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.
Petya once wrote a sad love song and shared it to Vasya. The song is a string consisting of lowercase English letters. Vasya made up q questions about this song. Each question is about a subsegment of the song starting from the l-th letter to the r-th letter. Vasya considers a substring made up from characters on this segment and repeats each letter in the subsegment k times, where k is the index of the corresponding letter in the alphabet. For example, if the question is about the substring "abbcb", then Vasya repeats letter 'a' once, each of the letters 'b' twice, letter 'c" three times, so that the resulting string is "abbbbcccbb", its length is 10. Vasya is interested about the length of the resulting string.
Help Petya find the length of each string obtained by Vasya.
Input
The first line contains two integers n and q (1≤ n≤ 100 000, 1≤ q ≤ 100 000) — the length of the song and the number of questions.
The second line contains one string s — the song, consisting of n lowercase letters of English letters.
Vasya's questions are contained in the next q lines. Each line contains two integers l and r (1 ≤ l ≤ r ≤ n) — the bounds of the question.
Output
Print q lines: for each question print the length of the string obtained by Vasya.
Examples
Input
7 3
abacaba
1 3
2 5
1 7
Output
4
7
11
Input
7 4
abbabaa
1 3
5 7
6 6
2 4
Output
5
4
1
5
Input
13 7
sonoshikumiwo
1 5
2 10
7 7
1 13
4 8
2 5
3 9
Output
82
125
9
191
62
63
97
Note
In the first example Vasya is interested in three questions. In the first question Vasya considers the substring "aba", that transforms to "abba", so the answer is equal to 4. In the second question Vasya considers "baca", that transforms to "bbaccca", so the answer is 7. In the third question Vasya considers the string "abacaba",that transforms to "abbacccabba" of length 11.
Submitted Solution:
```
def I():
return input()
def II():
return int(I())
def M():
return map(int,I().split())
def L():
return list(M())
# for _ in range(II())
d = {'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6, 'g': 7, 'h': 8, 'i': 9, 'j': 10, 'k': 11, 'l': 12, 'm': 13, 'n': 14, 'o': 15, 'p': 16, 'q': 17, 'r': 18, 's': 19, 't': 20, 'u': 21, 'v': 22, 'w': 23, 'x': 24, 'y': 25, 'z': 26}
n,q = M()
s = I()
a = [0]*n
a[0] = d.get(s[0])
for i in range(1,n):
a[i] = a[i-1] + d.get(s[i])
print(a)
for _ in range(q):
l,r = M()
if (l<2):
c = a[r-1]
else:
c=a[r-1]-a[l-2]
print(c)
``` | instruction | 0 | 101,464 | 18 | 202,928 |
No | output | 1 | 101,464 | 18 | 202,929 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya once wrote a sad love song and shared it to Vasya. The song is a string consisting of lowercase English letters. Vasya made up q questions about this song. Each question is about a subsegment of the song starting from the l-th letter to the r-th letter. Vasya considers a substring made up from characters on this segment and repeats each letter in the subsegment k times, where k is the index of the corresponding letter in the alphabet. For example, if the question is about the substring "abbcb", then Vasya repeats letter 'a' once, each of the letters 'b' twice, letter 'c" three times, so that the resulting string is "abbbbcccbb", its length is 10. Vasya is interested about the length of the resulting string.
Help Petya find the length of each string obtained by Vasya.
Input
The first line contains two integers n and q (1≤ n≤ 100 000, 1≤ q ≤ 100 000) — the length of the song and the number of questions.
The second line contains one string s — the song, consisting of n lowercase letters of English letters.
Vasya's questions are contained in the next q lines. Each line contains two integers l and r (1 ≤ l ≤ r ≤ n) — the bounds of the question.
Output
Print q lines: for each question print the length of the string obtained by Vasya.
Examples
Input
7 3
abacaba
1 3
2 5
1 7
Output
4
7
11
Input
7 4
abbabaa
1 3
5 7
6 6
2 4
Output
5
4
1
5
Input
13 7
sonoshikumiwo
1 5
2 10
7 7
1 13
4 8
2 5
3 9
Output
82
125
9
191
62
63
97
Note
In the first example Vasya is interested in three questions. In the first question Vasya considers the substring "aba", that transforms to "abba", so the answer is equal to 4. In the second question Vasya considers "baca", that transforms to "bbaccca", so the answer is 7. In the third question Vasya considers the string "abacaba",that transforms to "abbacccabba" of length 11.
Submitted Solution:
```
d={"a":1,"b":2,"c":3,"d":4,"e":5,"f":6,"g":7,"h":8,"i":9,"j":10,"k":11,"l":12,"m":13,"n":14,"o":15,"p":16,"q":17,"r":18,"s":19,"t":20,"u":21,"v":22,"w":23,"x":24,"y":25,"z":26}
n,q=map(int,input().split())
s=str(input())
arr=[]
arr.append(d[s[0]])
for i in range(1,n):
x=d[s[i]]
y=arr[-1]
arr.append(x+y)
for i in range(q):
a,b=map(int,input().split())
if a==1:
if b==n:
print(arr[-1])
else:
print(b-1)
``` | instruction | 0 | 101,465 | 18 | 202,930 |
No | output | 1 | 101,465 | 18 | 202,931 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya once wrote a sad love song and shared it to Vasya. The song is a string consisting of lowercase English letters. Vasya made up q questions about this song. Each question is about a subsegment of the song starting from the l-th letter to the r-th letter. Vasya considers a substring made up from characters on this segment and repeats each letter in the subsegment k times, where k is the index of the corresponding letter in the alphabet. For example, if the question is about the substring "abbcb", then Vasya repeats letter 'a' once, each of the letters 'b' twice, letter 'c" three times, so that the resulting string is "abbbbcccbb", its length is 10. Vasya is interested about the length of the resulting string.
Help Petya find the length of each string obtained by Vasya.
Input
The first line contains two integers n and q (1≤ n≤ 100 000, 1≤ q ≤ 100 000) — the length of the song and the number of questions.
The second line contains one string s — the song, consisting of n lowercase letters of English letters.
Vasya's questions are contained in the next q lines. Each line contains two integers l and r (1 ≤ l ≤ r ≤ n) — the bounds of the question.
Output
Print q lines: for each question print the length of the string obtained by Vasya.
Examples
Input
7 3
abacaba
1 3
2 5
1 7
Output
4
7
11
Input
7 4
abbabaa
1 3
5 7
6 6
2 4
Output
5
4
1
5
Input
13 7
sonoshikumiwo
1 5
2 10
7 7
1 13
4 8
2 5
3 9
Output
82
125
9
191
62
63
97
Note
In the first example Vasya is interested in three questions. In the first question Vasya considers the substring "aba", that transforms to "abba", so the answer is equal to 4. In the second question Vasya considers "baca", that transforms to "bbaccca", so the answer is 7. In the third question Vasya considers the string "abacaba",that transforms to "abbacccabba" of length 11.
Submitted Solution:
```
import sys
input = sys.stdin.readline
n, q = map(int, input().split())
s = input()
v = [0]
for i in range(n):
v.append(v[-1] + (ord(s[i])-96))
print(v)
for _ in range(q):
l, r = map(int, input().split())
print(v[r]-v[l-1])
``` | instruction | 0 | 101,466 | 18 | 202,932 |
No | output | 1 | 101,466 | 18 | 202,933 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya once wrote a sad love song and shared it to Vasya. The song is a string consisting of lowercase English letters. Vasya made up q questions about this song. Each question is about a subsegment of the song starting from the l-th letter to the r-th letter. Vasya considers a substring made up from characters on this segment and repeats each letter in the subsegment k times, where k is the index of the corresponding letter in the alphabet. For example, if the question is about the substring "abbcb", then Vasya repeats letter 'a' once, each of the letters 'b' twice, letter 'c" three times, so that the resulting string is "abbbbcccbb", its length is 10. Vasya is interested about the length of the resulting string.
Help Petya find the length of each string obtained by Vasya.
Input
The first line contains two integers n and q (1≤ n≤ 100 000, 1≤ q ≤ 100 000) — the length of the song and the number of questions.
The second line contains one string s — the song, consisting of n lowercase letters of English letters.
Vasya's questions are contained in the next q lines. Each line contains two integers l and r (1 ≤ l ≤ r ≤ n) — the bounds of the question.
Output
Print q lines: for each question print the length of the string obtained by Vasya.
Examples
Input
7 3
abacaba
1 3
2 5
1 7
Output
4
7
11
Input
7 4
abbabaa
1 3
5 7
6 6
2 4
Output
5
4
1
5
Input
13 7
sonoshikumiwo
1 5
2 10
7 7
1 13
4 8
2 5
3 9
Output
82
125
9
191
62
63
97
Note
In the first example Vasya is interested in three questions. In the first question Vasya considers the substring "aba", that transforms to "abba", so the answer is equal to 4. In the second question Vasya considers "baca", that transforms to "bbaccca", so the answer is 7. In the third question Vasya considers the string "abacaba",that transforms to "abbacccabba" of length 11.
Submitted Solution:
```
n, q=map(int, input().split())
s=input()
alp={'a':1, 'b':2, 'c':3, 'd':4, 'e':5, 'f':6, 'g':7, 'h':8, 'i':9, 'j':10, 'k':11, 'l':12, 'm':13, 'n':14, 'o':15, 'p':16, 'q':17, 'r':18, 's':19, 't':20, 'u':21, 'v':22, 'w':23, 'x':24, 'y':25, 'z':26}
li=[0]
sum=0
for i in range(n):
sum+=alp[s[i]]
li.append(sum)
print(li)
for i in range(q):
x, y=map(int, input().split())
print(li[y]-li[x-1])
``` | instruction | 0 | 101,467 | 18 | 202,934 |
No | output | 1 | 101,467 | 18 | 202,935 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, a start up by two students of a state university of city F gained incredible popularity. Now it's time to start a new company. But what do we call it?
The market analysts came up with a very smart plan: the name of the company should be identical to its reflection in a mirror! In other words, if we write out the name of the company on a piece of paper in a line (horizontally, from left to right) with large English letters, then put this piece of paper in front of the mirror, then the reflection of the name in the mirror should perfectly match the line written on the piece of paper.
There are many suggestions for the company name, so coming up to the mirror with a piece of paper for each name wouldn't be sensible. The founders of the company decided to automatize this process. They asked you to write a program that can, given a word, determine whether the word is a 'mirror' word or not.
Input
The first line contains a non-empty name that needs to be checked. The name contains at most 105 large English letters. The name will be written with the next sans serif font:
<image>
Output
Print 'YES' (without the quotes), if the given name matches its mirror reflection. Otherwise, print 'NO' (without the quotes).
Examples
Input
AHA
Output
YES
Input
Z
Output
NO
Input
XO
Output
NO
Submitted Solution:
```
t=list(input())
if t==t[::-1]:
f=set(t)
h=set(list('BCDEFGJKLNPQRSZ'))
if f&h==set():
print('YES')
else:
print('NO')
else:
print('NO')
``` | instruction | 0 | 101,537 | 18 | 203,074 |
Yes | output | 1 | 101,537 | 18 | 203,075 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, a start up by two students of a state university of city F gained incredible popularity. Now it's time to start a new company. But what do we call it?
The market analysts came up with a very smart plan: the name of the company should be identical to its reflection in a mirror! In other words, if we write out the name of the company on a piece of paper in a line (horizontally, from left to right) with large English letters, then put this piece of paper in front of the mirror, then the reflection of the name in the mirror should perfectly match the line written on the piece of paper.
There are many suggestions for the company name, so coming up to the mirror with a piece of paper for each name wouldn't be sensible. The founders of the company decided to automatize this process. They asked you to write a program that can, given a word, determine whether the word is a 'mirror' word or not.
Input
The first line contains a non-empty name that needs to be checked. The name contains at most 105 large English letters. The name will be written with the next sans serif font:
<image>
Output
Print 'YES' (without the quotes), if the given name matches its mirror reflection. Otherwise, print 'NO' (without the quotes).
Examples
Input
AHA
Output
YES
Input
Z
Output
NO
Input
XO
Output
NO
Submitted Solution:
```
n=input()
i=0
k=len(n)-1
_b=True
while i<=k and _b:
if n[i]==n[k] and n[i] in ['A','H','I','M','O','T','U','V','W','X','Y']:
_b=True
i+=1
k-=1
else:
_b=False
if _b: print('YES')
else: print('NO')
``` | instruction | 0 | 101,538 | 18 | 203,076 |
Yes | output | 1 | 101,538 | 18 | 203,077 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, a start up by two students of a state university of city F gained incredible popularity. Now it's time to start a new company. But what do we call it?
The market analysts came up with a very smart plan: the name of the company should be identical to its reflection in a mirror! In other words, if we write out the name of the company on a piece of paper in a line (horizontally, from left to right) with large English letters, then put this piece of paper in front of the mirror, then the reflection of the name in the mirror should perfectly match the line written on the piece of paper.
There are many suggestions for the company name, so coming up to the mirror with a piece of paper for each name wouldn't be sensible. The founders of the company decided to automatize this process. They asked you to write a program that can, given a word, determine whether the word is a 'mirror' word or not.
Input
The first line contains a non-empty name that needs to be checked. The name contains at most 105 large English letters. The name will be written with the next sans serif font:
<image>
Output
Print 'YES' (without the quotes), if the given name matches its mirror reflection. Otherwise, print 'NO' (without the quotes).
Examples
Input
AHA
Output
YES
Input
Z
Output
NO
Input
XO
Output
NO
Submitted Solution:
```
s = input()
a = set([chr(i + ord('A')) for i in range(26)])
b = set(s)
c = set(['A', 'H', 'I', 'M', 'O', 'T', 'U', 'V', 'W', 'X', 'Y'])
a &= b
if a<=c and s==s[::-1]:
print('YES')
else:
print('NO')
``` | instruction | 0 | 101,539 | 18 | 203,078 |
Yes | output | 1 | 101,539 | 18 | 203,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, a start up by two students of a state university of city F gained incredible popularity. Now it's time to start a new company. But what do we call it?
The market analysts came up with a very smart plan: the name of the company should be identical to its reflection in a mirror! In other words, if we write out the name of the company on a piece of paper in a line (horizontally, from left to right) with large English letters, then put this piece of paper in front of the mirror, then the reflection of the name in the mirror should perfectly match the line written on the piece of paper.
There are many suggestions for the company name, so coming up to the mirror with a piece of paper for each name wouldn't be sensible. The founders of the company decided to automatize this process. They asked you to write a program that can, given a word, determine whether the word is a 'mirror' word or not.
Input
The first line contains a non-empty name that needs to be checked. The name contains at most 105 large English letters. The name will be written with the next sans serif font:
<image>
Output
Print 'YES' (without the quotes), if the given name matches its mirror reflection. Otherwise, print 'NO' (without the quotes).
Examples
Input
AHA
Output
YES
Input
Z
Output
NO
Input
XO
Output
NO
Submitted Solution:
```
word = list(input())
def poop(word):
correct = ['A','H','I','M','O','T','U','V','W','X','Y']
for i in word:
if i not in correct:
return "NO"
if list(reversed(word)) == word:
return "YES"
return "NO"
print(poop(word))
``` | instruction | 0 | 101,540 | 18 | 203,080 |
Yes | output | 1 | 101,540 | 18 | 203,081 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, a start up by two students of a state university of city F gained incredible popularity. Now it's time to start a new company. But what do we call it?
The market analysts came up with a very smart plan: the name of the company should be identical to its reflection in a mirror! In other words, if we write out the name of the company on a piece of paper in a line (horizontally, from left to right) with large English letters, then put this piece of paper in front of the mirror, then the reflection of the name in the mirror should perfectly match the line written on the piece of paper.
There are many suggestions for the company name, so coming up to the mirror with a piece of paper for each name wouldn't be sensible. The founders of the company decided to automatize this process. They asked you to write a program that can, given a word, determine whether the word is a 'mirror' word or not.
Input
The first line contains a non-empty name that needs to be checked. The name contains at most 105 large English letters. The name will be written with the next sans serif font:
<image>
Output
Print 'YES' (without the quotes), if the given name matches its mirror reflection. Otherwise, print 'NO' (without the quotes).
Examples
Input
AHA
Output
YES
Input
Z
Output
NO
Input
XO
Output
NO
Submitted Solution:
```
def rev(n):
s=''
for i in range(len(n)-1,-1,-1):
s+=n[i]
return s
l=['A','H','I','M','O','T','U','V,','W','X','Y']
n=input()
i=0
while i<len(n):
if(n[i] in l):
i+=1
else:
i=0
break
if(i==0):
print('NO')
else:
if(len(n)%2==0):
j=len(n)//2
a=n[0:j]
b=rev(n[j:])
if(a==b):
print('YES')
else:
print('NO')
else:
j=len(n)//2
a=n[0:j]
b=rev(n[j+1:])
if(a==b):
print('YES')
else:
print('NO')
``` | instruction | 0 | 101,541 | 18 | 203,082 |
No | output | 1 | 101,541 | 18 | 203,083 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, a start up by two students of a state university of city F gained incredible popularity. Now it's time to start a new company. But what do we call it?
The market analysts came up with a very smart plan: the name of the company should be identical to its reflection in a mirror! In other words, if we write out the name of the company on a piece of paper in a line (horizontally, from left to right) with large English letters, then put this piece of paper in front of the mirror, then the reflection of the name in the mirror should perfectly match the line written on the piece of paper.
There are many suggestions for the company name, so coming up to the mirror with a piece of paper for each name wouldn't be sensible. The founders of the company decided to automatize this process. They asked you to write a program that can, given a word, determine whether the word is a 'mirror' word or not.
Input
The first line contains a non-empty name that needs to be checked. The name contains at most 105 large English letters. The name will be written with the next sans serif font:
<image>
Output
Print 'YES' (without the quotes), if the given name matches its mirror reflection. Otherwise, print 'NO' (without the quotes).
Examples
Input
AHA
Output
YES
Input
Z
Output
NO
Input
XO
Output
NO
Submitted Solution:
```
ch=input()
b=len(ch)
if (len(ch)==1):
print("NO")
elif(b%2==0):
if(ch[:int(b/2)]==ch[int(b/2):][::-1]):
print("YES")
else :
print("NO")
elif(b%2!=0):
if(ch[:int(b/2)]==ch[int(b/2)+1:][::-1]):
print("YES")
else :
print("NO")
``` | instruction | 0 | 101,542 | 18 | 203,084 |
No | output | 1 | 101,542 | 18 | 203,085 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, a start up by two students of a state university of city F gained incredible popularity. Now it's time to start a new company. But what do we call it?
The market analysts came up with a very smart plan: the name of the company should be identical to its reflection in a mirror! In other words, if we write out the name of the company on a piece of paper in a line (horizontally, from left to right) with large English letters, then put this piece of paper in front of the mirror, then the reflection of the name in the mirror should perfectly match the line written on the piece of paper.
There are many suggestions for the company name, so coming up to the mirror with a piece of paper for each name wouldn't be sensible. The founders of the company decided to automatize this process. They asked you to write a program that can, given a word, determine whether the word is a 'mirror' word or not.
Input
The first line contains a non-empty name that needs to be checked. The name contains at most 105 large English letters. The name will be written with the next sans serif font:
<image>
Output
Print 'YES' (without the quotes), if the given name matches its mirror reflection. Otherwise, print 'NO' (without the quotes).
Examples
Input
AHA
Output
YES
Input
Z
Output
NO
Input
XO
Output
NO
Submitted Solution:
```
# Name : Sachdev Hitesh
# College : GLSICA
user = input()
resu = user[::-1]
c = 0
if user == resu:
for i in user:
if i == 'A' or i == 'H' or i == 'I' or i == 'O' or i == 'T' or i == 'V' or \
i == 'W' or i == 'X' or i == 'Y' :
c = c + 1
if c == 0:
print("NO")
else:
print("YES")
``` | instruction | 0 | 101,543 | 18 | 203,086 |
No | output | 1 | 101,543 | 18 | 203,087 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, a start up by two students of a state university of city F gained incredible popularity. Now it's time to start a new company. But what do we call it?
The market analysts came up with a very smart plan: the name of the company should be identical to its reflection in a mirror! In other words, if we write out the name of the company on a piece of paper in a line (horizontally, from left to right) with large English letters, then put this piece of paper in front of the mirror, then the reflection of the name in the mirror should perfectly match the line written on the piece of paper.
There are many suggestions for the company name, so coming up to the mirror with a piece of paper for each name wouldn't be sensible. The founders of the company decided to automatize this process. They asked you to write a program that can, given a word, determine whether the word is a 'mirror' word or not.
Input
The first line contains a non-empty name that needs to be checked. The name contains at most 105 large English letters. The name will be written with the next sans serif font:
<image>
Output
Print 'YES' (without the quotes), if the given name matches its mirror reflection. Otherwise, print 'NO' (without the quotes).
Examples
Input
AHA
Output
YES
Input
Z
Output
NO
Input
XO
Output
NO
Submitted Solution:
```
ch=input()
b=len(ch)
if (len(ch)==1):
print("NO")
if(b%2==0):
if(ch[:int(b/2)]==ch[int(b/2):]):
print("YES")
else :
print("NO")
if(b%2!=0):
if(ch[:int(b/2)]==ch[int(b/2)+1:]):
print("YES")
else :
print("NO")
``` | instruction | 0 | 101,544 | 18 | 203,088 |
No | output | 1 | 101,544 | 18 | 203,089 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a boolean function of three variables which is defined by its truth table. You need to find an expression of minimum length that equals to this function. The expression may consist of:
* Operation AND ('&', ASCII code 38)
* Operation OR ('|', ASCII code 124)
* Operation NOT ('!', ASCII code 33)
* Variables x, y and z (ASCII codes 120-122)
* Parentheses ('(', ASCII code 40, and ')', ASCII code 41)
If more than one expression of minimum length exists, you should find the lexicographically smallest one.
Operations have standard priority. NOT has the highest priority, then AND goes, and OR has the lowest priority. The expression should satisfy the following grammar:
E ::= E '|' T | T
T ::= T '&' F | F
F ::= '!' F | '(' E ')' | 'x' | 'y' | 'z'
Input
The first line contains one integer n — the number of functions in the input (1 ≤ n ≤ 10 000).
The following n lines contain descriptions of functions, the i-th of them contains a string of length 8 that consists of digits 0 and 1 — the truth table of the i-th function. The digit on position j (0 ≤ j < 8) equals to the value of the function in case of <image>, <image> and <image>.
Output
You should output n lines, the i-th line should contain the expression of minimum length which equals to the i-th function. If there is more than one such expression, output the lexicographically smallest of them. Expressions should satisfy the given grammar and shouldn't contain white spaces.
Example
Input
4
00110011
00000111
11110000
00011111
Output
y
(y|z)&x
!x
x|y&z
Note
The truth table for the second function:
<image>
Submitted Solution:
```
ans = {'01100000': '!x&!y&z&(y|z)', '01110101': '!x&y|z', '11111011': '!x&z|y', '00010101': '(x|y)&z', '10111000': '!x&y&!z&!y', '01111011': '!x&z|!z&x|y', '01100001': '!x&!y&z&(y|z)|x&y&z', '10001001': '!y&!z|x&y&z', '10101001': '!x&y&!z|x&y&z', '10111011': '!z|y', '10111111': '!z|x|y', '11110000': '!x', '01101010': '!(x|y)&z&(x|y|z)', '00000101': 'x&z', '01010001': '!x&!y&z', '11011000': '!x&z&!y&!z', '11100111': '!x&!y&!y&z|x&z', '11101010': '!(x|y)&z', '11101111': '!y&z|x', '01000000': '!x&!y&z', '00100011': '!z&!x&y', '11001000': '!x&z&!y', '00100111': '!z&y|x&z', '01010011': '!x&z|x&y', '01110100': '!x&y|!y&z', '00100100': '!x&!z&y|!y&x&z', '10111010': '!x&y|!z', '10110001': '!x&!z|y&z', '10100000': '!x&!z', '00010001': 'y&z', '11010000': '!x&!y&!z', '11010010': '!x&!y&!z|!z&x&y', '01001100': '!y&(x|z)', '01100101': '!x&!z&y|!y&!x&z', '00110101': '!x&y|x&z', '10001000': '!y&!z', '10001010': '!y&!x&!z', '11000110': '!x&!z&!y|!z&x&y', '01101110': '!y&z&(x|y|z)', '00111100': '!x&y|!y&x', '01010101': 'z', '10100100': '!x&!z|!y&x&z', '01110011': '!x&z|y', '00011110': '!x&y&z|!y&z&x', '11101101': '!(x|z)&y|x&z', '00101000': '!x&y&!z&(x|y)', '11011010': '!x&!y&!z|!z&x', '00011101': '!y&x|y&z', '11101110': '!y&z', '11000001': '!x&!y|x&y&z', '10011001': '!y&!z|y&z', '01001110': '!y&z|!z&x', '10000001': '!(x|y|z)|x&y&z', '10110011': '!x&!z|y', '10101111': '!z|x', '00111110': '!x&y|!y&z&x', '01000001': '!x&!y&!y&!x&z', '11110011': '!x|y', '00110000': '!x&y', '10010100': '!(!x&z|!z&x|y)|!x&y&z', '00110110': '!x&z&y|!y&x&z', '01011111': 'x|z', '00011010': '!x&y&z|!z&x', '10010000': '!(!y&z|!z&y|x)', '01001111': '!y&z|x', '01100100': '!x&!z&y|!y&z', '00000001': 'x&y&z', '10100110': '!x&!y&!z|!y&x&z', '00000111': '(y|z)&x', '10011100': '!x&y&z|!y&!z&!x', '10000101': '!(x|y|z)|x&z', '11111010': '!x&z', '10010001': '!(x|y|z)|y&z', '10001011': '!y&!z|x&y', '00100101': '!x&!z&y|x&z', '11100100': '!x&!z&!y&z', '11000010': '!x&!y|!z&x&y', '11010001': '!x&!y|y&z', '01000110': '!y&z|!z&x&y', '11001111': '!y|x', '11100010': '!x&!y&!y&z', '10010010': '!(!x&y|!y&x|z)|!x&y&z', '11011011': '!x&!y|!z&x|y&z', '01011010': '!x&z|!z&x', '11000111': '!x&!z&!y|x&y', '10110100': '!x&!z&!y|!y&x&z', '00110010': '!x&z&y', '00011111': 'x|y&z', '01001101': '!y&!x&z|!y&x', '11111110': '!x&y&z', '10100010': '!x&!y&!z', '10011111': '!y&!z|x|y&z', '00100010': '!z&y', '00111010': '!x&y|!z&x', '00110011': 'y', '11100011': '!x&!y&z|x&y', '01001000': '!x&z&!y&(x|z)', '01011001': '!x&!y&z|!y&!z&x', '11110110': '!x|!y&z|!z&y', '01110001': '!x&!y&z|!x&y', '11010100': '!x&!y|!x&y&z', '00111001': '!x&!z&y|!y&!z&x', '10100001': '!x&!z|x&y&z', '10101101': '!x&y&!z|x&z', '00000100': '!y&x&z', '10110111': '!x&!z|x&z|y', '01100010': '!x&!y&z|!z&y', '00101010': '!z&(x|y)', '11110100': '!x|!y&z', '11111101': '!x&y|z', '01011011': '!x&!y&z|!z&x', '01011110': '!x&y&z|!z&x', '10001110': '!y&!z|!y&z&x', '01100011': '!x&!y&z|!z&!x&y', '11111111': '!x|x', '00010000': '!x&y&z', '11000100': '!x&!z&!y', '01001010': '!x&!y&z|!z&x', '11111001': '!(y|z)&x|y&z', '11101100': '!(x|z)&y', '01000011': '!x&!y&z|x&y', '00010111': '(x|y)&z|x&y', '00001000': '!y&!z&x', '10110101': '!x&!z&!y|x&z', '01001011': '!x&!y&z|!z&!y&x', '10110110': '!x&!z|!x&z&y|!y&x&z', '11001011': '!x&z&!y|x&y', '01111100': '!x&y&(x|y|z)', '00101111': '!z&y|x', '10111001': '!x&y&!z|y&z', '11001100': '!y', '10111100': '!x&!z&!y|!y&x', '10000110': '!(!x&y|!y&x|z)|!y&x&z', '10101000': '!x&y&!z', '11001010': '!x&!y|!z&x', '11011111': '!y|x|z', '00000110': '!y&z&(y|z)&x', '00111101': '!x&!z&y|!y&x', '01011101': '!y&x|z', '11010011': '!x&!y&!z|x&y', '01101100': '!(x|z)&y&(x|y|z)', '00001011': '!z&!y&x', '01101101': '!(x|z)&y&(x|y|z)|x&z', '11100000': '!x&!y&z', '01010010': '!x&z|!z&x&y', '10000100': '!(!x&z|!z&x|y)', '11010101': '!x&!y|z', '00000011': 'x&y', '10000011': '!(x|y|z)|x&y', '01110000': '!x&(y|z)', '00000010': '!z&x&y', '00110111': 'x&z|y', '00101011': '!z&!x&y|!z&x', '11110001': '!x|y&z', '01001001': '!x&z&!y&(x|z)|x&y&z', '11111100': '!x&y', '00111000': '!x&y|!y&!z&x', '11011100': '!x&z|!y', '10010111': '!(x|y|z)|(x|y)&z|x&y', '00001111': 'x', '11010111': '!x&!y|x&y|z', '11011101': '!y|z', '00010010': '!x&z&(x|z)&y', '01101000': '!(x|y)&z&!x&y&(x|y|z)', '11101000': '!(x|y)&z&!x&y', '10000000': '!(x|y|z)', '01110110': '!x&y&z|!z&y', '00010011': '(x|z)&y', '00101001': '!x&y&!z&(x|y)|x&y&z', '01000111': '!y&z|x&y', '01000100': '!y&z', '01111110': '!x&y&z&(x|y|z)', '11011110': '!x&z|!y|!z&x', '01100110': '!y&z|!z&y', '00010110': '!x&y&(x|y)&z|!z&x&y', '10100111': '!x&!y&!z|x&z', '10101011': '!z|x&y', '00101110': '!y&x|!z&y', '10101100': '!x&!z|!y&x', '01010000': '!x&z', '00001100': '!y&x', '00001101': '!y&!z&x', '00001001': '!y&!z&!z&!y&x', '11001001': '!x&z&!y|x&y&z', '10100101': '!x&!z|x&z', '10110000': '!x&!z&!y', '00001110': '!y&z&x', '00011000': '!x&y&z|!y&!z&x', '10010011': '!(x|y|z)|(x|z)&y', '10001101': '!y&!z|x&z', '01111000': '!(y|z)&x&(x|y|z)', '01010111': 'x&y|z', '10101010': '!z', '00111011': '!z&x|y', '11000011': '!x&!y|x&y', '10011110': '!x&y&z|!y&!z|!y&z&x', '11001101': '!y|x&z', '00111111': 'x|y', '01100111': '!y&!x&z|!z&y', '00110001': '!x&!z&y', '00100001': '!x&!z&!z&!x&y', '10001111': '!y&!z|x', '00101101': '!x&!z&y|!y&!z&x', '10011011': '!y&!x&!z|y&z', '00100000': '!x&!z&y', '10000111': '!(x|y|z)|(y|z)&x', '01011100': '!x&z|!y&x', '10011010': '!x&y&z|!y&!x&!z', '01110010': '!x&z|!z&y', '10010110': '!(!x&y|!y&x|z)|!x&y&(x|y)&z', '01111111': 'x|y|z', '11011001': '!x&z&!y|y&z', '11110111': '!x|y|z', '00100110': '!y&x&z|!z&y', '00011001': '!y&!z&x|y&z', '10010101': '!(x|y|z)|(x|y)&z', '01000101': '!y&!x&z', '10001100': '!y&!z&!x', '01101011': '!(x|y)&z&(x|y|z)|x&y', '01101001': '!(x|y)&!x&y&z&(!x&y|!y&x|z)', '10011000': '!x&y&z|!y&!z', '11100001': '!x&!y&z|x&y&z', '10101110': '!y&x|!z', '00110100': '!x&y|!y&x&z', '10000010': '!(!x&y|!y&x|z)', '10111110': '!x&y|!y&x|!z', '11111000': '!(y|z)&x', '01000010': '!x&!y&z|!z&x&y', '01111001': '!(y|z)&x&(x|y|z)|y&z', '10111101': '!x&!z|!y&x|y&z', '11010110': '!x&!y|!x&y&z|!z&x&y', '01111010': '!x&z&(x|y|z)', '11000000': '!x&!y', '00000000': '!x&x', '00101100': '!x&!z&y|!y&x', '11101001': '!(x|y)&z&!x&y|x&y&z', '11000101': '!x&!y|x&z', '11110101': '!x|z', '01010100': '!x&y&z', '10110010': '!x&!z|!x&z&y', '11001110': '!y|!z&x', '00011100': '!x&y&z|!y&x', '11101011': '!(x|y)&z|x&y', '10100011': '!x&!z|x&y', '01101111': '!y&z|!z&y|x', '00011011': '!z&x|y&z', '01111101': '!x&y|!y&x|z', '00001010': '!z&x', '11100110': '!x&!y&!z&!y&z', '10011101': '!y&!z&!x|y&z', '11110010': '!x|!z&y', '01010110': '!x&y&z|!z&x&y', '01011000': '!x&z|!y&!z&x', '11100101': '!x&!y&z|x&z', '00010100': '!x&y&(x|y)&z', '01110111': 'y|z'}
n=int(input())
for i in range(n):
print(ans[input()])
``` | instruction | 0 | 101,753 | 18 | 203,506 |
No | output | 1 | 101,753 | 18 | 203,507 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a boolean function of three variables which is defined by its truth table. You need to find an expression of minimum length that equals to this function. The expression may consist of:
* Operation AND ('&', ASCII code 38)
* Operation OR ('|', ASCII code 124)
* Operation NOT ('!', ASCII code 33)
* Variables x, y and z (ASCII codes 120-122)
* Parentheses ('(', ASCII code 40, and ')', ASCII code 41)
If more than one expression of minimum length exists, you should find the lexicographically smallest one.
Operations have standard priority. NOT has the highest priority, then AND goes, and OR has the lowest priority. The expression should satisfy the following grammar:
E ::= E '|' T | T
T ::= T '&' F | F
F ::= '!' F | '(' E ')' | 'x' | 'y' | 'z'
Input
The first line contains one integer n — the number of functions in the input (1 ≤ n ≤ 10 000).
The following n lines contain descriptions of functions, the i-th of them contains a string of length 8 that consists of digits 0 and 1 — the truth table of the i-th function. The digit on position j (0 ≤ j < 8) equals to the value of the function in case of <image>, <image> and <image>.
Output
You should output n lines, the i-th line should contain the expression of minimum length which equals to the i-th function. If there is more than one such expression, output the lexicographically smallest of them. Expressions should satisfy the given grammar and shouldn't contain white spaces.
Example
Input
4
00110011
00000111
11110000
00011111
Output
y
(y|z)&x
!x
x|y&z
Note
The truth table for the second function:
<image>
Submitted Solution:
```
s = '''00000000 !x&x
00000001 x&y&z
00000010 !x&y&z
00000011 y&z
00000100 !y&x&z
00000101 x&z
00000110 !x&y|!y&x&z
00000111 (x|y)&z
00001000 !x&!y&z
00001001 !x&!y|x&y&z
00001010 !x&z
00001011 !x&z|y&z
00001100 !y&z
00001101 !y&z|x&z
00001110 !(!z|x&y)
00001111 z
00010000 !z&x&y
00010001 x&y
00010010 !x&z|!z&x&y
00010011 (x|z)&y
00010100 !y&z|!z&y&x
00010101 (y|z)&x
00010110 !x&y&z|!y&z|!z&y&x
00010111 (x|y)&x&y|z
00011000 !x&!y&z|!z&x&y
00011001 !x&!y&z|x&y
00011010 !x&z|!z&x&y
00011011 !x&z|x&y
00011100 !y&z|!z&x&y
00011101 !y&z|x&y
00011110 !(x&y&z)&x&y|z
00011111 x&y|z
00100000 !x&!z&y
00100001 !x&!z|x&z&y
00100010 !x&y
00100011 !x&y|y&z
00100100 !x&!z&y|!y&x&z
00100101 !x&!z&y|x&z
00100110 !x&y|!y&x&z
00100111 !x&y|x&z
00101000 !x&!y&z|!z&y
00101001 !x&!y&z|!z&y|x&y&z
00101010 !x&(y|z)
00101011 !x&(y|z)|y&z
00101100 !x&!z&y|!y&z
00101101 !x&!y&z|!z&y|x&z
00101110 !x&y|!y&z
00101111 !x&y|z
00110000 !z&y
00110001 !z&y|x&y
00110010 !(!y|x&z)
00110011 y
00110100 !y&x&z|!z&y
00110101 !z&y|x&z
00110110 !(x&y&z)&x&z|y
00110111 x&z|y
00111000 !x&!y&z|!z&y
00111001 !x&!y&z|!z&y|x&y
00111010 !x&z|!z&y
00111011 !x&z|y
00111100 !y&z|!z&y
00111101 !y&z|!z&y|x&y
00111110 !(!y&!z|x&y&z)
00111111 y|z
01000000 !y&!z&x
01000001 !y&!z|y&z&x
01000010 !x&y&z|!y&!z&x
01000011 !y&!z&x|y&z
01000100 !y&x
01000101 !y&x|x&z
01000110 !x&y&z|!y&x
01000111 !y&x|y&z
01001000 !x&z|!z&x&!y
01001001 !x&z|!z&x&!y|x&y&z
01001010 !x&z|!y&!z&x
01001011 !x&z|!y&!z&x|y&z
01001100 !y&(x|z)
01001101 !y&(x|z)|x&z
01001110 !x&z|!y&x
01001111 !y&x|z
01010000 !z&x
01010001 !z&x|x&y
01010010 !x&y&z|!z&x
01010011 !z&x|y&z
01010100 !(!x|y&z)
01010101 x
01010110 x|y&z&!(x&y&z)
01010111 x|y&z
01011000 !x&!y&z|!z&x
01011001 !x&!y&z|!z&x|x&y
01011010 !x&z|!z&x
01011011 !x&z|!z&x|x&y
01011100 !y&z|!z&x
01011101 !y&z|x
01011110 !(!x&!z|x&y&z)
01011111 x|z
01100000 !x&y|!y&x&!z
01100001 !x&y|!y&x&!z|x&y&z
01100010 !x&y|!y&!z&x
01100011 !x&y|!y&!z&x|y&z
01100100 !x&!z&y|!y&x
01100101 !x&!z&y|!y&x|x&z
01100110 !x&y|!y&x
01100111 !x&y|!y&x|x&z
01101000 !x&!y&z|!x&y|!y&x&!z
01101001 !x&!y&z|!z&y|!y&!z|y&z&x
01101010 !x&(y|z)|!y&!z&x
01101011 !x&(y|z)|!y&!z&x|y&z
01101100 !x&!z&y|!y&(x|z)
01101101 !x&!z&y|!y&(x|z)|x&z
01101110 !x&(y|z)|!y&x
01101111 !x&y|!y&x|z
01110000 !z&(x|y)
01110001 !z&(x|y)|x&y
01110010 !x&y|!z&x
01110011 !z&x|y
01110100 !y&x|!z&y
01110101 !z&y|x
01110110 !(!x&!y|x&y&z)
01110111 x|y
01111000 !x&!y&z|!z&(x|y)
01111001 !x&!y&z|!z&(x|y)|x&y
01111010 !x&(y|z)|!z&x
01111011 !x&z|!z&x|y
01111100 !y&(x|z)|!z&y
01111101 !y&z|!z&y|x
01111110 !x&y|!y&z|!z&x
01111111 x|y|z
10000000 !(x|y|z)
10000001 !(x|y|z)|x&y&z
10000010 !x&!y&!z|y&z
10000011 !(x|y|z)|y&z
10000100 !x&!z|x&z&!y
10000101 !(x|y|z)|x&z
10000110 !x&!y&!z|y&z|!y&x&z
10000111 !(x|y|z)|(x|y)&z
10001000 !x&!y
10001001 !x&!y|x&y&z
10001010 !(!z&y|x)
10001011 !x&!y|y&z
10001100 !(!z&x|y)
10001101 !x&!y|x&z
10001110 !(!z&x|y)|!x&z
10001111 !x&!y|z
10010000 !x&!y|x&y&!z
10010001 !(x|y|z)|x&y
10010010 !x&!y&!z|y&z|!z&x&y
10010011 !(x|y|z)|(x|z)&y
10010100 !x&!y|x&y&!z|!y&x&z
10010101 !(x|y|z)|(y|z)&x
10010110 !x&!y&!z|y&z|!y&z|!z&y&x
10010111 !(x|y|z)|(x|y)&x&y|z
10011000 !x&!y|!z&x&y
10011001 !x&!y|x&y
10011010 !(!z&y|x)|!z&x&y
10011011 !(!z&y|x)|x&y
10011100 !(!z&x|y)|!z&x&y
10011101 !(!z&x|y)|x&y
10011110 !(x&y&z)&!x&!y|x&y|z
10011111 !x&!y|x&y|z
10100000 !x&!z
10100001 !x&!z|x&y&z
10100010 !(!y&z|x)
10100011 !x&!z|y&z
10100100 !x&!z|!y&x&z
10100101 !x&!z|x&z
10100110 !(!y&z|x)|!y&x&z
10100111 !(!y&z|x)|x&z
10101000 !(x|y&z)
10101001 !(x|y&z)|x&y&z
10101010 !x
10101011 !x|y&z
10101100 !x&!z|!y&z
10101101 !(x|y&z)|x&z
10101110 !x|!y&z
10101111 !x|z
10110000 !(!y&x|z)
10110001 !x&!z|x&y
10110010 !(!y&x|z)|!x&y
10110011 !x&!z|y
10110100 !(!y&x|z)|!y&x&z
10110101 !(!y&x|z)|x&z
10110110 !(x&y&z)&!x&!z|x&z|y
10110111 !x&!z|x&z|y
10111000 !x&!y|!z&y
10111001 !(x|y&z)|x&y
10111010 !x|!z&y
10111011 !x|y
10111100 !(!y&!z&x|y&z)
10111101 !x&!y|!z&y|x&z
10111110 !x|!y&z|!z&y
10111111 !x|y|z
11000000 !y&!z
11000001 !y&!z|x&y&z
11000010 !x&y&z|!y&!z
11000011 !y&!z|y&z
11000100 !(!x&z|y)
11000101 !y&!z|x&z
11000110 !(!x&z|y)|!x&y&z
11000111 !(!x&z|y)|y&z
11001000 !(x&z|y)
11001001 !(x&z|y)|x&y&z
11001010 !x&z|!y&!z
11001011 !(x&z|y)|y&z
11001100 !y
11001101 !y|x&z
11001110 !x&z|!y
11001111 !y|z
11010000 !(!x&y|z)
11010001 !y&!z|x&y
11010010 !(!x&y|z)|!x&y&z
11010011 !(!x&y|z)|y&z
11010100 !(!x&y|z)|!y&x
11010101 !y&!z|x
11010110 !(x&y&z)&!y&!z|x|y&z
11010111 !y&!z|x|y&z
11011000 !x&!y|!z&x
11011001 !(x&z|y)|x&y
11011010 !(!x&!z&y|x&z)
11011011 !x&!y|!z&x|y&z
11011100 !y|!z&x
11011101 !y|x
11011110 !x&z|!y|!z&x
11011111 !y|x|z
11100000 !(x&y|z)
11100001 !(x&y|z)|x&y&z
11100010 !x&y|!y&!z
11100011 !(x&y|z)|y&z
11100100 !x&!z|!y&x
11100101 !(x&y|z)|x&z
11100110 !(!x&!y&z|x&y)
11100111 !x&!z|!y&x|y&z
11101000 !((x|y)&x&y|z)
11101001 !((x|y)&x&y|z)|x&y&z
11101010 !x|!y&!z
11101011 !x|!y&!z|y&z
11101100 !x&!z|!y
11101101 !x&!z|!y|x&z
11101110 !x|!y
11101111 !x|!y|z
11110000 !z
11110001 !z|x&y
11110010 !x&y|!z
11110011 !z|y
11110100 !y&x|!z
11110101 !z|x
11110110 !x&y|!y&x|!z
11110111 !z|x|y
11111000 !x&!y|!z
11111001 !x&!y|!z|x&y
11111010 !x|!z
11111011 !x|!z|y
11111100 !y|!z
11111101 !y|!z|x
11111110 !(x&y&z)
11111111 !x|x'''
def vim(s):
rez = ''
for c in s:
if c == 'x':
rez += 'z'
elif c == 'z':
rez += 'x'
else:
rez += c
return rez
l = s.split('\n')
_d = dict()
for p in l:
_d[p.split()[0]] = p.split()[1]
n = int(input())
for i in range(n):
s = input().strip()
print(vim(_d[s]))
``` | instruction | 0 | 101,754 | 18 | 203,508 |
No | output | 1 | 101,754 | 18 | 203,509 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a boolean function of three variables which is defined by its truth table. You need to find an expression of minimum length that equals to this function. The expression may consist of:
* Operation AND ('&', ASCII code 38)
* Operation OR ('|', ASCII code 124)
* Operation NOT ('!', ASCII code 33)
* Variables x, y and z (ASCII codes 120-122)
* Parentheses ('(', ASCII code 40, and ')', ASCII code 41)
If more than one expression of minimum length exists, you should find the lexicographically smallest one.
Operations have standard priority. NOT has the highest priority, then AND goes, and OR has the lowest priority. The expression should satisfy the following grammar:
E ::= E '|' T | T
T ::= T '&' F | F
F ::= '!' F | '(' E ')' | 'x' | 'y' | 'z'
Input
The first line contains one integer n — the number of functions in the input (1 ≤ n ≤ 10 000).
The following n lines contain descriptions of functions, the i-th of them contains a string of length 8 that consists of digits 0 and 1 — the truth table of the i-th function. The digit on position j (0 ≤ j < 8) equals to the value of the function in case of <image>, <image> and <image>.
Output
You should output n lines, the i-th line should contain the expression of minimum length which equals to the i-th function. If there is more than one such expression, output the lexicographically smallest of them. Expressions should satisfy the given grammar and shouldn't contain white spaces.
Example
Input
4
00110011
00000111
11110000
00011111
Output
y
(y|z)&x
!x
x|y&z
Note
The truth table for the second function:
<image>
Submitted Solution:
```
base = [
"x", "(not x)", "y", "(not y)", "z", "(not z)",
]
expr = ["x", "(not x)", "y", "(not y)", "z", "(not z)"]
for x in base:
for y in base:
if x == y:
continue
for z in base:
if y == z:
continue
expr.append(x + "|" + y)
expr.append(x + "&" + y)
expr.append(x + "|" + y + "|" + z)
expr.append(x + "|" + "(" + y + "|" + z + ")")
expr.append(x + "&" + y + "|" + z)
expr.append(x + "&" + "(" + y + "|" + z + ")")
expr.append(x + "|" + y + "&" + z)
expr.append("(" + x + "|" + y + ")" + "&" + z)
expr.append(x + "&" + y + "&" + z)
expr.append(x + "&" + "(" + y + "&" + z + ")")
n = int(input())
for i in range(n):
s = input()
resp = "zzzzzzzzzzzzzzzzzz"
for exp in expr:
x = 0; y = 0; z = 0
if int(s[0]) != eval(exp):
continue
x = 0; y = 0; z = 1
if int(s[1]) != eval(exp):
continue
x = 0; y = 1; z = 0
if int(s[2]) != eval(exp):
continue
x = 0; y = 1; z = 1
if int(s[3]) != eval(exp):
continue
x = 1; y = 0; z = 0
if int(s[4]) != eval(exp):
continue
x = 1; y = 0; z = 1
if int(s[5]) != eval(exp):
continue
x = 1; y = 1; z = 0
if int(s[6]) != eval(exp):
continue
x = 1; y = 1; z = 1
if int(s[7]) != eval(exp):
continue
if (len(exp) < len(resp)):
resp = exp
elif (len(exp) < len(resp) and exp < resp):
resp = exp
i = resp.find("(not ")
while i != -1:
resp = resp[0:i] + "!" + resp[i+5] + resp[i+7:]
i = resp.find("(not ")
print(resp)
``` | instruction | 0 | 101,755 | 18 | 203,510 |
No | output | 1 | 101,755 | 18 | 203,511 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a boolean function of three variables which is defined by its truth table. You need to find an expression of minimum length that equals to this function. The expression may consist of:
* Operation AND ('&', ASCII code 38)
* Operation OR ('|', ASCII code 124)
* Operation NOT ('!', ASCII code 33)
* Variables x, y and z (ASCII codes 120-122)
* Parentheses ('(', ASCII code 40, and ')', ASCII code 41)
If more than one expression of minimum length exists, you should find the lexicographically smallest one.
Operations have standard priority. NOT has the highest priority, then AND goes, and OR has the lowest priority. The expression should satisfy the following grammar:
E ::= E '|' T | T
T ::= T '&' F | F
F ::= '!' F | '(' E ')' | 'x' | 'y' | 'z'
Input
The first line contains one integer n — the number of functions in the input (1 ≤ n ≤ 10 000).
The following n lines contain descriptions of functions, the i-th of them contains a string of length 8 that consists of digits 0 and 1 — the truth table of the i-th function. The digit on position j (0 ≤ j < 8) equals to the value of the function in case of <image>, <image> and <image>.
Output
You should output n lines, the i-th line should contain the expression of minimum length which equals to the i-th function. If there is more than one such expression, output the lexicographically smallest of them. Expressions should satisfy the given grammar and shouldn't contain white spaces.
Example
Input
4
00110011
00000111
11110000
00011111
Output
y
(y|z)&x
!x
x|y&z
Note
The truth table for the second function:
<image>
Submitted Solution:
```
f = ['!x&x', '!(x|y|z)', '!x&!y&z', '!x&!y', '!x&!z&y', '!x&!z', '!(!y&!z|x|y&z)', '!(x|y&z)', '!x&y&z', '!(!y&z|!z&y|x)', '!x&z', '!(!z&y|x)', '!x&y', '!(!y&z|x)', '!x&(y|z)', '!x', '!y&!z&x', '!y&!z', '!(!x&!z|x&z|y)', '!(x&z|y)', '!(!x&!y|x&y|z)', '!(x&y|z)', '!(!x&!y|x&y|z)|!x&!y&z', '!((x|y)&z|x&y)', '!x&y&z|!y&!z&x', '!x&y&z|!y&!z', '!x&z|!y&!z&x', '!x&z|!y&!z', '!x&y|!y&!z&x', '!x&y|!y&!z', '!x&(y|z)|!y&!z&x', '!x|!y&!z', '!y&x&z', '!(!x&z|!z&x|y)', '!y&z', '!(!z&x|y)', '!x&!z&y|!y&x&z', '!x&!z|!y&x&z', '!x&!z&y|!y&z', '!x&!z|!y&z', '!x&y&z|!y&x&z', '!(!x&z|!z&x|y)|!x&y&z', '!(!z|x&y)', '!(!z&x|y)|!x&z', '!x&y|!y&x&z', '!(!y&z|x)|!y&x&z', '!x&y|!y&z', '!x|!y&z', '!y&x', '!(!x&z|y)', '!y&(x|z)', '!y', '!x&!z&y|!y&x', '!x&!z|!y&x', '!x&!z&y|!y&(x|z)', '!x&!z|!y', '!x&y&z|!y&x', '!(!x&z|y)|!x&y&z', '!x&z|!y&x', '!x&z|!y', '!x&y|!y&x', '!(!x&!y&z|x&y)', '!x&(y|z)|!y&x', '!x|!y', '!z&x&y', '!(!x&y|!y&x|z)', '!x&!y&z|!z&x&y', '!x&!y|!z&x&y', '!z&y', '!(!y&x|z)', '!x&!y&z|!z&y', '!x&!y|!z&y', '!x&y&z|!z&x&y', '!(!x&y|!y&x|z)|!x&y&z', '!x&z|!z&x&y', '!(!z&y|x)|!z&x&y', '!(!y|x&z)', '!(!y&x|z)|!x&y', '!x&z|!z&y', '!x|!z&y', '!z&x', '!(!x&y|z)', '!x&!y&z|!z&x', '!x&!y|!z&x', '!z&(x|y)', '!z', '!x&!y&z|!z&(x|y)', '!x&!y|!z', '!x&y&z|!z&x', '!(!x&y|z)|!x&y&z', '!x&z|!z&x', '!(!x&!z&y|x&z)', '!x&y|!z&x', '!x&y|!z', '!x&(y|z)|!z&x', '!x|!z', '!y&x&z|!z&x&y', '!(!x&y|!y&x|z)|!y&x&z', '!y&z|!z&x&y', '!(!z&x|y)|!z&x&y', '!y&x&z|!z&y', '!(!y&x|z)|!y&x&z', '!y&z|!z&y', '!(!y&!z&x|y&z)', '!x&y&z|!y&x&z|!z&x&y', '!(!x&y|!y&x|z)|!x&y&z|!y&x&z', '!(!z|x&y)|!z&x&y', '!(!z&x|y)|!x&z|!z&x&y', '!(!y|x&z)|!y&x&z', '!(!y&x|z)|!x&y|!y&x&z', '!(!y&!z|x&y&z)', '!x|!y&z|!z&y', '!(!x|y&z)', '!(!x&y|z)|!y&x', '!y&z|!z&x', '!y|!z&x', '!y&x|!z&y', '!y&x|!z', '!y&(x|z)|!z&y', '!y|!z', '!(!x|y&z)|!x&y&z', '!(!x&y|z)|!x&y&z|!y&x', '!(!x&!z|x&y&z)', '!x&z|!y|!z&x', '!(!x&!y|x&y&z)', '!x&y|!y&x|!z', '!x&y|!y&z|!z&x', '!(x&y&z)', 'x&y&z', '!(x|y|z)|x&y&z', '!x&!y&z|x&y&z', '!x&!y|x&y&z', '!x&!z&y|x&y&z', '!x&!z|x&y&z', '!(!y&!z|x|y&z)|x&y&z', '!(x|y&z)|x&y&z', 'y&z', '!(x|y|z)|y&z', '!x&z|y&z', '!x&!y|y&z', '!x&y|y&z', '!x&!z|y&z', '!x&(y|z)|y&z', '!x|y&z', '!y&!z&x|x&y&z', '!y&!z|x&y&z', '!(!x&!z|x&z|y)|x&y&z', '!(x&z|y)|x&y&z', '!(!x&!y|x&y|z)|x&y&z', '!(x&y|z)|x&y&z', '!(!x&!y|x&y|z)|!x&!y&z|x&y&z', '!((x|y)&z|x&y)|x&y&z', '!y&!z&x|y&z', '!y&!z|y&z', '!x&z|!y&!z&x|y&z', '!(x&z|y)|y&z', '!x&y|!y&!z&x|y&z', '!(x&y|z)|y&z', '!x&(y|z)|!y&!z&x|y&z', '!x|!y&!z|y&z', 'x&z', '!(x|y|z)|x&z', '!y&z|x&z', '!x&!y|x&z', '!x&!z&y|x&z', '!x&!z|x&z', '!x&!z&y|!y&z|x&z', '!(x|y&z)|x&z', '(x|y)&z', '!(x|y|z)|(x|y)&z', 'z', '!x&!y|z', '!x&y|x&z', '!(!y&z|x)|x&z', '!x&y|z', '!x|z', '!y&x|x&z', '!y&!z|x&z', '!y&(x|z)|x&z', '!y|x&z', '!x&!z&y|!y&x|x&z', '!(x&y|z)|x&z', '!x&!z&y|!y&(x|z)|x&z', '!x&!z|!y|x&z', '!y&x|y&z', '!(!x&z|y)|y&z', '!y&x|z', '!y|z', '!x&y|!y&x|x&z', '!x&!z|!y&x|y&z', '!x&y|!y&x|z', '!x|!y|z', 'x&y', '!(x|y|z)|x&y', '!x&!y&z|x&y', '!x&!y|x&y', '!z&y|x&y', '!x&!z|x&y', '!x&!y&z|!z&y|x&y', '!(x|y&z)|x&y', '(x|z)&y', '!(x|y|z)|(x|z)&y', '!x&z|x&y', '!(!z&y|x)|x&y', 'y', '!x&!z|y', '!x&z|y', '!x|y', '!z&x|x&y', '!y&!z|x&y', '!x&!y&z|!z&x|x&y', '!(x&z|y)|x&y', '!z&(x|y)|x&y', '!z|x&y', '!x&!y&z|!z&(x|y)|x&y', '!x&!y|!z|x&y', '!z&x|y&z', '!(!x&y|z)|y&z', '!x&z|!z&x|x&y', '!x&!y|!z&x|y&z', '!z&x|y', '!z|y', '!x&z|!z&x|y', '!x|!z|y', '(y|z)&x', '!(x|y|z)|(y|z)&x', '!y&z|x&y', '!(!z&x|y)|x&y', '!z&y|x&z', '!(!y&x|z)|x&z', '!y&z|!z&y|x&y', '!x&!y|!z&y|x&z', '(x|y)&z|x&y', '!(x|y|z)|(x|y)&z|x&y', 'x&y|z', '!x&!y|x&y|z', 'x&z|y', '!x&!z|x&z|y', 'y|z', '!x|y|z', 'x', '!y&!z|x', '!y&z|x', '!y|x', '!z&y|x', '!z|x', '!y&z|!z&y|x', '!y|!z|x', 'x|y&z', '!y&!z|x|y&z', 'x|z', '!y|x|z', 'x|y', '!z|x|y', 'x|y|z', '!x|x']
n = int(input())
for i in range(n):
s = input()
a = 0
for j in range(8):
a += int(s[j] == '1') << j
print(f[a])
``` | instruction | 0 | 101,756 | 18 | 203,512 |
No | output | 1 | 101,756 | 18 | 203,513 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.
You are given two strings O and E. O contains the characters at the odd-numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
Constraints
* O and E consists of lowercase English letters (`a` - `z`).
* 1 \leq |O|,|E| \leq 50
* |O| - |E| is either 0 or 1.
Input
Input is given from Standard Input in the following format:
O
E
Output
Print the original password.
Examples
Input
xyz
abc
Output
xaybzc
Input
atcoderbeginnercontest
atcoderregularcontest
Output
aattccooddeerrbreeggiunlnaerrccoonntteesstt | instruction | 0 | 101,920 | 18 | 203,840 |
"Correct Solution:
```
I=input
for i,j in zip(I(),I()+" "):print(i+j,end="")
``` | output | 1 | 101,920 | 18 | 203,841 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.
You are given two strings O and E. O contains the characters at the odd-numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
Constraints
* O and E consists of lowercase English letters (`a` - `z`).
* 1 \leq |O|,|E| \leq 50
* |O| - |E| is either 0 or 1.
Input
Input is given from Standard Input in the following format:
O
E
Output
Print the original password.
Examples
Input
xyz
abc
Output
xaybzc
Input
atcoderbeginnercontest
atcoderregularcontest
Output
aattccooddeerrbreeggiunlnaerrccoonntteesstt | instruction | 0 | 101,921 | 18 | 203,842 |
"Correct Solution:
```
O = input()
E = input() + ' '
print(''.join(a + b for a, b in zip(O, E)))
``` | output | 1 | 101,921 | 18 | 203,843 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.
You are given two strings O and E. O contains the characters at the odd-numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
Constraints
* O and E consists of lowercase English letters (`a` - `z`).
* 1 \leq |O|,|E| \leq 50
* |O| - |E| is either 0 or 1.
Input
Input is given from Standard Input in the following format:
O
E
Output
Print the original password.
Examples
Input
xyz
abc
Output
xaybzc
Input
atcoderbeginnercontest
atcoderregularcontest
Output
aattccooddeerrbreeggiunlnaerrccoonntteesstt | instruction | 0 | 101,922 | 18 | 203,844 |
"Correct Solution:
```
print(''.join(map(''.join,zip(*open(0)))))
``` | output | 1 | 101,922 | 18 | 203,845 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.
You are given two strings O and E. O contains the characters at the odd-numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
Constraints
* O and E consists of lowercase English letters (`a` - `z`).
* 1 \leq |O|,|E| \leq 50
* |O| - |E| is either 0 or 1.
Input
Input is given from Standard Input in the following format:
O
E
Output
Print the original password.
Examples
Input
xyz
abc
Output
xaybzc
Input
atcoderbeginnercontest
atcoderregularcontest
Output
aattccooddeerrbreeggiunlnaerrccoonntteesstt | instruction | 0 | 101,923 | 18 | 203,846 |
"Correct Solution:
```
O = input()
E = input()
s = ''
for a in zip(O,E):s += a[0]+a[1]
if len(O) != len(E):s += O[-1]
print(s)
``` | output | 1 | 101,923 | 18 | 203,847 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.
You are given two strings O and E. O contains the characters at the odd-numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
Constraints
* O and E consists of lowercase English letters (`a` - `z`).
* 1 \leq |O|,|E| \leq 50
* |O| - |E| is either 0 or 1.
Input
Input is given from Standard Input in the following format:
O
E
Output
Print the original password.
Examples
Input
xyz
abc
Output
xaybzc
Input
atcoderbeginnercontest
atcoderregularcontest
Output
aattccooddeerrbreeggiunlnaerrccoonntteesstt | instruction | 0 | 101,924 | 18 | 203,848 |
"Correct Solution:
```
o = input()
e = input()
z =''
for i in range(len(e)):
z+=o[i]+e[i]
if len(o) > len(e):
z += o[-1]
print(z)
``` | output | 1 | 101,924 | 18 | 203,849 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.
You are given two strings O and E. O contains the characters at the odd-numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
Constraints
* O and E consists of lowercase English letters (`a` - `z`).
* 1 \leq |O|,|E| \leq 50
* |O| - |E| is either 0 or 1.
Input
Input is given from Standard Input in the following format:
O
E
Output
Print the original password.
Examples
Input
xyz
abc
Output
xaybzc
Input
atcoderbeginnercontest
atcoderregularcontest
Output
aattccooddeerrbreeggiunlnaerrccoonntteesstt | instruction | 0 | 101,925 | 18 | 203,850 |
"Correct Solution:
```
A=list(input())
B=list(input())
L=["0"]*(len(A)+len(B))
L[::2]=A
L[1::2]=B
print("".join(L))
``` | output | 1 | 101,925 | 18 | 203,851 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.
You are given two strings O and E. O contains the characters at the odd-numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
Constraints
* O and E consists of lowercase English letters (`a` - `z`).
* 1 \leq |O|,|E| \leq 50
* |O| - |E| is either 0 or 1.
Input
Input is given from Standard Input in the following format:
O
E
Output
Print the original password.
Examples
Input
xyz
abc
Output
xaybzc
Input
atcoderbeginnercontest
atcoderregularcontest
Output
aattccooddeerrbreeggiunlnaerrccoonntteesstt | instruction | 0 | 101,926 | 18 | 203,852 |
"Correct Solution:
```
*O,=open(0);print("".join(o+e for o,e in zip(*O)))
``` | output | 1 | 101,926 | 18 | 203,853 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.
You are given two strings O and E. O contains the characters at the odd-numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
Constraints
* O and E consists of lowercase English letters (`a` - `z`).
* 1 \leq |O|,|E| \leq 50
* |O| - |E| is either 0 or 1.
Input
Input is given from Standard Input in the following format:
O
E
Output
Print the original password.
Examples
Input
xyz
abc
Output
xaybzc
Input
atcoderbeginnercontest
atcoderregularcontest
Output
aattccooddeerrbreeggiunlnaerrccoonntteesstt | instruction | 0 | 101,927 | 18 | 203,854 |
"Correct Solution:
```
O=input();E=input()+' ';print(*[a+b for a,b in zip(O,E)],sep='')
``` | output | 1 | 101,927 | 18 | 203,855 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.
You are given two strings O and E. O contains the characters at the odd-numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
Constraints
* O and E consists of lowercase English letters (`a` - `z`).
* 1 \leq |O|,|E| \leq 50
* |O| - |E| is either 0 or 1.
Input
Input is given from Standard Input in the following format:
O
E
Output
Print the original password.
Examples
Input
xyz
abc
Output
xaybzc
Input
atcoderbeginnercontest
atcoderregularcontest
Output
aattccooddeerrbreeggiunlnaerrccoonntteesstt
Submitted Solution:
```
o=input()
e=input()
print(''.join(x+y for x,y in zip(list(o),list(e)))+('' if len(o)==len(e) else o[-1]))
``` | instruction | 0 | 101,928 | 18 | 203,856 |
Yes | output | 1 | 101,928 | 18 | 203,857 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.
You are given two strings O and E. O contains the characters at the odd-numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
Constraints
* O and E consists of lowercase English letters (`a` - `z`).
* 1 \leq |O|,|E| \leq 50
* |O| - |E| is either 0 or 1.
Input
Input is given from Standard Input in the following format:
O
E
Output
Print the original password.
Examples
Input
xyz
abc
Output
xaybzc
Input
atcoderbeginnercontest
atcoderregularcontest
Output
aattccooddeerrbreeggiunlnaerrccoonntteesstt
Submitted Solution:
```
O = input()
E = input() + " "
for i in range(len(O)): print(O[i]+E[i],end="",sep="")
``` | instruction | 0 | 101,929 | 18 | 203,858 |
Yes | output | 1 | 101,929 | 18 | 203,859 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.
You are given two strings O and E. O contains the characters at the odd-numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
Constraints
* O and E consists of lowercase English letters (`a` - `z`).
* 1 \leq |O|,|E| \leq 50
* |O| - |E| is either 0 or 1.
Input
Input is given from Standard Input in the following format:
O
E
Output
Print the original password.
Examples
Input
xyz
abc
Output
xaybzc
Input
atcoderbeginnercontest
atcoderregularcontest
Output
aattccooddeerrbreeggiunlnaerrccoonntteesstt
Submitted Solution:
```
s=input()
t=input()
ans=""
for i in range(len(t)):
ans+=s[i]+t[i]
if len(s)-len(t):ans+=s[-1]
print(ans)
``` | instruction | 0 | 101,930 | 18 | 203,860 |
Yes | output | 1 | 101,930 | 18 | 203,861 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.
You are given two strings O and E. O contains the characters at the odd-numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
Constraints
* O and E consists of lowercase English letters (`a` - `z`).
* 1 \leq |O|,|E| \leq 50
* |O| - |E| is either 0 or 1.
Input
Input is given from Standard Input in the following format:
O
E
Output
Print the original password.
Examples
Input
xyz
abc
Output
xaybzc
Input
atcoderbeginnercontest
atcoderregularcontest
Output
aattccooddeerrbreeggiunlnaerrccoonntteesstt
Submitted Solution:
```
O = list(input())
E = list(input())
for i in range(len(E)):
O.insert(2 * i + 1, E[i])
print("".join(O))
``` | instruction | 0 | 101,931 | 18 | 203,862 |
Yes | output | 1 | 101,931 | 18 | 203,863 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.
You are given two strings O and E. O contains the characters at the odd-numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
Constraints
* O and E consists of lowercase English letters (`a` - `z`).
* 1 \leq |O|,|E| \leq 50
* |O| - |E| is either 0 or 1.
Input
Input is given from Standard Input in the following format:
O
E
Output
Print the original password.
Examples
Input
xyz
abc
Output
xaybzc
Input
atcoderbeginnercontest
atcoderregularcontest
Output
aattccooddeerrbreeggiunlnaerrccoonntteesstt
Submitted Solution:
```
# -*- coding: utf-8 -*-
o = list(input()[::2])
e = list(input()[1::2])
ans = []
while len(o) and len(e):
if len(o):
ans.append(o[0])
del o[0]
if len(e):
ans.append(e[0])
del e[0]
print(''.join(ans))
``` | instruction | 0 | 101,932 | 18 | 203,864 |
No | output | 1 | 101,932 | 18 | 203,865 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.
You are given two strings O and E. O contains the characters at the odd-numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
Constraints
* O and E consists of lowercase English letters (`a` - `z`).
* 1 \leq |O|,|E| \leq 50
* |O| - |E| is either 0 or 1.
Input
Input is given from Standard Input in the following format:
O
E
Output
Print the original password.
Examples
Input
xyz
abc
Output
xaybzc
Input
atcoderbeginnercontest
atcoderregularcontest
Output
aattccooddeerrbreeggiunlnaerrccoonntteesstt
Submitted Solution:
```
O = input()
E = input()
a = ''
for i in O:
a += i
for j in E:
a += j
E = E.lstrip(j)
break
print(a)
``` | instruction | 0 | 101,933 | 18 | 203,866 |
No | output | 1 | 101,933 | 18 | 203,867 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.
You are given two strings O and E. O contains the characters at the odd-numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
Constraints
* O and E consists of lowercase English letters (`a` - `z`).
* 1 \leq |O|,|E| \leq 50
* |O| - |E| is either 0 or 1.
Input
Input is given from Standard Input in the following format:
O
E
Output
Print the original password.
Examples
Input
xyz
abc
Output
xaybzc
Input
atcoderbeginnercontest
atcoderregularcontest
Output
aattccooddeerrbreeggiunlnaerrccoonntteesstt
Submitted Solution:
```
o=list(input())
e=list(input())
for x,y in zip(x,y):print(x,y,end="")
``` | instruction | 0 | 101,934 | 18 | 203,868 |
No | output | 1 | 101,934 | 18 | 203,869 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.
You are given two strings O and E. O contains the characters at the odd-numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
Constraints
* O and E consists of lowercase English letters (`a` - `z`).
* 1 \leq |O|,|E| \leq 50
* |O| - |E| is either 0 or 1.
Input
Input is given from Standard Input in the following format:
O
E
Output
Print the original password.
Examples
Input
xyz
abc
Output
xaybzc
Input
atcoderbeginnercontest
atcoderregularcontest
Output
aattccooddeerrbreeggiunlnaerrccoonntteesstt
Submitted Solution:
```
O = input()
E = input()
S = ""
for i,j in zip(O,E):
S += i + j
print(S)
``` | instruction | 0 | 101,935 | 18 | 203,870 |
No | output | 1 | 101,935 | 18 | 203,871 |
Provide a correct Python 3 solution for this coding contest problem.
My arm is stuck and I can't pull it out.
I tried to pick up something that had rolled over to the back of my desk, and I inserted my arm while illuminating it with the screen of my cell phone instead of a flashlight, but I accidentally got caught in something and couldn't pull it out. It hurts if you move it forcibly.
What should I do. How can I escape? Do you call for help out loud? I'm embarrassed to dismiss it, but it seems difficult to escape on my own. No, no one should have been within reach. Hold down your impatience and look around. If you can reach that computer, you can call for help by email, but unfortunately it's too far away. After all, do we have to wait for someone to pass by by chance?
No, wait, there is a machine that sends emails. It's already in your hands, rather than within reach. The screen at the tip of my deeply extended hand is blurry and hard to see, but it's a mobile phone I've been using for many years. You should be able to send emails even if you can't see the screen.
Carefully press the button while drawing the screen in your head. Rely on the feel of your finger and type in a message by pressing the number keys many times to avoid making typos. It doesn't convert to kanji, and it should be transmitted even in hiragana. I'm glad I didn't switch to the mainstream smartphones for the touch panel.
Take a deep breath and put your finger on the send button. After a while, my cell phone quivered lightly. It is a signal that the transmission is completed. sigh. A friend must come to help after a while.
I think my friends are smart enough to interpret it properly, but the message I just sent may be different from the one I really wanted to send. The character input method of this mobile phone is the same as the widely used one, for example, use the "1" button to input the hiragana of the "A" line. "1" stands for "a", "11" stands for "i", and so on, "11111" stands for "o". Hiragana loops, that is, "111111" also becomes "a". The "ka" line uses "2", the "sa" line uses "3", ..., and the behavior when the same button is pressed in succession is the same as in the example of "1".
However, this character input method has some annoying behavior. If you do not operate for a while after pressing the number button, the conversion to hiragana will be forced. In other words, if you press "1" three times in a row, it will become "U", but if you press "1" three times after a while, it will become "Ah". To give another example, when you press "111111", the character string entered depends on the interval between the presses, which can be either "a" or "ah ah ah". "111111111111" may be "yes", "yeah ah", or twelve "a". Note that even if you press a different number button, it will be converted to hiragana, so "12345" can only be the character string "Akasatana".
Well, I didn't mean to press the wrong button because I typed it carefully, but I'm not confident that I could press the buttons at the right intervals. It's very possible that you've sent something different than the message you really wanted to send. Now, how many strings may have been sent? Oh, this might be a good way to kill time until a friend comes to help. They will come to help within 5 hours at the latest. I hope it can be solved by then.
Input
The input consists of multiple cases.
Each case is given in the following format.
string
The end of the input is given by the line where the input consists of "#"
string contains up to 100,000 numbers between 0 and 9.
No more than 50 inputs have a string length greater than 10,000.
The test case file size is guaranteed to be 5MB or less.
Also, the number of test cases does not exceed 100.
The characters that can be entered with each number key are as shown in the table below.
Numbers | Enterable characters
--- | ---
1 | Aiueo
2 | Kakikukeko
3 |
4 |
5 | What is it?
6 | Hahifuheho
7 | Mamimumemo
8 | Yayuyo
9 | Larry Lero
0 | Won
Output
Divide how to interpret the sentence by 1000000007 and output the remainder on one line.
Examples
Input
1
11
111111
111111111111
12345
11111111119999999999
11111111113333333333
11111111118888888888
11111111112222222222111111111
11111111110000000000444444444
11224111122411
888888888888999999999999888888888888999999999999999999
666666666666666777333333333338888888888
1111114444441111111444499999931111111222222222222888111111115555
#
Output
1
2
32
1856
1
230400
230400
156480
56217600
38181120
128
26681431
61684293
40046720
Input
1
11
111111
111111111111
12345
11111111119999999999
11111111113333333333
11111111118888888888
11111111112222222222111111111
11111111110000000000444444444
11224111122411
888888888888999999999999888888888888999999999999999999
666666666666666777333333333338888888888
1111114444441111111444499999931111111222222222222888111111115555
Output
1
2
32
1856
1
230400
230400
156480
56217600
38181120
128
26681431
61684293
40046720 | instruction | 0 | 101,991 | 18 | 203,982 |
"Correct Solution:
```
def f3(n):
a=b=c=0
for _ in range(n):
a,b,c=(a+b+c+1)%100000007,a,b
return a
def f5(n):
a=b=c=d=e=0
for _ in range(n):
a,b,c,d,e=(a+b+c+d+e+1)%100000007,a,b,c,d
return a
while 1:
s=input()
if s=="#":break
ans=1
num="_"
cnt=1
for n in s+"_":
if n==num:
cnt+=1
else:
if num in "80":
ans=ans*f3(cnt)%100000007
else:
ans=ans*f5(cnt)%100000007
num=n
cnt=1
print(ans)
``` | output | 1 | 101,991 | 18 | 203,983 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
My arm is stuck and I can't pull it out.
I tried to pick up something that had rolled over to the back of my desk, and I inserted my arm while illuminating it with the screen of my cell phone instead of a flashlight, but I accidentally got caught in something and couldn't pull it out. It hurts if you move it forcibly.
What should I do. How can I escape? Do you call for help out loud? I'm embarrassed to dismiss it, but it seems difficult to escape on my own. No, no one should have been within reach. Hold down your impatience and look around. If you can reach that computer, you can call for help by email, but unfortunately it's too far away. After all, do we have to wait for someone to pass by by chance?
No, wait, there is a machine that sends emails. It's already in your hands, rather than within reach. The screen at the tip of my deeply extended hand is blurry and hard to see, but it's a mobile phone I've been using for many years. You should be able to send emails even if you can't see the screen.
Carefully press the button while drawing the screen in your head. Rely on the feel of your finger and type in a message by pressing the number keys many times to avoid making typos. It doesn't convert to kanji, and it should be transmitted even in hiragana. I'm glad I didn't switch to the mainstream smartphones for the touch panel.
Take a deep breath and put your finger on the send button. After a while, my cell phone quivered lightly. It is a signal that the transmission is completed. sigh. A friend must come to help after a while.
I think my friends are smart enough to interpret it properly, but the message I just sent may be different from the one I really wanted to send. The character input method of this mobile phone is the same as the widely used one, for example, use the "1" button to input the hiragana of the "A" line. "1" stands for "a", "11" stands for "i", and so on, "11111" stands for "o". Hiragana loops, that is, "111111" also becomes "a". The "ka" line uses "2", the "sa" line uses "3", ..., and the behavior when the same button is pressed in succession is the same as in the example of "1".
However, this character input method has some annoying behavior. If you do not operate for a while after pressing the number button, the conversion to hiragana will be forced. In other words, if you press "1" three times in a row, it will become "U", but if you press "1" three times after a while, it will become "Ah". To give another example, when you press "111111", the character string entered depends on the interval between the presses, which can be either "a" or "ah ah ah". "111111111111" may be "yes", "yeah ah", or twelve "a". Note that even if you press a different number button, it will be converted to hiragana, so "12345" can only be the character string "Akasatana".
Well, I didn't mean to press the wrong button because I typed it carefully, but I'm not confident that I could press the buttons at the right intervals. It's very possible that you've sent something different than the message you really wanted to send. Now, how many strings may have been sent? Oh, this might be a good way to kill time until a friend comes to help. They will come to help within 5 hours at the latest. I hope it can be solved by then.
Input
The input consists of multiple cases.
Each case is given in the following format.
string
The end of the input is given by the line where the input consists of "#"
string contains up to 100,000 numbers between 0 and 9.
No more than 50 inputs have a string length greater than 10,000.
The test case file size is guaranteed to be 5MB or less.
Also, the number of test cases does not exceed 100.
The characters that can be entered with each number key are as shown in the table below.
Numbers | Enterable characters
--- | ---
1 | Aiueo
2 | Kakikukeko
3 |
4 |
5 | What is it?
6 | Hahifuheho
7 | Mamimumemo
8 | Yayuyo
9 | Larry Lero
0 | Won
Output
Divide how to interpret the sentence by 1000000007 and output the remainder on one line.
Examples
Input
1
11
111111
111111111111
12345
11111111119999999999
11111111113333333333
11111111118888888888
11111111112222222222111111111
11111111110000000000444444444
11224111122411
888888888888999999999999888888888888999999999999999999
666666666666666777333333333338888888888
1111114444441111111444499999931111111222222222222888111111115555
#
Output
1
2
32
1856
1
230400
230400
156480
56217600
38181120
128
26681431
61684293
40046720
Input
1
11
111111
111111111111
12345
11111111119999999999
11111111113333333333
11111111118888888888
11111111112222222222111111111
11111111110000000000444444444
11224111122411
888888888888999999999999888888888888999999999999999999
666666666666666777333333333338888888888
1111114444441111111444499999931111111222222222222888111111115555
Output
1
2
32
1856
1
230400
230400
156480
56217600
38181120
128
26681431
61684293
40046720
Submitted Solution:
```
def f3(n):
a=b=c=0
for _ in range(n):
a,b,c=(a+b+c+1)%100000007,a,b
return a
def f5(n):
a=b=c=d=e=0
for _ in range(n):
a,b,c,d,e=(a+b+c+d+e+1)%100000007,a,b,c,d
return a
while 1:
s=input()
if s=="#":break
ans=1
num="_"
cnt=1
for n in s+"_":
if n==num:
cnt+=1
else:
if num in "80":
ans*=f3(cnt)
else:
ans*=f5(cnt)
num=n
cnt=1
print(ans)
``` | instruction | 0 | 101,992 | 18 | 203,984 |
No | output | 1 | 101,992 | 18 | 203,985 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
My arm is stuck and I can't pull it out.
I tried to pick up something that had rolled over to the back of my desk, and I inserted my arm while illuminating it with the screen of my cell phone instead of a flashlight, but I accidentally got caught in something and couldn't pull it out. It hurts if you move it forcibly.
What should I do. How can I escape? Do you call for help out loud? I'm embarrassed to dismiss it, but it seems difficult to escape on my own. No, no one should have been within reach. Hold down your impatience and look around. If you can reach that computer, you can call for help by email, but unfortunately it's too far away. After all, do we have to wait for someone to pass by by chance?
No, wait, there is a machine that sends emails. It's already in your hands, rather than within reach. The screen at the tip of my deeply extended hand is blurry and hard to see, but it's a mobile phone I've been using for many years. You should be able to send emails even if you can't see the screen.
Carefully press the button while drawing the screen in your head. Rely on the feel of your finger and type in a message by pressing the number keys many times to avoid making typos. It doesn't convert to kanji, and it should be transmitted even in hiragana. I'm glad I didn't switch to the mainstream smartphones for the touch panel.
Take a deep breath and put your finger on the send button. After a while, my cell phone quivered lightly. It is a signal that the transmission is completed. sigh. A friend must come to help after a while.
I think my friends are smart enough to interpret it properly, but the message I just sent may be different from the one I really wanted to send. The character input method of this mobile phone is the same as the widely used one, for example, use the "1" button to input the hiragana of the "A" line. "1" stands for "a", "11" stands for "i", and so on, "11111" stands for "o". Hiragana loops, that is, "111111" also becomes "a". The "ka" line uses "2", the "sa" line uses "3", ..., and the behavior when the same button is pressed in succession is the same as in the example of "1".
However, this character input method has some annoying behavior. If you do not operate for a while after pressing the number button, the conversion to hiragana will be forced. In other words, if you press "1" three times in a row, it will become "U", but if you press "1" three times after a while, it will become "Ah". To give another example, when you press "111111", the character string entered depends on the interval between the presses, which can be either "a" or "ah ah ah". "111111111111" may be "yes", "yeah ah", or twelve "a". Note that even if you press a different number button, it will be converted to hiragana, so "12345" can only be the character string "Akasatana".
Well, I didn't mean to press the wrong button because I typed it carefully, but I'm not confident that I could press the buttons at the right intervals. It's very possible that you've sent something different than the message you really wanted to send. Now, how many strings may have been sent? Oh, this might be a good way to kill time until a friend comes to help. They will come to help within 5 hours at the latest. I hope it can be solved by then.
Input
The input consists of multiple cases.
Each case is given in the following format.
string
The end of the input is given by the line where the input consists of "#"
string contains up to 100,000 numbers between 0 and 9.
No more than 50 inputs have a string length greater than 10,000.
The test case file size is guaranteed to be 5MB or less.
Also, the number of test cases does not exceed 100.
The characters that can be entered with each number key are as shown in the table below.
Numbers | Enterable characters
--- | ---
1 | Aiueo
2 | Kakikukeko
3 |
4 |
5 | What is it?
6 | Hahifuheho
7 | Mamimumemo
8 | Yayuyo
9 | Larry Lero
0 | Won
Output
Divide how to interpret the sentence by 1000000007 and output the remainder on one line.
Examples
Input
1
11
111111
111111111111
12345
11111111119999999999
11111111113333333333
11111111118888888888
11111111112222222222111111111
11111111110000000000444444444
11224111122411
888888888888999999999999888888888888999999999999999999
666666666666666777333333333338888888888
1111114444441111111444499999931111111222222222222888111111115555
#
Output
1
2
32
1856
1
230400
230400
156480
56217600
38181120
128
26681431
61684293
40046720
Input
1
11
111111
111111111111
12345
11111111119999999999
11111111113333333333
11111111118888888888
11111111112222222222111111111
11111111110000000000444444444
11224111122411
888888888888999999999999888888888888999999999999999999
666666666666666777333333333338888888888
1111114444441111111444499999931111111222222222222888111111115555
Output
1
2
32
1856
1
230400
230400
156480
56217600
38181120
128
26681431
61684293
40046720
Submitted Solution:
```
def f3(n):
a=b=c=0
for _ in range(n):
a,b,c=a+b+c+1,a,b
return a
def f5(n):
a=b=c=d=e=0
for _ in range(n):
a,b,c,d,e=a+b+c+d+e+1,a,b,c,d
return a
while 1:
s=input()
if s=="#":break
ans=1
num="_"
cnt=1
for n in s+"_":
if n==num:
cnt+=1
else:
if num in "80":
ans*=f3(cnt)
else:
ans*=f5(cnt)
num=n
cnt=1
print(ans%100000007)
``` | instruction | 0 | 101,993 | 18 | 203,986 |
No | output | 1 | 101,993 | 18 | 203,987 |
Provide a correct Python 3 solution for this coding contest problem.
Find places where a string P is found within a text T. Print all indices of T where P found. The indices of T start with 0.
Constraints
* 1 ≤ length of T ≤ 1000000
* 1 ≤ length of P ≤ 10000
* The input consists of alphabetical characters and digits
Input
In the first line, a text T is given. In the second line, a string P is given.
Output
Print an index of T where P found in a line. Print the indices in ascending order.
Examples
Input
aabaaa
aa
Output
0
3
4
Input
xyzz
yz
Output
1
Input
abc
xyz
Output | instruction | 0 | 102,000 | 18 | 204,000 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
MOD = 10 ** 9 + 7
T = input()
P = input()
N = len(T)
M = len(P)
ans = []
p = P
for i in range(N-M+1):
# Tの区間[i,i+M)の文字列を取得
t = T[i:i+M]
if t == p:
ans.append(i)
if len(ans):
[print(i) for i in ans]
``` | output | 1 | 102,000 | 18 | 204,001 |
Provide a correct Python 3 solution for this coding contest problem.
Find places where a string P is found within a text T. Print all indices of T where P found. The indices of T start with 0.
Constraints
* 1 ≤ length of T ≤ 1000000
* 1 ≤ length of P ≤ 10000
* The input consists of alphabetical characters and digits
Input
In the first line, a text T is given. In the second line, a string P is given.
Output
Print an index of T where P found in a line. Print the indices in ascending order.
Examples
Input
aabaaa
aa
Output
0
3
4
Input
xyzz
yz
Output
1
Input
abc
xyz
Output | instruction | 0 | 102,001 | 18 | 204,002 |
"Correct Solution:
```
def main():
text = input()
len_text = len(text)
pattern = input()
len_pattern = len(pattern)
for i in range(len_text - len_pattern + 1):
if text[i:i+len_pattern] == pattern:
print(i)
return
main()
``` | output | 1 | 102,001 | 18 | 204,003 |
Provide a correct Python 3 solution for this coding contest problem.
Find places where a string P is found within a text T. Print all indices of T where P found. The indices of T start with 0.
Constraints
* 1 ≤ length of T ≤ 1000000
* 1 ≤ length of P ≤ 10000
* The input consists of alphabetical characters and digits
Input
In the first line, a text T is given. In the second line, a string P is given.
Output
Print an index of T where P found in a line. Print the indices in ascending order.
Examples
Input
aabaaa
aa
Output
0
3
4
Input
xyzz
yz
Output
1
Input
abc
xyz
Output | instruction | 0 | 102,002 | 18 | 204,004 |
"Correct Solution:
```
def kmpTable(w):
lw = len(w)
nx = [-1] * (lw + 1)
j = -1
for i in range(lw):
while j >= 0 and w[i] != w[j]:
j = nx[j]
j += 1
nx[i + 1] = j
return tuple(nx)
def kmpSearch(s, w):
ls = len(s)
start = 0
w_idx = 0
ret = []
nx = kmpTable(w)
while start + w_idx < ls:
if s[start + w_idx] == w[w_idx]:
w_idx += 1
if w_idx == len(w):
ret.append(start)
start = start + w_idx - nx[w_idx]
w_idx = nx[w_idx]
else:
if w_idx == 0:
start += 1
else:
start = start + w_idx - nx[w_idx]
w_idx = nx[w_idx]
return tuple(ret)
T = input()
P = input()
ans = kmpSearch(T, P)
if ans:
print("\n".join(map(str, ans)))
``` | output | 1 | 102,002 | 18 | 204,005 |
Provide a correct Python 3 solution for this coding contest problem.
Find places where a string P is found within a text T. Print all indices of T where P found. The indices of T start with 0.
Constraints
* 1 ≤ length of T ≤ 1000000
* 1 ≤ length of P ≤ 10000
* The input consists of alphabetical characters and digits
Input
In the first line, a text T is given. In the second line, a string P is given.
Output
Print an index of T where P found in a line. Print the indices in ascending order.
Examples
Input
aabaaa
aa
Output
0
3
4
Input
xyzz
yz
Output
1
Input
abc
xyz
Output | instruction | 0 | 102,004 | 18 | 204,008 |
"Correct Solution:
```
def f():
T,P=input(),input()
for i in range(len(T)):P!=T[i:i+len(P)]or print(i)
f()
``` | output | 1 | 102,004 | 18 | 204,009 |
Provide a correct Python 3 solution for this coding contest problem.
Find places where a string P is found within a text T. Print all indices of T where P found. The indices of T start with 0.
Constraints
* 1 ≤ length of T ≤ 1000000
* 1 ≤ length of P ≤ 10000
* The input consists of alphabetical characters and digits
Input
In the first line, a text T is given. In the second line, a string P is given.
Output
Print an index of T where P found in a line. Print the indices in ascending order.
Examples
Input
aabaaa
aa
Output
0
3
4
Input
xyzz
yz
Output
1
Input
abc
xyz
Output | instruction | 0 | 102,005 | 18 | 204,010 |
"Correct Solution:
```
if __name__ == '__main__':
# ??????????????\???
# T = 'abc'
# P = 'xyz'
T = input()
P = input()
# ???????????????
results = []
for i in range(len(T) - len(P) + 1):
temp = T[i:i+len(P)]
if temp == P:
results.append(i)
# ???????????¨???
for r in results:
print(r)
``` | output | 1 | 102,005 | 18 | 204,011 |
Provide a correct Python 3 solution for this coding contest problem.
Find places where a string P is found within a text T. Print all indices of T where P found. The indices of T start with 0.
Constraints
* 1 ≤ length of T ≤ 1000000
* 1 ≤ length of P ≤ 10000
* The input consists of alphabetical characters and digits
Input
In the first line, a text T is given. In the second line, a string P is given.
Output
Print an index of T where P found in a line. Print the indices in ascending order.
Examples
Input
aabaaa
aa
Output
0
3
4
Input
xyzz
yz
Output
1
Input
abc
xyz
Output | instruction | 0 | 102,007 | 18 | 204,014 |
"Correct Solution:
```
#とりあえず
t=input()
p=input()
l=len(t)-len(p)
lp=len(p)
ans=[]
for i in range(l+1):
if t[i:i+len(p)]==p:
print(i)
``` | output | 1 | 102,007 | 18 | 204,015 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find places where a string P is found within a text T. Print all indices of T where P found. The indices of T start with 0.
Constraints
* 1 ≤ length of T ≤ 1000000
* 1 ≤ length of P ≤ 10000
* The input consists of alphabetical characters and digits
Input
In the first line, a text T is given. In the second line, a string P is given.
Output
Print an index of T where P found in a line. Print the indices in ascending order.
Examples
Input
aabaaa
aa
Output
0
3
4
Input
xyzz
yz
Output
1
Input
abc
xyz
Output
Submitted Solution:
```
# AOJ ALDS1_14_B "String Search"
# returns i s.t. S[i+j] = T[j] for 0 <= j < len(T)
def RollingHash(S, T):
if len(S) < len(T):
return []
# gcd(h, b) = 1
h = 10**11+7
b = 10**7+7
L = len(T)
bL = 1
for i in range(L):
bL = bL * b % h
hashS = 0
for i in range(L):
hashS = (hashS * b + ord(S[i])) % h
hashT = 0
for i in range(L):
hashT = (hashT * b + ord(T[i])) % h
correctIndexes = []
if hashS == hashT:
correctIndexes.append(0)
for j in range(len(S)-L):
hashS = (hashS * b - ord(S[j])*bL + ord(S[L+j])) % h
if hashS == hashT:
correctIndexes.append(j+1)
return correctIndexes
if __name__ == "__main__":
import sys
input = sys.stdin.readline
S = input().rstrip()
T = input().rstrip()
ans = RollingHash(S, T)
for a in ans:
print(a)
``` | instruction | 0 | 102,008 | 18 | 204,016 |
Yes | output | 1 | 102,008 | 18 | 204,017 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find places where a string P is found within a text T. Print all indices of T where P found. The indices of T start with 0.
Constraints
* 1 ≤ length of T ≤ 1000000
* 1 ≤ length of P ≤ 10000
* The input consists of alphabetical characters and digits
Input
In the first line, a text T is given. In the second line, a string P is given.
Output
Print an index of T where P found in a line. Print the indices in ascending order.
Examples
Input
aabaaa
aa
Output
0
3
4
Input
xyzz
yz
Output
1
Input
abc
xyz
Output
Submitted Solution:
```
def make_kmp_table(t):
i = 2
j = 0
m = len(t)
tbl = [0] * (m + 1)
tbl[0] = -1
while i <= m:
if t[i - 1] == t[j]:
tbl[i] = j + 1
i += 1
j += 1
elif j > 0:
j = tbl[j]
else:
tbl[i] = 0
i += 1
return tbl
def kmp(s, t):
matched_indices = []
tbl = make_kmp_table(t)
i = 0
j = 0
n = len(s)
m = len(t)
while i + j < n:
if t[j] == s[i + j]:
j += 1
if j == m:
matched_indices.append(i)
i += j - tbl[j]
j = tbl[j]
else:
i += j - tbl[j]
if j > 0:
j = tbl[j]
return matched_indices
t = input()
p = input()
result = kmp(t, p)
if result:
print(*result, sep='\n')
``` | instruction | 0 | 102,009 | 18 | 204,018 |
Yes | output | 1 | 102,009 | 18 | 204,019 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find places where a string P is found within a text T. Print all indices of T where P found. The indices of T start with 0.
Constraints
* 1 ≤ length of T ≤ 1000000
* 1 ≤ length of P ≤ 10000
* The input consists of alphabetical characters and digits
Input
In the first line, a text T is given. In the second line, a string P is given.
Output
Print an index of T where P found in a line. Print the indices in ascending order.
Examples
Input
aabaaa
aa
Output
0
3
4
Input
xyzz
yz
Output
1
Input
abc
xyz
Output
Submitted Solution:
```
def kmpTable(w):
lw = len(w)
nx = [-1] * (lw + 1)
j = -1
for i in range(lw):
while j >= 0 and w[i] != w[j]:
j = nx[j]
j += 1
nx[i + 1] = j
return tuple(nx)
def kmpSearch(s, w):
ls = len(s)
start = 0
w_idx = 0
ret = []
nx = kmpTable(w)
while start + w_idx < ls:
if s[start + w_idx] == w[w_idx]:
w_idx += 1
if w_idx == len(w):
ret.append(start)
start = start + w_idx - nx[w_idx]
w_idx = nx[w_idx]
else:
if w_idx == 0:
start += 1
else:
start = start + w_idx - nx[w_idx]
w_idx = nx[w_idx]
return ret
T = input()
P = input()
ans = kmpSearch(T, P)
if ans:
print(*ans, sep="\n")
# print("\n".join(map(str, ans)))
``` | instruction | 0 | 102,010 | 18 | 204,020 |
Yes | output | 1 | 102,010 | 18 | 204,021 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find places where a string P is found within a text T. Print all indices of T where P found. The indices of T start with 0.
Constraints
* 1 ≤ length of T ≤ 1000000
* 1 ≤ length of P ≤ 10000
* The input consists of alphabetical characters and digits
Input
In the first line, a text T is given. In the second line, a string P is given.
Output
Print an index of T where P found in a line. Print the indices in ascending order.
Examples
Input
aabaaa
aa
Output
0
3
4
Input
xyzz
yz
Output
1
Input
abc
xyz
Output
Submitted Solution:
```
import sys
input = sys.stdin.readline
def make_table(s):
n = len(s)
res = [-1]*(n+1)
j = -1
for i in range(n):
while j>=0 and s[i]!=s[j]:
j = res[j]
j += 1
res[i+1] = j
return res
def kmp(s, w): #s中のwと一致する箇所の先頭インデックスのリストを作成
table = make_table(w)
res = []
m, i, n = 0, 0, len(s)
while m+i<n:
if w[i]==s[m+i]:
i += 1
if i==len(w):
res.append(m)
m = m+i-table[i]
i = table[i]
else:
m = m+i-table[i]
if i>0:
i = table[i]
return res
T = input()[:-1]
P = input()[:-1]
res = kmp(T, P)
for res_i in res:
print(res_i)
``` | instruction | 0 | 102,011 | 18 | 204,022 |
Yes | output | 1 | 102,011 | 18 | 204,023 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find places where a string P is found within a text T. Print all indices of T where P found. The indices of T start with 0.
Constraints
* 1 ≤ length of T ≤ 1000000
* 1 ≤ length of P ≤ 10000
* The input consists of alphabetical characters and digits
Input
In the first line, a text T is given. In the second line, a string P is given.
Output
Print an index of T where P found in a line. Print the indices in ascending order.
Examples
Input
aabaaa
aa
Output
0
3
4
Input
xyzz
yz
Output
1
Input
abc
xyz
Output
Submitted Solution:
```
T = list(input())
P = list(input())
for i in range(len(T)-len(P)+1):
if P[0] == T[i]:
for j in range(len(P)):
if P[j] == T[i + j]:
pass
else: break
else: print(i)
``` | instruction | 0 | 102,012 | 18 | 204,024 |
No | output | 1 | 102,012 | 18 | 204,025 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find places where a string P is found within a text T. Print all indices of T where P found. The indices of T start with 0.
Constraints
* 1 ≤ length of T ≤ 1000000
* 1 ≤ length of P ≤ 10000
* The input consists of alphabetical characters and digits
Input
In the first line, a text T is given. In the second line, a string P is given.
Output
Print an index of T where P found in a line. Print the indices in ascending order.
Examples
Input
aabaaa
aa
Output
0
3
4
Input
xyzz
yz
Output
1
Input
abc
xyz
Output
Submitted Solution:
```
t = list(input())
p = list(input())
n = len(t) - len(p) + 1
for i in range(n):
if p[0] == t[i]:
for j in range(len(p)):
flag = True
if not p[j] == t[i+j]:
flag = False
break
if flag:
print(i)
``` | instruction | 0 | 102,013 | 18 | 204,026 |
No | output | 1 | 102,013 | 18 | 204,027 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find places where a string P is found within a text T. Print all indices of T where P found. The indices of T start with 0.
Constraints
* 1 ≤ length of T ≤ 1000000
* 1 ≤ length of P ≤ 10000
* The input consists of alphabetical characters and digits
Input
In the first line, a text T is given. In the second line, a string P is given.
Output
Print an index of T where P found in a line. Print the indices in ascending order.
Examples
Input
aabaaa
aa
Output
0
3
4
Input
xyzz
yz
Output
1
Input
abc
xyz
Output
Submitted Solution:
```
## SA-IS
from collections import defaultdict
def l_or_s_classification(target):
length = len(target)
l_or_s = ["" for n in range(length)]
for i in range(length):
if i == length - 1 or target[i] < target[i + 1]:
l_or_s[i] = "S"
else:
l_or_s[i] = "L"
return l_or_s
def left_most_s(target, l_or_s):
length = len(target)
lms = []
for i in range(1, length):
if l_or_s[i - 1] == "L" and l_or_s[i] == "S":
lms.append(i)
return lms
def get_bucket(target):
characters = set(target)
bucket_size = []
bucket_capacity = defaultdict(int)
bucket_index = {}
for t in target:
bucket_capacity[t] += 1
for i, c in enumerate(sorted(characters)):
bucket_size.append(bucket_capacity[c])
bucket_index[c] = i
return (bucket_size, bucket_capacity, bucket_index)
def get_bucket_start_end(bucket_size):
length = len(bucket_size)
temp_start_sum = 0
temp_end_sum = -1
start_index = [0]
end_index = []
for i in range(length):
temp_start_sum += bucket_size[i]
temp_end_sum += bucket_size[i]
start_index.append(temp_start_sum)
end_index.append(temp_end_sum)
start_index.pop()
return (start_index, end_index)
def lms_bucket_sort(target, bucket_index, end_index, suffix_index, choice):
length = len(target)
for ch in choice:
c = target[ch]
this_index = end_index[bucket_index[c]]
suffix_index[this_index] = ch
end_index[bucket_index[c]] -= 1
def make_suffix_array(target):
target += "$"
l_or_s = l_or_s_classification(target)
lms = left_most_s(target, l_or_s)
bucket_size, bucket_capacity, bucket_index = get_bucket(target)
start_index, end_index = get_bucket_start_end(bucket_size)
lms_end_index = end_index[:]
length = len(target)
suffix_string = [target[n:] for n in range(length)]
suffix_index = [-1 for n in range(length)]
lms_bucket_sort(target, bucket_index, lms_end_index, suffix_index, lms)
# print(suffix_string)
# print(suffix_index, lms, l_or_s)
for suf in suffix_index:
if suf != -1 and l_or_s[suf - 1] == "L":
c = target[suf - 1]
suffix_index[start_index[bucket_index[c]]] = suf - 1
start_index[bucket_index[c]] += 1
# print(suffix_index)
for i in range(length - 1, -1, -1):
suf = suffix_index[i]
if suf != -1 and l_or_s[suf - 1] == "S":
c = target[suf - 1]
suffix_index[end_index[bucket_index[c]]] = suf - 1
end_index[bucket_index[c]] -= 1
suffix_array = [suffix_string[n] for n in suffix_index]
# print(suffix_index)
return (suffix_array, suffix_index)
def binary_range_search(suffix_array, pattern):
length = len(suffix_array)
left = 0
temp_r = length - 1
pat_len = len(pattern)
while left < temp_r:
mid = (left + temp_r) // 2
comp_len = min(len(suffix_array[mid]), pat_len)
if pattern[:comp_len] > suffix_array[mid][:comp_len]:
left = mid + 1
else:
temp_r = mid
right = length - 1
temp_l = 0
while temp_l < right:
mid = (temp_l + right + 1) // 2
comp_len = min(len(suffix_array[mid]), pat_len)
if pattern[:comp_len] < suffix_array[mid][:comp_len]:
right = mid - 1
else:
temp_l = mid
if left == right and suffix_array[left][:pat_len] != pattern:
return []
else:
return [left, right]
target = input().strip()
pattern = input().strip()
if len(target) >= len(pattern):
suffix_array, suffix_index = make_suffix_array(target)
match_range = binary_range_search(suffix_array, pattern)
for i in sorted(suffix_index[match_range[0]:match_range[1] + 1]):
print(i)
``` | instruction | 0 | 102,014 | 18 | 204,028 |
No | output | 1 | 102,014 | 18 | 204,029 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find places where a string P is found within a text T. Print all indices of T where P found. The indices of T start with 0.
Constraints
* 1 ≤ length of T ≤ 1000000
* 1 ≤ length of P ≤ 10000
* The input consists of alphabetical characters and digits
Input
In the first line, a text T is given. In the second line, a string P is given.
Output
Print an index of T where P found in a line. Print the indices in ascending order.
Examples
Input
aabaaa
aa
Output
0
3
4
Input
xyzz
yz
Output
1
Input
abc
xyz
Output
Submitted Solution:
```
import random, math
random.seed()
def gen(a, b, num):
result = set()
while 1:
while 1:
v = random.randint(a, b)//2*2+1
if v not in result:
break
for x in range(3, int(math.sqrt(v))+1, 2):
if v % x == 0:
break
else:
result.add(v)
if len(result) == num:
break
return result
class RH():
def __init__(self, s, base, mod):
self.base = base
self.mod = mod
self.rev = pow(base, mod-2, mod)
l = len(s)
self.h = h = [0]*(l+1)
tmp = 0
for i in range(l):
num = ord(s[i])
tmp = (tmp*base + num) % mod
h[i+1] = tmp
def calc(self, l, r):
return (self.h[r] - self.h[l] * pow(self.base, r-l, self.mod) + self.mod) % self.mod
class RRH():
def __init__(self, s, num=10, primes=None):
primes = primes or gen(2, 10**3, num)
MOD = 10**9+7
self.rhs = [RH(s, p, MOD) for p in primes]
def calc(self, l, r):
return [rh.calc(l, r) for rh in self.rhs]
T = input(); P = input()
primes = gen(2, 10**3, 3)
HT = RRH(T, primes=primes)
HP = RRH(P, primes=primes)
pv = HP.calc(0, len(P))
for i in range(len(T)-len(P)+1):
if HT.calc(i, i+len(P)) == pv:
print(i)
``` | instruction | 0 | 102,015 | 18 | 204,030 |
No | output | 1 | 102,015 | 18 | 204,031 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which replace all the lower-case letters of a given text with the corresponding captital letters.
Input
A text including lower-case letters, periods, and space is given in a line. The number of characters in the text is less than or equal to 200.
Output
Print the converted text.
Example
Input
this is a pen.
Output
THIS IS A PEN. | instruction | 0 | 102,803 | 18 | 205,606 |
"Correct Solution:
```
str=input()
print(str.upper())
``` | output | 1 | 102,803 | 18 | 205,607 |
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