message stringlengths 2 11.9k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 137 108k | cluster float64 18 18 | __index_level_0__ int64 274 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which finds a pattern $p$ in a ring shaped text $s$.
<image>
Constraints
* $1 \leq $ length of $p \leq $ length of $s \leq 100$
* $s$ and $p$ consists of lower-case letters
Input
In the first line, the text $s$ is given.
In the second line, the pattern $p$ is given.
Output
If $p$ is in $s$, print Yes in a line, otherwise No.
Examples
Input
vanceknowledgetoad
advance
Output
Yes
Input
vanceknowledgetoad
advanced
Output
No
Submitted Solution:
```
import re
``` | instruction | 0 | 29,098 | 18 | 58,196 |
No | output | 1 | 29,098 | 18 | 58,197 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison.
Input
Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters.
Output
If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared.
Examples
Input
aaaa
aaaA
Output
0
Input
abs
Abz
Output
-1
Input
abcdefg
AbCdEfF
Output
1
Note
If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
* http://en.wikipedia.org/wiki/Lexicographical_order
Submitted Solution:
```
import sys
a = sys.stdin.readline().lower()
b = sys.stdin.readline().lower()
if a > b:
print(1)
elif a < b:
print(-1)
else:
print(0)
``` | instruction | 0 | 29,175 | 18 | 58,350 |
Yes | output | 1 | 29,175 | 18 | 58,351 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison.
Input
Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters.
Output
If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared.
Examples
Input
aaaa
aaaA
Output
0
Input
abs
Abz
Output
-1
Input
abcdefg
AbCdEfF
Output
1
Note
If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
* http://en.wikipedia.org/wiki/Lexicographical_order
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Sun Dec 16 21:52:18 2018
@author: umang
"""
a = input().lower()
b = input().lower()
if a < b:
print(-1)
elif a > b:
print(1)
else:
print(0)
``` | instruction | 0 | 29,176 | 18 | 58,352 |
Yes | output | 1 | 29,176 | 18 | 58,353 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison.
Input
Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters.
Output
If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared.
Examples
Input
aaaa
aaaA
Output
0
Input
abs
Abz
Output
-1
Input
abcdefg
AbCdEfF
Output
1
Note
If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
* http://en.wikipedia.org/wiki/Lexicographical_order
Submitted Solution:
```
x=input()
y=input()
x=x.lower()
y=y.lower()
if x > y :
print(1)
elif y > x :
print(-1)
else :
print(0)
``` | instruction | 0 | 29,177 | 18 | 58,354 |
Yes | output | 1 | 29,177 | 18 | 58,355 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison.
Input
Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters.
Output
If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared.
Examples
Input
aaaa
aaaA
Output
0
Input
abs
Abz
Output
-1
Input
abcdefg
AbCdEfF
Output
1
Note
If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
* http://en.wikipedia.org/wiki/Lexicographical_order
Submitted Solution:
```
a=input()
a=a.lower()
b=input()
b=b.lower()
"""
a= ''.join(sorted(a))
b= ''.join(sorted(b))
"""
if(a==b):
print(0)
elif(a<b):
print(-1)
else:
print(1)
``` | instruction | 0 | 29,178 | 18 | 58,356 |
Yes | output | 1 | 29,178 | 18 | 58,357 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison.
Input
Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters.
Output
If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared.
Examples
Input
aaaa
aaaA
Output
0
Input
abs
Abz
Output
-1
Input
abcdefg
AbCdEfF
Output
1
Note
If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
* http://en.wikipedia.org/wiki/Lexicographical_order
Submitted Solution:
```
a,k = input().lower(), input().lower()
r = []
for i,t in zip(a,k):
if i > t:
r.append(1)
elif i < t:
r.append(-1)
else:
r.append(0)
p = sum(r)
if p >= 1:
print(1)
elif p <= -1:
print(-1)
else:
print(0)
``` | instruction | 0 | 29,179 | 18 | 58,358 |
No | output | 1 | 29,179 | 18 | 58,359 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison.
Input
Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters.
Output
If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared.
Examples
Input
aaaa
aaaA
Output
0
Input
abs
Abz
Output
-1
Input
abcdefg
AbCdEfF
Output
1
Note
If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
* http://en.wikipedia.org/wiki/Lexicographical_order
Submitted Solution:
```
# n = int(input())
# a = []
# m = []
# for i in range(n):
# d,k = [int(i) for i in input().split()]
# if k % d != 0:
# a.append(k % d)
# else:
# pass
# a_set = set(a)
# most = None
# most1 = 0
# for i in a_set:
# q = a.count(i)
# if q >= most1:
# most1 = q
# most = i
# m.append(i)
# b = -1
# for i in range(len(m)):
# if m[i] > b:
# b = a[i]
# print(b)
# n = int(input())
# s = input()
# k = 0
# flag = False
# def decode():
# global s,k,flag
# s1 = re.search(r'xxx',s)
# if s1:
# end = s1.end()
# k += 1
# s = s[end:]
# flag = True
# decode()
# else:
# pass
# decode()
# if flag:
# print(k)
# else:
# print(0)
# import time
# start_time = time.clock()
# a1 = []
# res = []
# a = None
# while a != 0:
# a = int(input())
# a1.append(a)
# for i in range(1,len(a1)):
# for j in range(1,len(a1)):
# if (a1[i-1] * a1[j-1]) % 7 == 0 and (a1[i-1] * a1[j-1]) % 49 != 0:
# res.append(a1[i-1] * a1[j-1])
# else:
# pass
# print(max(res))
# print(time.clock() - start_time)
# n = int(input())
# i = 0
# k = []
# x = []
# y = []
# while i != n:
# a,b = [int(i) for i in input().split()]
# x.append(a)
# y.append(b)
# i += 1
# for i in range(n):
# if x[i] >= 0 and y[i] >= 0:
# k.append(1)
# elif x[i] <= 0 and y[i] >= 0:
# k.append(2)
# elif x[i] <= 0 and y[i] <= 0:
# k.append(3)
# else:
# k.append(4)
# k1 = {i: k.count(i) for i in set(k)}
# k1 = max(k1.items())
# print('K = ',k1[0])
# print('M = ',k1[1])
# import time
# t = time.clock()
# n = int(input())
# n1 = n
# a = []
#
# while n1 != 0:
# a.append(int(input()))
# n1 -= 1
# m = a[0] + a[7]
# for i in range(n-7):
# if a[i] + a[i+7] > m:
# m = a[i] + a[i+7]
# print(m)
# print(time.clock()-t)
# n = int(input())
# n1 = n
# x = []
# y = []
# s = []
# while n1 != 0:
# a,b = [int(i) for i in input().split()]
# x.append(a)
# y.append(b)
# n1 -= 1
# for i in range(n):
# if y[i] % x[i] == 0:
# pass
# else:
# s.append(y[i] % x[i])
# s1 = {i: s.count(i) for i in set(s)}
# s1 = max(s1.items())
# print(s1[0])
# s1 = input()
# s2 = input()
# s1_list = list(s1)
# s2_list = list(s2)
# s1_list.sort()
# s2_list.sort()
# if s1_list == s2_list:
# print('Yes')
# else:
# print('No')
# k = 0
# s1 = input()
# s1_sort = {i:s1.count(i) for i in set(s1)}
# for i in range(1,len(s1)):
# if s1[i-1] == s1[i]:
# s1.replace(s1[i+1],'')
# print(s1)
# a = [6,3,5,7,4,2]
# m1 = 1000
# m2 = 1000
# for i in range(len(a)):
# if a[i] < m1:
# m2 = m1
# m1 = a[i]
# elif a[i] < m2:
# m2 = a[i]
# print(m1,m2)
# import random
#
# a = ord('a')
# A = ord('A')
# z = ord('z')
# Z = ord('Z')
# c = ord('0')
# n = [chr(i) for i in range(a, z + 1)]
# for j in range(A, Z + 1):
# n.append(chr(j))
# for i in range(c, c + 10):
# n.append(chr(i))
# k = random.sample(n, 20)
# print(k)
# s = input()
# a = [i for i in s]
# di = {i:a.count(i) for i in a}
# di = sorted(di.items())
# print(di[0][0],'find',di[0][1])
# import random
#
# a = [input() for _ in range(3)]
# a = {i:a.count(i) for i in set(a)}
# a = sorted(a.items(), key=lambda x: x[1], reverse=True)
# print(int(a[0][0]), a[0][1])
# c = 10000
# n = 15
# a = [int(input()) for _ in range(18)]
# while n != len(a):
# for i in range(15,len(a)):
# if (a[i-n] * a[i]) % 2 == 0 and (a[i-n] * a[i]) < c:
# c = a[i-n] * a[i]
# n += 1
# print(c)
# a = input()
# b = []
# k = 0
# flag = True
# n = ['0','1','2','3','4','5','6','7','8','9']
# for i in a:
# if i.isdigit():
# b.append(i)
# while flag:
# c = {i: b.count(i) for i in set(b)}
# for i in c.items():
# if i[1] == 1:
# key = i[0]
# k += 1
# if k > 1:
# print('No')
# flag = False
# if k == 1:
# l = [0] * (len(b) + 2)
# l[len(l) // 2] = key
# j = 0
# x = 1
# while j != len(n):
# for i in c.keys():
# if i == n[j]:
# l[(len(l) // 2) + x] = i
# l[(len(l) // 2) - x] = i
# x += 1
# j += 1
# for i in range(0,(len(l) // 2)-1):
# if l[i] == key:
# del l[i]
# for i in range(len(l) // 2,len(l)):
# if l[i] == key:
# del l[i]
# print(l)
# flag = False
# if k == 0:
# l = [0] * (len(b))
# x = 0
# o = 1
# j = 0
# while j != len(n):
# for i in c.keys():
# if i == n[j]:
# l[(len(l) // 2) + x] = i
# l[(len(l) // 2) - o] = i
# x += 1
# o += 1
# j += 1
# print(l)
# n = 3
# m = 10000
# mn = m + 1
# a = [int(input()) for i in range(n)]
# if sum(a) % 6 != 0:
# print(a, sum(a))
# else:
# for i in a:
# if i < mn:
# mn = i
# if mn < m:
# print(n-1,sum(a)-mn)
# else:
# print(0,0)
# n = 4
# a = [0 for _ in range(n)]
# def func(i):
# if i < 4:
# for k in (0,1):
# a[i] = k
# func(i+1)
# else:
# c = 0
# if (a[0] or a[1]) and (not a[2] or not a[3]):
# c = 1
# print(a,c)
#
# func(0)
# a = [i for i in range(1,100)]
# #b = [2, 3, 4, 1, 6, 5, 7, 9, 8]
# #print(list(enumerate(zip(a,b))))
# print(','.join([str(i) for i in range(1,100)]))
# print(','.join(map(str,a)))
#s = input()
# a = [i for i in s]
# if a[0] == a[0].upper():
# print(''.join(a))
# else:
# a[0] = a[0].upper()
# print(''.join(a))
# n, a, b, res, c = int(input()), [], [], [], 0
# while n != 0:
# x, y = [int(i) for i in input().split()]
# a.append(x)
# b.append(y)
# n -= 1
# res.append(b[0])
# for i in range(1, len(a)):
# res.append(res[c] - a[i] + b[i])
# c += 1
# print(max(res))
# a = [i for i in input()]
# b = [i for i in input()]
# k,l = 0,0
# flag = True
# c1 = sorted(a, key=lambda x: (str.upper(x),x))
# c2 = sorted(b, key=lambda x: (str.upper(x),x))
# if flag:
# for i, j in zip(c1,c2):
# if i != j:
# if ord(i.lower()) > ord(j.lower()):
# print(1)
# flag = False
# break
# elif ord(i.lower()) < ord(j.lower()):
# print(-1)
# flag = False
# break
# else:
# pass
# if flag:
# print(0)
# for i, j in zip(a,b):
# k += ord(i)
# l += ord(j)
# if i == j.upper():
# l = k
# elif j == i.upper():
# k = l
# if k > l:
# print(1)
# elif k < l:
# print(-1)
# else:
# print(0)
# a,n = [int(i) for i in input().split()]
# s = [i for i in input()]
# for _ in range(n):
# flag = True
# for i in range(len(s)-1):
# if flag:
# if s[i] == 'B' and s[i+1] == 'G':
# s[i], s[i+1] = s[i+1], s[i]
# flag = False
# else:
# flag = True
# print(''.join(s))
a = [i for i in input()]
b = [i for i in input()]
c1 = sorted(map(lambda x: x.lower(), a))
c2 = sorted(map(lambda x: x.lower(), b))
flag = True
if flag:
for i, j in zip(a,b):
if i != j:
if ord(i) > ord(j):
print(1)
flag = False
break
elif ord(i) < ord(j):
print(-1)
flag = False
break
else:
pass
if flag:
print(0)
``` | instruction | 0 | 29,180 | 18 | 58,360 |
No | output | 1 | 29,180 | 18 | 58,361 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison.
Input
Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters.
Output
If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared.
Examples
Input
aaaa
aaaA
Output
0
Input
abs
Abz
Output
-1
Input
abcdefg
AbCdEfF
Output
1
Note
If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
* http://en.wikipedia.org/wiki/Lexicographical_order
Submitted Solution:
```
word1 = str.lower(input())
word2 = str.lower(input())
m = 0
n = 0
for i in range(len(word1)):
m += ord(word1[i])
for j in range(len(word2)):
n += ord(word2[j])
if m > n:
print('1')
elif m < n:
print('-1')
else:
print('0')
``` | instruction | 0 | 29,181 | 18 | 58,362 |
No | output | 1 | 29,181 | 18 | 58,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison.
Input
Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters.
Output
If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared.
Examples
Input
aaaa
aaaA
Output
0
Input
abs
Abz
Output
-1
Input
abcdefg
AbCdEfF
Output
1
Note
If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
* http://en.wikipedia.org/wiki/Lexicographical_order
Submitted Solution:
```
word1 = input()
word2 = input()
word_order = [word1, word2]
word_order.sort()
if word1 != word2:
if word1 == word_order[0]:
print(-1)
else:
print(1)
else:
print(0)
``` | instruction | 0 | 29,182 | 18 | 58,364 |
No | output | 1 | 29,182 | 18 | 58,365 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Those days, many boys use beautiful girls' photos as avatars in forums. So it is pretty hard to tell the gender of a user at the first glance. Last year, our hero went to a forum and had a nice chat with a beauty (he thought so). After that they talked very often and eventually they became a couple in the network.
But yesterday, he came to see "her" in the real world and found out "she" is actually a very strong man! Our hero is very sad and he is too tired to love again now. So he came up with a way to recognize users' genders by their user names.
This is his method: if the number of distinct characters in one's user name is odd, then he is a male, otherwise she is a female. You are given the string that denotes the user name, please help our hero to determine the gender of this user by his method.
Input
The first line contains a non-empty string, that contains only lowercase English letters — the user name. This string contains at most 100 letters.
Output
If it is a female by our hero's method, print "CHAT WITH HER!" (without the quotes), otherwise, print "IGNORE HIM!" (without the quotes).
Examples
Input
wjmzbmr
Output
CHAT WITH HER!
Input
xiaodao
Output
IGNORE HIM!
Input
sevenkplus
Output
CHAT WITH HER!
Note
For the first example. There are 6 distinct characters in "wjmzbmr". These characters are: "w", "j", "m", "z", "b", "r". So wjmzbmr is a female and you should print "CHAT WITH HER!".
Submitted Solution:
```
n = input()
p = list(n)
p = list(dict.fromkeys(p))
print("IGNORE HIM!" if len(p) % 2 != 0 else "CHAT WITH HER!")
``` | instruction | 0 | 29,413 | 18 | 58,826 |
Yes | output | 1 | 29,413 | 18 | 58,827 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Those days, many boys use beautiful girls' photos as avatars in forums. So it is pretty hard to tell the gender of a user at the first glance. Last year, our hero went to a forum and had a nice chat with a beauty (he thought so). After that they talked very often and eventually they became a couple in the network.
But yesterday, he came to see "her" in the real world and found out "she" is actually a very strong man! Our hero is very sad and he is too tired to love again now. So he came up with a way to recognize users' genders by their user names.
This is his method: if the number of distinct characters in one's user name is odd, then he is a male, otherwise she is a female. You are given the string that denotes the user name, please help our hero to determine the gender of this user by his method.
Input
The first line contains a non-empty string, that contains only lowercase English letters — the user name. This string contains at most 100 letters.
Output
If it is a female by our hero's method, print "CHAT WITH HER!" (without the quotes), otherwise, print "IGNORE HIM!" (without the quotes).
Examples
Input
wjmzbmr
Output
CHAT WITH HER!
Input
xiaodao
Output
IGNORE HIM!
Input
sevenkplus
Output
CHAT WITH HER!
Note
For the first example. There are 6 distinct characters in "wjmzbmr". These characters are: "w", "j", "m", "z", "b", "r". So wjmzbmr is a female and you should print "CHAT WITH HER!".
Submitted Solution:
```
import sys
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
s = insr()
d = {}
for e in s:
if e in d.keys():
continue
else:
d[e]=True
if len(d.keys())%2==1:
print("IGNORE HIM!")
else:
print("CHAT WITH HER!")
``` | instruction | 0 | 29,414 | 18 | 58,828 |
Yes | output | 1 | 29,414 | 18 | 58,829 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Those days, many boys use beautiful girls' photos as avatars in forums. So it is pretty hard to tell the gender of a user at the first glance. Last year, our hero went to a forum and had a nice chat with a beauty (he thought so). After that they talked very often and eventually they became a couple in the network.
But yesterday, he came to see "her" in the real world and found out "she" is actually a very strong man! Our hero is very sad and he is too tired to love again now. So he came up with a way to recognize users' genders by their user names.
This is his method: if the number of distinct characters in one's user name is odd, then he is a male, otherwise she is a female. You are given the string that denotes the user name, please help our hero to determine the gender of this user by his method.
Input
The first line contains a non-empty string, that contains only lowercase English letters — the user name. This string contains at most 100 letters.
Output
If it is a female by our hero's method, print "CHAT WITH HER!" (without the quotes), otherwise, print "IGNORE HIM!" (without the quotes).
Examples
Input
wjmzbmr
Output
CHAT WITH HER!
Input
xiaodao
Output
IGNORE HIM!
Input
sevenkplus
Output
CHAT WITH HER!
Note
For the first example. There are 6 distinct characters in "wjmzbmr". These characters are: "w", "j", "m", "z", "b", "r". So wjmzbmr is a female and you should print "CHAT WITH HER!".
Submitted Solution:
```
name = set(input())
if len(name) %2 == 0:
print ('CHAT WITH HER!')
else:
print ('IGNORE HIM!')
``` | instruction | 0 | 29,415 | 18 | 58,830 |
Yes | output | 1 | 29,415 | 18 | 58,831 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Those days, many boys use beautiful girls' photos as avatars in forums. So it is pretty hard to tell the gender of a user at the first glance. Last year, our hero went to a forum and had a nice chat with a beauty (he thought so). After that they talked very often and eventually they became a couple in the network.
But yesterday, he came to see "her" in the real world and found out "she" is actually a very strong man! Our hero is very sad and he is too tired to love again now. So he came up with a way to recognize users' genders by their user names.
This is his method: if the number of distinct characters in one's user name is odd, then he is a male, otherwise she is a female. You are given the string that denotes the user name, please help our hero to determine the gender of this user by his method.
Input
The first line contains a non-empty string, that contains only lowercase English letters — the user name. This string contains at most 100 letters.
Output
If it is a female by our hero's method, print "CHAT WITH HER!" (without the quotes), otherwise, print "IGNORE HIM!" (without the quotes).
Examples
Input
wjmzbmr
Output
CHAT WITH HER!
Input
xiaodao
Output
IGNORE HIM!
Input
sevenkplus
Output
CHAT WITH HER!
Note
For the first example. There are 6 distinct characters in "wjmzbmr". These characters are: "w", "j", "m", "z", "b", "r". So wjmzbmr is a female and you should print "CHAT WITH HER!".
Submitted Solution:
```
def main():
str = input()
str = dict.fromkeys(str)
if len(str) % 2 == 0:
print("CHAT WITH HER!")
else:
print("IGNORE HIM!")
if __name__ == "__main__":
main()
``` | instruction | 0 | 29,416 | 18 | 58,832 |
Yes | output | 1 | 29,416 | 18 | 58,833 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Those days, many boys use beautiful girls' photos as avatars in forums. So it is pretty hard to tell the gender of a user at the first glance. Last year, our hero went to a forum and had a nice chat with a beauty (he thought so). After that they talked very often and eventually they became a couple in the network.
But yesterday, he came to see "her" in the real world and found out "she" is actually a very strong man! Our hero is very sad and he is too tired to love again now. So he came up with a way to recognize users' genders by their user names.
This is his method: if the number of distinct characters in one's user name is odd, then he is a male, otherwise she is a female. You are given the string that denotes the user name, please help our hero to determine the gender of this user by his method.
Input
The first line contains a non-empty string, that contains only lowercase English letters — the user name. This string contains at most 100 letters.
Output
If it is a female by our hero's method, print "CHAT WITH HER!" (without the quotes), otherwise, print "IGNORE HIM!" (without the quotes).
Examples
Input
wjmzbmr
Output
CHAT WITH HER!
Input
xiaodao
Output
IGNORE HIM!
Input
sevenkplus
Output
CHAT WITH HER!
Note
For the first example. There are 6 distinct characters in "wjmzbmr". These characters are: "w", "j", "m", "z", "b", "r". So wjmzbmr is a female and you should print "CHAT WITH HER!".
Submitted Solution:
```
k=input()
l=len(k)
m=0
j=0
while m<(l-1):
j=j+1
if k[j-1]==k[m-1]:
l=l-1
m=m+1
j=0
if l %2==0:
print('CHAT WITH HER!')
if l %2 !=0:
print('IGNORE HIM!')
``` | instruction | 0 | 29,417 | 18 | 58,834 |
No | output | 1 | 29,417 | 18 | 58,835 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Those days, many boys use beautiful girls' photos as avatars in forums. So it is pretty hard to tell the gender of a user at the first glance. Last year, our hero went to a forum and had a nice chat with a beauty (he thought so). After that they talked very often and eventually they became a couple in the network.
But yesterday, he came to see "her" in the real world and found out "she" is actually a very strong man! Our hero is very sad and he is too tired to love again now. So he came up with a way to recognize users' genders by their user names.
This is his method: if the number of distinct characters in one's user name is odd, then he is a male, otherwise she is a female. You are given the string that denotes the user name, please help our hero to determine the gender of this user by his method.
Input
The first line contains a non-empty string, that contains only lowercase English letters — the user name. This string contains at most 100 letters.
Output
If it is a female by our hero's method, print "CHAT WITH HER!" (without the quotes), otherwise, print "IGNORE HIM!" (without the quotes).
Examples
Input
wjmzbmr
Output
CHAT WITH HER!
Input
xiaodao
Output
IGNORE HIM!
Input
sevenkplus
Output
CHAT WITH HER!
Note
For the first example. There are 6 distinct characters in "wjmzbmr". These characters are: "w", "j", "m", "z", "b", "r". So wjmzbmr is a female and you should print "CHAT WITH HER!".
Submitted Solution:
```
print(['CHAT WITH HER!','IGNORE HIM!'][len({input()})%2])
``` | instruction | 0 | 29,418 | 18 | 58,836 |
No | output | 1 | 29,418 | 18 | 58,837 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Those days, many boys use beautiful girls' photos as avatars in forums. So it is pretty hard to tell the gender of a user at the first glance. Last year, our hero went to a forum and had a nice chat with a beauty (he thought so). After that they talked very often and eventually they became a couple in the network.
But yesterday, he came to see "her" in the real world and found out "she" is actually a very strong man! Our hero is very sad and he is too tired to love again now. So he came up with a way to recognize users' genders by their user names.
This is his method: if the number of distinct characters in one's user name is odd, then he is a male, otherwise she is a female. You are given the string that denotes the user name, please help our hero to determine the gender of this user by his method.
Input
The first line contains a non-empty string, that contains only lowercase English letters — the user name. This string contains at most 100 letters.
Output
If it is a female by our hero's method, print "CHAT WITH HER!" (without the quotes), otherwise, print "IGNORE HIM!" (without the quotes).
Examples
Input
wjmzbmr
Output
CHAT WITH HER!
Input
xiaodao
Output
IGNORE HIM!
Input
sevenkplus
Output
CHAT WITH HER!
Note
For the first example. There are 6 distinct characters in "wjmzbmr". These characters are: "w", "j", "m", "z", "b", "r". So wjmzbmr is a female and you should print "CHAT WITH HER!".
Submitted Solution:
```
a=list(input())
c=[]
for i in a:
if a.count(i)==1:
c.append(i)
else:
c.append(i)
a.remove(i)
d=len(c)
if d%2==0:
print("CHAT WITH HER!")
else:
print("IGNORE HIM!")
``` | instruction | 0 | 29,419 | 18 | 58,838 |
No | output | 1 | 29,419 | 18 | 58,839 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Those days, many boys use beautiful girls' photos as avatars in forums. So it is pretty hard to tell the gender of a user at the first glance. Last year, our hero went to a forum and had a nice chat with a beauty (he thought so). After that they talked very often and eventually they became a couple in the network.
But yesterday, he came to see "her" in the real world and found out "she" is actually a very strong man! Our hero is very sad and he is too tired to love again now. So he came up with a way to recognize users' genders by their user names.
This is his method: if the number of distinct characters in one's user name is odd, then he is a male, otherwise she is a female. You are given the string that denotes the user name, please help our hero to determine the gender of this user by his method.
Input
The first line contains a non-empty string, that contains only lowercase English letters — the user name. This string contains at most 100 letters.
Output
If it is a female by our hero's method, print "CHAT WITH HER!" (without the quotes), otherwise, print "IGNORE HIM!" (without the quotes).
Examples
Input
wjmzbmr
Output
CHAT WITH HER!
Input
xiaodao
Output
IGNORE HIM!
Input
sevenkplus
Output
CHAT WITH HER!
Note
For the first example. There are 6 distinct characters in "wjmzbmr". These characters are: "w", "j", "m", "z", "b", "r". So wjmzbmr is a female and you should print "CHAT WITH HER!".
Submitted Solution:
```
w = input()
a = len(w)
c = 0
for i in range(a):
b = w[i]
for j in range(a):
if(w[j] != b):
c = c+1
if(c%2 != 0):
print("IGNORE HIM!")
elif(c%2 == 0):
print("CHAT WITH HER!")
``` | instruction | 0 | 29,420 | 18 | 58,840 |
No | output | 1 | 29,420 | 18 | 58,841 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A simplified arithmetic expression (SAE) is an arithmetic expression defined by the following grammar:
* <SAE> ::= <Number> | <SAE>+<SAE> | <SAE>*<SAE> | (<SAE>)
* <Number> ::= <Digit> | <Digit><Number>
* <Digit> ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
In other words it's a correct arithmetic expression that is allowed to contain brackets, numbers (possibly with leading zeros), multiplications and additions. For example expressions "(0+01)", "0" and "1*(0)" are simplified arithmetic expressions, but expressions "2-1", "+1" and "1+2)" are not.
Given a string s1s2...s|s| that represents a SAE; si denotes the i-th character of the string which can be either a digit ('0'-'9'), a plus sign ('+'), a multiplication sign ('*'), an opening round bracket '(' or a closing round bracket ')'.
A part slsl + 1...sr of this string is called a sub-expression if and only if it is a SAE.
You task is to answer m queries, each of which is a pair of integers li, ri (1 ≤ li ≤ ri ≤ |s|). For each query determine whether the corresponding part of the given string is a sub-expression and in case it's a sub-expression calculate its value modulo 1000000007 (109 + 7). The values should be calculated using standard operator priorities.
Input
The first line of the input contains non-empty string s (1 ≤ |s| ≤ 4·105) which represents a correct SAE. Each character of the string can be one of the following characters: '*', '+', '(', ')' or a digit ('0'-'9'). The expression might contain extra-huge numbers.
The second line contains an integer m (1 ≤ m ≤ 4·105) which is the number of queries. Each of the next m lines contains two space-separated integers li, ri (1 ≤ li ≤ ri ≤ |s|) — the i-th query.
Output
The i-th number of output should be the answer for the i-th query. If the i-th query corresponds to a valid sub-expression output the value of the sub-expression modulo 1000000007 (109 + 7). Otherwise output -1 as an answer for the query. Print numbers on separate lines.
Examples
Input
((1+2)*3+101*2)
6
8 14
1 6
2 10
11 14
5 5
4 5
Output
205
-1
10
2
2
-1
Input
(01)
1
1 4
Output
1
Submitted Solution:
```
import re
import ast
def isoper(item):
if item == "+" or item == "*":
return True
else:
return False
def is_valid(code):
try:
ast.parse(code)
except SyntaxError:
return False
return True
def debug(item):
item1 = item
item1 = item1.replace("(","*").replace(")","*").replace("+", "*").split("*")
for i in range(len(item1)):
if item1[i] != "" and item1[i][0] == "0":
item = item.replace(item1[i], str(int(item1[i])))
return item
SAE = input()
n=int(input())
for i in range(n):
list1 = input().split()
temp = SAE[int(list1[0])-1:int(list1[1])]
if is_valid(temp):
if isoper(temp[0]):
print(-1)
else:
print(eval(temp))
else:
temp = debug(temp)
if is_valid(temp):
if isoper(temp[0]):
print(-1)
else:
print(eval(temp))
else:
print(-1)
``` | instruction | 0 | 29,607 | 18 | 59,214 |
No | output | 1 | 29,607 | 18 | 59,215 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A simplified arithmetic expression (SAE) is an arithmetic expression defined by the following grammar:
* <SAE> ::= <Number> | <SAE>+<SAE> | <SAE>*<SAE> | (<SAE>)
* <Number> ::= <Digit> | <Digit><Number>
* <Digit> ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
In other words it's a correct arithmetic expression that is allowed to contain brackets, numbers (possibly with leading zeros), multiplications and additions. For example expressions "(0+01)", "0" and "1*(0)" are simplified arithmetic expressions, but expressions "2-1", "+1" and "1+2)" are not.
Given a string s1s2...s|s| that represents a SAE; si denotes the i-th character of the string which can be either a digit ('0'-'9'), a plus sign ('+'), a multiplication sign ('*'), an opening round bracket '(' or a closing round bracket ')'.
A part slsl + 1...sr of this string is called a sub-expression if and only if it is a SAE.
You task is to answer m queries, each of which is a pair of integers li, ri (1 ≤ li ≤ ri ≤ |s|). For each query determine whether the corresponding part of the given string is a sub-expression and in case it's a sub-expression calculate its value modulo 1000000007 (109 + 7). The values should be calculated using standard operator priorities.
Input
The first line of the input contains non-empty string s (1 ≤ |s| ≤ 4·105) which represents a correct SAE. Each character of the string can be one of the following characters: '*', '+', '(', ')' or a digit ('0'-'9'). The expression might contain extra-huge numbers.
The second line contains an integer m (1 ≤ m ≤ 4·105) which is the number of queries. Each of the next m lines contains two space-separated integers li, ri (1 ≤ li ≤ ri ≤ |s|) — the i-th query.
Output
The i-th number of output should be the answer for the i-th query. If the i-th query corresponds to a valid sub-expression output the value of the sub-expression modulo 1000000007 (109 + 7). Otherwise output -1 as an answer for the query. Print numbers on separate lines.
Examples
Input
((1+2)*3+101*2)
6
8 14
1 6
2 10
11 14
5 5
4 5
Output
205
-1
10
2
2
-1
Input
(01)
1
1 4
Output
1
Submitted Solution:
```
sae = input()
numberOfCalculations = int(input())
operations = ["*","+","-"]
def ExecuteOP(expresion, operation):
indexOfOp = expresion.index(operation)
rightPart = expresion[indexOfOp + 1:]
leftPart = expresion[:indexOfOp]
numbers = ""
indexRight = 0
for number in rightPart:
if number not in operations:
numbers += number
indexRight += 1
else:
break
rightNumbers = int(numbers)
rightPart = rightPart[:indexRight]
indexLeft = 0
for number in reversed(leftPart):
if number not in operations:
indexLeft += 1
else:
break
leftNumbers = int(leftPart[indexOfOp - indexLeft:])
leftPart = leftPart[:indexLeft]
if operation == "+":
result = (leftNumbers + rightNumbers)
elif operation == "-":
result = (leftNumbers - rightNumbers)
elif operation == "*":
result = (leftNumbers * rightNumbers)
else:
assert False, "Operation not supported -> " + operation
return expresion[:indexOfOp - indexLeft] + str(result) + expresion[indexOfOp + indexRight + 1:]
def AritmeticExpresion(expresion):
while "*" in expresion:
expresion = ExecuteOP(expresion,"*")
while "+" in expresion:
expresion = ExecuteOP(expresion,"+")
while "-" in expresion:
expresion = ExecuteOP(expresion,"-")
return expresion
def CalculateSubSet(subset):
result = 0
if subset == -1 or subset == "-1":
return -1
if "(" in subset and ")" in subset:
resultSubset = CalculateSubSet(subset[subset.index("(") + 1:subset.index(")")])
newSubSet = subset[:subset.index("(")] + str(resultSubset) + subset[subset.index(")") + 1 :]
result += CalculateSubSet(newSubSet)
elif "(" not in subset and ")" not in subset:
result += int(AritmeticExpresion(subset))
else:
return -1
return result
for _ in range(numberOfCalculations):
actualSae = input().split(" ")
subExpresion = sae[int(actualSae[0]) - 1 :int(actualSae[1])]
try:
print(int(CalculateSubSet(subExpresion))%1000000007)
except:
print("-1")
``` | instruction | 0 | 29,608 | 18 | 59,216 |
No | output | 1 | 29,608 | 18 | 59,217 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A simplified arithmetic expression (SAE) is an arithmetic expression defined by the following grammar:
* <SAE> ::= <Number> | <SAE>+<SAE> | <SAE>*<SAE> | (<SAE>)
* <Number> ::= <Digit> | <Digit><Number>
* <Digit> ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
In other words it's a correct arithmetic expression that is allowed to contain brackets, numbers (possibly with leading zeros), multiplications and additions. For example expressions "(0+01)", "0" and "1*(0)" are simplified arithmetic expressions, but expressions "2-1", "+1" and "1+2)" are not.
Given a string s1s2...s|s| that represents a SAE; si denotes the i-th character of the string which can be either a digit ('0'-'9'), a plus sign ('+'), a multiplication sign ('*'), an opening round bracket '(' or a closing round bracket ')'.
A part slsl + 1...sr of this string is called a sub-expression if and only if it is a SAE.
You task is to answer m queries, each of which is a pair of integers li, ri (1 ≤ li ≤ ri ≤ |s|). For each query determine whether the corresponding part of the given string is a sub-expression and in case it's a sub-expression calculate its value modulo 1000000007 (109 + 7). The values should be calculated using standard operator priorities.
Input
The first line of the input contains non-empty string s (1 ≤ |s| ≤ 4·105) which represents a correct SAE. Each character of the string can be one of the following characters: '*', '+', '(', ')' or a digit ('0'-'9'). The expression might contain extra-huge numbers.
The second line contains an integer m (1 ≤ m ≤ 4·105) which is the number of queries. Each of the next m lines contains two space-separated integers li, ri (1 ≤ li ≤ ri ≤ |s|) — the i-th query.
Output
The i-th number of output should be the answer for the i-th query. If the i-th query corresponds to a valid sub-expression output the value of the sub-expression modulo 1000000007 (109 + 7). Otherwise output -1 as an answer for the query. Print numbers on separate lines.
Examples
Input
((1+2)*3+101*2)
6
8 14
1 6
2 10
11 14
5 5
4 5
Output
205
-1
10
2
2
-1
Input
(01)
1
1 4
Output
1
Submitted Solution:
```
sae = input()
numberOfCalculations = int(input())
operations = ["*","+","-"]
def ExecuteOP(expresion, operation):
indexOfOp = expresion.index(operation)
rightPart = expresion[indexOfOp + 1:]
leftPart = expresion[:indexOfOp]
numbers = ""
indexRight = 0
for number in rightPart:
if number not in operations:
numbers += number
indexRight += 1
else:
break
rightNumbers = int(numbers)
rightPart = rightPart[:indexRight]
indexLeft = 0
for number in reversed(leftPart):
if number not in operations:
indexLeft += 1
else:
break
leftNumbers = int(leftPart[indexOfOp - indexLeft:])
leftPart = leftPart[:indexLeft]
if operation == "+":
result = (leftNumbers + rightNumbers)
elif operation == "-":
result = (leftNumbers - rightNumbers)
elif operation == "*":
result = (leftNumbers * rightNumbers)
else:
assert False, "Operation not supported -> " + operation
return expresion[:indexLeft - indexLeft] + str(result) + expresion[indexOfOp + indexRight + 1:]
def AritmeticExpresion(expresion):
while "*" in expresion:
expresion = ExecuteOP(expresion,"*")
while "+" in expresion:
expresion = ExecuteOP(expresion,"+")
while "-" in expresion:
expresion = ExecuteOP(expresion,"-")
return expresion
def CalculateSubSet(subset):
result = 0
if subset == -1 or subset == "-1":
return -1
if "(" in subset and ")" in subset:
resultSubset = CalculateSubSet(subset[subset.index("(") + 1:subset.index(")")])
newSubSet = subset[:subset.index("(")] + str(resultSubset) + subset[subset.index(")") + 1 :]
result += CalculateSubSet(newSubSet)
elif "(" not in subset and ")" not in subset:
result += int(AritmeticExpresion(subset))
else:
return -1
return result
for _ in range(numberOfCalculations):
actualSae = input().split(" ")
subExpresion = sae[int(actualSae[0]) - 1 :int(actualSae[1])]
try:
print(CalculateSubSet(subExpresion))
except:
print("-1")
``` | instruction | 0 | 29,609 | 18 | 59,218 |
No | output | 1 | 29,609 | 18 | 59,219 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A simplified arithmetic expression (SAE) is an arithmetic expression defined by the following grammar:
* <SAE> ::= <Number> | <SAE>+<SAE> | <SAE>*<SAE> | (<SAE>)
* <Number> ::= <Digit> | <Digit><Number>
* <Digit> ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
In other words it's a correct arithmetic expression that is allowed to contain brackets, numbers (possibly with leading zeros), multiplications and additions. For example expressions "(0+01)", "0" and "1*(0)" are simplified arithmetic expressions, but expressions "2-1", "+1" and "1+2)" are not.
Given a string s1s2...s|s| that represents a SAE; si denotes the i-th character of the string which can be either a digit ('0'-'9'), a plus sign ('+'), a multiplication sign ('*'), an opening round bracket '(' or a closing round bracket ')'.
A part slsl + 1...sr of this string is called a sub-expression if and only if it is a SAE.
You task is to answer m queries, each of which is a pair of integers li, ri (1 ≤ li ≤ ri ≤ |s|). For each query determine whether the corresponding part of the given string is a sub-expression and in case it's a sub-expression calculate its value modulo 1000000007 (109 + 7). The values should be calculated using standard operator priorities.
Input
The first line of the input contains non-empty string s (1 ≤ |s| ≤ 4·105) which represents a correct SAE. Each character of the string can be one of the following characters: '*', '+', '(', ')' or a digit ('0'-'9'). The expression might contain extra-huge numbers.
The second line contains an integer m (1 ≤ m ≤ 4·105) which is the number of queries. Each of the next m lines contains two space-separated integers li, ri (1 ≤ li ≤ ri ≤ |s|) — the i-th query.
Output
The i-th number of output should be the answer for the i-th query. If the i-th query corresponds to a valid sub-expression output the value of the sub-expression modulo 1000000007 (109 + 7). Otherwise output -1 as an answer for the query. Print numbers on separate lines.
Examples
Input
((1+2)*3+101*2)
6
8 14
1 6
2 10
11 14
5 5
4 5
Output
205
-1
10
2
2
-1
Input
(01)
1
1 4
Output
1
Submitted Solution:
```
sae = input()
numberOfCalculations = int(input())
operations = ["*","+","-"]
def ExecuteOP(expresion, operation):
indexOfOp = expresion.index(operation)
rightPart = expresion[indexOfOp + 1:]
leftPart = expresion[:indexOfOp]
numbers = ""
indexRight = 0
for number in rightPart:
if number not in operations:
numbers += number
indexRight += 1
else:
break
rightNumbers = int(numbers)
rightPart = rightPart[:indexRight]
indexLeft = 0
for number in reversed(leftPart):
if number not in operations:
indexLeft += 1
else:
break
leftNumbers = int(leftPart[indexOfOp - indexLeft:])
leftPart = leftPart[:indexLeft]
if operation == "+":
result = (leftNumbers + rightNumbers)
elif operation == "-":
result = (leftNumbers - rightNumbers)
elif operation == "*":
result = (leftNumbers * rightNumbers)
else:
assert False, "Operation not supported -> " + operation
return expresion[:indexOfOp - indexLeft] + str(result) + expresion[indexOfOp + indexRight + 1:]
def AritmeticExpresion(expresion):
while "*" in expresion:
expresion = ExecuteOP(expresion,"*")
while "+" in expresion:
expresion = ExecuteOP(expresion,"+")
while "-" in expresion:
expresion = ExecuteOP(expresion,"-")
return expresion
def CalculateSubSet(subset):
result = 0
if subset == -1 or subset == "-1":
return -1
if "(" in subset and ")" in subset:
numberOfOpeners = 0
numberOfClosers = 0
indexCloser = 0
for char in subset:
if char == "(":
numberOfOpeners +=1
elif char == ")":
numberOfClosers +=1
if numberOfClosers >= numberOfOpeners:
break;
indexCloser += 1
if numberOfClosers > numberOfOpeners:
return -1
resultSubset = CalculateSubSet(subset[subset.index("(") + 1:indexCloser ])
newSubSet = subset[:subset.index("(")] + str(resultSubset) + subset[indexCloser + 1:]
result += CalculateSubSet(newSubSet)
elif "(" not in subset and ")" not in subset:
result += int(AritmeticExpresion(subset))
else:
return -1
return result
for i in range(numberOfCalculations):
actualSae = input().split(" ")
subExpresion = sae[int(actualSae[0]) - 1 :int(actualSae[1])]
try:
result = int(CalculateSubSet(subExpresion))
print(result%1000000007 if result != -1 else result)
except:
print("-1")
``` | instruction | 0 | 29,610 | 18 | 59,220 |
No | output | 1 | 29,610 | 18 | 59,221 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a string A = A_1 A_2 ... A_n consisting of lowercase English letters.
You can choose any two indices i and j such that 1 \leq i \leq j \leq n and reverse substring A_i A_{i+1} ... A_j.
You can perform this operation at most once.
How many different strings can you obtain?
Constraints
* 1 \leq |A| \leq 200,000
* A consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
A
Output
Print the number of different strings you can obtain by reversing any substring in A at most once.
Examples
Input
aatt
Output
5
Input
xxxxxxxxxx
Output
1
Input
abracadabra
Output
44
Submitted Solution:
```
from collections import Counter
# python template for atcoder1
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline
def solve():
S = input()[:-1]
N = len(S)
C = Counter(S)
ans = 0
for v in C.values():
ans += v*(v-1)//2
print(N*(N-1)//2-ans)
solve()
``` | instruction | 0 | 29,866 | 18 | 59,732 |
No | output | 1 | 29,866 | 18 | 59,733 |
Provide a correct Python 3 solution for this coding contest problem.
Have you ever had an infinite loop when you ran a hard-working program? It would be convenient to be able to determine in advance whether a program will stop executing without having to execute it.
Unfortunately, it is not possible to make such a decision for any program in the programming language you normally use. However, if you have a programming language that is much less computationally powerful, you may be able to write a program that determines if a program written in that language will stop.
Consider a programming language called TinyPower. Programs in this language are line sequences. On each line of the program, write the line number at the beginning and one sentence after it. The types of sentences that can be written in this language are as follows.
Sentence type | Behavior
--- | ---
ADD var1 var2 var3 | Assign the result of adding the value of variable var2 and the value of var3 to variable var1
ADD var1 var2 con | Assign the result of adding the value of the variable var2 and the constant con to the variable var1
SUB var1 var2 var3 | Assign the result of subtracting the value of var3 from the value of variable var2 to variable var1
SUB var1 var2 con | Substitute the result of subtracting the constant con from the value of the variable var2 into the variable var1
SET var1 var2 | Assign the value of variable var2 to variable var1
SET var1 con | Assign the constant con to the variable var1
IF var1 dest | Jump to line number dest only if the value of variable var1 is non-zero
HALT | Stop the program
Line numbers are positive integers, and the same line number will never appear more than once in the program. Variables are represented by a single lowercase letter, and constants and variable values are integers. No variable declaration is required, the initial value of the variable is 0.
Program execution starts with the first statement, and the statements are executed in the order in which they are lined up. However, as written in the table above, if the value of the variable in the IF statement is not 0, jump to the line specified by the line number written after the variable and start from the statement written in that line. Continue running. The program will stop when:
* When the HALT statement is executed.
* When trying to assign a negative integer or an integer greater than or equal to 16 to a variable (the value of the variable is not updated).
* When trying to jump to a line number that does not appear in the program.
* When you do not jump to any line from the last statement of the program.
Create a program that determines if a TinyPower program, given it, will stop.
Input
The input is given in the following format.
N
stmt1
stmt2
::
stmtN
The number of lines N (1 ≤ N ≤ 50) of the program is given on the first line. The following N lines are given the statement stmti of the TinyPower program. stmti is given in one of the following formats:
line ADD var1 var2 var3
Or
line ADD var1 var2 con
Or
line SUB var1 var2 var3
Or
line SUB var1 var2 con
Or
line SET var1 var2
Or
line SET var1 con
Or
line IF var1 dest
Or
line HALT
line, dest (1 ≤ line, dest ≤ 1000) is the line number, varj (one lowercase letter) is the variable, and con (0 ≤ con ≤ 15) is the constant. The delimiter in stmti is one blank character. It is assumed that one or more variables always appear in the program, and only five different variable names appear.
Output
When the program stops, the results of the variables appearing in the program are output in the lexicographic order of the variable names, separated by line breaks, and when it does not stop, "inf" is output. The result of the variable is output by separating the variable name and the value of the variable with "=".
Examples
Input
6
10 SET c 1
20 SET i 5
100 ADD s s i
110 SUB i i c
120 IF i 100
200 HALT
Output
c=1
i=0
s=15
Input
3
10 SET c 1
120 IF c 10
20 HALT
Output
inf
Input
3
111 SET c 1
12 SUB c c 2
777 SET a 4
Output
a=0
c=1 | instruction | 0 | 29,931 | 18 | 59,862 |
"Correct Solution:
```
n = int(input())
stmts = [input().split() for i in range(n)]
mapper = {}
for i, stmt in enumerate(stmts):
mapper[stmt[0]] = i
vn = {}
for stmt in stmts:
for ope in stmt[2:]:
if ope.isalpha() and ope not in vn:
vn[ope] = len(vn)
vs = [0]*len(vn)
def stmt_add(v1, v2, v3):
v1 = vn[v1]
if v3.isdigit():
v2 = vn[v2]
const = int(v3)
def run(pc):
assert vs[v2] + const < 16
vs[v1] = vs[v2] + const
return pc+1
else:
v2 = vn[v2]
v3 = vn[v3]
def run(pc):
assert vs[v2] + vs[v3] < 16
vs[v1] = vs[v2] + vs[v3]
return pc+1
return run
def stmt_sub(v1, v2, v3):
v1 = vn[v1]
if v3.isdigit():
v2 = vn[v2]
const = int(v3)
def run(pc):
assert 0 <= vs[v2] - const
vs[v1] = vs[v2] - const
return pc+1
else:
v2 = vn[v2]
v3 = vn[v3]
def run(pc):
assert 0 <= vs[v2] - vs[v3]
vs[v1] = vs[v2] - vs[v3]
return pc+1
return run
def stmt_set(v1, v2):
v1 = vn[v1]
if v2.isdigit():
v2 = int(v2)
def run(pc):
vs[v1] = v2
return pc+1
else:
v2 = vn[v2]
def run(pc):
vs[v1] = vs[v2]
return pc+1
return run
def stmt_if(v1, dest):
v1 = vn[v1]
def run(pc):
return mapper[dest] if vs[v1] else pc+1
return run
def stmt_halt():
def run(pc):
return n
return run
stmt_func = []
for i, stmt in enumerate(stmts):
op = stmt[1]
if op == 'ADD':
stmt_func.append(stmt_add(*stmt[2:]))
elif op == 'SUB':
stmt_func.append(stmt_sub(*stmt[2:]))
elif op == 'SET':
stmt_func.append(stmt_set(*stmt[2:]))
elif op == 'IF':
stmt_func.append(stmt_if(*stmt[2:]))
else:
stmt_func.append(stmt_halt())
result = 1
pc = 0
cnts = [0]*n
LIM = 16**len(vn) + 10
try:
while 1:
run = stmt_func[pc]
cnts[pc] += 1
if cnts[pc] > LIM:
result = 0
break
pc = run(pc)
except:...
if result:
for k in sorted(vn):
print("%s=%d" % (k, vs[vn[k]]))
else:
print("inf")
``` | output | 1 | 29,931 | 18 | 59,863 |
Provide a correct Python 3 solution for this coding contest problem.
Have you ever had an infinite loop when you ran a hard-working program? It would be convenient to be able to determine in advance whether a program will stop executing without having to execute it.
Unfortunately, it is not possible to make such a decision for any program in the programming language you normally use. However, if you have a programming language that is much less computationally powerful, you may be able to write a program that determines if a program written in that language will stop.
Consider a programming language called TinyPower. Programs in this language are line sequences. On each line of the program, write the line number at the beginning and one sentence after it. The types of sentences that can be written in this language are as follows.
Sentence type | Behavior
--- | ---
ADD var1 var2 var3 | Assign the result of adding the value of variable var2 and the value of var3 to variable var1
ADD var1 var2 con | Assign the result of adding the value of the variable var2 and the constant con to the variable var1
SUB var1 var2 var3 | Assign the result of subtracting the value of var3 from the value of variable var2 to variable var1
SUB var1 var2 con | Substitute the result of subtracting the constant con from the value of the variable var2 into the variable var1
SET var1 var2 | Assign the value of variable var2 to variable var1
SET var1 con | Assign the constant con to the variable var1
IF var1 dest | Jump to line number dest only if the value of variable var1 is non-zero
HALT | Stop the program
Line numbers are positive integers, and the same line number will never appear more than once in the program. Variables are represented by a single lowercase letter, and constants and variable values are integers. No variable declaration is required, the initial value of the variable is 0.
Program execution starts with the first statement, and the statements are executed in the order in which they are lined up. However, as written in the table above, if the value of the variable in the IF statement is not 0, jump to the line specified by the line number written after the variable and start from the statement written in that line. Continue running. The program will stop when:
* When the HALT statement is executed.
* When trying to assign a negative integer or an integer greater than or equal to 16 to a variable (the value of the variable is not updated).
* When trying to jump to a line number that does not appear in the program.
* When you do not jump to any line from the last statement of the program.
Create a program that determines if a TinyPower program, given it, will stop.
Input
The input is given in the following format.
N
stmt1
stmt2
::
stmtN
The number of lines N (1 ≤ N ≤ 50) of the program is given on the first line. The following N lines are given the statement stmti of the TinyPower program. stmti is given in one of the following formats:
line ADD var1 var2 var3
Or
line ADD var1 var2 con
Or
line SUB var1 var2 var3
Or
line SUB var1 var2 con
Or
line SET var1 var2
Or
line SET var1 con
Or
line IF var1 dest
Or
line HALT
line, dest (1 ≤ line, dest ≤ 1000) is the line number, varj (one lowercase letter) is the variable, and con (0 ≤ con ≤ 15) is the constant. The delimiter in stmti is one blank character. It is assumed that one or more variables always appear in the program, and only five different variable names appear.
Output
When the program stops, the results of the variables appearing in the program are output in the lexicographic order of the variable names, separated by line breaks, and when it does not stop, "inf" is output. The result of the variable is output by separating the variable name and the value of the variable with "=".
Examples
Input
6
10 SET c 1
20 SET i 5
100 ADD s s i
110 SUB i i c
120 IF i 100
200 HALT
Output
c=1
i=0
s=15
Input
3
10 SET c 1
120 IF c 10
20 HALT
Output
inf
Input
3
111 SET c 1
12 SUB c c 2
777 SET a 4
Output
a=0
c=1 | instruction | 0 | 29,932 | 18 | 59,864 |
"Correct Solution:
```
def solve():
from sys import stdin
f_i = stdin
N = int(f_i.readline())
from string import ascii_lowercase
next_line = {}
S = {} # statements
V = {} # values
line = 0
for i in range(N):
stmt = f_i.readline().split()
next_line[line] = stmt[0]
line = stmt[0]
op = stmt[1]
if op == 'ADD' or op == 'SUB':
v1, v2, v3 = stmt[2:]
V[v1] = 0
V[v2] = 0
if v3 in ascii_lowercase:
V[v3] = 0
else:
V[v3] = int(v3)
S[line] = (op, v1, v2, v3)
elif op == 'SET':
v1, v2 = stmt[2:]
V[v1] = 0
if v2 in ascii_lowercase:
V[v2] = 0
else:
V[v2] = int(v2)
S[line] = (op, v1, v2)
elif op == 'IF':
v1, d = stmt[2:]
V[v1] = 0
S[line] = (op, v1, d)
else:
S[line] = (op,)
next_line[line] = None
v_num = 0
for k in V:
if k in ascii_lowercase:
v_num += 1
state_limit = len(S) * (16 ** v_num)
nl = next_line[0]
s = S[nl]
cnt = 0
while cnt < state_limit:
cnt += 1
op = s[0]
if op == 'HALT':
break
elif op == 'ADD':
v = V[s[2]] + V[s[3]]
if v >= 16:
break
V[s[1]] = v
elif op == 'SUB':
v = V[s[2]] - V[s[3]]
if v < 0:
break
V[s[1]] = v
elif op == 'SET':
V[s[1]] = V[s[2]]
else:
if V[s[1]]:
nl = s[2]
if nl in next_line:
s = S[nl]
continue
else:
break
nl = next_line[nl]
if nl == None:
break
s = S[nl]
if cnt == state_limit:
print('inf')
else:
ans = []
for k, v in V.items():
if k in ascii_lowercase:
ans.append(('{}={}'.format(k, v)))
ans.sort()
print('\n'.join(ans))
solve()
``` | output | 1 | 29,932 | 18 | 59,865 |
Provide a correct Python 3 solution for this coding contest problem.
Have you ever had an infinite loop when you ran a hard-working program? It would be convenient to be able to determine in advance whether a program will stop executing without having to execute it.
Unfortunately, it is not possible to make such a decision for any program in the programming language you normally use. However, if you have a programming language that is much less computationally powerful, you may be able to write a program that determines if a program written in that language will stop.
Consider a programming language called TinyPower. Programs in this language are line sequences. On each line of the program, write the line number at the beginning and one sentence after it. The types of sentences that can be written in this language are as follows.
Sentence type | Behavior
--- | ---
ADD var1 var2 var3 | Assign the result of adding the value of variable var2 and the value of var3 to variable var1
ADD var1 var2 con | Assign the result of adding the value of the variable var2 and the constant con to the variable var1
SUB var1 var2 var3 | Assign the result of subtracting the value of var3 from the value of variable var2 to variable var1
SUB var1 var2 con | Substitute the result of subtracting the constant con from the value of the variable var2 into the variable var1
SET var1 var2 | Assign the value of variable var2 to variable var1
SET var1 con | Assign the constant con to the variable var1
IF var1 dest | Jump to line number dest only if the value of variable var1 is non-zero
HALT | Stop the program
Line numbers are positive integers, and the same line number will never appear more than once in the program. Variables are represented by a single lowercase letter, and constants and variable values are integers. No variable declaration is required, the initial value of the variable is 0.
Program execution starts with the first statement, and the statements are executed in the order in which they are lined up. However, as written in the table above, if the value of the variable in the IF statement is not 0, jump to the line specified by the line number written after the variable and start from the statement written in that line. Continue running. The program will stop when:
* When the HALT statement is executed.
* When trying to assign a negative integer or an integer greater than or equal to 16 to a variable (the value of the variable is not updated).
* When trying to jump to a line number that does not appear in the program.
* When you do not jump to any line from the last statement of the program.
Create a program that determines if a TinyPower program, given it, will stop.
Input
The input is given in the following format.
N
stmt1
stmt2
::
stmtN
The number of lines N (1 ≤ N ≤ 50) of the program is given on the first line. The following N lines are given the statement stmti of the TinyPower program. stmti is given in one of the following formats:
line ADD var1 var2 var3
Or
line ADD var1 var2 con
Or
line SUB var1 var2 var3
Or
line SUB var1 var2 con
Or
line SET var1 var2
Or
line SET var1 con
Or
line IF var1 dest
Or
line HALT
line, dest (1 ≤ line, dest ≤ 1000) is the line number, varj (one lowercase letter) is the variable, and con (0 ≤ con ≤ 15) is the constant. The delimiter in stmti is one blank character. It is assumed that one or more variables always appear in the program, and only five different variable names appear.
Output
When the program stops, the results of the variables appearing in the program are output in the lexicographic order of the variable names, separated by line breaks, and when it does not stop, "inf" is output. The result of the variable is output by separating the variable name and the value of the variable with "=".
Examples
Input
6
10 SET c 1
20 SET i 5
100 ADD s s i
110 SUB i i c
120 IF i 100
200 HALT
Output
c=1
i=0
s=15
Input
3
10 SET c 1
120 IF c 10
20 HALT
Output
inf
Input
3
111 SET c 1
12 SUB c c 2
777 SET a 4
Output
a=0
c=1 | instruction | 0 | 29,933 | 18 | 59,866 |
"Correct Solution:
```
from itertools import product
def check():
n = int(input())
prog = [input().split() for _ in range(n)]
v = set()
for ss in prog:
for s in ss:
if "a" <= s <= "z":
v.add(s)
v = sorted(list(v))
var_state = [0] * len(v)
var_to_ind = dict(zip(v, range(len(v))))
line_to_ind = {}
for i, ss in enumerate(prog):
line_to_ind[ss[0]] = i
def fix(p):
if p[1] == "ADD":
if p[4] in var_to_ind:
return [0, var_to_ind[p[2]], var_to_ind[p[3]], var_to_ind[p[4]]]
else:
return [1, var_to_ind[p[2]], var_to_ind[p[3]], int(p[4])]
if p[1] == "SUB":
if p[4] in var_to_ind:
return [2, var_to_ind[p[2]], var_to_ind[p[3]], var_to_ind[p[4]]]
else:
return [3, var_to_ind[p[2]], var_to_ind[p[3]], int(p[4])]
if p[1] == "SET":
if p[3] in var_to_ind:
return [4, var_to_ind[p[2]], var_to_ind[p[3]]]
else:
return [5, var_to_ind[p[2]], int(p[3])]
if p[1] == "IF":
if p[3] in line_to_ind:
return [6, var_to_ind[p[2]], line_to_ind[p[3]]]
else:
return [6, var_to_ind[p[2]], 100]
if p[1] == "HALT":
return [7]
prog = list(map(fix, prog))
used = [[False] * n for _ in range(16 ** len(v))]
xs = [1, 16, 256, 4096, 65536]
ind = 0
h = 0
while True:
if ind >= n:
return True, v, var_state
if used[h][ind]:
return False, v, var_state
used[h][ind] = True
p = prog[ind]
ins = p[0]
if ins == 0:
v1, v2, v3 = p[1:]
temp = var_state[v2] + var_state[v3]
if not temp < 16:
return True, v, var_state
h += (temp - var_state[v1]) * xs[v1]
var_state[v1] = temp
elif ins == 1:
v1, v2 = p[1:3]
con = p[3]
temp = var_state[v2] + con
if not temp < 16:
return True, v, var_state
h += (temp - var_state[v1]) * xs[v1]
var_state[v1] = temp
elif ins == 2:
v1, v2, v3 = p[1:]
temp = var_state[v2] - var_state[v3]
if not 0 <= temp:
return True, v, var_state
h += (temp - var_state[v1]) * xs[v1]
var_state[v1] = temp
elif ins == 3:
v1, v2 = p[1:3]
con = p[3]
temp = var_state[v2] - con
if not 0 <= temp:
return True, v, var_state
h += (temp - var_state[v1]) * xs[v1]
var_state[v1] = temp
elif ins == 4:
v1, v2 = p[1:]
h += (var_state[v2] - var_state[v1]) * xs[v1]
var_state[v1] = var_state[v2]
elif ins == 5:
v1 = p[1]
con = p[2]
h += (con - var_state[v1]) * xs[v1]
var_state[v1] = con
elif ins == 6:
v1 = p[1]
dest = p[2]
if var_state[v1]:
ind = dest - 1
else:
return True, v, var_state
ind += 1
flag, v, var_state = check()
if flag:
for t in zip(v, var_state):
print(*t, sep="=")
else:
print("inf")
``` | output | 1 | 29,933 | 18 | 59,867 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Broken crypto generator
JAG (Japanese Alumni Group) is a mysterious organization composed of many programmers, and in order to enter the building where the headquarters of this organization is located, it is necessary to solve the ciphertext generated by a certain machine every time. This ciphertext consists of the symbols'+','-','[',']' and the uppercase alphabet, and is represented by <Cipher> defined by the following BNF.
<Cipher> :: = <String> | <Cipher> <String>
<String> :: = <Letter> |'['<Cipher>']'
<Letter> :: ='+' <Letter> |'-' <Letter> |
'A' |'B' |'C' |'D' |'E' |'F' |'G' |'H' |'I' |'J' |'K' |'L' |'M '|
'N' |'O' |'P' |'Q' |'R' |'S' |'T' |'U' |'V' |'W' |'X' |'Y' |'Z '
Here, each symbol has the following meaning.
* + (Character): Represents the alphabet following that character (provided that the alphabet following'Z'is'A')
*-(Character): Represents the alphabet before that character (provided that the alphabet before'A'is'Z')
* [(Character string)]: Represents a character string that is horizontally inverted.
However, the machine that generates this ciphertext is currently out of order, and some letters of the alphabet in the ciphertext may be broken and unreadable. Unreadable characters are tentatively represented as'?'. As a result of the investigation, it was found that the method of filling the broken characters is such that the decrypted character string is the smallest in the dictionary order among the possible character strings after decoding. Your job is to decrypt this ciphertext correctly.
Input
The input consists of multiple datasets. Each dataset consists of one line containing a string in which some uppercase letters have been replaced with ‘?’ In the ciphertext defined by BNF above. You can assume that the length of each string is less than $ 80 $. You can also assume that the number of'?' In each dataset is greater than or equal to $ 0 $ and less than or equal to $ 3 $.
The end of the input is represented by a line containing only one character,'.'.
Output
For each data set, output the decrypted character string when the ciphertext is decrypted so that the decrypted character string is the smallest in the dictionary order.
Sample Input
A + A ++ A
Z-Z--Z + -Z
[ESREVER]
J ---? --- J
++++++++ A +++ Z ----------- A +++ Z
[[++-+-? [-++-? ++-+++ L]] [-+ ----- + -O]] ++++ --- + L
..
Output for Sample Input
ABC
ZYXZ
REVERSE
JAG
ICPC
JAPAN
Example
Input
A+A++A
Z-Z--Z+-Z
[ESREVER]
J---?---J
++++++++A+++Z-----------A+++Z
[[++-+--?[--++-?++-+++L]][-+-----+-O]]++++---+L
.
Output
ABC
ZYXZ
REVERSE
JAG
ICPC
JAPAN
Submitted Solution:
```
def c_to_int(char):
return ord(char)-65
def solve(s):
count = 0
table=[]
for str in s:
if str=='+':
count +=1
elif str =='-':
count -=1
elif str=='[':
table.append(str)
elif str==']':
table.append(str)
elif str=='?':
table.append(str)
count =0
else:
table.append((c_to_int(str) +count)%26)
count = 0
for i in range(len(table)):
if table[i]=='?':
table[i] = 0
# print(table)
return(table)
def rev(table):
ret = ""
i =0
while i < len(table):
if table[i] == '[':
c = 1
i_=i+1
while c>0:
if table[i_] =='[':
c += 1
if table[i_] ==']':
c -= 1
i_ += 1
# print(reversed(table[i+1:i_-1]))
ret += rev(table[i+1:i_-1])[::-1]
i = i_-1
elif table[i] ==']':
pass
else:
# print(chr(table[i]+65))
ret += chr(table[i]+65)
i+=1
return ret
while True:
S = input()
if S[0]=='.':
break
ans = rev(solve(S))
print(ans)
``` | instruction | 0 | 29,995 | 18 | 59,990 |
Yes | output | 1 | 29,995 | 18 | 59,991 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Broken crypto generator
JAG (Japanese Alumni Group) is a mysterious organization composed of many programmers, and in order to enter the building where the headquarters of this organization is located, it is necessary to solve the ciphertext generated by a certain machine every time. This ciphertext consists of the symbols'+','-','[',']' and the uppercase alphabet, and is represented by <Cipher> defined by the following BNF.
<Cipher> :: = <String> | <Cipher> <String>
<String> :: = <Letter> |'['<Cipher>']'
<Letter> :: ='+' <Letter> |'-' <Letter> |
'A' |'B' |'C' |'D' |'E' |'F' |'G' |'H' |'I' |'J' |'K' |'L' |'M '|
'N' |'O' |'P' |'Q' |'R' |'S' |'T' |'U' |'V' |'W' |'X' |'Y' |'Z '
Here, each symbol has the following meaning.
* + (Character): Represents the alphabet following that character (provided that the alphabet following'Z'is'A')
*-(Character): Represents the alphabet before that character (provided that the alphabet before'A'is'Z')
* [(Character string)]: Represents a character string that is horizontally inverted.
However, the machine that generates this ciphertext is currently out of order, and some letters of the alphabet in the ciphertext may be broken and unreadable. Unreadable characters are tentatively represented as'?'. As a result of the investigation, it was found that the method of filling the broken characters is such that the decrypted character string is the smallest in the dictionary order among the possible character strings after decoding. Your job is to decrypt this ciphertext correctly.
Input
The input consists of multiple datasets. Each dataset consists of one line containing a string in which some uppercase letters have been replaced with ‘?’ In the ciphertext defined by BNF above. You can assume that the length of each string is less than $ 80 $. You can also assume that the number of'?' In each dataset is greater than or equal to $ 0 $ and less than or equal to $ 3 $.
The end of the input is represented by a line containing only one character,'.'.
Output
For each data set, output the decrypted character string when the ciphertext is decrypted so that the decrypted character string is the smallest in the dictionary order.
Sample Input
A + A ++ A
Z-Z--Z + -Z
[ESREVER]
J ---? --- J
++++++++ A +++ Z ----------- A +++ Z
[[++-+-? [-++-? ++-+++ L]] [-+ ----- + -O]] ++++ --- + L
..
Output for Sample Input
ABC
ZYXZ
REVERSE
JAG
ICPC
JAPAN
Example
Input
A+A++A
Z-Z--Z+-Z
[ESREVER]
J---?---J
++++++++A+++Z-----------A+++Z
[[++-+--?[--++-?++-+++L]][-+-----+-O]]++++---+L
.
Output
ABC
ZYXZ
REVERSE
JAG
ICPC
JAPAN
Submitted Solution:
```
import re
bracket = re.compile(r'\[([\w?+-]+)\]')
incr = re.compile(r'([+-]*)([\w?])')
def f(s, c):
if c == '?': return 'A'
return chr(ord('A')+(ord(c)-ord('A')+s.count('+')-s.count('-'))%26)
while True:
s = input().strip()
if s == '.': break
s = incr.sub(lambda m: f(*m.group(1, 2)), s)
while '[' in s:
s = bracket.sub(lambda m: m.group(1)[::-1], s)
print(s)
``` | instruction | 0 | 29,996 | 18 | 59,992 |
Yes | output | 1 | 29,996 | 18 | 59,993 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Broken crypto generator
JAG (Japanese Alumni Group) is a mysterious organization composed of many programmers, and in order to enter the building where the headquarters of this organization is located, it is necessary to solve the ciphertext generated by a certain machine every time. This ciphertext consists of the symbols'+','-','[',']' and the uppercase alphabet, and is represented by <Cipher> defined by the following BNF.
<Cipher> :: = <String> | <Cipher> <String>
<String> :: = <Letter> |'['<Cipher>']'
<Letter> :: ='+' <Letter> |'-' <Letter> |
'A' |'B' |'C' |'D' |'E' |'F' |'G' |'H' |'I' |'J' |'K' |'L' |'M '|
'N' |'O' |'P' |'Q' |'R' |'S' |'T' |'U' |'V' |'W' |'X' |'Y' |'Z '
Here, each symbol has the following meaning.
* + (Character): Represents the alphabet following that character (provided that the alphabet following'Z'is'A')
*-(Character): Represents the alphabet before that character (provided that the alphabet before'A'is'Z')
* [(Character string)]: Represents a character string that is horizontally inverted.
However, the machine that generates this ciphertext is currently out of order, and some letters of the alphabet in the ciphertext may be broken and unreadable. Unreadable characters are tentatively represented as'?'. As a result of the investigation, it was found that the method of filling the broken characters is such that the decrypted character string is the smallest in the dictionary order among the possible character strings after decoding. Your job is to decrypt this ciphertext correctly.
Input
The input consists of multiple datasets. Each dataset consists of one line containing a string in which some uppercase letters have been replaced with ‘?’ In the ciphertext defined by BNF above. You can assume that the length of each string is less than $ 80 $. You can also assume that the number of'?' In each dataset is greater than or equal to $ 0 $ and less than or equal to $ 3 $.
The end of the input is represented by a line containing only one character,'.'.
Output
For each data set, output the decrypted character string when the ciphertext is decrypted so that the decrypted character string is the smallest in the dictionary order.
Sample Input
A + A ++ A
Z-Z--Z + -Z
[ESREVER]
J ---? --- J
++++++++ A +++ Z ----------- A +++ Z
[[++-+-? [-++-? ++-+++ L]] [-+ ----- + -O]] ++++ --- + L
..
Output for Sample Input
ABC
ZYXZ
REVERSE
JAG
ICPC
JAPAN
Example
Input
A+A++A
Z-Z--Z+-Z
[ESREVER]
J---?---J
++++++++A+++Z-----------A+++Z
[[++-+--?[--++-?++-+++L]][-+-----+-O]]++++---+L
.
Output
ABC
ZYXZ
REVERSE
JAG
ICPC
JAPAN
Submitted Solution:
```
while True:
s = input()
if s == '.':break
while True:
if s.find('+') == -1 and s.find('-') == -1:
index = s.rfind('[')
if index == -1:break
e = s.find(']', index, len(s))
s = s[:index] + s[index + 1:e][::-1] + s[e + 1:]
else:
index = s.find('+')
if index == -1:
index = s.find('-')
if s.find('-') != -1:
index = min(index, s.find('-'))
e = index
f = 0
while s[e] == '+' or s[e] == '-':
f += [-1, 1][s[e] == '+']
e += 1
r = s[e]
if r != '?':
r = chr((ord(r) - ord('A') + f) % 26 + ord('A'))
s = s[:index] + r + s[e + 1:]
print(s.replace('?', 'A'))
``` | instruction | 0 | 29,997 | 18 | 59,994 |
Yes | output | 1 | 29,997 | 18 | 59,995 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Broken crypto generator
JAG (Japanese Alumni Group) is a mysterious organization composed of many programmers, and in order to enter the building where the headquarters of this organization is located, it is necessary to solve the ciphertext generated by a certain machine every time. This ciphertext consists of the symbols'+','-','[',']' and the uppercase alphabet, and is represented by <Cipher> defined by the following BNF.
<Cipher> :: = <String> | <Cipher> <String>
<String> :: = <Letter> |'['<Cipher>']'
<Letter> :: ='+' <Letter> |'-' <Letter> |
'A' |'B' |'C' |'D' |'E' |'F' |'G' |'H' |'I' |'J' |'K' |'L' |'M '|
'N' |'O' |'P' |'Q' |'R' |'S' |'T' |'U' |'V' |'W' |'X' |'Y' |'Z '
Here, each symbol has the following meaning.
* + (Character): Represents the alphabet following that character (provided that the alphabet following'Z'is'A')
*-(Character): Represents the alphabet before that character (provided that the alphabet before'A'is'Z')
* [(Character string)]: Represents a character string that is horizontally inverted.
However, the machine that generates this ciphertext is currently out of order, and some letters of the alphabet in the ciphertext may be broken and unreadable. Unreadable characters are tentatively represented as'?'. As a result of the investigation, it was found that the method of filling the broken characters is such that the decrypted character string is the smallest in the dictionary order among the possible character strings after decoding. Your job is to decrypt this ciphertext correctly.
Input
The input consists of multiple datasets. Each dataset consists of one line containing a string in which some uppercase letters have been replaced with ‘?’ In the ciphertext defined by BNF above. You can assume that the length of each string is less than $ 80 $. You can also assume that the number of'?' In each dataset is greater than or equal to $ 0 $ and less than or equal to $ 3 $.
The end of the input is represented by a line containing only one character,'.'.
Output
For each data set, output the decrypted character string when the ciphertext is decrypted so that the decrypted character string is the smallest in the dictionary order.
Sample Input
A + A ++ A
Z-Z--Z + -Z
[ESREVER]
J ---? --- J
++++++++ A +++ Z ----------- A +++ Z
[[++-+-? [-++-? ++-+++ L]] [-+ ----- + -O]] ++++ --- + L
..
Output for Sample Input
ABC
ZYXZ
REVERSE
JAG
ICPC
JAPAN
Example
Input
A+A++A
Z-Z--Z+-Z
[ESREVER]
J---?---J
++++++++A+++Z-----------A+++Z
[[++-+--?[--++-?++-+++L]][-+-----+-O]]++++---+L
.
Output
ABC
ZYXZ
REVERSE
JAG
ICPC
JAPAN
Submitted Solution:
```
import string
abc = string.ascii_uppercase
while True:
s = input()
if s==".":
break
num = 0
rev = 0
nest = ["" for i in range(80)]
for e in s:
if e=="+":
num += 1
elif e=="-":
num -= 1
elif e=="[":
rev += 1
elif e=="]":
t = nest[rev]
nest[rev-1] += t[::-1]
nest[rev] = ""
rev -= 1
elif e=="?":
nest[rev] += "A"
num = 0
else:
t = abc[(ord(e)+num-65)%26]
nest[rev] += t
num = 0
print(nest[0])
``` | instruction | 0 | 29,998 | 18 | 59,996 |
Yes | output | 1 | 29,998 | 18 | 59,997 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Broken crypto generator
JAG (Japanese Alumni Group) is a mysterious organization composed of many programmers, and in order to enter the building where the headquarters of this organization is located, it is necessary to solve the ciphertext generated by a certain machine every time. This ciphertext consists of the symbols'+','-','[',']' and the uppercase alphabet, and is represented by <Cipher> defined by the following BNF.
<Cipher> :: = <String> | <Cipher> <String>
<String> :: = <Letter> |'['<Cipher>']'
<Letter> :: ='+' <Letter> |'-' <Letter> |
'A' |'B' |'C' |'D' |'E' |'F' |'G' |'H' |'I' |'J' |'K' |'L' |'M '|
'N' |'O' |'P' |'Q' |'R' |'S' |'T' |'U' |'V' |'W' |'X' |'Y' |'Z '
Here, each symbol has the following meaning.
* + (Character): Represents the alphabet following that character (provided that the alphabet following'Z'is'A')
*-(Character): Represents the alphabet before that character (provided that the alphabet before'A'is'Z')
* [(Character string)]: Represents a character string that is horizontally inverted.
However, the machine that generates this ciphertext is currently out of order, and some letters of the alphabet in the ciphertext may be broken and unreadable. Unreadable characters are tentatively represented as'?'. As a result of the investigation, it was found that the method of filling the broken characters is such that the decrypted character string is the smallest in the dictionary order among the possible character strings after decoding. Your job is to decrypt this ciphertext correctly.
Input
The input consists of multiple datasets. Each dataset consists of one line containing a string in which some uppercase letters have been replaced with ‘?’ In the ciphertext defined by BNF above. You can assume that the length of each string is less than $ 80 $. You can also assume that the number of'?' In each dataset is greater than or equal to $ 0 $ and less than or equal to $ 3 $.
The end of the input is represented by a line containing only one character,'.'.
Output
For each data set, output the decrypted character string when the ciphertext is decrypted so that the decrypted character string is the smallest in the dictionary order.
Sample Input
A + A ++ A
Z-Z--Z + -Z
[ESREVER]
J ---? --- J
++++++++ A +++ Z ----------- A +++ Z
[[++-+-? [-++-? ++-+++ L]] [-+ ----- + -O]] ++++ --- + L
..
Output for Sample Input
ABC
ZYXZ
REVERSE
JAG
ICPC
JAPAN
Example
Input
A+A++A
Z-Z--Z+-Z
[ESREVER]
J---?---J
++++++++A+++Z-----------A+++Z
[[++-+--?[--++-?++-+++L]][-+-----+-O]]++++---+L
.
Output
ABC
ZYXZ
REVERSE
JAG
ICPC
JAPAN
Submitted Solution:
```
while True:
s = input()
if s == '.':break
while True:
index = min(max(0, s.find('+')), max(0, s.find('-')))
if index == 0 and s[0] != '+' and s[0] != '-':
index = s.rfind('[')
if index == -1:
break
e = s[index:].find(']')
s = s[:index] + s[index + 1:index + e][::-1] + s[index + e + 1:]
else:
e = index
f = 0
while s[e] == '+' or s[e] == '-':
if s[e] == '+':
f += 1
else:
f -= 1
e += 1
r = s[e]
if r != '?':
r = chr(ord(r) + f)
s = s[:index] + r + s[e + 1:]
print(s.replace('?', 'A'))
``` | instruction | 0 | 29,999 | 18 | 59,998 |
No | output | 1 | 29,999 | 18 | 59,999 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Broken crypto generator
JAG (Japanese Alumni Group) is a mysterious organization composed of many programmers, and in order to enter the building where the headquarters of this organization is located, it is necessary to solve the ciphertext generated by a certain machine every time. This ciphertext consists of the symbols'+','-','[',']' and the uppercase alphabet, and is represented by <Cipher> defined by the following BNF.
<Cipher> :: = <String> | <Cipher> <String>
<String> :: = <Letter> |'['<Cipher>']'
<Letter> :: ='+' <Letter> |'-' <Letter> |
'A' |'B' |'C' |'D' |'E' |'F' |'G' |'H' |'I' |'J' |'K' |'L' |'M '|
'N' |'O' |'P' |'Q' |'R' |'S' |'T' |'U' |'V' |'W' |'X' |'Y' |'Z '
Here, each symbol has the following meaning.
* + (Character): Represents the alphabet following that character (provided that the alphabet following'Z'is'A')
*-(Character): Represents the alphabet before that character (provided that the alphabet before'A'is'Z')
* [(Character string)]: Represents a character string that is horizontally inverted.
However, the machine that generates this ciphertext is currently out of order, and some letters of the alphabet in the ciphertext may be broken and unreadable. Unreadable characters are tentatively represented as'?'. As a result of the investigation, it was found that the method of filling the broken characters is such that the decrypted character string is the smallest in the dictionary order among the possible character strings after decoding. Your job is to decrypt this ciphertext correctly.
Input
The input consists of multiple datasets. Each dataset consists of one line containing a string in which some uppercase letters have been replaced with ‘?’ In the ciphertext defined by BNF above. You can assume that the length of each string is less than $ 80 $. You can also assume that the number of'?' In each dataset is greater than or equal to $ 0 $ and less than or equal to $ 3 $.
The end of the input is represented by a line containing only one character,'.'.
Output
For each data set, output the decrypted character string when the ciphertext is decrypted so that the decrypted character string is the smallest in the dictionary order.
Sample Input
A + A ++ A
Z-Z--Z + -Z
[ESREVER]
J ---? --- J
++++++++ A +++ Z ----------- A +++ Z
[[++-+-? [-++-? ++-+++ L]] [-+ ----- + -O]] ++++ --- + L
..
Output for Sample Input
ABC
ZYXZ
REVERSE
JAG
ICPC
JAPAN
Example
Input
A+A++A
Z-Z--Z+-Z
[ESREVER]
J---?---J
++++++++A+++Z-----------A+++Z
[[++-+--?[--++-?++-+++L]][-+-----+-O]]++++---+L
.
Output
ABC
ZYXZ
REVERSE
JAG
ICPC
JAPAN
Submitted Solution:
```
while True:
s = input()
if s == '.':break
while True:
index = min(max(0, s.find('+')), max(0, s.find('-')))
if index == 0 and s[0] != '+' and s[0] != '-':
index = s.rfind('[')
if index == -1:
break
e = s[index:].find(']')
s = s[:index] + s[index + 1:e][::-1] + s[e + 1:]
else:
e = index
f = 0
while s[e] == '+' or s[e] == '-':
e += 1
if s[e] == '+':
f += 1
else:
f -= 1
r = s[e]
if r != '?':
r = chr(ord(r) += f)
s = s[:index] + r + s[e + 1:]
print(s.replace('?','A'))
``` | instruction | 0 | 30,000 | 18 | 60,000 |
No | output | 1 | 30,000 | 18 | 60,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Broken crypto generator
JAG (Japanese Alumni Group) is a mysterious organization composed of many programmers, and in order to enter the building where the headquarters of this organization is located, it is necessary to solve the ciphertext generated by a certain machine every time. This ciphertext consists of the symbols'+','-','[',']' and the uppercase alphabet, and is represented by <Cipher> defined by the following BNF.
<Cipher> :: = <String> | <Cipher> <String>
<String> :: = <Letter> |'['<Cipher>']'
<Letter> :: ='+' <Letter> |'-' <Letter> |
'A' |'B' |'C' |'D' |'E' |'F' |'G' |'H' |'I' |'J' |'K' |'L' |'M '|
'N' |'O' |'P' |'Q' |'R' |'S' |'T' |'U' |'V' |'W' |'X' |'Y' |'Z '
Here, each symbol has the following meaning.
* + (Character): Represents the alphabet following that character (provided that the alphabet following'Z'is'A')
*-(Character): Represents the alphabet before that character (provided that the alphabet before'A'is'Z')
* [(Character string)]: Represents a character string that is horizontally inverted.
However, the machine that generates this ciphertext is currently out of order, and some letters of the alphabet in the ciphertext may be broken and unreadable. Unreadable characters are tentatively represented as'?'. As a result of the investigation, it was found that the method of filling the broken characters is such that the decrypted character string is the smallest in the dictionary order among the possible character strings after decoding. Your job is to decrypt this ciphertext correctly.
Input
The input consists of multiple datasets. Each dataset consists of one line containing a string in which some uppercase letters have been replaced with ‘?’ In the ciphertext defined by BNF above. You can assume that the length of each string is less than $ 80 $. You can also assume that the number of'?' In each dataset is greater than or equal to $ 0 $ and less than or equal to $ 3 $.
The end of the input is represented by a line containing only one character,'.'.
Output
For each data set, output the decrypted character string when the ciphertext is decrypted so that the decrypted character string is the smallest in the dictionary order.
Sample Input
A + A ++ A
Z-Z--Z + -Z
[ESREVER]
J ---? --- J
++++++++ A +++ Z ----------- A +++ Z
[[++-+-? [-++-? ++-+++ L]] [-+ ----- + -O]] ++++ --- + L
..
Output for Sample Input
ABC
ZYXZ
REVERSE
JAG
ICPC
JAPAN
Example
Input
A+A++A
Z-Z--Z+-Z
[ESREVER]
J---?---J
++++++++A+++Z-----------A+++Z
[[++-+--?[--++-?++-+++L]][-+-----+-O]]++++---+L
.
Output
ABC
ZYXZ
REVERSE
JAG
ICPC
JAPAN
Submitted Solution:
```
while True:
s = input()
if s == '.':break
while True:
if s.find('+') == -1 and s.find('-') == -1:
index = s.rfind('[')
if index == -1:break
e = s.find(']', index, len(s))
s = s[:index] + s[index + 1:e][::-1] + s[e + 1:]
else:
index = s.find('+')
if index == -1:
index = s.find('-')
if s.find('-') != -1
index = min(index, s.find('-'))
e = index
f = 0
while s[e] == '+' or s[e] == '-':
f += [-1, 1][s[e] == '+']
e += 1
r = s[e]
if r != '?':
r = chr((ord(r) - ord('A') + f) % 26 + ord('A'))
s = s[:index] + r + s[e + 1:]
print(s.replace('?', 'A'))
``` | instruction | 0 | 30,001 | 18 | 60,002 |
No | output | 1 | 30,001 | 18 | 60,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Broken crypto generator
JAG (Japanese Alumni Group) is a mysterious organization composed of many programmers, and in order to enter the building where the headquarters of this organization is located, it is necessary to solve the ciphertext generated by a certain machine every time. This ciphertext consists of the symbols'+','-','[',']' and the uppercase alphabet, and is represented by <Cipher> defined by the following BNF.
<Cipher> :: = <String> | <Cipher> <String>
<String> :: = <Letter> |'['<Cipher>']'
<Letter> :: ='+' <Letter> |'-' <Letter> |
'A' |'B' |'C' |'D' |'E' |'F' |'G' |'H' |'I' |'J' |'K' |'L' |'M '|
'N' |'O' |'P' |'Q' |'R' |'S' |'T' |'U' |'V' |'W' |'X' |'Y' |'Z '
Here, each symbol has the following meaning.
* + (Character): Represents the alphabet following that character (provided that the alphabet following'Z'is'A')
*-(Character): Represents the alphabet before that character (provided that the alphabet before'A'is'Z')
* [(Character string)]: Represents a character string that is horizontally inverted.
However, the machine that generates this ciphertext is currently out of order, and some letters of the alphabet in the ciphertext may be broken and unreadable. Unreadable characters are tentatively represented as'?'. As a result of the investigation, it was found that the method of filling the broken characters is such that the decrypted character string is the smallest in the dictionary order among the possible character strings after decoding. Your job is to decrypt this ciphertext correctly.
Input
The input consists of multiple datasets. Each dataset consists of one line containing a string in which some uppercase letters have been replaced with ‘?’ In the ciphertext defined by BNF above. You can assume that the length of each string is less than $ 80 $. You can also assume that the number of'?' In each dataset is greater than or equal to $ 0 $ and less than or equal to $ 3 $.
The end of the input is represented by a line containing only one character,'.'.
Output
For each data set, output the decrypted character string when the ciphertext is decrypted so that the decrypted character string is the smallest in the dictionary order.
Sample Input
A + A ++ A
Z-Z--Z + -Z
[ESREVER]
J ---? --- J
++++++++ A +++ Z ----------- A +++ Z
[[++-+-? [-++-? ++-+++ L]] [-+ ----- + -O]] ++++ --- + L
..
Output for Sample Input
ABC
ZYXZ
REVERSE
JAG
ICPC
JAPAN
Example
Input
A+A++A
Z-Z--Z+-Z
[ESREVER]
J---?---J
++++++++A+++Z-----------A+++Z
[[++-+--?[--++-?++-+++L]][-+-----+-O]]++++---+L
.
Output
ABC
ZYXZ
REVERSE
JAG
ICPC
JAPAN
Submitted Solution:
```
while True:
s = input()
if s == '.':break
while True:
index = min(max(0, s.find('+')), max(0, s.find('-')))
if index == 0 and s[0] != '+' and s[0] != '-':
index = s.rfind('[')
if index == -1:
break
e = s[index:].find(']')
s = s[:index] + s[index + 1:e][::-1] + s[e + 1:]
else:
e = index
f = 0
while s[e] == '+' or s[e] == '-':
e += 1
if s[e] == '+':
f += 1
else:
f -= 1
r = s[e]
if s[e] != '?':
r = chr(ord(r) += f)
s = s[:index] + r + s[e + 1:]
print(s.replace('?','A'))
``` | instruction | 0 | 30,002 | 18 | 60,004 |
No | output | 1 | 30,002 | 18 | 60,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Gogol continue their fierce competition. Crucial moment is just around the corner: Gogol is ready to release it's new tablet Lastus 3000.
This new device is equipped with specially designed artificial intelligence (AI). Employees of Pineapple did their best to postpone the release of Lastus 3000 as long as possible. Finally, they found out, that the name of the new artificial intelligence is similar to the name of the phone, that Pineapple released 200 years ago. As all rights on its name belong to Pineapple, they stand on changing the name of Gogol's artificial intelligence.
Pineapple insists, that the name of their phone occurs in the name of AI as a substring. Because the name of technology was already printed on all devices, the Gogol's director decided to replace some characters in AI name with "#". As this operation is pretty expensive, you should find the minimum number of characters to replace with "#", such that the name of AI doesn't contain the name of the phone as a substring.
Substring is a continuous subsequence of a string.
Input
The first line of the input contains the name of AI designed by Gogol, its length doesn't exceed 100 000 characters. Second line contains the name of the phone released by Pineapple 200 years ago, its length doesn't exceed 30. Both string are non-empty and consist of only small English letters.
Output
Print the minimum number of characters that must be replaced with "#" in order to obtain that the name of the phone doesn't occur in the name of AI as a substring.
Examples
Input
intellect
tell
Output
1
Input
google
apple
Output
0
Input
sirisiri
sir
Output
2
Note
In the first sample AI's name may be replaced with "int#llect".
In the second sample Gogol can just keep things as they are.
In the third sample one of the new possible names of AI may be "s#ris#ri".
Submitted Solution:
```
t = input()
p = input()
s = p + '&' + t
n = len(s)
k = len(p)
z = [0] * n
l = 0
r = 0
for i in range(1, n):
if i <= r:
z[i] = min(r - i + 1, z[i - l])
while i +z[i] < n and s[z[i]] == s[i + z[i]]:
z[i] += 1
if i + z[i] - 1 > r:
r = i + z[i] - 1
l = i
ans = 0
i = len(p) - 1
while i < n:
if z[i] == len(p):
ans += 1
i += k
else:
i += 1
print(ans)
``` | instruction | 0 | 31,392 | 18 | 62,784 |
Yes | output | 1 | 31,392 | 18 | 62,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Gogol continue their fierce competition. Crucial moment is just around the corner: Gogol is ready to release it's new tablet Lastus 3000.
This new device is equipped with specially designed artificial intelligence (AI). Employees of Pineapple did their best to postpone the release of Lastus 3000 as long as possible. Finally, they found out, that the name of the new artificial intelligence is similar to the name of the phone, that Pineapple released 200 years ago. As all rights on its name belong to Pineapple, they stand on changing the name of Gogol's artificial intelligence.
Pineapple insists, that the name of their phone occurs in the name of AI as a substring. Because the name of technology was already printed on all devices, the Gogol's director decided to replace some characters in AI name with "#". As this operation is pretty expensive, you should find the minimum number of characters to replace with "#", such that the name of AI doesn't contain the name of the phone as a substring.
Substring is a continuous subsequence of a string.
Input
The first line of the input contains the name of AI designed by Gogol, its length doesn't exceed 100 000 characters. Second line contains the name of the phone released by Pineapple 200 years ago, its length doesn't exceed 30. Both string are non-empty and consist of only small English letters.
Output
Print the minimum number of characters that must be replaced with "#" in order to obtain that the name of the phone doesn't occur in the name of AI as a substring.
Examples
Input
intellect
tell
Output
1
Input
google
apple
Output
0
Input
sirisiri
sir
Output
2
Note
In the first sample AI's name may be replaced with "int#llect".
In the second sample Gogol can just keep things as they are.
In the third sample one of the new possible names of AI may be "s#ris#ri".
Submitted Solution:
```
def get_next(a,next_a):
i = 0; k = -1; next_a[i] = k
while (i < len(a)):
if (k == -1 or a[i] == a[k]):
i += 1
k += 1
next_a[i] = k
else:
k = next_a[k]
def KMP_sa(s,a,next_a,count):
i = 0; j = 0;
while (i < len(s) and j < len(a)):
while (j > 0 and s[i] != a[j]):
j = next_a[j]
if (s[i] == a[j]):
j += 1
if (j == len(a)):
count.append(i-len(a)+1)
# print(i)
j = next_a[j]
i += 1
s = str(input())
a = str(input())
count = []
next_a = [0 for i in range(0,len(a)+1)]
get_next(a,next_a)
KMP_sa(s,a,next_a,count)
# print(count)
# print(next_a)
i = 0
len_c = len(count)
len_a = len(a)-1
tot = 0
while (i<len_c):
pos = count[i]
tot += 1
while (i+1 < len_c and pos+len_a >= count[i+1]):
i += 1
i += 1
print(tot)
``` | instruction | 0 | 31,393 | 18 | 62,786 |
Yes | output | 1 | 31,393 | 18 | 62,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Gogol continue their fierce competition. Crucial moment is just around the corner: Gogol is ready to release it's new tablet Lastus 3000.
This new device is equipped with specially designed artificial intelligence (AI). Employees of Pineapple did their best to postpone the release of Lastus 3000 as long as possible. Finally, they found out, that the name of the new artificial intelligence is similar to the name of the phone, that Pineapple released 200 years ago. As all rights on its name belong to Pineapple, they stand on changing the name of Gogol's artificial intelligence.
Pineapple insists, that the name of their phone occurs in the name of AI as a substring. Because the name of technology was already printed on all devices, the Gogol's director decided to replace some characters in AI name with "#". As this operation is pretty expensive, you should find the minimum number of characters to replace with "#", such that the name of AI doesn't contain the name of the phone as a substring.
Substring is a continuous subsequence of a string.
Input
The first line of the input contains the name of AI designed by Gogol, its length doesn't exceed 100 000 characters. Second line contains the name of the phone released by Pineapple 200 years ago, its length doesn't exceed 30. Both string are non-empty and consist of only small English letters.
Output
Print the minimum number of characters that must be replaced with "#" in order to obtain that the name of the phone doesn't occur in the name of AI as a substring.
Examples
Input
intellect
tell
Output
1
Input
google
apple
Output
0
Input
sirisiri
sir
Output
2
Note
In the first sample AI's name may be replaced with "int#llect".
In the second sample Gogol can just keep things as they are.
In the third sample one of the new possible names of AI may be "s#ris#ri".
Submitted Solution:
```
n=input()
n1=input()
x=n.count(n1)
print(x)
``` | instruction | 0 | 31,394 | 18 | 62,788 |
Yes | output | 1 | 31,394 | 18 | 62,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Gogol continue their fierce competition. Crucial moment is just around the corner: Gogol is ready to release it's new tablet Lastus 3000.
This new device is equipped with specially designed artificial intelligence (AI). Employees of Pineapple did their best to postpone the release of Lastus 3000 as long as possible. Finally, they found out, that the name of the new artificial intelligence is similar to the name of the phone, that Pineapple released 200 years ago. As all rights on its name belong to Pineapple, they stand on changing the name of Gogol's artificial intelligence.
Pineapple insists, that the name of their phone occurs in the name of AI as a substring. Because the name of technology was already printed on all devices, the Gogol's director decided to replace some characters in AI name with "#". As this operation is pretty expensive, you should find the minimum number of characters to replace with "#", such that the name of AI doesn't contain the name of the phone as a substring.
Substring is a continuous subsequence of a string.
Input
The first line of the input contains the name of AI designed by Gogol, its length doesn't exceed 100 000 characters. Second line contains the name of the phone released by Pineapple 200 years ago, its length doesn't exceed 30. Both string are non-empty and consist of only small English letters.
Output
Print the minimum number of characters that must be replaced with "#" in order to obtain that the name of the phone doesn't occur in the name of AI as a substring.
Examples
Input
intellect
tell
Output
1
Input
google
apple
Output
0
Input
sirisiri
sir
Output
2
Note
In the first sample AI's name may be replaced with "int#llect".
In the second sample Gogol can just keep things as they are.
In the third sample one of the new possible names of AI may be "s#ris#ri".
Submitted Solution:
```
s = input()
s = list(s)
t = input()
t = list(t)
l = len(s)
lt = len(t)
res = 0
for i in range(l):
#print(last)
if i + lt <= l and t == s[i:i+lt]:
s[i + lt - 1] = '#'
res += 1
print(res)
``` | instruction | 0 | 31,395 | 18 | 62,790 |
Yes | output | 1 | 31,395 | 18 | 62,791 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Gogol continue their fierce competition. Crucial moment is just around the corner: Gogol is ready to release it's new tablet Lastus 3000.
This new device is equipped with specially designed artificial intelligence (AI). Employees of Pineapple did their best to postpone the release of Lastus 3000 as long as possible. Finally, they found out, that the name of the new artificial intelligence is similar to the name of the phone, that Pineapple released 200 years ago. As all rights on its name belong to Pineapple, they stand on changing the name of Gogol's artificial intelligence.
Pineapple insists, that the name of their phone occurs in the name of AI as a substring. Because the name of technology was already printed on all devices, the Gogol's director decided to replace some characters in AI name with "#". As this operation is pretty expensive, you should find the minimum number of characters to replace with "#", such that the name of AI doesn't contain the name of the phone as a substring.
Substring is a continuous subsequence of a string.
Input
The first line of the input contains the name of AI designed by Gogol, its length doesn't exceed 100 000 characters. Second line contains the name of the phone released by Pineapple 200 years ago, its length doesn't exceed 30. Both string are non-empty and consist of only small English letters.
Output
Print the minimum number of characters that must be replaced with "#" in order to obtain that the name of the phone doesn't occur in the name of AI as a substring.
Examples
Input
intellect
tell
Output
1
Input
google
apple
Output
0
Input
sirisiri
sir
Output
2
Note
In the first sample AI's name may be replaced with "int#llect".
In the second sample Gogol can just keep things as they are.
In the third sample one of the new possible names of AI may be "s#ris#ri".
Submitted Solution:
```
entrada1 = input("")
palavra = input("")
qntd = 0
word = ""
letra = 0
for i in range(len(entrada1) + 1):
if(word == palavra):
qntd += 1
letra = 0
word = ""
if(i < len(entrada1) and letra < len(palavra)):
if (entrada1[i] == palavra[letra]):
letra += 1
word += entrada1[i]
else :
word = ""
letra = 0
if(qntd == 75):
qntd = 76
print (qntd)
``` | instruction | 0 | 31,396 | 18 | 62,792 |
No | output | 1 | 31,396 | 18 | 62,793 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Gogol continue their fierce competition. Crucial moment is just around the corner: Gogol is ready to release it's new tablet Lastus 3000.
This new device is equipped with specially designed artificial intelligence (AI). Employees of Pineapple did their best to postpone the release of Lastus 3000 as long as possible. Finally, they found out, that the name of the new artificial intelligence is similar to the name of the phone, that Pineapple released 200 years ago. As all rights on its name belong to Pineapple, they stand on changing the name of Gogol's artificial intelligence.
Pineapple insists, that the name of their phone occurs in the name of AI as a substring. Because the name of technology was already printed on all devices, the Gogol's director decided to replace some characters in AI name with "#". As this operation is pretty expensive, you should find the minimum number of characters to replace with "#", such that the name of AI doesn't contain the name of the phone as a substring.
Substring is a continuous subsequence of a string.
Input
The first line of the input contains the name of AI designed by Gogol, its length doesn't exceed 100 000 characters. Second line contains the name of the phone released by Pineapple 200 years ago, its length doesn't exceed 30. Both string are non-empty and consist of only small English letters.
Output
Print the minimum number of characters that must be replaced with "#" in order to obtain that the name of the phone doesn't occur in the name of AI as a substring.
Examples
Input
intellect
tell
Output
1
Input
google
apple
Output
0
Input
sirisiri
sir
Output
2
Note
In the first sample AI's name may be replaced with "int#llect".
In the second sample Gogol can just keep things as they are.
In the third sample one of the new possible names of AI may be "s#ris#ri".
Submitted Solution:
```
s, t = input(), input()
ls, lt = len(s), len(t)
if ls < lt:
print(0)
else:
ans = 0
for i in range(ls-lt+1):
if s[i:i+lt] == t:
ans += 1
print(ans)
``` | instruction | 0 | 31,397 | 18 | 62,794 |
No | output | 1 | 31,397 | 18 | 62,795 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Gogol continue their fierce competition. Crucial moment is just around the corner: Gogol is ready to release it's new tablet Lastus 3000.
This new device is equipped with specially designed artificial intelligence (AI). Employees of Pineapple did their best to postpone the release of Lastus 3000 as long as possible. Finally, they found out, that the name of the new artificial intelligence is similar to the name of the phone, that Pineapple released 200 years ago. As all rights on its name belong to Pineapple, they stand on changing the name of Gogol's artificial intelligence.
Pineapple insists, that the name of their phone occurs in the name of AI as a substring. Because the name of technology was already printed on all devices, the Gogol's director decided to replace some characters in AI name with "#". As this operation is pretty expensive, you should find the minimum number of characters to replace with "#", such that the name of AI doesn't contain the name of the phone as a substring.
Substring is a continuous subsequence of a string.
Input
The first line of the input contains the name of AI designed by Gogol, its length doesn't exceed 100 000 characters. Second line contains the name of the phone released by Pineapple 200 years ago, its length doesn't exceed 30. Both string are non-empty and consist of only small English letters.
Output
Print the minimum number of characters that must be replaced with "#" in order to obtain that the name of the phone doesn't occur in the name of AI as a substring.
Examples
Input
intellect
tell
Output
1
Input
google
apple
Output
0
Input
sirisiri
sir
Output
2
Note
In the first sample AI's name may be replaced with "int#llect".
In the second sample Gogol can just keep things as they are.
In the third sample one of the new possible names of AI may be "s#ris#ri".
Submitted Solution:
```
a = input()
b = input()
count = 0
while True:
n = a.find(b)
if n==-1:
break
else:
count+=1
a = a.replace(b, b[:len(b)-1]+"0", 1)
print(a, n)
print(count)
``` | instruction | 0 | 31,398 | 18 | 62,796 |
No | output | 1 | 31,398 | 18 | 62,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Gogol continue their fierce competition. Crucial moment is just around the corner: Gogol is ready to release it's new tablet Lastus 3000.
This new device is equipped with specially designed artificial intelligence (AI). Employees of Pineapple did their best to postpone the release of Lastus 3000 as long as possible. Finally, they found out, that the name of the new artificial intelligence is similar to the name of the phone, that Pineapple released 200 years ago. As all rights on its name belong to Pineapple, they stand on changing the name of Gogol's artificial intelligence.
Pineapple insists, that the name of their phone occurs in the name of AI as a substring. Because the name of technology was already printed on all devices, the Gogol's director decided to replace some characters in AI name with "#". As this operation is pretty expensive, you should find the minimum number of characters to replace with "#", such that the name of AI doesn't contain the name of the phone as a substring.
Substring is a continuous subsequence of a string.
Input
The first line of the input contains the name of AI designed by Gogol, its length doesn't exceed 100 000 characters. Second line contains the name of the phone released by Pineapple 200 years ago, its length doesn't exceed 30. Both string are non-empty and consist of only small English letters.
Output
Print the minimum number of characters that must be replaced with "#" in order to obtain that the name of the phone doesn't occur in the name of AI as a substring.
Examples
Input
intellect
tell
Output
1
Input
google
apple
Output
0
Input
sirisiri
sir
Output
2
Note
In the first sample AI's name may be replaced with "int#llect".
In the second sample Gogol can just keep things as they are.
In the third sample one of the new possible names of AI may be "s#ris#ri".
Submitted Solution:
```
s1 = list(input())
s2 = list(input())
cont = 0
for i in range(len(s1) - len(s2)):
aux = s1[i:i+len(s2)]
if aux == s2:
#meio = (i+i+len(s2)) / 2
#s1[meio] = "#"
cont += 1
print(cont)
``` | instruction | 0 | 31,399 | 18 | 62,798 |
No | output | 1 | 31,399 | 18 | 62,799 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input
The first line contains an integer n (1 ≤ n ≤ 100). Each of the following n lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output
Print n lines. The i-th line should contain the result of replacing of the i-th word from the input data.
Examples
Input
4
word
localization
internationalization
pneumonoultramicroscopicsilicovolcanoconiosis
Output
word
l10n
i18n
p43s
Submitted Solution:
```
n=input()
n=int(n)
i=0
while i<n:
word=input()
if len(word)<=10:
print(word)
else:
word=word[0]+str(len(word)-2)+word[len(word)-1]
print(word)
i=i+1
``` | instruction | 0 | 31,425 | 18 | 62,850 |
Yes | output | 1 | 31,425 | 18 | 62,851 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input
The first line contains an integer n (1 ≤ n ≤ 100). Each of the following n lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output
Print n lines. The i-th line should contain the result of replacing of the i-th word from the input data.
Examples
Input
4
word
localization
internationalization
pneumonoultramicroscopicsilicovolcanoconiosis
Output
word
l10n
i18n
p43s
Submitted Solution:
```
T = int(input())
for i in range (T):
a = str(input())
if len(a) > 10 :
print(a[0],end='')
print(int(len(a)-2),end='')
print(a[-1])
else :
print(a)
``` | instruction | 0 | 31,426 | 18 | 62,852 |
Yes | output | 1 | 31,426 | 18 | 62,853 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input
The first line contains an integer n (1 ≤ n ≤ 100). Each of the following n lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output
Print n lines. The i-th line should contain the result of replacing of the i-th word from the input data.
Examples
Input
4
word
localization
internationalization
pneumonoultramicroscopicsilicovolcanoconiosis
Output
word
l10n
i18n
p43s
Submitted Solution:
```
# cook your dish here
n=int(input())
while n>0:
s=input()
if(len(s)>10):
s1=s[0]+str(len(s)-2)+s[len(s)-1]
print(s1)
else:
print(s)
n-=1
``` | instruction | 0 | 31,427 | 18 | 62,854 |
Yes | output | 1 | 31,427 | 18 | 62,855 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input
The first line contains an integer n (1 ≤ n ≤ 100). Each of the following n lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output
Print n lines. The i-th line should contain the result of replacing of the i-th word from the input data.
Examples
Input
4
word
localization
internationalization
pneumonoultramicroscopicsilicovolcanoconiosis
Output
word
l10n
i18n
p43s
Submitted Solution:
```
word = str(input("give me a word"))
if len(word) < 10 :
print (word)
if 10 < len(word) :
print( word[0],(len(word)-2),word[-1])
``` | instruction | 0 | 31,428 | 18 | 62,856 |
No | output | 1 | 31,428 | 18 | 62,857 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input
The first line contains an integer n (1 ≤ n ≤ 100). Each of the following n lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output
Print n lines. The i-th line should contain the result of replacing of the i-th word from the input data.
Examples
Input
4
word
localization
internationalization
pneumonoultramicroscopicsilicovolcanoconiosis
Output
word
l10n
i18n
p43s
Submitted Solution:
```
a=int(input())
z=[]
for i in range(a):
x=input()
z.append(x)
for i in z:
if len(i)>10:
print(i[0],str(len(i)-2),i[-1])
else:
print(i)
``` | instruction | 0 | 31,429 | 18 | 62,858 |
No | output | 1 | 31,429 | 18 | 62,859 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input
The first line contains an integer n (1 ≤ n ≤ 100). Each of the following n lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output
Print n lines. The i-th line should contain the result of replacing of the i-th word from the input data.
Examples
Input
4
word
localization
internationalization
pneumonoultramicroscopicsilicovolcanoconiosis
Output
word
l10n
i18n
p43s
Submitted Solution:
```
import sys
n = int(input())
while n != 0:
word = input()
l = int(len(word))
if len(word) < 10:
print(word)
else:
first = word[0]
mid = l - 2
last = word[l-1]
sys.stdout.write(str(first))
sys.stdout.write(str(mid))
sys.stdout.write(str(last))
print("\n")
n -= 1
``` | instruction | 0 | 31,430 | 18 | 62,860 |
No | output | 1 | 31,430 | 18 | 62,861 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input
The first line contains an integer n (1 ≤ n ≤ 100). Each of the following n lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output
Print n lines. The i-th line should contain the result of replacing of the i-th word from the input data.
Examples
Input
4
word
localization
internationalization
pneumonoultramicroscopicsilicovolcanoconiosis
Output
word
l10n
i18n
p43s
Submitted Solution:
```
import sys
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
# --------------------------------------------------
num = inp()
for x in range(num):
s = input()
if len(s)>10:
count = len(s)-2
print(s[0] + str(count) + s[-1])
else:
print(s)
``` | instruction | 0 | 31,431 | 18 | 62,862 |
No | output | 1 | 31,431 | 18 | 62,863 |
Provide a correct Python 3 solution for this coding contest problem.
Relative B man came to A child's house. He is 3 years old and loves singing. He is singing the song "Kobutanuki Tsuneko" (written and composed by Naozumi Yamamoto), which he learned from kindergarten. In this song, the four words "kobuta," "raccoon dog," "fox," and "cat" are arranged in order, and the last and first sounds are the same. Mr. B was asked by Mr. A to tell him if he could make a similar shiritori from the words that Mr. B said.
So, in order to help Ako, from the given words, use all the words to make a shiritori in order, and then the first letter of the first word and the last letter of the last word are the same. Let's write a program that determines whether or not it can be done.
Create a program that takes n words as input, determines whether or not a shiritori can be created from those word pairs, and outputs OK if possible and NG if not.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
word1
word2
::
wordn
The number of words n (2 ≤ n ≤ 10000) is given on the first line. The next n lines are given n words wordi (a string of up to 32 single-byte lowercase letters).
The number of datasets does not exceed 50.
Output
The judgment result is output to one line for each input data set.
Example
Input
5
apple
yellow
georgia
king
email
7
apple
yellow
georgia
king
email
wink
lucky
0
Output
NG
OK | instruction | 0 | 31,702 | 18 | 63,404 |
"Correct Solution:
```
from collections import deque
ca = ord('a')
while 1:
N = int(input())
if N == 0:
break
G = [[] for i in range(26)]
S = [0]*26; T = [0]*26
for i in range(N):
word = input()
s = ord(word[0]) - ca; t = ord(word[-1]) - ca
S[s] += 1; T[t] += 1
G[s].append(t)
ok = 1
for i in range(26):
if S[i] != T[i]:
ok = 0
if ok:
que = deque()
U = [0]*26
for i in range(26):
if S[i]:
que.append(i)
U[i] = 1
break
while que:
v = que.popleft()
for w in G[v]:
if U[w]:
continue
U[w] = 1
que.append(w)
for i in range(26):
if S[i] and not U[i]:
ok = 0
print("OK" if ok else "NG")
``` | output | 1 | 31,702 | 18 | 63,405 |
Provide a correct Python 3 solution for this coding contest problem.
Relative B man came to A child's house. He is 3 years old and loves singing. He is singing the song "Kobutanuki Tsuneko" (written and composed by Naozumi Yamamoto), which he learned from kindergarten. In this song, the four words "kobuta," "raccoon dog," "fox," and "cat" are arranged in order, and the last and first sounds are the same. Mr. B was asked by Mr. A to tell him if he could make a similar shiritori from the words that Mr. B said.
So, in order to help Ako, from the given words, use all the words to make a shiritori in order, and then the first letter of the first word and the last letter of the last word are the same. Let's write a program that determines whether or not it can be done.
Create a program that takes n words as input, determines whether or not a shiritori can be created from those word pairs, and outputs OK if possible and NG if not.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
word1
word2
::
wordn
The number of words n (2 ≤ n ≤ 10000) is given on the first line. The next n lines are given n words wordi (a string of up to 32 single-byte lowercase letters).
The number of datasets does not exceed 50.
Output
The judgment result is output to one line for each input data set.
Example
Input
5
apple
yellow
georgia
king
email
7
apple
yellow
georgia
king
email
wink
lucky
0
Output
NG
OK | instruction | 0 | 31,703 | 18 | 63,406 |
"Correct Solution:
```
from collections import deque
while True:
n = int(input())
if n == 0:
break
ins = {}
outs = {}
node = set()
base = ord("a")
edges = [[0] * 26 for _ in range(26)]
for i in range(26):
ins[i] = 0
outs[i] = 0
for _ in range(n):
s = input()
fr = ord(s[0]) - base
to = ord(s[-1]) - base
ins[to] += 1
outs[fr] += 1
edges[fr][to] = 1
node.add(fr)
node.add(to)
for i in range(26):
if ins[i] != outs[i]:
print("NG")
break
else:
node = sorted(list(node))
visited = [False] * 26
visited[node[0]] = True
que = deque()
que.append(node[0])
while que:
p = que.popleft()
for i in range(26):
if edges[p][i] and not visited[i]:
visited[i] = True
que.append(i)
for p in node:
if not visited[p]:
print("NG")
break
else:
print("OK")
``` | output | 1 | 31,703 | 18 | 63,407 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.