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Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays the sleuth with his friends. The rules of the game are as follows: those who play for the first time, that is Vasya is the sleuth, he should investigate a "crime" and find out what is happening. He can ask any questions whatsoever that can be answered with "Yes" or "No". All the rest agree beforehand to answer the questions like that: if the question’s last letter is a vowel, they answer "Yes" and if the last letter is a consonant, they answer "No". Of course, the sleuth knows nothing about it and his task is to understand that.
Unfortunately, Vasya is not very smart. After 5 hours of endless stupid questions everybody except Vasya got bored. That’s why Vasya’s friends ask you to write a program that would give answers instead of them.
The English alphabet vowels are: A, E, I, O, U, Y
The English alphabet consonants are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Z
Input
The single line contains a question represented by a non-empty line consisting of large and small Latin letters, spaces and a question mark. The line length does not exceed 100. It is guaranteed that the question mark occurs exactly once in the line — as the last symbol and that the line contains at least one letter.
Output
Print answer for the question in a single line: YES if the answer is "Yes", NO if the answer is "No".
Remember that in the reply to the question the last letter, not the last character counts. I. e. the spaces and the question mark do not count as letters.
Examples
Input
Is it a melon?
Output
NO
Input
Is it an apple?
Output
YES
Input
Is it a banana ?
Output
YES
Input
Is it an apple and a banana simultaneouSLY?
Output
YES
|
instruction
| 0
| 107,343
| 6
| 214,686
|
Tags: implementation
Correct Solution:
```
def isvowel(c):
return c.lower() in ['a', 'e', 'i', 'o', 'u', 'y']
s = input()
for x in s[-2::-1]:
if x.isalpha():
if isvowel(x):
print('YES')
else:
print('NO')
break
```
|
output
| 1
| 107,343
| 6
| 214,687
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays the sleuth with his friends. The rules of the game are as follows: those who play for the first time, that is Vasya is the sleuth, he should investigate a "crime" and find out what is happening. He can ask any questions whatsoever that can be answered with "Yes" or "No". All the rest agree beforehand to answer the questions like that: if the question’s last letter is a vowel, they answer "Yes" and if the last letter is a consonant, they answer "No". Of course, the sleuth knows nothing about it and his task is to understand that.
Unfortunately, Vasya is not very smart. After 5 hours of endless stupid questions everybody except Vasya got bored. That’s why Vasya’s friends ask you to write a program that would give answers instead of them.
The English alphabet vowels are: A, E, I, O, U, Y
The English alphabet consonants are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Z
Input
The single line contains a question represented by a non-empty line consisting of large and small Latin letters, spaces and a question mark. The line length does not exceed 100. It is guaranteed that the question mark occurs exactly once in the line — as the last symbol and that the line contains at least one letter.
Output
Print answer for the question in a single line: YES if the answer is "Yes", NO if the answer is "No".
Remember that in the reply to the question the last letter, not the last character counts. I. e. the spaces and the question mark do not count as letters.
Examples
Input
Is it a melon?
Output
NO
Input
Is it an apple?
Output
YES
Input
Is it a banana ?
Output
YES
Input
Is it an apple and a banana simultaneouSLY?
Output
YES
|
instruction
| 0
| 107,344
| 6
| 214,688
|
Tags: implementation
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Tue Mar 30 19:32:51 2021
@author: nehas
"""
s=input()
s=s.replace(" ","")
l=['a','e','i','o','u','y','A','I','E','O','U','Y']
s1=""
for char in s:
if(char.isalpha()==True):
s1=s1+char
if s1[len(s1)-1] in l:
print("YES")
else:
print("NO")
```
|
output
| 1
| 107,344
| 6
| 214,689
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays the sleuth with his friends. The rules of the game are as follows: those who play for the first time, that is Vasya is the sleuth, he should investigate a "crime" and find out what is happening. He can ask any questions whatsoever that can be answered with "Yes" or "No". All the rest agree beforehand to answer the questions like that: if the question’s last letter is a vowel, they answer "Yes" and if the last letter is a consonant, they answer "No". Of course, the sleuth knows nothing about it and his task is to understand that.
Unfortunately, Vasya is not very smart. After 5 hours of endless stupid questions everybody except Vasya got bored. That’s why Vasya’s friends ask you to write a program that would give answers instead of them.
The English alphabet vowels are: A, E, I, O, U, Y
The English alphabet consonants are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Z
Input
The single line contains a question represented by a non-empty line consisting of large and small Latin letters, spaces and a question mark. The line length does not exceed 100. It is guaranteed that the question mark occurs exactly once in the line — as the last symbol and that the line contains at least one letter.
Output
Print answer for the question in a single line: YES if the answer is "Yes", NO if the answer is "No".
Remember that in the reply to the question the last letter, not the last character counts. I. e. the spaces and the question mark do not count as letters.
Examples
Input
Is it a melon?
Output
NO
Input
Is it an apple?
Output
YES
Input
Is it a banana ?
Output
YES
Input
Is it an apple and a banana simultaneouSLY?
Output
YES
|
instruction
| 0
| 107,345
| 6
| 214,690
|
Tags: implementation
Correct Solution:
```
s=input()
s=s.replace(" ","")
if(s[-2].lower()=="a" or s[-2].lower()=="e" or s[-2].lower()=="i" or s[-2].lower()=="o" or s[-2].lower()=="u" or s[-2].lower()=="y"):
print("YES")
else:
print("NO")
```
|
output
| 1
| 107,345
| 6
| 214,691
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays the sleuth with his friends. The rules of the game are as follows: those who play for the first time, that is Vasya is the sleuth, he should investigate a "crime" and find out what is happening. He can ask any questions whatsoever that can be answered with "Yes" or "No". All the rest agree beforehand to answer the questions like that: if the question’s last letter is a vowel, they answer "Yes" and if the last letter is a consonant, they answer "No". Of course, the sleuth knows nothing about it and his task is to understand that.
Unfortunately, Vasya is not very smart. After 5 hours of endless stupid questions everybody except Vasya got bored. That’s why Vasya’s friends ask you to write a program that would give answers instead of them.
The English alphabet vowels are: A, E, I, O, U, Y
The English alphabet consonants are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Z
Input
The single line contains a question represented by a non-empty line consisting of large and small Latin letters, spaces and a question mark. The line length does not exceed 100. It is guaranteed that the question mark occurs exactly once in the line — as the last symbol and that the line contains at least one letter.
Output
Print answer for the question in a single line: YES if the answer is "Yes", NO if the answer is "No".
Remember that in the reply to the question the last letter, not the last character counts. I. e. the spaces and the question mark do not count as letters.
Examples
Input
Is it a melon?
Output
NO
Input
Is it an apple?
Output
YES
Input
Is it a banana ?
Output
YES
Input
Is it an apple and a banana simultaneouSLY?
Output
YES
|
instruction
| 0
| 107,346
| 6
| 214,692
|
Tags: implementation
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Tue Mar 10 22:10:44 2020
@author: DELL
"""
o=input()
v='aeiouyAEIOUY'
for i in range(len(o)-1,-1,-1):
k=o[i]
if k.isalpha():
if k in v:
print('YES')
else:
print('NO')
break
```
|
output
| 1
| 107,346
| 6
| 214,693
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays the sleuth with his friends. The rules of the game are as follows: those who play for the first time, that is Vasya is the sleuth, he should investigate a "crime" and find out what is happening. He can ask any questions whatsoever that can be answered with "Yes" or "No". All the rest agree beforehand to answer the questions like that: if the question’s last letter is a vowel, they answer "Yes" and if the last letter is a consonant, they answer "No". Of course, the sleuth knows nothing about it and his task is to understand that.
Unfortunately, Vasya is not very smart. After 5 hours of endless stupid questions everybody except Vasya got bored. That’s why Vasya’s friends ask you to write a program that would give answers instead of them.
The English alphabet vowels are: A, E, I, O, U, Y
The English alphabet consonants are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Z
Input
The single line contains a question represented by a non-empty line consisting of large and small Latin letters, spaces and a question mark. The line length does not exceed 100. It is guaranteed that the question mark occurs exactly once in the line — as the last symbol and that the line contains at least one letter.
Output
Print answer for the question in a single line: YES if the answer is "Yes", NO if the answer is "No".
Remember that in the reply to the question the last letter, not the last character counts. I. e. the spaces and the question mark do not count as letters.
Examples
Input
Is it a melon?
Output
NO
Input
Is it an apple?
Output
YES
Input
Is it a banana ?
Output
YES
Input
Is it an apple and a banana simultaneouSLY?
Output
YES
|
instruction
| 0
| 107,347
| 6
| 214,694
|
Tags: implementation
Correct Solution:
```
s=list(map(str,input()))
for i in range(len(s)):
if s[-2]==' ':
s.remove(s[-2])
k=s[-2]
if k=='A' or k=='E' or k=='I' or k=='O' or k=='U' or k=='Y' or k=='a' or k=='e' or k=='i' or k=='o' or k=='u' or k=='y':
print('YES')
else:
print('NO')
```
|
output
| 1
| 107,347
| 6
| 214,695
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays the sleuth with his friends. The rules of the game are as follows: those who play for the first time, that is Vasya is the sleuth, he should investigate a "crime" and find out what is happening. He can ask any questions whatsoever that can be answered with "Yes" or "No". All the rest agree beforehand to answer the questions like that: if the question’s last letter is a vowel, they answer "Yes" and if the last letter is a consonant, they answer "No". Of course, the sleuth knows nothing about it and his task is to understand that.
Unfortunately, Vasya is not very smart. After 5 hours of endless stupid questions everybody except Vasya got bored. That’s why Vasya’s friends ask you to write a program that would give answers instead of them.
The English alphabet vowels are: A, E, I, O, U, Y
The English alphabet consonants are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Z
Input
The single line contains a question represented by a non-empty line consisting of large and small Latin letters, spaces and a question mark. The line length does not exceed 100. It is guaranteed that the question mark occurs exactly once in the line — as the last symbol and that the line contains at least one letter.
Output
Print answer for the question in a single line: YES if the answer is "Yes", NO if the answer is "No".
Remember that in the reply to the question the last letter, not the last character counts. I. e. the spaces and the question mark do not count as letters.
Examples
Input
Is it a melon?
Output
NO
Input
Is it an apple?
Output
YES
Input
Is it a banana ?
Output
YES
Input
Is it an apple and a banana simultaneouSLY?
Output
YES
|
instruction
| 0
| 107,348
| 6
| 214,696
|
Tags: implementation
Correct Solution:
```
s = input().replace(' ', '')
vowels, ln = 'aAeEiIoOuUyY', len(s)
print('YES' if s[ln-2] in vowels else 'NO')
```
|
output
| 1
| 107,348
| 6
| 214,697
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays the sleuth with his friends. The rules of the game are as follows: those who play for the first time, that is Vasya is the sleuth, he should investigate a "crime" and find out what is happening. He can ask any questions whatsoever that can be answered with "Yes" or "No". All the rest agree beforehand to answer the questions like that: if the question’s last letter is a vowel, they answer "Yes" and if the last letter is a consonant, they answer "No". Of course, the sleuth knows nothing about it and his task is to understand that.
Unfortunately, Vasya is not very smart. After 5 hours of endless stupid questions everybody except Vasya got bored. That’s why Vasya’s friends ask you to write a program that would give answers instead of them.
The English alphabet vowels are: A, E, I, O, U, Y
The English alphabet consonants are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Z
Input
The single line contains a question represented by a non-empty line consisting of large and small Latin letters, spaces and a question mark. The line length does not exceed 100. It is guaranteed that the question mark occurs exactly once in the line — as the last symbol and that the line contains at least one letter.
Output
Print answer for the question in a single line: YES if the answer is "Yes", NO if the answer is "No".
Remember that in the reply to the question the last letter, not the last character counts. I. e. the spaces and the question mark do not count as letters.
Examples
Input
Is it a melon?
Output
NO
Input
Is it an apple?
Output
YES
Input
Is it a banana ?
Output
YES
Input
Is it an apple and a banana simultaneouSLY?
Output
YES
|
instruction
| 0
| 107,349
| 6
| 214,698
|
Tags: implementation
Correct Solution:
```
n = input().lower().split("?")
a = list(n[0].strip())[-1]
if(a == "a" or a == "e" or a == "o" or a == "i" or a == "u" or a == "y"):
print("YES")
else:
print("NO")
```
|
output
| 1
| 107,349
| 6
| 214,699
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays the sleuth with his friends. The rules of the game are as follows: those who play for the first time, that is Vasya is the sleuth, he should investigate a "crime" and find out what is happening. He can ask any questions whatsoever that can be answered with "Yes" or "No". All the rest agree beforehand to answer the questions like that: if the question’s last letter is a vowel, they answer "Yes" and if the last letter is a consonant, they answer "No". Of course, the sleuth knows nothing about it and his task is to understand that.
Unfortunately, Vasya is not very smart. After 5 hours of endless stupid questions everybody except Vasya got bored. That’s why Vasya’s friends ask you to write a program that would give answers instead of them.
The English alphabet vowels are: A, E, I, O, U, Y
The English alphabet consonants are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Z
Input
The single line contains a question represented by a non-empty line consisting of large and small Latin letters, spaces and a question mark. The line length does not exceed 100. It is guaranteed that the question mark occurs exactly once in the line — as the last symbol and that the line contains at least one letter.
Output
Print answer for the question in a single line: YES if the answer is "Yes", NO if the answer is "No".
Remember that in the reply to the question the last letter, not the last character counts. I. e. the spaces and the question mark do not count as letters.
Examples
Input
Is it a melon?
Output
NO
Input
Is it an apple?
Output
YES
Input
Is it a banana ?
Output
YES
Input
Is it an apple and a banana simultaneouSLY?
Output
YES
|
instruction
| 0
| 107,350
| 6
| 214,700
|
Tags: implementation
Correct Solution:
```
s=input()
s=s[::-1]
s=s[1:]
s=s.strip()
#print(s)
v=['a','e','i','o','u','y','A','E','I','O','U','Y']
if(s[0] in v):
print("YES")
else:
print("NO")
```
|
output
| 1
| 107,350
| 6
| 214,701
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A Martian boy is named s — he has got this name quite recently from his parents for his coming of age birthday. Now he enjoys looking for his name everywhere. If he sees that he can obtain his name from some string by removing zero or more letters (at that, the remaining letters remain in the same order), he gets happy. For example, if s=«aba», then strings «baobab», «aabbaa», «helloabahello» make him very happy and strings «aab», «baaa» and «helloabhello» do not.
However rather than being happy once, he loves twice as much being happy twice! So, when he got string t as a present, he wanted to cut it in two parts (the left part and the right part) so that each part made him happy.
Help s determine the number of distinct ways to cut the given string t into two parts in the required manner.
Input
The first line contains string s, consisting of lowercase English letters. The length of string s is from 1 to 1000 letters.
The second line contains string t, that also consists of lowercase English letters. The length of string t is from 1 to 106 letters.
Output
Print the sought number of ways to cut string t in two so that each part made s happy.
Examples
Input
aba
baobababbah
Output
2
Input
mars
sunvenusearthmarsjupitersaturnuranusneptune
Output
0
|
instruction
| 0
| 107,359
| 6
| 214,718
|
Tags: *special, greedy
Correct Solution:
```
from collections import Counter
import string
import math
import sys
from fractions import Fraction
def array_int():
return [int(i) for i in sys.stdin.readline().split()]
def vary(arrber_of_variables):
if arrber_of_variables==1:
return int(sys.stdin.readline())
if arrber_of_variables>=2:
return map(int,sys.stdin.readline().split())
def makedict(var):
return dict(Counter(var))
# i am noob wanted to be better and trying hard for that
def printDivisors(n):
divisors=[]
# Note that this loop runs till square root
i = 1
while i <= math.sqrt(n):
if (n % i == 0) :
# If divisors are equal, print only one
if (n//i == i) :
divisors.append(i)
else :
# Otherwise print both
divisors.extend((i,n//i))
i = i + 1
return divisors
def countTotalBits(num):
# convert number into it's binary and
# remove first two characters 0b.
binary = bin(num)[2:]
return(len(binary))
def isPrime(n) :
# Corner cases
if (n <= 1) :
return False
if (n <= 3) :
return True
# This is checked so that we can skip
# middle five numbers in below loop
if (n % 2 == 0 or n % 3 == 0) :
return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
# print(math.gcd(3,2))
""" def dfs(node,val):
global tree,visited
visited[node]=1
ans[node]=val
val^=1
for i in tree[node]:
if visited[i]==-1:
dfs(i,val) """
s=input()
t=input()
m=len(s)
n=len(t)
j=0
flag=0
for i in range(n):
if j>=m:
break
if t[i]==s[j]:
j+=1
continue
if j!=m:
print(0)
else:
# print(i)
p=m-1
flag=0
for k in range(n-1,i,-1):
if t[k]==s[p]:
if p==0:
flag=1
break
p-=1
# print(k)
# print(flag)
if flag!=1:
print(0)
else:
print(k-(i-1))
```
|
output
| 1
| 107,359
| 6
| 214,719
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A Martian boy is named s — he has got this name quite recently from his parents for his coming of age birthday. Now he enjoys looking for his name everywhere. If he sees that he can obtain his name from some string by removing zero or more letters (at that, the remaining letters remain in the same order), he gets happy. For example, if s=«aba», then strings «baobab», «aabbaa», «helloabahello» make him very happy and strings «aab», «baaa» and «helloabhello» do not.
However rather than being happy once, he loves twice as much being happy twice! So, when he got string t as a present, he wanted to cut it in two parts (the left part and the right part) so that each part made him happy.
Help s determine the number of distinct ways to cut the given string t into two parts in the required manner.
Input
The first line contains string s, consisting of lowercase English letters. The length of string s is from 1 to 1000 letters.
The second line contains string t, that also consists of lowercase English letters. The length of string t is from 1 to 106 letters.
Output
Print the sought number of ways to cut string t in two so that each part made s happy.
Examples
Input
aba
baobababbah
Output
2
Input
mars
sunvenusearthmarsjupitersaturnuranusneptune
Output
0
|
instruction
| 0
| 107,360
| 6
| 214,720
|
Tags: *special, greedy
Correct Solution:
```
s = input()
t = input()
m = len(s)
n = len(t)
first = 0
last = n
j = 0
flag = 0
i = 0
while i < n and j < m:
if t[i] == s[j]:
j += 1
i += 1
first = i - 1
if j == m:
flag = 1
i = n - 1
j = m - 1
while i >= 0 and j >= 0:
if t[i] == s[j]:
j -= 1
i -= 1
last = i + 1
# print(first,last)
if flag == 0 or last <= first:
ans = 0
else:
ans = last - first
print(ans)
```
|
output
| 1
| 107,360
| 6
| 214,721
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A Martian boy is named s — he has got this name quite recently from his parents for his coming of age birthday. Now he enjoys looking for his name everywhere. If he sees that he can obtain his name from some string by removing zero or more letters (at that, the remaining letters remain in the same order), he gets happy. For example, if s=«aba», then strings «baobab», «aabbaa», «helloabahello» make him very happy and strings «aab», «baaa» and «helloabhello» do not.
However rather than being happy once, he loves twice as much being happy twice! So, when he got string t as a present, he wanted to cut it in two parts (the left part and the right part) so that each part made him happy.
Help s determine the number of distinct ways to cut the given string t into two parts in the required manner.
Input
The first line contains string s, consisting of lowercase English letters. The length of string s is from 1 to 1000 letters.
The second line contains string t, that also consists of lowercase English letters. The length of string t is from 1 to 106 letters.
Output
Print the sought number of ways to cut string t in two so that each part made s happy.
Examples
Input
aba
baobababbah
Output
2
Input
mars
sunvenusearthmarsjupitersaturnuranusneptune
Output
0
|
instruction
| 0
| 107,361
| 6
| 214,722
|
Tags: *special, greedy
Correct Solution:
```
#! /usr/bin/python3
def solve():
s = input()
t = input()
ls = len(s)
if s[ls - 1] == '\n':
ls = ls - 1
lt = len(t)
if t[lt - 1] == '\n':
lt = lt - 1
#print("ls = ",ls," lt = ", lt)
i = 0
j = 0
pos = -1
while 1:
if s[i] == t[j]:
i = i + 1
j = j + 1
else:
j = j + 1
if i == ls:
pos = j
if i == ls or j == lt:
break
if pos == -1:
print(0)
return
i = ls - 1
j = lt - 1
reverse = -1
while 1:
if s[i] == t[j]:
i = i - 1
j = j - 1
else:
j = j - 1
if i == -1:
reverse = j
if i == -1 or j == -1:
break
# print("pos = ",pos," reverse = ", reverse)
reverse = reverse + 1
pos = pos - 1
if reverse == -1 or reverse < pos:
print(0)
else:
print(reverse - pos)
solve()
```
|
output
| 1
| 107,361
| 6
| 214,723
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A Martian boy is named s — he has got this name quite recently from his parents for his coming of age birthday. Now he enjoys looking for his name everywhere. If he sees that he can obtain his name from some string by removing zero or more letters (at that, the remaining letters remain in the same order), he gets happy. For example, if s=«aba», then strings «baobab», «aabbaa», «helloabahello» make him very happy and strings «aab», «baaa» and «helloabhello» do not.
However rather than being happy once, he loves twice as much being happy twice! So, when he got string t as a present, he wanted to cut it in two parts (the left part and the right part) so that each part made him happy.
Help s determine the number of distinct ways to cut the given string t into two parts in the required manner.
Input
The first line contains string s, consisting of lowercase English letters. The length of string s is from 1 to 1000 letters.
The second line contains string t, that also consists of lowercase English letters. The length of string t is from 1 to 106 letters.
Output
Print the sought number of ways to cut string t in two so that each part made s happy.
Examples
Input
aba
baobababbah
Output
2
Input
mars
sunvenusearthmarsjupitersaturnuranusneptune
Output
0
|
instruction
| 0
| 107,362
| 6
| 214,724
|
Tags: *special, greedy
Correct Solution:
```
import sys
count = 0
matrix = []
for line in sys.stdin:
if count == 0:
pattern = line.strip()
count += 1
else:
s = line.strip()
dp1 = [0 for i in range(len(s))]
dp2 = [0 for i in range(len(s))]
i = 0
j = 0
while i < len(pattern) and j < len(s):
if pattern[i] == s[j]:
i += 1
j += 1
left = len(s) + 1
if i == len(pattern):
left = j - 1
i = len(pattern) - 1
j = len(s) - 1
while i >= 0 and j >= 0:
if pattern[i] == s[j]:
i -= 1
j -= 1
right = -1
if i < 0:
right = j + 1
res = right - left if right -left > 0 else 0
print(res)
```
|
output
| 1
| 107,362
| 6
| 214,725
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A Martian boy is named s — he has got this name quite recently from his parents for his coming of age birthday. Now he enjoys looking for his name everywhere. If he sees that he can obtain his name from some string by removing zero or more letters (at that, the remaining letters remain in the same order), he gets happy. For example, if s=«aba», then strings «baobab», «aabbaa», «helloabahello» make him very happy and strings «aab», «baaa» and «helloabhello» do not.
However rather than being happy once, he loves twice as much being happy twice! So, when he got string t as a present, he wanted to cut it in two parts (the left part and the right part) so that each part made him happy.
Help s determine the number of distinct ways to cut the given string t into two parts in the required manner.
Input
The first line contains string s, consisting of lowercase English letters. The length of string s is from 1 to 1000 letters.
The second line contains string t, that also consists of lowercase English letters. The length of string t is from 1 to 106 letters.
Output
Print the sought number of ways to cut string t in two so that each part made s happy.
Examples
Input
aba
baobababbah
Output
2
Input
mars
sunvenusearthmarsjupitersaturnuranusneptune
Output
0
|
instruction
| 0
| 107,363
| 6
| 214,726
|
Tags: *special, greedy
Correct Solution:
```
s1 = input().rstrip()
s2 = input().rstrip()
len1 = len(s1)
len2 = len(s2)
l = 1
r = 0
got = 0
for i in range(0,len2):
if s2[ i ] == s1[ got ] :
got = got + 1
if got == len1 :
l = i
break
got = len1 - 1
for j in range(0,len2):
i = len2 - j - 1
if s2[ i ] == s1[ got ] :
got = got - 1
if got < 0 :
r = i - 1
break
if l <= r :
print( r - l + 1 )
else :
print( "0" )
```
|
output
| 1
| 107,363
| 6
| 214,727
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A Martian boy is named s — he has got this name quite recently from his parents for his coming of age birthday. Now he enjoys looking for his name everywhere. If he sees that he can obtain his name from some string by removing zero or more letters (at that, the remaining letters remain in the same order), he gets happy. For example, if s=«aba», then strings «baobab», «aabbaa», «helloabahello» make him very happy and strings «aab», «baaa» and «helloabhello» do not.
However rather than being happy once, he loves twice as much being happy twice! So, when he got string t as a present, he wanted to cut it in two parts (the left part and the right part) so that each part made him happy.
Help s determine the number of distinct ways to cut the given string t into two parts in the required manner.
Input
The first line contains string s, consisting of lowercase English letters. The length of string s is from 1 to 1000 letters.
The second line contains string t, that also consists of lowercase English letters. The length of string t is from 1 to 106 letters.
Output
Print the sought number of ways to cut string t in two so that each part made s happy.
Examples
Input
aba
baobababbah
Output
2
Input
mars
sunvenusearthmarsjupitersaturnuranusneptune
Output
0
|
instruction
| 0
| 107,364
| 6
| 214,728
|
Tags: *special, greedy
Correct Solution:
```
import sys
# def solution(a, b, c, d, k):
def solution(s, l):
char_pos = dict()
charset = set(s)
for i in range(len(l)):
c = l[i]
if c in charset:
if c not in char_pos:
char_pos[c] = [i]
else:
char_pos[c].append(i)
cursors = {c: 0 for c in charset}
end = -1
for c in s:
if c not in char_pos:
return 0
while cursors[c] < len(char_pos[c]) and char_pos[c][cursors[c]] <= end:
cursors[c] += 1
if cursors[c] >= len(char_pos[c]):
return 0
end = char_pos[c][cursors[c]]
cursors[c] += 1
cursors = {c: len(char_pos[c]) - 1 for c in charset}
begin = len(l)
for c in s[::-1]:
while cursors[c] >= 0 and char_pos[c][cursors[c]] >= begin:
cursors[c] -= 1
if cursors[c] < 0:
return 0
begin = char_pos[c][cursors[c]]
cursors[c] -= 1
if begin - end <= 0:
return 0
return begin - end
# def solution(s, l):
# if len(s) == 0 or len(l) == 0:
# print(0)
# return
# intervals = []
# cursor_s = 0
# cursor_l = 0
# begin = 0
# while cursor_l < len(l):
# if s[cursor_s] == l[cursor_l]:
# if cursor_s == 0:
# begin = cursor_l
# cursor_s += 1
# if cursor_s == len(s):
# intervals.append((begin, cursor_l))
# cursor_s = 0
# cursor_l += 1
# # print(intervals)
# if len(intervals) < 2:
# print(0)
# return
# res = intervals[-1][0] - intervals[0][1]
# print(res)
s = sys.stdin.readline().strip()
l = sys.stdin.readline().strip()
print(solution(s, l))
# input_lst = []
# for i in range(m):
# a, b = sys.stdin.readline().strip().split()
# a = int(a)
# b = int(b)
# input_lst.append((a, b))
# print(get_B(input_lst))
# print('abc'[::-1])
```
|
output
| 1
| 107,364
| 6
| 214,729
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A Martian boy is named s — he has got this name quite recently from his parents for his coming of age birthday. Now he enjoys looking for his name everywhere. If he sees that he can obtain his name from some string by removing zero or more letters (at that, the remaining letters remain in the same order), he gets happy. For example, if s=«aba», then strings «baobab», «aabbaa», «helloabahello» make him very happy and strings «aab», «baaa» and «helloabhello» do not.
However rather than being happy once, he loves twice as much being happy twice! So, when he got string t as a present, he wanted to cut it in two parts (the left part and the right part) so that each part made him happy.
Help s determine the number of distinct ways to cut the given string t into two parts in the required manner.
Input
The first line contains string s, consisting of lowercase English letters. The length of string s is from 1 to 1000 letters.
The second line contains string t, that also consists of lowercase English letters. The length of string t is from 1 to 106 letters.
Output
Print the sought number of ways to cut string t in two so that each part made s happy.
Examples
Input
aba
baobababbah
Output
2
Input
mars
sunvenusearthmarsjupitersaturnuranusneptune
Output
0
|
instruction
| 0
| 107,365
| 6
| 214,730
|
Tags: *special, greedy
Correct Solution:
```
t = input()
s = input()
l_t = len(t)
l_s = len(s)
x = 0
f = l_s+1
for i in range(l_s):
if t[x] == s[i]: x += 1
if x == l_t:
f = i
break
c = -1
y = l_t-1
for i in range(l_s-1,-1,-1):
if t[y] == s[i]: y -= 1
if y == -1:
c = i
break
if c - f <= 0: print(0)
else: print(c - f)
```
|
output
| 1
| 107,365
| 6
| 214,731
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A Martian boy is named s — he has got this name quite recently from his parents for his coming of age birthday. Now he enjoys looking for his name everywhere. If he sees that he can obtain his name from some string by removing zero or more letters (at that, the remaining letters remain in the same order), he gets happy. For example, if s=«aba», then strings «baobab», «aabbaa», «helloabahello» make him very happy and strings «aab», «baaa» and «helloabhello» do not.
However rather than being happy once, he loves twice as much being happy twice! So, when he got string t as a present, he wanted to cut it in two parts (the left part and the right part) so that each part made him happy.
Help s determine the number of distinct ways to cut the given string t into two parts in the required manner.
Input
The first line contains string s, consisting of lowercase English letters. The length of string s is from 1 to 1000 letters.
The second line contains string t, that also consists of lowercase English letters. The length of string t is from 1 to 106 letters.
Output
Print the sought number of ways to cut string t in two so that each part made s happy.
Examples
Input
aba
baobababbah
Output
2
Input
mars
sunvenusearthmarsjupitersaturnuranusneptune
Output
0
|
instruction
| 0
| 107,366
| 6
| 214,732
|
Tags: *special, greedy
Correct Solution:
```
def search_podstroki(t,s):
c1 = 0
for i in range(len(t)):
if (t[i] == s[0]):
s = s[1::]
if (len(s) == 0):
c1 = -1
return i+1
if (c1 == 0):
return -1
s = input()
t = input()
e = 0
if (len(t) <= len(s)):
print('0')
else:
s1 = s
h = search_podstroki(t,s1)
s1 = s1[::-1]
t = t[::-1]
g = search_podstroki(t,s1)
if ((g == -1) or (h == -1)):
print('0')
else:
g = (len(t) - g) + 1
if (g < h):
print('0')
else:
print(g-h)
```
|
output
| 1
| 107,366
| 6
| 214,733
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
|
instruction
| 0
| 107,876
| 6
| 215,752
|
Tags: constructive algorithms, hashing, strings
Correct Solution:
```
def solve(s):
n = len(s)
for i in range(n):
s2 = s[i:] + s[:i]
# print(s2)
if s != s2 and s2[::-1] == s2:
return 1
for i in range( (n // 2) + 1, n):
if s[i] != s[0]:
return 2
# print(s[i])
return "Impossible"
s = input()
print(solve(s))
```
|
output
| 1
| 107,876
| 6
| 215,753
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
|
instruction
| 0
| 107,877
| 6
| 215,754
|
Tags: constructive algorithms, hashing, strings
Correct Solution:
```
import sys
s = input()
n = len(s)
for i in range(n):
t = s[i:] + s[:i]
if t != s and t == t[::-1]:
print(1)
sys.exit(0)
if s[:n//2] != s[n-n//2:]:
print(2)
sys.exit(0)
is4 = True
for i in range(n):
if not (n % 2 == 1 and i == n//2):
if s[i] != s[0]:
is4 = False
if is4 == False:
print(2)
else:
print("Impossible")
```
|
output
| 1
| 107,877
| 6
| 215,755
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
|
instruction
| 0
| 107,878
| 6
| 215,756
|
Tags: constructive algorithms, hashing, strings
Correct Solution:
```
def is_palindrome(S):
a = S[::-1]
return S == a
s = input()
if len(s) == 1:
print('Impossible')
exit(0)
hop = 0
if len(s) % 2:
if len(set(s[:len(s)//2])) == 1:
print('Impossible')
else:
print(2)
else:
if len(set(s)) == 1:
print('Impossible')
else:
if s[:len(s)//2] != s[len(s)//2:]:
print(1)
else:
fl = 2
while len(s)%2==0:
s = s[:len(s)//2]
if (len(s) % 2 == 0 and not is_palindrome(s[:len(s)//2])):
fl = 1
break
print(fl)
```
|
output
| 1
| 107,878
| 6
| 215,757
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
|
instruction
| 0
| 107,879
| 6
| 215,758
|
Tags: constructive algorithms, hashing, strings
Correct Solution:
```
st=input()
le=len(st)
if(le==1):
print('Impossible')
else:
flag=False
for i in range(1,le//2):
if(st[i]!=st[i-1]):
flag=True
break
if(not(flag)):
print('Impossible')
else:
flag=True
for i in range(le):
ne=st[i:]+st[:i]
if(ne==ne[::-1] and ne!=st):
flag=False
break
if(flag):
print(2)
else:
print(1)
```
|
output
| 1
| 107,879
| 6
| 215,759
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
|
instruction
| 0
| 107,880
| 6
| 215,760
|
Tags: constructive algorithms, hashing, strings
Correct Solution:
```
# cook your dish here
s=input()
lz=len(s)
found=0
ans=-1
for i in range(0,lz//2):
k=s[0:i+1]
if k!=k[::-1]:
found=1
ans=2
for i in range(0,lz-1):
k=s[i+1:lz]+s[0:i+1]
if k==k[::-1] and k!=s:
found=1
ans=1
if not found:
print("Impossible")
else:
print(ans)
```
|
output
| 1
| 107,880
| 6
| 215,761
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
|
instruction
| 0
| 107,881
| 6
| 215,762
|
Tags: constructive algorithms, hashing, strings
Correct Solution:
```
s = input()
if len(set(s[:len(s) // 2])) <= 1:
print("Impossible")
exit()
if len(s) % 2 == 0:
for i in range(1, len(s)):
n = s[i:] + s[:i]
if (n == n[::-1]) and (n != s):
print("1")
exit()
print("2")
```
|
output
| 1
| 107,881
| 6
| 215,763
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
|
instruction
| 0
| 107,882
| 6
| 215,764
|
Tags: constructive algorithms, hashing, strings
Correct Solution:
```
s=input()
hai=False
if(len(s)%2==0):
for i in range(1,len(s)):
s1=s[i:]+s[:i]
if(s1!=s and s1==s1[::-1]):
print(1)
exit()
cur=len(s)//2
cur2=cur
cur-=1
while(cur>0):
if(s[:cur+1]!=s[cur2:]):
print(2)
exit()
cur-=1
cur2+=1
print("Impossible")
else:
cur=len(s)//2
cur2=cur+1
cur-=1
while(cur>0):
if(s[:cur+1]!=s[cur2:]):
print(2)
exit()
cur-=1
cur2+=1
print("Impossible")
```
|
output
| 1
| 107,882
| 6
| 215,765
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
|
instruction
| 0
| 107,883
| 6
| 215,766
|
Tags: constructive algorithms, hashing, strings
Correct Solution:
```
s = input()
if len(set(s[:len(s) // 2])) <= 1:
print("Impossible");exit()
if len(s) % 2 == 0:
for i in range(1, len(s)):
n = s[i:] + s[:i]
if (n == n[::-1]) and (n != s):
print(1);exit()
print(2)
```
|
output
| 1
| 107,883
| 6
| 215,767
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
Submitted Solution:
```
# |
# _` | __ \ _` | __| _ \ __ \ _` | _` |
# ( | | | ( | ( ( | | | ( | ( |
# \__,_| _| _| \__,_| \___| \___/ _| _| \__,_| \__,_|
import sys
import math
def read_line():
return sys.stdin.readline()[:-1]
def read_int():
return int(sys.stdin.readline())
def read_int_line():
return [int(v) for v in sys.stdin.readline().split()]
def reverse(s):
return s[::-1]
def isPalindrome(s):
# Calling reverse function
rev = reverse(s)
# Checking if both string are equal or not
if (s == rev):
return True
return False
s = read_line()
n = len(s)
d = {}
for i in s:
if i not in d:
d[i] = 1
else:
d[i] +=1
f = True
if n&1==1 and len(list(d.keys())) <= 2 :
if len(list(d.keys())) == 1:
f = False
else:
if 1 in list(d.values()):
f = False
elif n&1 != 1 and len(list(d.keys())) == 1:
f = False
ans = 2
for i in range(1,n):
pre = s[:i]
post = s[i:]
if isPalindrome(post+pre):
if post+pre != s:
ans = 1
break
if f:
print(ans)
else:
print("Impossible")
```
|
instruction
| 0
| 107,884
| 6
| 215,768
|
Yes
|
output
| 1
| 107,884
| 6
| 215,769
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
Submitted Solution:
```
s = input()
n = len(s)
if n <= 2:
print('Impossible')
exit()
if len(s) % 2 == 1:
s = s[:n//2] + s[n//2 + 1:]
if len(set(s)) == 1:
print('Impossible')
exit()
print(2)
exit()
if len(set(s)) == 1:
print('Impossible')
exit()
for i in range(0, n):
cut = s[:i]
orig = s[i:]
pal = orig + cut
if pal == pal[::-1] and pal != s:
print(1)
exit()
print(2)
exit()
```
|
instruction
| 0
| 107,885
| 6
| 215,770
|
Yes
|
output
| 1
| 107,885
| 6
| 215,771
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
Submitted Solution:
```
from collections import defaultdict as dd
import math
def nn():
return int(input())
def li():
return list(input())
def mi():
return map(int, input().split())
def lm():
return list(map(int, input().split()))
def ispal(s):
for i in range(len(s)//2):
if not s[i]==s[-i-1]:
return False
return True
s=li()
if len(s)%2==0:
bad=1
for i in range(1,len(s)//2):
if not s[i]==s[i-1]:
bad=0
if bad==1:
print('Impossible')
else:
seg=s
while ispal(seg[0:len(seg)//2]):
if (len(seg)//2)%2==1:
print(2)
bad=1
break
seg=seg[0:len(seg)//2]
if bad==1:
pass
else:
print(1)
if len(s)%2==1:
bad=1
for i in range(1,len(s)//2):
if not s[i]==s[i-1]:
bad=0
if bad==1:
print('Impossible')
else:
print(2)
```
|
instruction
| 0
| 107,886
| 6
| 215,772
|
Yes
|
output
| 1
| 107,886
| 6
| 215,773
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
Submitted Solution:
```
import sys
a = input()
def check(x,y):
return True if x != y and x == x[::-1] else False
n = len(a)
for i in range(1,len(a)):
if check(a[i:]+a[:i],a):
print(1)
sys.exit(0)
for i in range(1, len(a)//2 + (len(a)%2)):
t = a[-i:] + a[i:-i] + a[:i]
if check(t,a):
print("2")
sys.exit(0)
print("Impossible")
```
|
instruction
| 0
| 107,887
| 6
| 215,774
|
Yes
|
output
| 1
| 107,887
| 6
| 215,775
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
Submitted Solution:
```
def isequal(s):
a=True
i=0
while(a and i<(len(s)-1)) :
if(s[i]==s[i+1]):
i=i+1
else :
a=False
return a
s=str(input())
if (len(s) == 1 or isequal(s)) :
print("Impossible")
elif (len(s)%2==0) :
print(1)
else :
print (2)
```
|
instruction
| 0
| 107,888
| 6
| 215,776
|
No
|
output
| 1
| 107,888
| 6
| 215,777
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
Submitted Solution:
```
def checkPali(s):
for i in range(len(s)):
if s[i] != s[len(s) - i - 1]:
return False
return True
def check(s):
for i in range(len(s)):
s2 = s[i : ] + s[ : i]
if checkPali(s2) and s != s2:
return i
def checkOdd(s):
for i in range(1, len(s) // 2):
if s[i] != s[i - 1]:
return False
return True
s = input()
flag = True
if len(s) == 1:
flag = False
elif len(s) % 2 != 0 and checkOdd(s):
flag = True
else:
for i in range(1, len(s)):
if s[i] != s[i - 1]:
flag = False
if flag:
print("Impossible")
elif check(s):
print(1)
else:
print(2)
```
|
instruction
| 0
| 107,889
| 6
| 215,778
|
No
|
output
| 1
| 107,889
| 6
| 215,779
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
Submitted Solution:
```
def isequal(s):
a=True
i=0
while(a and i<(len(s)-1)) :
if(s[i]==s[i+1]):
i=i+1
else :
a=False
return a
s=str(input())
if (len(s) == 1 or isequal(s)) :
print("Impossible")
elif (len(s)%2==0) :
if (s[0:(len(s)//2)]==s[(len(s)//2):len(s)]) :
print(2)
else:
print(1)
else :
print (2)
```
|
instruction
| 0
| 107,890
| 6
| 215,780
|
No
|
output
| 1
| 107,890
| 6
| 215,781
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
Submitted Solution:
```
def solve(s):
if len(s)%2:
print(s[0])
s = s[:len(s)//2]+s[0]+s[len(s)//2+1:]
# print(s)
if len(set(list(s))) == 1:
return 'Impossible'
if len(s)%2:
return 2
n = len(s)
mid = n//2
if s[:mid] != s[mid:]:
return 1
if n%4:
return 2
if s[:mid//2] != s[mid//2:mid]:
return 1
return 2
s = input()
print(solve(s))
# print(solve('nolon'))
# print(solve('otto'))
# print(solve('qqqq'))
# print(solve('kinnikkinnik'))
```
|
instruction
| 0
| 107,891
| 6
| 215,782
|
No
|
output
| 1
| 107,891
| 6
| 215,783
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's analyze a program written on some strange programming language. The variables in this language have names consisting of 1 to 4 characters, and each character is a lowercase or an uppercase Latin letter, or a digit. There is an extra constraint that the first character should not be a digit.
There are four types of operations in the program, each denoted by one of the characters: $, ^, # or &.
Each line of the program has one of the following formats:
* <lvalue>=<rvalue>, where <lvalue> and <rvalue> are valid variable names;
* <lvalue>=<arg1><op><arg2>, where <lvalue>, <arg1> and <arg2> are valid variable names, and <op> is an operation character.
The program is executed line-by-line, and the result of execution is stored in a variable having the name res. If res is never assigned in the program, then the result will be equal to the value of res before running the program.
Two programs are called equivalent if no matter which operations do characters $, ^, # and & denote (but, obviously, performing the same operation on the same arguments gives the same result) and which values do variables have before execution of program, the value of res after running the first program is equal to the value of res after running the second program (the programs are executed independently).
You are given a program consisting of n lines. Your task is to write a program consisting of minimum possible number of lines that is equivalent to the program you are given.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of lines in the program.
Then n lines follow — the program itself. Each line corresponds to the format described in the statement and has no extra whitespaces.
Output
In the first line print k — the minimum number of lines in the equivalent program.
Then print k lines without any whitespaces — an equivalent program having exactly k lines, in the same format it is described in the statement.
Examples
Input
4
c=aa#bb
d12=c
res=c^d12
tmp=aa$c
Output
2
aaaa=aa#bb
res=aaaa^aaaa
Input
2
max=aaaa$bbbb
min=bbbb^aaaa
Output
0
|
instruction
| 0
| 107,892
| 6
| 215,784
|
Tags: graphs, greedy, hashing, implementation
Correct Solution:
```
#!/usr/bin/env python3
import sys
sys.setrecursionlimit(65536)
vars, reserved = {}, set()
rules, irules, R = [], {}, 0
def wwj():
n__ = 0
def dak():
nonlocal n__
while 'a' + str(n__) in reserved:
n__ += 1
n__ += 1
return 'a' + str(n__ - 1)
return dak
next_var = wwj()
L = int(input())
for i in range(L):
k, v = input().split('=')
if v.isidentifier():
if v in vars:
vars[k] = vars[v]
else:
vars[k] = v
reserved.add(v)
else:
j = [i for i in range(len(v)) if not v[i].isalnum()][0]
v1, op, v2 = vars.get(v[:j], v[:j]), v[j], vars.get(v[j + 1:], v[j + 1:])
r = (op, v1, v2)
if r not in irules:
rules.append(r)
irules[r] = R
R += 1
vars[k] = irules[r]
if vars.get('res', 'res') == 'res':
print(0)
exit()
resolve = {}
output = []
def interprete(v, name = None):
if isinstance(v, str):
return v
if v in resolve:
return resolve[v]
op, v1, v2 = rules[v]
u1, u2 = interprete(v1), interprete(v2)
if name is None:
name = next_var()
resolve[v] = name
output.append('{}={}{}{}'.format(name, u1, op, u2))
return name
R = interprete(vars['res'], 'res')
if R != 'res':
output.append('res={}'.format(R))
print(len(output), '\n'.join(output), sep = '\n')
```
|
output
| 1
| 107,892
| 6
| 215,785
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's analyze a program written on some strange programming language. The variables in this language have names consisting of 1 to 4 characters, and each character is a lowercase or an uppercase Latin letter, or a digit. There is an extra constraint that the first character should not be a digit.
There are four types of operations in the program, each denoted by one of the characters: $, ^, # or &.
Each line of the program has one of the following formats:
* <lvalue>=<rvalue>, where <lvalue> and <rvalue> are valid variable names;
* <lvalue>=<arg1><op><arg2>, where <lvalue>, <arg1> and <arg2> are valid variable names, and <op> is an operation character.
The program is executed line-by-line, and the result of execution is stored in a variable having the name res. If res is never assigned in the program, then the result will be equal to the value of res before running the program.
Two programs are called equivalent if no matter which operations do characters $, ^, # and & denote (but, obviously, performing the same operation on the same arguments gives the same result) and which values do variables have before execution of program, the value of res after running the first program is equal to the value of res after running the second program (the programs are executed independently).
You are given a program consisting of n lines. Your task is to write a program consisting of minimum possible number of lines that is equivalent to the program you are given.
Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of lines in the program.
Then n lines follow — the program itself. Each line corresponds to the format described in the statement and has no extra whitespaces.
Output
In the first line print k — the minimum number of lines in the equivalent program.
Then print k lines without any whitespaces — an equivalent program having exactly k lines, in the same format it is described in the statement.
Examples
Input
4
c=aa#bb
d12=c
res=c^d12
tmp=aa$c
Output
2
aaaa=aa#bb
res=aaaa^aaaa
Input
2
max=aaaa$bbbb
min=bbbb^aaaa
Output
0
|
instruction
| 0
| 107,893
| 6
| 215,786
|
Tags: graphs, greedy, hashing, implementation
Correct Solution:
```
from random import seed, randint
import sys
sys.setrecursionlimit(10000)
opr = ['#', '^', '&', '$']
namespace = { "res" : (False, "res") }
rules = dict()
lookup = dict()
cnt = -1
def get_tag(var):
if var in namespace:
return namespace[var][1]
else:
return var
N = int(input())
for _ in range(N):
lval, rval = input().split('=')
for c in opr:
if c in rval:
arg1, arg2 = map(get_tag, rval.split(c))
rule = (arg1, arg2, c)
if rule in rules:
namespace[lval] = (True, rules[rule])
else:
cnt += 1
namespace[lval] = (True, cnt)
rules[rule] = cnt
lookup[cnt] = rule
break
else:
if rval in namespace:
namespace[lval] = namespace[rval]
else:
namespace[lval] = (False, rval)
if namespace["res"] == (False, "res"):
print("0")
exit()
program = []
myvars = dict()
def reserve():
return ''.join(chr(randint(0, 25) + ord('a')) for _ in range(4))
def implement(rule, final):
if type(rule) == str:
return rule
elif rule in myvars:
return myvars[rule]
else:
if final:
name = "res"
else:
name = reserve()
myvars[rule] = name
arg1, arg2, op = lookup[rule]
var1, var2 = implement(arg1, False), implement(arg2, False)
program.append(name + "=" + var1 + op + var2)
return name
seed(123)
if namespace["res"][0]:
implement(namespace["res"][1], True)
else:
program.append("res=" + namespace["res"][1])
print(len(program))
print("\n".join(program))
#print(namespace)
#print(rules)
```
|
output
| 1
| 107,893
| 6
| 215,787
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya trains to compose crossword puzzles. He can only compose crosswords of a very simplе type so far. All of them consist of exactly six words; the words can be read only from top to bottom vertically and from the left to the right horizontally. The words are arranged in the form of a rectangular "eight" or infinity sign, not necessarily symmetrical.
The top-left corner of the crossword coincides with the top-left corner of the rectangle. The same thing is correct for the right-bottom corners. The crossword can't degrade, i.e. it always has exactly four blank areas, two of which are surrounded by letters. Look into the output for the samples for clarification.
Help Vasya — compose a crossword of the described type using the given six words. It is allowed to use the words in any order.
Input
Six lines contain the given words. Every word consists of no more than 30 and no less than 3 uppercase Latin letters.
Output
If it is impossible to solve the problem, print Impossible. Otherwise, print the sought crossword. All the empty squares should be marked as dots.
If there can be several solutions to that problem, print the lexicographically minimum one. I.e. the solution where the first line is less than the first line of other solutions should be printed. If the two lines are equal, compare the second lines and so on. The lexicographical comparison of lines is realized by the < operator in the modern programming languages.
Examples
Input
NOD
BAA
YARD
AIRWAY
NEWTON
BURN
Output
BAA...
U.I...
R.R...
NEWTON
..A..O
..YARD
Input
AAA
AAA
AAAAA
AAA
AAA
AAAAA
Output
AAA..
A.A..
AAAAA
..A.A
..AAA
Input
PTC
JYNYFDSGI
ZGPPC
IXEJNDOP
JJFS
SSXXQOFGJUZ
Output
JJFS....
Y..S....
N..X....
Y..X....
F..Q....
D..O....
S..F....
G..G....
IXEJNDOP
...U...T
...ZGPPC
|
instruction
| 0
| 108,217
| 6
| 216,434
|
Tags: implementation
Correct Solution:
```
ar = [input() for i in range(6)]
ar.sort()
ans = []
for i in range(6):
for j in range(6):
if(i == j): continue
if(ar[i][0] != ar[j][0]): continue
for k in range(6):
if(i==k or j==k): continue
if(len(ar[k]) < 5): continue
if(len(ar[j])+2 > len(ar[k])): continue
if(ar[i][-1] != ar[k][0]): continue
for l in range(6):
if(i==l or j==l or k==l): continue
if(len(ar[l]) < 5): continue
if(len(ar[i])+2 > len(ar[l])): continue
if(ar[j][-1] != ar[l][0]): continue
if(ar[k][len(ar[j])-1] != ar[l][len(ar[i])-1]): continue
for m in range(6):
if(i==m or j==m or k==m or l==m): continue
if((len(ar[j])+len(ar[m]) - 1) != len(ar[k])): continue
if(ar[l][-1] != ar[m][0]): continue
for n in range(6):
if(i==n or j==n or k==n or l==n or m==n): continue
if((len(ar[i])+len(ar[n]) - 1) != len(ar[l])): continue
if(ar[k][-1] != ar[n][0]): continue
if(ar[m][-1] != ar[n][-1]): continue
tmp = []
s = ar[i] + ('.' * (len(ar[n])-1))
tmp.append(s)
for a in range(1, len(ar[j])-1):
s = ar[j][a] + ('.' * (len(ar[i])-2)) + ar[k][a] + ('.' * (len(ar[n])-1))
tmp.append(s)
s = ar[l]
tmp.append(s)
for a in range(1, len(ar[m])-1):
s = ('.' * (len(ar[i])-1)) + ar[k][len(ar[j])-1 + a] + ('.' * (len(ar[n])-2)) + ar[m][a]
tmp.append(s)
s = ('.' * (len(ar[i])-1)) + ar[n]
tmp.append(s)
ans.append(tmp)
if(len(ans) == 0):
print("Impossible")
else:
tmp = min(ans)
for t in tmp:
print(t)
```
|
output
| 1
| 108,217
| 6
| 216,435
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya trains to compose crossword puzzles. He can only compose crosswords of a very simplе type so far. All of them consist of exactly six words; the words can be read only from top to bottom vertically and from the left to the right horizontally. The words are arranged in the form of a rectangular "eight" or infinity sign, not necessarily symmetrical.
The top-left corner of the crossword coincides with the top-left corner of the rectangle. The same thing is correct for the right-bottom corners. The crossword can't degrade, i.e. it always has exactly four blank areas, two of which are surrounded by letters. Look into the output for the samples for clarification.
Help Vasya — compose a crossword of the described type using the given six words. It is allowed to use the words in any order.
Input
Six lines contain the given words. Every word consists of no more than 30 and no less than 3 uppercase Latin letters.
Output
If it is impossible to solve the problem, print Impossible. Otherwise, print the sought crossword. All the empty squares should be marked as dots.
If there can be several solutions to that problem, print the lexicographically minimum one. I.e. the solution where the first line is less than the first line of other solutions should be printed. If the two lines are equal, compare the second lines and so on. The lexicographical comparison of lines is realized by the < operator in the modern programming languages.
Examples
Input
NOD
BAA
YARD
AIRWAY
NEWTON
BURN
Output
BAA...
U.I...
R.R...
NEWTON
..A..O
..YARD
Input
AAA
AAA
AAAAA
AAA
AAA
AAAAA
Output
AAA..
A.A..
AAAAA
..A.A
..AAA
Input
PTC
JYNYFDSGI
ZGPPC
IXEJNDOP
JJFS
SSXXQOFGJUZ
Output
JJFS....
Y..S....
N..X....
Y..X....
F..Q....
D..O....
S..F....
G..G....
IXEJNDOP
...U...T
...ZGPPC
|
instruction
| 0
| 108,218
| 6
| 216,436
|
Tags: implementation
Correct Solution:
```
from itertools import permutations
v = []
for p in permutations(input() for i in range(6)):
if len(p[1]) != len(p[0]) + len(p[2]) - 1 or len(p[4]) != len(p[3]) + len(p[5]) - 1:
continue
elif p[0][0] != p[3][0] or p[0][-1] != p[4][0]:
continue
elif p[1][0] != p[3][-1] or p[1][len(p[0]) - 1] != p[4][len(p[3]) - 1] or p[1][-1] != p[5][0]:
continue
elif p[2][0] != p[4][-1] or p[2][-1] != p[5][-1]:
continue
else:
x, y, c = '.' * (len(p[1]) - len(p[0])), '.' * \
(len(p[1]) - len(p[2])), []
c.append(p[0] + x)
for i in range(1, len(p[3]) - 1):
c.append(p[3][i] + '.' * (len(p[0]) - 2) + p[4][i] + x)
c.append(p[1])
for i in range(1, len(p[5]) - 1):
c.append(y + p[4][len(p[3]) + i - 1] +
'.' * (len(p[2]) - 2) + p[5][i])
c.append(y + p[2])
v.append(c)
print('\n'.join(sorted(v)[0]) if v else 'Impossible')
```
|
output
| 1
| 108,218
| 6
| 216,437
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya trains to compose crossword puzzles. He can only compose crosswords of a very simplе type so far. All of them consist of exactly six words; the words can be read only from top to bottom vertically and from the left to the right horizontally. The words are arranged in the form of a rectangular "eight" or infinity sign, not necessarily symmetrical.
The top-left corner of the crossword coincides with the top-left corner of the rectangle. The same thing is correct for the right-bottom corners. The crossword can't degrade, i.e. it always has exactly four blank areas, two of which are surrounded by letters. Look into the output for the samples for clarification.
Help Vasya — compose a crossword of the described type using the given six words. It is allowed to use the words in any order.
Input
Six lines contain the given words. Every word consists of no more than 30 and no less than 3 uppercase Latin letters.
Output
If it is impossible to solve the problem, print Impossible. Otherwise, print the sought crossword. All the empty squares should be marked as dots.
If there can be several solutions to that problem, print the lexicographically minimum one. I.e. the solution where the first line is less than the first line of other solutions should be printed. If the two lines are equal, compare the second lines and so on. The lexicographical comparison of lines is realized by the < operator in the modern programming languages.
Examples
Input
NOD
BAA
YARD
AIRWAY
NEWTON
BURN
Output
BAA...
U.I...
R.R...
NEWTON
..A..O
..YARD
Input
AAA
AAA
AAAAA
AAA
AAA
AAAAA
Output
AAA..
A.A..
AAAAA
..A.A
..AAA
Input
PTC
JYNYFDSGI
ZGPPC
IXEJNDOP
JJFS
SSXXQOFGJUZ
Output
JJFS....
Y..S....
N..X....
Y..X....
F..Q....
D..O....
S..F....
G..G....
IXEJNDOP
...U...T
...ZGPPC
|
instruction
| 0
| 108,219
| 6
| 216,438
|
Tags: implementation
Correct Solution:
```
from itertools import permutations
v = []
for p in permutations(input() for i in range(6)):
if len(p[1]) != len(p[0]) + len(p[2]) - 1 or len(p[4]) != len(p[3]) + len(p[5]) - 1:
continue
elif p[0][0] != p[3][0] or p[0][-1] != p[4][0]:
continue
elif p[1][0] != p[3][-1] or p[1][len(p[0]) - 1] != p[4][len(p[3]) - 1] or p[1][-1] != p[5][0]:
continue
elif p[2][0] != p[4][-1] or p[2][-1] != p[5][-1]:
continue
else:
x, y, c = '.' * (len(p[1]) - len(p[0])), '.' * (len(p[1]) - len(p[2])), []
c.append(p[0] + x)
for i in range(1, len(p[3]) - 1):
c.append(p[3][i] + '.' * (len(p[0]) - 2) + p[4][i] + x)
c.append(p[1])
for i in range(1, len(p[5]) - 1):
c.append(y + p[4][len(p[3]) + i - 1] + '.' * (len(p[2]) - 2) + p[5][i])
c.append(y + p[2])
v.append(c)
print('\n'.join(sorted(v)[0]) if v else 'Impossible')
```
|
output
| 1
| 108,219
| 6
| 216,439
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya trains to compose crossword puzzles. He can only compose crosswords of a very simplе type so far. All of them consist of exactly six words; the words can be read only from top to bottom vertically and from the left to the right horizontally. The words are arranged in the form of a rectangular "eight" or infinity sign, not necessarily symmetrical.
The top-left corner of the crossword coincides with the top-left corner of the rectangle. The same thing is correct for the right-bottom corners. The crossword can't degrade, i.e. it always has exactly four blank areas, two of which are surrounded by letters. Look into the output for the samples for clarification.
Help Vasya — compose a crossword of the described type using the given six words. It is allowed to use the words in any order.
Input
Six lines contain the given words. Every word consists of no more than 30 and no less than 3 uppercase Latin letters.
Output
If it is impossible to solve the problem, print Impossible. Otherwise, print the sought crossword. All the empty squares should be marked as dots.
If there can be several solutions to that problem, print the lexicographically minimum one. I.e. the solution where the first line is less than the first line of other solutions should be printed. If the two lines are equal, compare the second lines and so on. The lexicographical comparison of lines is realized by the < operator in the modern programming languages.
Examples
Input
NOD
BAA
YARD
AIRWAY
NEWTON
BURN
Output
BAA...
U.I...
R.R...
NEWTON
..A..O
..YARD
Input
AAA
AAA
AAAAA
AAA
AAA
AAAAA
Output
AAA..
A.A..
AAAAA
..A.A
..AAA
Input
PTC
JYNYFDSGI
ZGPPC
IXEJNDOP
JJFS
SSXXQOFGJUZ
Output
JJFS....
Y..S....
N..X....
Y..X....
F..Q....
D..O....
S..F....
G..G....
IXEJNDOP
...U...T
...ZGPPC
|
instruction
| 0
| 108,220
| 6
| 216,440
|
Tags: implementation
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
from itertools import permutations
def main():
s = [input().strip() for _ in range(6)]
ans = ['a'*30 for _ in range(30)]
dx = [0,1,0,-1,0,-1]
dy = [1,0,1,0,-1,0]
for i in enumerate(permutations(s)):
i = list(i[1])
i[3],i[4],i[5] = i[3][::-1],i[4][::-1],i[5][::-1]
fl = 0
for j in range(1,7):
if i[j%6][0] != i[j-1][-1]:
fl = 1
break
if fl:
continue
ans1 = [['a']*30 for _ in range(30)]
x,y,fl,s = 0,0,0,-1
for ind,j in enumerate(i):
for ch in j:
if x < 0 or y < 0 or x >= 30 or y >= 30:
fl = 1
break
ans1[x][y] = ch
x += dx[ind]
y += dy[ind]
if fl:
break
x -= dx[ind]
y -= dy[ind]
if s == -1:
s = y
if s != -1:
ii = 0
for xx in i[1]:
if ans1[ii][s] != xx:
fl = 1
break
ii += 1
if fl or x or y:
continue
ma = 0
for ii in range(30):
ma = max(ma,30-ans1[ii].count('a'))
fin = []
for ii in range(30):
fin.append(''.join(ans1[ii][:ma]))
fin[-1] = fin[-1].replace('a','.')
if fin[-1] == '.'*ma:
fin.pop()
break
fll = 0
for ii in range(min(len(fin),len(ans))):
if fin[ii] < ans[ii]:
fll = 1
ans = fin
break
elif fin[ii] > ans[ii]:
fll = 1
break
if not fll:
if len(fin) < len(ans):
ans = fin
if ans == ['a'*30 for _ in range(30)]:
print('Impossible')
exit()
print(*ans,sep='\n')
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
```
|
output
| 1
| 108,220
| 6
| 216,441
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya trains to compose crossword puzzles. He can only compose crosswords of a very simplе type so far. All of them consist of exactly six words; the words can be read only from top to bottom vertically and from the left to the right horizontally. The words are arranged in the form of a rectangular "eight" or infinity sign, not necessarily symmetrical.
The top-left corner of the crossword coincides with the top-left corner of the rectangle. The same thing is correct for the right-bottom corners. The crossword can't degrade, i.e. it always has exactly four blank areas, two of which are surrounded by letters. Look into the output for the samples for clarification.
Help Vasya — compose a crossword of the described type using the given six words. It is allowed to use the words in any order.
Input
Six lines contain the given words. Every word consists of no more than 30 and no less than 3 uppercase Latin letters.
Output
If it is impossible to solve the problem, print Impossible. Otherwise, print the sought crossword. All the empty squares should be marked as dots.
If there can be several solutions to that problem, print the lexicographically minimum one. I.e. the solution where the first line is less than the first line of other solutions should be printed. If the two lines are equal, compare the second lines and so on. The lexicographical comparison of lines is realized by the < operator in the modern programming languages.
Examples
Input
NOD
BAA
YARD
AIRWAY
NEWTON
BURN
Output
BAA...
U.I...
R.R...
NEWTON
..A..O
..YARD
Input
AAA
AAA
AAAAA
AAA
AAA
AAAAA
Output
AAA..
A.A..
AAAAA
..A.A
..AAA
Input
PTC
JYNYFDSGI
ZGPPC
IXEJNDOP
JJFS
SSXXQOFGJUZ
Output
JJFS....
Y..S....
N..X....
Y..X....
F..Q....
D..O....
S..F....
G..G....
IXEJNDOP
...U...T
...ZGPPC
|
instruction
| 0
| 108,221
| 6
| 216,442
|
Tags: implementation
Correct Solution:
```
import sys
from array import array # noqa: F401
from itertools import permutations
def input():
return sys.stdin.buffer.readline().decode('utf-8')
a = [input().rstrip() for _ in range(6)]
ans = ['z' * 100 for _ in range(6)]
inf = ['z' * 100 for _ in range(6)]
for words in permutations(a):
if (
len(words[0]) + len(words[4]) - 1 == len(words[2])
and len(words[1]) + len(words[5]) - 1 == len(words[3])
and words[0][0] == words[1][0]
and words[1][-1] == words[2][0]
and words[0][-1] == words[3][0]
and words[2][len(words[0]) - 1] == words[3][len(words[1]) - 1]
and words[2][-1] == words[5][0]
and words[3][-1] == words[4][0]
and words[4][-1] == words[5][-1]
):
res = [['.'] * len(words[2]) for _ in range(len(words[3]))]
for i in range(len(words[0])):
res[0][i] = words[0][i]
for i in range(len(words[1])):
res[i][0] = words[1][i]
for i in range(len(words[2])):
res[len(words[1]) - 1][i] = words[2][i]
for i in range(len(words[3])):
res[i][len(words[0]) - 1] = words[3][i]
for i in range(len(words[4])):
res[len(words[3]) - 1][len(words[0]) - 1 + i] = words[4][i]
for i in range(len(words[5])):
res[len(words[1]) - 1 + i][len(words[2]) - 1] = words[5][i]
res = [''.join(row) for row in res]
if ans > res:
ans = res
if ans == inf:
print('Impossible')
else:
print('\n'.join(ans))
```
|
output
| 1
| 108,221
| 6
| 216,443
|
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