash(s,base,M)
dic={}
dic[0]=[0]*26
for k,s in enumerate(S):
Set=set()
l=len(s)
for j in range(l):
Hash=RH[k].get(j,l)
if not Hash in dic:
dic[Hash]=[0]*26
for c in Set:
dic[Hash][ord(c)-97]+=1
Set.add(s[j])
for c in Set:
dic[0][ord(c)-97]+=1
ans=0
for i,s in enumerate(S):
Hash=RH[i].get(1,len(s))
ans+=dic[Hash][ord(s[0])-97]-1
print(ans)
if __name__=='__main__':
main()
``` | instruction | 0 | 45,167 | 18 | 90,334 |
Yes | output | 1 | 45,167 | 18 | 90,335 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak can repeatedly remove one of the first two characters of a string, for example abcxyx \rightarrow acxyx \rightarrow cxyx \rightarrow cyx.
You are given N different strings S_1, S_2, \ldots, S_N. Among N \cdot (N-1) / 2 pairs (S_i, S_j), in how many pairs could Limak obtain one string from the other?
Constraints
* 2 \leq N \leq 200\,000
* S_i consists of lowercase English letters `a`-`z`.
* S_i \neq S_j
* 1 \leq |S_i|
* |S_1| + |S_2| + \ldots + |S_N| \leq 10^6
Input
Input is given from Standard Input in the following format.
N
S_1
S_2
\vdots
S_N
Output
Print the number of unordered pairs (S_i, S_j) where i \neq j and Limak can obtain one string from the other.
Examples
Input
3
abcxyx
cyx
abc
Output
1
Input
6
b
a
abc
c
d
ab
Output
5
Submitted Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
max2 = lambda x,y: x if x > y else y
min2 = lambda x,y: x if x < y else y
class TrieNode:
def __init__(self, n_alphabets):
self.children = [None]*n_alphabets
self.end = False
class Trie:
def __init__(self, n_alphabets):
self.n_alphabets = n_alphabets
self.root = TrieNode(n_alphabets)
def add(self, s):
n = self.root
for c in s:
if n.children[c] is None:
n.children[c] = TrieNode(self.n_alphabets)
n = n.children[c]
n.end = True
def solve(L):
L = [tuple(ord(c)-ord('a') for c in reversed(s)) for s in L]
trie = Trie(26)
for s in L:
trie.add(s)
cnt = 0
for s in L:
app = [0]*26
for c in s:
app[c] += 1
node = trie.root
for c in s:
if node is None:
break
for d,n in enumerate(node.children):
if n is not None and app[d] > 0 and n.end:
cnt += 1
app[c] -= 1
node = node.children[c]
return cnt - len(L)
if __name__ == '__main__':
N = int(readline())
L = [s.decode().rstrip() for s in read().split()]
print(solve(L))
``` | instruction | 0 | 45,168 | 18 | 90,336 |
Yes | output | 1 | 45,168 | 18 | 90,337 |
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