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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 β€ n_0, n_1, n_2 β€ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Submitted Solution:
```
for _ in range(int(input())):
n0,n1,n2=map(int,input().split())
if n1%2==1:
print('0'*n0+'01'*(n1//2+1)+'1'*n2)
else:
if n1==0 and n2==0:
print('0'*(n0+1))
else:
print('10'*(n1//2)+'0'*n0+'1'*(n2+1))
```
|
instruction
| 0
| 5,934
| 0
| 11,868
|
Yes
|
output
| 1
| 5,934
| 0
| 11,869
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 β€ n_0, n_1, n_2 β€ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Submitted Solution:
```
def solve():
n0, n1, n2 = [int(x) for x in input().split()]
res = []
if not n1:
if n0:
res = ['0' for _ in range(n0 + 1)]
else:
res = ['1' for _ in range(n2 + 1)]
else:
res = ['0' if i & 1 else '1' for i in range(n1 + 1)]
res[1:1] = ['0' for _ in range(n0)]
res[0:0] = ['1' for _ in range(n2)]
print(''.join(res))
for _ in range(int(input())):
solve()
```
|
instruction
| 0
| 5,935
| 0
| 11,870
|
Yes
|
output
| 1
| 5,935
| 0
| 11,871
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 β€ n_0, n_1, n_2 β€ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Submitted Solution:
```
t = int(input())
for _ in range(t):
a, b, c = [int(x) for x in input().split()]
ans = ""
for i in range(b):
if i % 2 == 0:
ans += "1"
else:
ans += "0"
if b % 2 != 0:
ans = "1" * c + ans
ans += "0" * (a + 1)
if b % 2 == 0:
if b == 0:
if a > 0:
ans += "0" * (a + 1)
if c > 0:
ans += "1" * (c + 1)
else:
ans= "1" * c + ans
ans += "0" * a
ans += "1"
print(ans)
```
|
instruction
| 0
| 5,936
| 0
| 11,872
|
Yes
|
output
| 1
| 5,936
| 0
| 11,873
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 β€ n_0, n_1, n_2 β€ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Submitted Solution:
```
# t = int(input())
# n, k = map(int, input().split())
# a = list(map(int, input().split()))
# from collections import deque
#import copy
#import collections
#import sys
t = int(input())
for T in range(t):
n0, n1, n2 = map(int, input().split())
ans = ''
if n1==0:
if n0>0:
ans = '0'*(n0+1)
else:
ans = '1'*(n2+1)
else:
ans = '0'*(n0+1) + '1'*(n2+1)
flag = 0
for i in range(n1-1):
ans += str(flag)
flag ^= 1
print(ans)
```
|
instruction
| 0
| 5,937
| 0
| 11,874
|
Yes
|
output
| 1
| 5,937
| 0
| 11,875
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 β€ n_0, n_1, n_2 β€ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Submitted Solution:
```
n0 = '0'
n2 = '1'
t = int(input())
for _ in range(t):
output = ''
a, b, c = map(int, input().split())
if c: output += n2*(c+1)
if output:
for i in range(b):
output += str(i%2)
else:
if b:
for i in range(b+1):
output += str(i%2)
if b and output[-1] == '0':
output += n0*a
else:
if a: output += n0*(a+1)
print(output)
```
|
instruction
| 0
| 5,938
| 0
| 11,876
|
No
|
output
| 1
| 5,938
| 0
| 11,877
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 β€ n_0, n_1, n_2 β€ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Submitted Solution:
```
for _ in range((int)(input())):
zero,one,two=list(map(int,input().split()))
res=[]
if one%2==0:
if zero!=0:
res.append('0'*(zero+1))
one-=1
if two!=0:
res.append('1'*(two+1))
res.append('01'*(one))
else:
res.append('1'*(two))
if one!=0:
res.append('10'*(one//2+1))
res.append('0'*(zero))
print(''.join([i for i in res]))
```
|
instruction
| 0
| 5,939
| 0
| 11,878
|
No
|
output
| 1
| 5,939
| 0
| 11,879
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 β€ n_0, n_1, n_2 β€ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Submitted Solution:
```
for _ in range(int(input())):
n0, n1, n2 = map(int, input().split(' '))
if n0 != 0:
print(0, end='')
for i in range(n0):
print(0, end='')
if n2 != 0:
print(1, end='')
for i in range(n2):
print(1, end='')
start = 1
if n2 != 0:
start = 0
if n0 == 0 and n2 == 0:
print(0, end='')
for i in range(n1):
print(start, end='')
start = (start + 1) % 2
print()
```
|
instruction
| 0
| 5,940
| 0
| 11,880
|
No
|
output
| 1
| 5,940
| 0
| 11,881
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For some binary string s (i.e. each character s_i is either '0' or '1'), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of '1' (ones) in it was calculated.
You are given three numbers:
* n_0 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
* n_1 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
* n_2 β the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s="1110011110", the following substrings would be written: "11", "11", "10", "00", "01", "11", "11", "11", "10". Thus, n_0=1, n_1=3, n_2=5.
Your task is to restore any suitable binary string s from the given values n_0, n_1, n_2. It is guaranteed that at least one of the numbers n_0, n_1, n_2 is greater than 0. Also, it is guaranteed that a solution exists.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then test cases follow.
Each test case consists of one line which contains three integers n_0, n_1, n_2 (0 β€ n_0, n_1, n_2 β€ 100; n_0 + n_1 + n_2 > 0). It is guaranteed that the answer for given n_0, n_1, n_2 exists.
Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.
Example
Input
7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0
Output
1110011110
0011
0110001100101011
10
0000111
1111
000
Submitted Solution:
```
n = int(input())
def check(x,y,z):
s=''
for _ in range(z):
s += ''.join('1')
for i in range(y):
if i%2!=0:
s+=''.join('1')
else:
s+=''.join('0')
for _ in range(x):
s+=''.join('0')
print(s)
for _ in range(n):
n0,n1,n2 = map(int,input().split())
check(n0,n1,n2)
```
|
instruction
| 0
| 5,941
| 0
| 11,882
|
No
|
output
| 1
| 5,941
| 0
| 11,883
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You're given a list of n strings a1, a2, ..., an. You'd like to concatenate them together in some order such that the resulting string would be lexicographically smallest.
Given the list of strings, output the lexicographically smallest concatenation.
Input
The first line contains integer n β the number of strings (1 β€ n β€ 5Β·104).
Each of the next n lines contains one string ai (1 β€ |ai| β€ 50) consisting of only lowercase English letters. The sum of string lengths will not exceed 5Β·104.
Output
Print the only string a β the lexicographically smallest string concatenation.
Examples
Input
4
abba
abacaba
bcd
er
Output
abacabaabbabcder
Input
5
x
xx
xxa
xxaa
xxaaa
Output
xxaaaxxaaxxaxxx
Input
3
c
cb
cba
Output
cbacbc
|
instruction
| 0
| 6,240
| 0
| 12,480
|
Tags: sortings, strings
Correct Solution:
```
from functools import cmp_to_key as ctk
def comp(a, b):
if a + b < b + a:
return -1
elif a + b > b + a:
return 1
else:
return 0
n = int(input())
string = list()
for i in range(n):
strtem = input()
string.append(strtem)
string.sort(key = ctk(comp))
print(''.join(string))
```
|
output
| 1
| 6,240
| 0
| 12,481
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given a list of n strings a1, a2, ..., an. You'd like to concatenate them together in some order such that the resulting string would be lexicographically smallest.
Given the list of strings, output the lexicographically smallest concatenation.
Input
The first line contains integer n β the number of strings (1 β€ n β€ 5Β·104).
Each of the next n lines contains one string ai (1 β€ |ai| β€ 50) consisting of only lowercase English letters. The sum of string lengths will not exceed 5Β·104.
Output
Print the only string a β the lexicographically smallest string concatenation.
Examples
Input
4
abba
abacaba
bcd
er
Output
abacabaabbabcder
Input
5
x
xx
xxa
xxaa
xxaaa
Output
xxaaaxxaaxxaxxx
Input
3
c
cb
cba
Output
cbacbc
Submitted Solution:
```
n = int(input())
a = []
for i in range(n):
a.append(input())
from functools import cmp_to_key
# a = ['abc', 'abcabb', 'abcabcabb', 'abcabcabbabc']
a_p = sorted(a, key=cmp_to_key(lambda x, y: -1 if x + y < y + x else 1))
print(''.join(a_p))
```
|
instruction
| 0
| 6,245
| 0
| 12,490
|
Yes
|
output
| 1
| 6,245
| 0
| 12,491
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given a list of n strings a1, a2, ..., an. You'd like to concatenate them together in some order such that the resulting string would be lexicographically smallest.
Given the list of strings, output the lexicographically smallest concatenation.
Input
The first line contains integer n β the number of strings (1 β€ n β€ 5Β·104).
Each of the next n lines contains one string ai (1 β€ |ai| β€ 50) consisting of only lowercase English letters. The sum of string lengths will not exceed 5Β·104.
Output
Print the only string a β the lexicographically smallest string concatenation.
Examples
Input
4
abba
abacaba
bcd
er
Output
abacabaabbabcder
Input
5
x
xx
xxa
xxaa
xxaaa
Output
xxaaaxxaaxxaxxx
Input
3
c
cb
cba
Output
cbacbc
Submitted Solution:
```
from functools import cmp_to_key
n = int(input())
a=[]
for i in range(n):
a.append(input())
def cmp(x,y):
if x+y < y+x :
return -1
elif x+y > y+x:
return 1
else:
return 0
print(''.join(sorted(a,key=cmp_to_key(cmp))))
#print(a.sort(key = lambda x,y: cmp(x,y)))
```
|
instruction
| 0
| 6,246
| 0
| 12,492
|
Yes
|
output
| 1
| 6,246
| 0
| 12,493
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given a list of n strings a1, a2, ..., an. You'd like to concatenate them together in some order such that the resulting string would be lexicographically smallest.
Given the list of strings, output the lexicographically smallest concatenation.
Input
The first line contains integer n β the number of strings (1 β€ n β€ 5Β·104).
Each of the next n lines contains one string ai (1 β€ |ai| β€ 50) consisting of only lowercase English letters. The sum of string lengths will not exceed 5Β·104.
Output
Print the only string a β the lexicographically smallest string concatenation.
Examples
Input
4
abba
abacaba
bcd
er
Output
abacabaabbabcder
Input
5
x
xx
xxa
xxaa
xxaaa
Output
xxaaaxxaaxxaxxx
Input
3
c
cb
cba
Output
cbacbc
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Thu Apr 14 15:41:54 2016
@author: kebl4230
Too slow. Possibly lots of unnecessary list manipulation.
"""
from functools import cmp_to_key
n = int(input())
strings = list()
for i in range(n):
strings.append(input())
def cmpfunc(x, y):
a = x + y
b = y + x
if a < b:
return -1
elif a == b:
return 0
else:
return 1
bb = sorted(strings, key=cmp_to_key(cmpfunc))
print("".join(bb))
"""
n = int(input())
strings = list()
for i in range(n):
strings.append(input())
aa = "abcdefghijklmnopqrstuvwxyz"
positions = [strings.copy()]
def myfunc(mylist, index, pos):
scores = [aa.find(bb[index]) if index < len(bb) else 27 for bb in mylist]
r = 0
mins = min(scores)
while mins <= 27:
if r == 0:
for i in range(len(scores)):
if scores[i] == mins:
scores[i] = 100
else:
group = [mylist[n] for n in range(len(scores)) if scores[n] == mins]
positions.insert(pos + r, group)
for string in group:
mylist.remove(string)
while any(s == mins for s in scores):
scores.remove(mins)
r += 1
mins = min(scores)
index = 0
maxlen = len(positions)
pos = 0
while any(len(pos) > 1 for pos in positions):
myfunc(positions[pos], index, pos)
pos += 1
if pos == maxlen:
pos = 0
maxlen = len(positions)
index += 1
result = ''
for aa in positions:
result += aa[0]
print(result)
"""
```
|
instruction
| 0
| 6,247
| 0
| 12,494
|
Yes
|
output
| 1
| 6,247
| 0
| 12,495
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given a list of n strings a1, a2, ..., an. You'd like to concatenate them together in some order such that the resulting string would be lexicographically smallest.
Given the list of strings, output the lexicographically smallest concatenation.
Input
The first line contains integer n β the number of strings (1 β€ n β€ 5Β·104).
Each of the next n lines contains one string ai (1 β€ |ai| β€ 50) consisting of only lowercase English letters. The sum of string lengths will not exceed 5Β·104.
Output
Print the only string a β the lexicographically smallest string concatenation.
Examples
Input
4
abba
abacaba
bcd
er
Output
abacabaabbabcder
Input
5
x
xx
xxa
xxaa
xxaaa
Output
xxaaaxxaaxxaxxx
Input
3
c
cb
cba
Output
cbacbc
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
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 Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 10**9+7
from functools import cmp_to_key
Ri = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
n = int(ri())
s = []
for i in range(n):
s.append(ri())
s.sort(key = cmp_to_key(lambda x,y : -1 if x+y < y+x else 1 ))
print("".join(s))
```
|
instruction
| 0
| 6,248
| 0
| 12,496
|
Yes
|
output
| 1
| 6,248
| 0
| 12,497
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given a list of n strings a1, a2, ..., an. You'd like to concatenate them together in some order such that the resulting string would be lexicographically smallest.
Given the list of strings, output the lexicographically smallest concatenation.
Input
The first line contains integer n β the number of strings (1 β€ n β€ 5Β·104).
Each of the next n lines contains one string ai (1 β€ |ai| β€ 50) consisting of only lowercase English letters. The sum of string lengths will not exceed 5Β·104.
Output
Print the only string a β the lexicographically smallest string concatenation.
Examples
Input
4
abba
abacaba
bcd
er
Output
abacabaabbabcder
Input
5
x
xx
xxa
xxaa
xxaaa
Output
xxaaaxxaaxxaxxx
Input
3
c
cb
cba
Output
cbacbc
Submitted Solution:
```
n = int(input())
ans = []
for i in range(n):
a = input()
ans.append(a)
ans.sort(reverse = True)
ans = ''.join(ans)
print(ans)
```
|
instruction
| 0
| 6,249
| 0
| 12,498
|
No
|
output
| 1
| 6,249
| 0
| 12,499
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given a list of n strings a1, a2, ..., an. You'd like to concatenate them together in some order such that the resulting string would be lexicographically smallest.
Given the list of strings, output the lexicographically smallest concatenation.
Input
The first line contains integer n β the number of strings (1 β€ n β€ 5Β·104).
Each of the next n lines contains one string ai (1 β€ |ai| β€ 50) consisting of only lowercase English letters. The sum of string lengths will not exceed 5Β·104.
Output
Print the only string a β the lexicographically smallest string concatenation.
Examples
Input
4
abba
abacaba
bcd
er
Output
abacabaabbabcder
Input
5
x
xx
xxa
xxaa
xxaaa
Output
xxaaaxxaaxxaxxx
Input
3
c
cb
cba
Output
cbacbc
Submitted Solution:
```
n=int(input())
m=list()
for i in range(n):
m.append(str(input()))
for i in range(n-1):
if m[i]+m[i+1]>m[i+1]+m[i]:
m[i+1]=m[i+1]+m[i]
else:
m[i+1]=m[i]+m[i+1]
print(m[i+1])
```
|
instruction
| 0
| 6,250
| 0
| 12,500
|
No
|
output
| 1
| 6,250
| 0
| 12,501
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given a list of n strings a1, a2, ..., an. You'd like to concatenate them together in some order such that the resulting string would be lexicographically smallest.
Given the list of strings, output the lexicographically smallest concatenation.
Input
The first line contains integer n β the number of strings (1 β€ n β€ 5Β·104).
Each of the next n lines contains one string ai (1 β€ |ai| β€ 50) consisting of only lowercase English letters. The sum of string lengths will not exceed 5Β·104.
Output
Print the only string a β the lexicographically smallest string concatenation.
Examples
Input
4
abba
abacaba
bcd
er
Output
abacabaabbabcder
Input
5
x
xx
xxa
xxaa
xxaaa
Output
xxaaaxxaaxxaxxx
Input
3
c
cb
cba
Output
cbacbc
Submitted Solution:
```
from functools import cmp_to_key
def c(x,y):
if x+y < y+x:
return 1
else:
return -1
n = int(input())
l = []
for _ in range(n):
l.append(input())
l.sort(key = cmp_to_key(c))
print(''.join(l))
```
|
instruction
| 0
| 6,251
| 0
| 12,502
|
No
|
output
| 1
| 6,251
| 0
| 12,503
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given a list of n strings a1, a2, ..., an. You'd like to concatenate them together in some order such that the resulting string would be lexicographically smallest.
Given the list of strings, output the lexicographically smallest concatenation.
Input
The first line contains integer n β the number of strings (1 β€ n β€ 5Β·104).
Each of the next n lines contains one string ai (1 β€ |ai| β€ 50) consisting of only lowercase English letters. The sum of string lengths will not exceed 5Β·104.
Output
Print the only string a β the lexicographically smallest string concatenation.
Examples
Input
4
abba
abacaba
bcd
er
Output
abacabaabbabcder
Input
5
x
xx
xxa
xxaa
xxaaa
Output
xxaaaxxaaxxaxxx
Input
3
c
cb
cba
Output
cbacbc
Submitted Solution:
```
from functools import cmp_to_key
def f(x, y):
return -1 if x + y <= y + x else 1
n = int(input())
A = []
for i in range(n):
s = input()
A.append(s)
print(A)
print(''.join(sorted(A, key = cmp_to_key(f))))
```
|
instruction
| 0
| 6,252
| 0
| 12,504
|
No
|
output
| 1
| 6,252
| 0
| 12,505
|
Provide a correct Python 3 solution for this coding contest problem.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
|
instruction
| 0
| 6,381
| 0
| 12,762
|
"Correct Solution:
```
S = input()
N = len(S)
A = (len(S) - 1) // 2
B = (len(S) + 3) // 2
if S[0:A] == S[B - 1:N]:
print('Yes')
else:
print('No')
```
|
output
| 1
| 6,381
| 0
| 12,763
|
Provide a correct Python 3 solution for this coding contest problem.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
|
instruction
| 0
| 6,382
| 0
| 12,764
|
"Correct Solution:
```
s = input()
n = len(s)
if s[:int(n//2)] == s[int(n//2)+1:]:
print("Yes")
else:
print("No")
```
|
output
| 1
| 6,382
| 0
| 12,765
|
Provide a correct Python 3 solution for this coding contest problem.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
|
instruction
| 0
| 6,383
| 0
| 12,766
|
"Correct Solution:
```
S=list(input())
N=len(S)
if S==S[::-1] and S[:(N-1)//2]==S[(N+3)//2-1:]:
print("Yes")
else:print("No")
```
|
output
| 1
| 6,383
| 0
| 12,767
|
Provide a correct Python 3 solution for this coding contest problem.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
|
instruction
| 0
| 6,384
| 0
| 12,768
|
"Correct Solution:
```
s = input()
if s[len(s) // 2 + 1:] == s[:len(s) // 2]:
print('Yes')
else:
print('No')
```
|
output
| 1
| 6,384
| 0
| 12,769
|
Provide a correct Python 3 solution for this coding contest problem.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
|
instruction
| 0
| 6,385
| 0
| 12,770
|
"Correct Solution:
```
s = input()
print("Yes" if s == s[::-1] and s[:len(s)//2] == s[len(s)//2-1::-1] else "No")
```
|
output
| 1
| 6,385
| 0
| 12,771
|
Provide a correct Python 3 solution for this coding contest problem.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
|
instruction
| 0
| 6,386
| 0
| 12,772
|
"Correct Solution:
```
S = str(input())
N = len(S)
if S[:(N-1)//2] == S[(N+3)//2 - 1:] :
print('Yes')
else:
print('No')
```
|
output
| 1
| 6,386
| 0
| 12,773
|
Provide a correct Python 3 solution for this coding contest problem.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
|
instruction
| 0
| 6,387
| 0
| 12,774
|
"Correct Solution:
```
S = input()
N = len(S)
N1s = (N-1)//2
N2s = (N+3)//2-1
if S[:N1s] == S[N2s:]:
print("Yes")
else:
print("No")
```
|
output
| 1
| 6,387
| 0
| 12,775
|
Provide a correct Python 3 solution for this coding contest problem.
A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:
* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.
Determine whether S is a strong palindrome.
Constraints
* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
If S is a strong palindrome, print `Yes`; otherwise, print `No`.
Examples
Input
akasaka
Output
Yes
Input
level
Output
No
Input
atcoder
Output
No
|
instruction
| 0
| 6,388
| 0
| 12,776
|
"Correct Solution:
```
s=input();h=len(s)//2;print(' YNeos'[s[:h]==s[h-1::-1]==s[-h:]::2])
```
|
output
| 1
| 6,388
| 0
| 12,777
|
Provide a correct Python 3 solution for this coding contest problem.
For a given sequence A = {a0, a1, ... , an-1}, find the length of the longest increasing subsequnece (LIS) in A.
An increasing subsequence of A is defined by a subsequence {ai0, ai1, ... , aik} where 0 β€ i0 < i1 < ... < ik < n and ai0 < ai1 < ... < aik.
Constraints
* 1 β€ n β€ 100000
* 0 β€ ai β€ 109
Input
n
a0
a1
:
an-1
In the first line, an integer n is given. In the next n lines, elements of A are given.
Output
The length of the longest increasing subsequence of A.
Examples
Input
5
5
1
3
2
4
Output
3
Input
3
1
1
1
Output
1
|
instruction
| 0
| 7,508
| 0
| 15,016
|
"Correct Solution:
```
#!/usr/bin/env python3
# n,m = map(int,sys.stdin.readline().split())
# a = list(map(int,sys.stdin.readline().split()))
# a = [sys.stdin.readline() for _ in range(n)]
# s = sys.stdin.readline().rstrip()
# n = int(sys.stdin.readline())
INF = float("inf")
import sys,bisect
sys.setrecursionlimit(15000)
n = int(sys.stdin.readline())
a = [int(sys.stdin.readline()) for _ in range(n)]
# naive O(n**2)
# dp = [0]*n
# for i in range(n):
# m = 0
# for j in range(0,i):
# if a[j] < a[i] and m <= dp[j]:
# m = dp[j]
# dp[i] = m+1
# print(max(dp))
import bisect
dp = [float("inf")]*n
dp[0] = a[0]
l = 0
for i in range(1,n):
if dp[l] < a[i]:
l += 1
dp[l] = a[i]
else:
j = bisect.bisect_left(dp,a[i])
dp[j] = a[i]
print(bisect.bisect_left(dp,float("inf")))
```
|
output
| 1
| 7,508
| 0
| 15,017
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Nezzar has a binary string s of length n that he wants to share with his best friend, Nanako. Nanako will spend q days inspecting the binary string. At the same time, Nezzar wants to change the string s into string f during these q days, because it looks better.
It is known that Nanako loves consistency so much. On the i-th day, Nanako will inspect a segment of string s from position l_i to position r_i inclusive. If the segment contains both characters '0' and '1', Nanako becomes unhappy and throws away the string.
After this inspection, at the i-th night, Nezzar can secretly change strictly less than half of the characters in the segment from l_i to r_i inclusive, otherwise the change will be too obvious.
Now Nezzar wonders, if it is possible to avoid Nanako being unhappy and at the same time have the string become equal to the string f at the end of these q days and nights.
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β the number of test cases.
The first line of each test case contains two integers n,q (1 β€ n β€ 2 β
10^5, 0 β€ q β€ 2 β
10^5).
The second line of each test case contains a binary string s of length n.
The third line of each test case contains a binary string f of length n.
Then q lines follow, i-th of them contains two integers l_i,r_i (1 β€ l_i β€ r_i β€ n) β bounds of the segment, that Nanako will inspect on the i-th day.
It is guaranteed that the sum of n for all test cases doesn't exceed 2 β
10^5, and the sum of q for all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print "YES" on the single line if it is possible to avoid Nanako being unhappy and have the string f at the end of q days and nights. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
4
5 2
00000
00111
1 5
1 3
2 1
00
01
1 2
10 6
1111111111
0110001110
1 10
5 9
7 10
1 7
3 5
6 10
5 2
10000
11000
2 5
1 3
Output
YES
NO
YES
NO
Note
In the first test case, \underline{00000} β \underline{000}11 β 00111 is one of the possible sequences of string changes.
In the second test case, it can be shown that it is impossible to have the string f at the end.
|
instruction
| 0
| 7,783
| 0
| 15,566
|
Tags: data structures, greedy
Correct Solution:
```
import sys
readline = sys.stdin.readline
"""
import io,os
readline = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
"""
from operator import add
class Lazysegtree:
def __init__(self, A, fx, ex, fm, em, fa, initialize = True):
self.N = len(A)
self.N0 = 2**(self.N-1).bit_length()
self.dep = (self.N-1).bit_length()
self.fx = fx
self.ex = ex
self.em = em
self.lazy = [em]*(2*self.N0)
if initialize:
self.data = [self.ex]*self.N0 + A + [self.ex]*(self.N0 - self.N)
for i in range(self.N0-1, -1, -1):
self.data[i] = self.fx(self.data[2*i], self.data[2*i+1])
else:
self.data = [self.ex]*(2*self.N0)
def fa(self, ope, idx):
if ope is not None:
return ope*(1<<(self.dep - idx.bit_length() + 1))
return self.data[idx]
def fm(self, ope1, ope2):
return ope1
def __repr__(self):
s = 'data'
l = 'lazy'
cnt = 1
for i in range(self.N0.bit_length()):
s = '\n'.join((s, ' '.join(map(str, self.data[cnt:cnt+(1<<i)]))))
l = '\n'.join((l, ' '.join(map(str, self.lazy[cnt:cnt+(1<<i)]))))
cnt += 1<<i
return '\n'.join((s, l))
def _ascend(self, k):
k = k >> 1
c = k.bit_length()
for j in range(c):
idx = k >> j
self.data[idx] = self.fx(self.data[2*idx], self.data[2*idx+1])
def _descend(self, k):
k = k >> 1
idx = 1
c = k.bit_length()
for j in range(1, c+1):
idx = k >> (c - j)
if self.lazy[idx] == self.em:
continue
self.data[2*idx] = self.fa(self.lazy[idx], 2*idx)
self.data[2*idx+1] = self.fa(self.lazy[idx], 2*idx+1)
self.lazy[2*idx] = self.fm(self.lazy[idx], self.lazy[2*idx])
self.lazy[2*idx+1] = self.fm(self.lazy[idx], self.lazy[2*idx+1])
self.lazy[idx] = self.em
def query(self, l, r):
L = l+self.N0
R = r+self.N0
self._descend(L//(L & -L))
self._descend(R//(R & -R)-1)
sl = self.ex
sr = self.ex
while L < R:
if R & 1:
R -= 1
sr = self.fx(self.data[R], sr)
if L & 1:
sl = self.fx(sl, self.data[L])
L += 1
L >>= 1
R >>= 1
return self.fx(sl, sr)
def operate(self, l, r, x):
L = l+self.N0
R = r+self.N0
Li = L//(L & -L)
Ri = R//(R & -R)
self._descend(Li)
self._descend(Ri-1)
while L < R :
if R & 1:
R -= 1
self.data[R] = self.fa(x, R)
self.lazy[R] = self.fm(x, self.lazy[R])
if L & 1:
self.data[L] = self.fa(x, L)
self.lazy[L] = self.fm(x, self.lazy[L])
L += 1
L >>= 1
R >>= 1
self._ascend(Li)
self._ascend(Ri-1)
def propall(self):
for idx in range(1, self.N0):
self.data[2*idx] = self.fa(self.lazy[idx], 2*idx)
self.data[2*idx+1] = self.fa(self.lazy[idx], 2*idx+1)
self.lazy[2*idx] = self.fm(self.lazy[idx], self.lazy[2*idx])
self.lazy[2*idx+1] = self.fm(self.lazy[idx], self.lazy[2*idx+1])
T = int(readline())
Ans = ['NO']*T
for qu in range(T):
N, Q = map(int, readline().split())
S = list(map(int, readline().strip()))
T = list(map(int, readline().strip()))
"""
S = list(map(lambda x: int(x)-48, readline().strip()))
T = list(map(lambda x: int(x)-48, readline().strip()))
"""
Qs = [tuple(map(int, readline().split())) for _ in range(Q)]
K = Lazysegtree(T, add, 0, None, None, None)
for l, r in Qs[::-1]:
l -= 1
leng = r-l
x = K.query(l, r)
if 2*x > leng:
K.operate(l, r, 1)
elif 2*x == leng:
break
else:
K.operate(l, r, 0)
else:
if [K.query(i, i+1) for i in range(N)] == S:
Ans[qu] = 'YES'
print('\n'.join(Ans))
```
|
output
| 1
| 7,783
| 0
| 15,567
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Nezzar has a binary string s of length n that he wants to share with his best friend, Nanako. Nanako will spend q days inspecting the binary string. At the same time, Nezzar wants to change the string s into string f during these q days, because it looks better.
It is known that Nanako loves consistency so much. On the i-th day, Nanako will inspect a segment of string s from position l_i to position r_i inclusive. If the segment contains both characters '0' and '1', Nanako becomes unhappy and throws away the string.
After this inspection, at the i-th night, Nezzar can secretly change strictly less than half of the characters in the segment from l_i to r_i inclusive, otherwise the change will be too obvious.
Now Nezzar wonders, if it is possible to avoid Nanako being unhappy and at the same time have the string become equal to the string f at the end of these q days and nights.
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β the number of test cases.
The first line of each test case contains two integers n,q (1 β€ n β€ 2 β
10^5, 0 β€ q β€ 2 β
10^5).
The second line of each test case contains a binary string s of length n.
The third line of each test case contains a binary string f of length n.
Then q lines follow, i-th of them contains two integers l_i,r_i (1 β€ l_i β€ r_i β€ n) β bounds of the segment, that Nanako will inspect on the i-th day.
It is guaranteed that the sum of n for all test cases doesn't exceed 2 β
10^5, and the sum of q for all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print "YES" on the single line if it is possible to avoid Nanako being unhappy and have the string f at the end of q days and nights. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
4
5 2
00000
00111
1 5
1 3
2 1
00
01
1 2
10 6
1111111111
0110001110
1 10
5 9
7 10
1 7
3 5
6 10
5 2
10000
11000
2 5
1 3
Output
YES
NO
YES
NO
Note
In the first test case, \underline{00000} β \underline{000}11 β 00111 is one of the possible sequences of string changes.
In the second test case, it can be shown that it is impossible to have the string f at the end.
|
instruction
| 0
| 7,784
| 0
| 15,568
|
Tags: data structures, greedy
Correct Solution:
```
import io
import os
from math import floor
# https://github.com/atcoder/ac-library/blob/master/atcoder/lazysegtree.hpp
class LazySegTree:
def __init__(self, _op, _e, _mapping, _composition, _id, v):
def set(p, x):
assert 0 <= p < _n
p += _size
for i in range(_log, 0, -1):
_push(p >> i)
_d[p] = x
for i in range(1, _log + 1):
_update(p >> i)
def get(p):
assert 0 <= p < _n
p += _size
for i in range(_log, 0, -1):
_push(p >> i)
return _d[p]
def prod(l, r):
assert 0 <= l <= r <= _n
if l == r:
return _e
l += _size
r += _size
for i in range(_log, 0, -1):
if ((l >> i) << i) != l:
_push(l >> i)
if ((r >> i) << i) != r:
_push(r >> i)
sml = _e
smr = _e
while l < r:
if l & 1:
sml = _op(sml, _d[l])
l += 1
if r & 1:
r -= 1
smr = _op(_d[r], smr)
l >>= 1
r >>= 1
return _op(sml, smr)
def apply(l, r, f):
assert 0 <= l <= r <= _n
if l == r:
return
l += _size
r += _size
for i in range(_log, 0, -1):
if ((l >> i) << i) != l:
_push(l >> i)
if ((r >> i) << i) != r:
_push((r - 1) >> i)
l2 = l
r2 = r
while l < r:
if l & 1:
_all_apply(l, f)
l += 1
if r & 1:
r -= 1
_all_apply(r, f)
l >>= 1
r >>= 1
l = l2
r = r2
for i in range(1, _log + 1):
if ((l >> i) << i) != l:
_update(l >> i)
if ((r >> i) << i) != r:
_update((r - 1) >> i)
def _update(k):
_d[k] = _op(_d[2 * k], _d[2 * k + 1])
def _all_apply(k, f):
_d[k] = _mapping(f, _d[k])
if k < _size:
_lz[k] = _composition(f, _lz[k])
def _push(k):
_all_apply(2 * k, _lz[k])
_all_apply(2 * k + 1, _lz[k])
_lz[k] = _id
_n = len(v)
_log = _n.bit_length()
_size = 1 << _log
_d = [_e] * (2 * _size)
_lz = [_id] * _size
for i in range(_n):
_d[_size + i] = v[i]
for i in range(_size - 1, 0, -1):
_update(i)
self.set = set
self.get = get
self.prod = prod
self.apply = apply
MIL = 1 << 20
def makeNode(total, count):
# Pack a pair into a float
return (total * MIL) + count
def getTotal(node):
return floor(node / MIL)
def getCount(node):
return node - getTotal(node) * MIL
nodeIdentity = makeNode(0.0, 0.0)
def nodeOp(node1, node2):
return node1 + node2
# Equivalent to the following:
return makeNode(
getTotal(node1) + getTotal(node2), getCount(node1) + getCount(node2)
)
identityMapping = -1
def mapping(tag, node):
if tag == identityMapping:
return node
# If assigned, new total is the number assigned times count
count = getCount(node)
return makeNode(tag * count, count)
def composition(mapping1, mapping2):
# If assigned multiple times, take first non-identity assignment
return mapping1 if mapping1 != identityMapping else mapping2
def solve(N, Q, S, F, LR):
segTree = LazySegTree(
nodeOp,
nodeIdentity,
mapping,
composition,
identityMapping,
[makeNode(float(d), 1.0) for d in F],
)
for l, r in LR[::-1]:
l -= 1
countOnes = getTotal(segTree.prod(l, r))
countZeroes = r - l - countOnes
if countZeroes > countOnes:
# Majority were zeroes
segTree.apply(l, r, 0)
elif countZeroes < countOnes:
# Majority were ones
segTree.apply(l, r, 1)
else:
# No majority
return "NO"
for i in range(N):
if getTotal(segTree.get(i)) != S[i]:
return "NO"
return "YES"
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
TC = int(input())
for tc in range(1, TC + 1):
N, Q = [int(x) for x in input().split()]
S = [int(d) for d in input().decode().rstrip()]
F = [int(d) for d in input().decode().rstrip()]
LR = [[int(x) for x in input().split()] for i in range(Q)]
ans = solve(N, Q, S, F, LR)
print(ans)
```
|
output
| 1
| 7,784
| 0
| 15,569
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Nezzar has a binary string s of length n that he wants to share with his best friend, Nanako. Nanako will spend q days inspecting the binary string. At the same time, Nezzar wants to change the string s into string f during these q days, because it looks better.
It is known that Nanako loves consistency so much. On the i-th day, Nanako will inspect a segment of string s from position l_i to position r_i inclusive. If the segment contains both characters '0' and '1', Nanako becomes unhappy and throws away the string.
After this inspection, at the i-th night, Nezzar can secretly change strictly less than half of the characters in the segment from l_i to r_i inclusive, otherwise the change will be too obvious.
Now Nezzar wonders, if it is possible to avoid Nanako being unhappy and at the same time have the string become equal to the string f at the end of these q days and nights.
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β the number of test cases.
The first line of each test case contains two integers n,q (1 β€ n β€ 2 β
10^5, 0 β€ q β€ 2 β
10^5).
The second line of each test case contains a binary string s of length n.
The third line of each test case contains a binary string f of length n.
Then q lines follow, i-th of them contains two integers l_i,r_i (1 β€ l_i β€ r_i β€ n) β bounds of the segment, that Nanako will inspect on the i-th day.
It is guaranteed that the sum of n for all test cases doesn't exceed 2 β
10^5, and the sum of q for all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print "YES" on the single line if it is possible to avoid Nanako being unhappy and have the string f at the end of q days and nights. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
4
5 2
00000
00111
1 5
1 3
2 1
00
01
1 2
10 6
1111111111
0110001110
1 10
5 9
7 10
1 7
3 5
6 10
5 2
10000
11000
2 5
1 3
Output
YES
NO
YES
NO
Note
In the first test case, \underline{00000} β \underline{000}11 β 00111 is one of the possible sequences of string changes.
In the second test case, it can be shown that it is impossible to have the string f at the end.
|
instruction
| 0
| 7,785
| 0
| 15,570
|
Tags: data structures, greedy
Correct Solution:
```
import sys
input = sys.stdin.readline
def indexes(L,R):
INDLIST=[]
R-=1
L>>=1
R>>=1
while L!=R:
if L>R:
INDLIST.append(L)
L>>=1
else:
INDLIST.append(R)
R>>=1
while L!=0:
INDLIST.append(L)
L>>=1
return INDLIST
def updates(l,r,x):
L=l+seg_el
R=r+seg_el
L//=(L & (-L))
R//=(R & (-R))
UPIND=indexes(L,R)
for ind in UPIND[::-1]:
if LAZY[ind]!=None:
update_lazy = LAZY[ind] *(1<<(seg_height - 1 - (ind.bit_length())))
LAZY[ind<<1]=LAZY[1+(ind<<1)]=LAZY[ind]
SEG[ind<<1]=SEG[1+(ind<<1)]=update_lazy
LAZY[ind]=None
while L!=R:
if L > R:
SEG[L]=x * (1<<(seg_height - (L.bit_length())))
LAZY[L]=x
L+=1
L//=(L & (-L))
else:
R-=1
SEG[R]=x * (1<<(seg_height - (R.bit_length())))
LAZY[R]=x
R//=(R & (-R))
for ind in UPIND:
SEG[ind]=SEG[ind<<1]+SEG[1+(ind<<1)]
def getvalues(l,r):
L=l+seg_el
R=r+seg_el
L//=(L & (-L))
R//=(R & (-R))
UPIND=indexes(L,R)
for ind in UPIND[::-1]:
if LAZY[ind]!=None:
update_lazy = LAZY[ind] *(1<<(seg_height - 1 - (ind.bit_length())))
LAZY[ind<<1]=LAZY[1+(ind<<1)]=LAZY[ind]
SEG[ind<<1]=SEG[1+(ind<<1)]=update_lazy
LAZY[ind]=None
ANS=0
while L!=R:
if L > R:
ANS+=SEG[L]
L+=1
L//=(L & (-L))
else:
R-=1
ANS+=SEG[R]
R//=(R & (-R))
return ANS
t=int(input())
for tests in range(t):
n,q=map(int,input().split())
S=input().strip()
F=input().strip()
Q=[tuple(map(int,input().split())) for i in range(q)]
seg_el=1<<(n.bit_length())
seg_height=1+n.bit_length()
SEG=[0]*(2*seg_el)
LAZY=[None]*(2*seg_el)
for i in range(n):
SEG[i+seg_el]=int(F[i])
for i in range(seg_el-1,0,-1):
SEG[i]=SEG[i*2]+SEG[i*2+1]
for l,r in Q[::-1]:
SUM=r-l+1
xx=getvalues(l-1,r)
if xx*2==SUM:
print("NO")
break
else:
if xx*2>SUM:
updates(l-1,r,1)
else:
updates(l-1,r,0)
else:
for i in range(n):
if getvalues(i,i+1)==int(S[i]):
True
else:
print("NO")
break
else:
print("YES")
```
|
output
| 1
| 7,785
| 0
| 15,571
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Nezzar has a binary string s of length n that he wants to share with his best friend, Nanako. Nanako will spend q days inspecting the binary string. At the same time, Nezzar wants to change the string s into string f during these q days, because it looks better.
It is known that Nanako loves consistency so much. On the i-th day, Nanako will inspect a segment of string s from position l_i to position r_i inclusive. If the segment contains both characters '0' and '1', Nanako becomes unhappy and throws away the string.
After this inspection, at the i-th night, Nezzar can secretly change strictly less than half of the characters in the segment from l_i to r_i inclusive, otherwise the change will be too obvious.
Now Nezzar wonders, if it is possible to avoid Nanako being unhappy and at the same time have the string become equal to the string f at the end of these q days and nights.
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β the number of test cases.
The first line of each test case contains two integers n,q (1 β€ n β€ 2 β
10^5, 0 β€ q β€ 2 β
10^5).
The second line of each test case contains a binary string s of length n.
The third line of each test case contains a binary string f of length n.
Then q lines follow, i-th of them contains two integers l_i,r_i (1 β€ l_i β€ r_i β€ n) β bounds of the segment, that Nanako will inspect on the i-th day.
It is guaranteed that the sum of n for all test cases doesn't exceed 2 β
10^5, and the sum of q for all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print "YES" on the single line if it is possible to avoid Nanako being unhappy and have the string f at the end of q days and nights. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
4
5 2
00000
00111
1 5
1 3
2 1
00
01
1 2
10 6
1111111111
0110001110
1 10
5 9
7 10
1 7
3 5
6 10
5 2
10000
11000
2 5
1 3
Output
YES
NO
YES
NO
Note
In the first test case, \underline{00000} β \underline{000}11 β 00111 is one of the possible sequences of string changes.
In the second test case, it can be shown that it is impossible to have the string f at the end.
|
instruction
| 0
| 7,786
| 0
| 15,572
|
Tags: data structures, greedy
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
def __init__(self, file):
self.newlines = 0
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline()
# --------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().split())
def RLL(): return list(map(int, sys.stdin.readline().split()))
def N(): return int(input())
def S(): return input().strip()
def print_list(l): print(' '.join(map(str, l)))
# sys.setrecursionlimit(200000)
# import random
# from functools import reduce
# from functools import lru_cache
# from heapq import *
# from collections import deque as dq
# import math
# import bisect as bs
# from collections import Counter
# from collections import defaultdict as dc
def set_range(lx, rx, v, now):
if l <= lx and r >= rx:
label[now] = v
node[now] = rx - lx if v else 0
return
mx = (lx + rx) >> 1
nl = now << 1
nr = nl + 1
if label[now] != -1: # ιε―Ήη§°ζδ½ε―η¨θ―₯η―θ
node[nl] = node[nr] = node[now] >> 1
label[nl] = label[nr] = label[now]
label[now] = -1
if l < mx: set_range(lx, mx, v, nl)
if r > mx: set_range(mx, rx, v, nr)
node[now] = node[nl] + node[nr]
def get_range(lx, rx, now):
if l <= lx and r >= rx: return node[now]
mx = (lx + rx) >> 1
nl = now << 1
nr = nl + 1
if label[now] != -1: return min(rx, r) - max(lx, l) if label[now] else 0
m1 = get_range(lx, mx, nl) if l < mx else 0
m2 = get_range(mx, rx, nr) if r > mx else 0
return m1 + m2
def match(lx, rx, now):
if lx >= n: return False
if now >= sn: return node[now] != s[lx]
if label[now] != -1: return any(v != label[now] for v in s[lx:rx])
mx = (rx + lx) >> 1
return match(lx, mx, now << 1) or match(mx, rx, now << 1 | 1)
ans = []
for _ in range(N()):
n, q = RL()
s, t = [int(c) for c in S()], [int(c) for c in S()]
p = [tuple(RL()) for _ in range(q)]
sn = 1 << n.bit_length()
node = [0] * sn + t + [0] * (sn - n)
label = [-1] * (2 * sn)
for i in range(sn - 1, 0, -1): node[i] = node[i << 1] + node[i << 1 | 1]
f = True
for l, r in p[::-1]:
l -= 1
count = get_range(0, sn, 1) * 2
if count == r - l:
f = False
break
set_range(0, sn, 1 if count > r - l else 0, 1)
ans.append('YES' if f and not match(0, sn, 1) else 'NO')
print('\n'.join(ans))
```
|
output
| 1
| 7,786
| 0
| 15,573
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Nezzar has a binary string s of length n that he wants to share with his best friend, Nanako. Nanako will spend q days inspecting the binary string. At the same time, Nezzar wants to change the string s into string f during these q days, because it looks better.
It is known that Nanako loves consistency so much. On the i-th day, Nanako will inspect a segment of string s from position l_i to position r_i inclusive. If the segment contains both characters '0' and '1', Nanako becomes unhappy and throws away the string.
After this inspection, at the i-th night, Nezzar can secretly change strictly less than half of the characters in the segment from l_i to r_i inclusive, otherwise the change will be too obvious.
Now Nezzar wonders, if it is possible to avoid Nanako being unhappy and at the same time have the string become equal to the string f at the end of these q days and nights.
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β the number of test cases.
The first line of each test case contains two integers n,q (1 β€ n β€ 2 β
10^5, 0 β€ q β€ 2 β
10^5).
The second line of each test case contains a binary string s of length n.
The third line of each test case contains a binary string f of length n.
Then q lines follow, i-th of them contains two integers l_i,r_i (1 β€ l_i β€ r_i β€ n) β bounds of the segment, that Nanako will inspect on the i-th day.
It is guaranteed that the sum of n for all test cases doesn't exceed 2 β
10^5, and the sum of q for all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print "YES" on the single line if it is possible to avoid Nanako being unhappy and have the string f at the end of q days and nights. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
4
5 2
00000
00111
1 5
1 3
2 1
00
01
1 2
10 6
1111111111
0110001110
1 10
5 9
7 10
1 7
3 5
6 10
5 2
10000
11000
2 5
1 3
Output
YES
NO
YES
NO
Note
In the first test case, \underline{00000} β \underline{000}11 β 00111 is one of the possible sequences of string changes.
In the second test case, it can be shown that it is impossible to have the string f at the end.
|
instruction
| 0
| 7,787
| 0
| 15,574
|
Tags: data structures, greedy
Correct Solution:
```
import sys, io, os
if os.environ['USERNAME']=='kissz':
inp1=inp2=open('in.txt','r').readline
def debug(*args):
print(*args,file=sys.stderr)
else:
#inp1=input
inp1=inp2=sys.stdin.readline
#inp2=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def debug(*args):
pass
# SCRIPT STARTS HERE
class lazytree:
def __init__(self,array):
n=len(array)-1
i=1
while n:
n>>=1
i+=1
self.tree=[0]*(2**(i-1)-1)+array+[0]*(2**(i-1)-len(array))
self.l=0
self.r=2**(i-1)-1
self.lazy=[0]*(2**i-1)
self.lazyval=[0]*(2**i-1)
for i in range(2**(i-1)-2,-1,-1):
self.tree[i]=self.tree[2*i+1]+self.tree[2*i+2]
def query(self,l,r):
if l==self.l and r==self.r:
parents=[]
children=[(0,l,r)]
else:
parents=[(0,self.l,self.r)]
children=[]
i=0
while i<len(parents):
loc,L,R=parents[i]
left=(loc*2+1,L,L+(R-L+1)//2-1)
right=(loc*2+2,L+(R-L+1)//2,R)
if self.lazy[loc]:
self.lazy[loc]=0
self.tree[left[0]]=self.tree[right[0]]=(left[2]-left[1]+1)*self.lazyval[loc]
self.lazy[left[0]]=self.lazy[right[0]]=1
self.lazyval[left[0]]=self.lazyval[right[0]]=self.lazyval[loc]
if l<=left[1] and left[2]<=r:
children.append(left)
elif left[1]<=l<=left[2] or left[1]<=r<=left[2]:
parents.append(left)
if l<=right[1] and right[2]<=r:
children.append(right)
elif right[1]<=l<=right[2] or right[1]<=r<=right[2]:
parents.append(right)
i+=1
ones=sum(self.tree[child[0]] for child in children)
zeros=r-l+1-ones
if ones==zeros:
return -1
elif ones>zeros:
for child in children:
self.tree[child[0]]=child[2]-child[1]+1
self.lazy[child[0]]=1
self.lazyval[child[0]]=1
else:
for child in children:
self.tree[child[0]]=0
self.lazy[child[0]]=1
self.lazyval[child[0]]=0
for i,_,_ in reversed(parents):
self.tree[i]=self.tree[2*i+1]+self.tree[2*i+2]
return self.tree[child[0]]
def get_all(self):
n=self.r-self.l
for i in range(n):
if self.lazy[i]:
self.lazy[i]=0
self.tree[2*i+1]=self.tree[2*i+2]=self.lazyval[i] # destroying tree
self.lazy[2*i+1]=self.lazy[2*i+2]=1
self.lazyval[2*i+1]=self.lazyval[2*i+2]=self.lazyval[i]
return self.tree[n:]
for _ in range(int(inp2())):
n,q=map(int,inp2().split())
s=[int(c) for c in inp1().strip()]
f=[int(c) for c in inp1().strip()]
moves=[]
ans=True
for _ in range(q):
l,r=map(int,inp2().split())
moves.append((l-1,r-1))
tr=lazytree(f)
for l,r in reversed(moves):
success=tr.query(l,r)
if success==-1:
break
else:
f=tr.get_all()[:n]
if f!=s: success=-1
if success>=0:
print('YES')
else:
print('NO')
```
|
output
| 1
| 7,787
| 0
| 15,575
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Nezzar has a binary string s of length n that he wants to share with his best friend, Nanako. Nanako will spend q days inspecting the binary string. At the same time, Nezzar wants to change the string s into string f during these q days, because it looks better.
It is known that Nanako loves consistency so much. On the i-th day, Nanako will inspect a segment of string s from position l_i to position r_i inclusive. If the segment contains both characters '0' and '1', Nanako becomes unhappy and throws away the string.
After this inspection, at the i-th night, Nezzar can secretly change strictly less than half of the characters in the segment from l_i to r_i inclusive, otherwise the change will be too obvious.
Now Nezzar wonders, if it is possible to avoid Nanako being unhappy and at the same time have the string become equal to the string f at the end of these q days and nights.
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β the number of test cases.
The first line of each test case contains two integers n,q (1 β€ n β€ 2 β
10^5, 0 β€ q β€ 2 β
10^5).
The second line of each test case contains a binary string s of length n.
The third line of each test case contains a binary string f of length n.
Then q lines follow, i-th of them contains two integers l_i,r_i (1 β€ l_i β€ r_i β€ n) β bounds of the segment, that Nanako will inspect on the i-th day.
It is guaranteed that the sum of n for all test cases doesn't exceed 2 β
10^5, and the sum of q for all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print "YES" on the single line if it is possible to avoid Nanako being unhappy and have the string f at the end of q days and nights. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
4
5 2
00000
00111
1 5
1 3
2 1
00
01
1 2
10 6
1111111111
0110001110
1 10
5 9
7 10
1 7
3 5
6 10
5 2
10000
11000
2 5
1 3
Output
YES
NO
YES
NO
Note
In the first test case, \underline{00000} β \underline{000}11 β 00111 is one of the possible sequences of string changes.
In the second test case, it can be shown that it is impossible to have the string f at the end.
|
instruction
| 0
| 7,788
| 0
| 15,576
|
Tags: data structures, greedy
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
def __init__(self, file):
self.newlines = 0
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = sys.stdin.readline
# --------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().split())
def RLL(): return list(map(int, sys.stdin.readline().split()))
def N(): return int(input())
def S(): return input().strip()
def print_list(l): print(' '.join(map(str, l)))
# sys.setrecursionlimit(200000)
# import random
# from functools import reduce
# from functools import lru_cache
# from heapq import *
# from collections import deque as dq
# import math
# import bisect as bs
# from collections import Counter
# from collections import defaultdict as dc
def set_range(lx, rx, v, now):
if l <= lx and r >= rx:
label[now] = v
node[now] = rx - lx if v else 0
return
mx = (lx + rx) >> 1
nl = now << 1
nr = nl + 1
if label[now] != -1: # ιε―Ήη§°ζδ½ε―η¨θ―₯η―θ
node[nl] = node[nr] = node[now] >> 1
label[nl] = label[nr] = label[now]
label[now] = -1
if l < mx: set_range(lx, mx, v, nl)
if r > mx: set_range(mx, rx, v, nr)
node[now] = node[nl] + node[nr]
def get_range(lx, rx, now):
if l <= lx and r >= rx: return node[now]
mx = (lx + rx) >> 1
nl = now << 1
nr = nl + 1
if label[now] != -1: return min(rx, r) - max(lx, l) if label[now] else 0
m1 = get_range(lx, mx, nl) if l < mx else 0
m2 = get_range(mx, rx, nr) if r > mx else 0
return m1 + m2
def match(lx, rx, now):
if lx >= n: return False
if now >= sn: return node[now] != s[lx]
if label[now] != -1: return any(v != label[now] for v in s[lx:rx])
mx = (rx + lx) >> 1
return match(lx, mx, now << 1) or match(mx, rx, now << 1 | 1)
for _ in range(N()):
n, q = RL()
s, t = [int(c) for c in S()], [int(c) for c in S()]
p = [tuple(RL()) for _ in range(q)]
sn = 1 << n.bit_length()
node = [0] * sn + t + [0] * (sn - n)
label = [-1] * (2 * sn)
for i in range(sn - 1, 0, -1): node[i] = node[i << 1] + node[i << 1 | 1]
f = True
for l, r in p[::-1]:
l -= 1
count = get_range(0, sn, 1) * 2
if count == r - l:
f = False
break
set_range(0, sn, 1 if count > r - l else 0, 1)
print('YES' if f and not match(0, sn, 1) else 'NO')
```
|
output
| 1
| 7,788
| 0
| 15,577
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Nezzar has a binary string s of length n that he wants to share with his best friend, Nanako. Nanako will spend q days inspecting the binary string. At the same time, Nezzar wants to change the string s into string f during these q days, because it looks better.
It is known that Nanako loves consistency so much. On the i-th day, Nanako will inspect a segment of string s from position l_i to position r_i inclusive. If the segment contains both characters '0' and '1', Nanako becomes unhappy and throws away the string.
After this inspection, at the i-th night, Nezzar can secretly change strictly less than half of the characters in the segment from l_i to r_i inclusive, otherwise the change will be too obvious.
Now Nezzar wonders, if it is possible to avoid Nanako being unhappy and at the same time have the string become equal to the string f at the end of these q days and nights.
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β the number of test cases.
The first line of each test case contains two integers n,q (1 β€ n β€ 2 β
10^5, 0 β€ q β€ 2 β
10^5).
The second line of each test case contains a binary string s of length n.
The third line of each test case contains a binary string f of length n.
Then q lines follow, i-th of them contains two integers l_i,r_i (1 β€ l_i β€ r_i β€ n) β bounds of the segment, that Nanako will inspect on the i-th day.
It is guaranteed that the sum of n for all test cases doesn't exceed 2 β
10^5, and the sum of q for all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print "YES" on the single line if it is possible to avoid Nanako being unhappy and have the string f at the end of q days and nights. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
4
5 2
00000
00111
1 5
1 3
2 1
00
01
1 2
10 6
1111111111
0110001110
1 10
5 9
7 10
1 7
3 5
6 10
5 2
10000
11000
2 5
1 3
Output
YES
NO
YES
NO
Note
In the first test case, \underline{00000} β \underline{000}11 β 00111 is one of the possible sequences of string changes.
In the second test case, it can be shown that it is impossible to have the string f at the end.
|
instruction
| 0
| 7,789
| 0
| 15,578
|
Tags: data structures, greedy
Correct Solution:
```
# lazy segment tree
t=[0]*(4*200000)
lazy=[-1]*(4*200000)
def build(a,v,tl,tr):
if (tl==tr):
t[v]=a[tl]
else:
tm=(tl+tr)//2
build(a,v*2,tl,tm)
build(a,v*2+1,tm+1,tr)
t[v]=t[v*2]+t[v*2+1]
return
def push(v,tl,tr):
if lazy[v]==-1:
return
tm=(tl+tr)//2
t[v*2]=lazy[v]*(tm-tl+1)
lazy[v*2]=lazy[v];
t[v*2+1]=lazy[v]*(tr-tm)
lazy[v*2+1]=lazy[v]
lazy[v]=-1
return
def update(v,tl,tr,l,r,setval):
# print('v:{} tl:{} tr:{} l:{} r:{}'.format(v,tl,tr,l,r))
if (r<tl or l>tr):
return
if (l<=tl and tr<=r):
t[v]=(tr-tl+1)*setval
lazy[v]=setval
else:
push(v,tl,tr)
tm=(tl+tr)//2
update(v*2,tl,tm,l,r,setval)
update(v*2+1,tm+1,tr,l,r,setval)
t[v]=t[v*2]+t[v*2+1]
return
def query(v,tl,tr,l,r):
if (r<tl or l>tr):
return 0
if l<=tl and tr<=r:
return t[v]
push(v,tl,tr)
tm=(tl+tr)//2
returnVal=query(v*2,tl,tm,l,r)+query(v*2+1,tm+1,tr,l,r)
return returnVal
def reconstruct(v,tl,tr,f2):
if tl==tr:
f2.append(t[v])
else:
push(v,tl,tr)
tm=(tl+tr)//2
reconstruct(v*2,tl,tm,f2)
reconstruct(v*2+1,tm+1,tr,f2)
return
def main():
z=int(input())
allans=[]
for _ in range(z):
n,q=readIntArr()
sf=input()
ss=input()
lr=[None for __ in range(q)]
for i in range(q-1,-1,-1):
l,r=readIntArr()
l-=1
r-=1
lr[i]=[l,r]
s=[c-ord('0') for c in ss]
f=[c-ord('0') for c in sf]
for i in range(4*n):
t[i]=0
lazy[i]=-1
build(s,1,0,n-1)
ok=True
for i in range(q):
l,r=lr[i]
nElems=r-l+1
total=query(1,0,n-1,l,r)
if total*2<nElems: updateVal=0
elif total*2>nElems: updateVal=1
else:
ok=False
break
update(1,0,n-1,l,r,updateVal)
if not ok:
allans.append('NO')
continue
f2=[]
reconstruct(1,0,n-1,f2)
ok=True
for i in range(n):
if f[i]!=f2[i]:
ok=False
break
if ok:
allans.append('YES')
else:
allans.append('NO')
multiLineArrayPrint(allans)
return
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
# input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
def readIntArr():
return [int(x) for x in input().split()]
# def readFloatArr():
# return [float(x) for x in input().split()]
def makeArr(defaultValFactory,dimensionArr): # eg. makeArr(lambda:0,[n,m])
dv=defaultValFactory;da=dimensionArr
if len(da)==1:return [dv() for _ in range(da[0])]
else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
def queryInteractive(x,y):
print('? {} {}'.format(x,y))
sys.stdout.flush()
return int(input())
def answerInteractive(ans):
print('! {}'.format(ans))
sys.stdout.flush()
inf=float('inf')
MOD=10**9+7
# MOD=998244353
for _abc in range(1):
main()
```
|
output
| 1
| 7,789
| 0
| 15,579
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Nezzar has a binary string s of length n that he wants to share with his best friend, Nanako. Nanako will spend q days inspecting the binary string. At the same time, Nezzar wants to change the string s into string f during these q days, because it looks better.
It is known that Nanako loves consistency so much. On the i-th day, Nanako will inspect a segment of string s from position l_i to position r_i inclusive. If the segment contains both characters '0' and '1', Nanako becomes unhappy and throws away the string.
After this inspection, at the i-th night, Nezzar can secretly change strictly less than half of the characters in the segment from l_i to r_i inclusive, otherwise the change will be too obvious.
Now Nezzar wonders, if it is possible to avoid Nanako being unhappy and at the same time have the string become equal to the string f at the end of these q days and nights.
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β the number of test cases.
The first line of each test case contains two integers n,q (1 β€ n β€ 2 β
10^5, 0 β€ q β€ 2 β
10^5).
The second line of each test case contains a binary string s of length n.
The third line of each test case contains a binary string f of length n.
Then q lines follow, i-th of them contains two integers l_i,r_i (1 β€ l_i β€ r_i β€ n) β bounds of the segment, that Nanako will inspect on the i-th day.
It is guaranteed that the sum of n for all test cases doesn't exceed 2 β
10^5, and the sum of q for all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print "YES" on the single line if it is possible to avoid Nanako being unhappy and have the string f at the end of q days and nights. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
4
5 2
00000
00111
1 5
1 3
2 1
00
01
1 2
10 6
1111111111
0110001110
1 10
5 9
7 10
1 7
3 5
6 10
5 2
10000
11000
2 5
1 3
Output
YES
NO
YES
NO
Note
In the first test case, \underline{00000} β \underline{000}11 β 00111 is one of the possible sequences of string changes.
In the second test case, it can be shown that it is impossible to have the string f at the end.
|
instruction
| 0
| 7,790
| 0
| 15,580
|
Tags: data structures, greedy
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
def __init__(self, file):
self.newlines = 0
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline()
# --------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().split())
def RLL(): return list(map(int, sys.stdin.readline().split()))
def N(): return int(input())
def S(): return input().strip()
def print_list(l): print(' '.join(map(str, l)))
# sys.setrecursionlimit(200000)
# import random
# from functools import reduce
# from functools import lru_cache
# from heapq import *
# from collections import deque as dq
# import math
# import bisect as bs
# from collections import Counter
# from collections import defaultdict as dc
def set_range(lx, rx, v, now):
if l <= lx and r >= rx:
label[now] = v
node[now] = rx - lx if v else 0
return
mx = (lx + rx) >> 1
nl = now << 1
nr = nl + 1
if label[now] != -1: # ιε―Ήη§°ζδ½ε―η¨θ―₯η―θ
node[nl] = node[nr] = node[now] >> 1
label[nl] = label[nr] = label[now]
label[now] = -1
if l < mx:
set_range(lx, mx, v, nl)
if r > mx:
set_range(mx, rx, v, nr)
node[now] = node[nl] + node[nr]
def get_range(lx, rx, now):
if l <= lx and r >= rx:
return node[now]
mx = (lx + rx) >> 1
nl = now << 1
nr = nl + 1
if label[now] != -1: # ιε―Ήη§°ζδ½ε―η¨θ―₯η―θ
return min(rx, r) - max(lx, l) if label[now] else 0
m1 = get_range(lx, mx, nl) if l < mx else 0
m2 = get_range(mx, rx, nr) if r > mx else 0
return m1 + m2
def match(lx, rx, now):
if lx >= n: return False
if now >= sn:
return node[now] != s[lx]
if label[now] != -1:
return any(v != label[now] for v in s[lx:rx])
mx = (rx + lx) >> 1
return match(lx, mx, now << 1) or match(mx, rx, now << 1 | 1)
ans = []
for _ in range(N()):
n, q = RL()
s = [int(c) for c in S()]
t = [int(c) for c in S()]
p = [tuple(RL()) for _ in range(q)]
sn = 1 << n.bit_length()
node = [0] * sn + t + [0] * (sn - n)
label = [-1] * (2 * sn)
for i in range(sn - 1, 0, -1): node[i] = node[i << 1] + node[i << 1 | 1]
f = True
for l, r in p[::-1]:
l -= 1
count = get_range(0, sn, 1) * 2
if count == r - l:
f = False
break
if count > r - l:
set_range(0, sn, 1, 1)
else:
set_range(0, sn, 0, 1)
ans.append('YES' if f and not match(0, sn, 1) else 'NO')
print('\n'.join(ans))
```
|
output
| 1
| 7,790
| 0
| 15,581
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nezzar has a binary string s of length n that he wants to share with his best friend, Nanako. Nanako will spend q days inspecting the binary string. At the same time, Nezzar wants to change the string s into string f during these q days, because it looks better.
It is known that Nanako loves consistency so much. On the i-th day, Nanako will inspect a segment of string s from position l_i to position r_i inclusive. If the segment contains both characters '0' and '1', Nanako becomes unhappy and throws away the string.
After this inspection, at the i-th night, Nezzar can secretly change strictly less than half of the characters in the segment from l_i to r_i inclusive, otherwise the change will be too obvious.
Now Nezzar wonders, if it is possible to avoid Nanako being unhappy and at the same time have the string become equal to the string f at the end of these q days and nights.
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β the number of test cases.
The first line of each test case contains two integers n,q (1 β€ n β€ 2 β
10^5, 0 β€ q β€ 2 β
10^5).
The second line of each test case contains a binary string s of length n.
The third line of each test case contains a binary string f of length n.
Then q lines follow, i-th of them contains two integers l_i,r_i (1 β€ l_i β€ r_i β€ n) β bounds of the segment, that Nanako will inspect on the i-th day.
It is guaranteed that the sum of n for all test cases doesn't exceed 2 β
10^5, and the sum of q for all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print "YES" on the single line if it is possible to avoid Nanako being unhappy and have the string f at the end of q days and nights. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
4
5 2
00000
00111
1 5
1 3
2 1
00
01
1 2
10 6
1111111111
0110001110
1 10
5 9
7 10
1 7
3 5
6 10
5 2
10000
11000
2 5
1 3
Output
YES
NO
YES
NO
Note
In the first test case, \underline{00000} β \underline{000}11 β 00111 is one of the possible sequences of string changes.
In the second test case, it can be shown that it is impossible to have the string f at the end.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
def __init__(self, file):
self.newlines = 0
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline()
# --------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().split())
def RLL(): return list(map(int, sys.stdin.readline().split()))
def N(): return int(input())
def S(): return input().strip()
def print_list(l): print(' '.join(map(str, l)))
# sys.setrecursionlimit(200000)
# import random
# from functools import reduce
# from functools import lru_cache
# from heapq import *
# from collections import deque as dq
# import math
# import bisect as bs
# from collections import Counter
# from collections import defaultdict as dc
def set_range(lx, rx, v, now):
if l <= lx and r >= rx:
label[now] = v
node[now] = rx - lx if v else 0
return
mx = (lx + rx) >> 1
nl = now << 1
nr = nl + 1
if label[now] != -1: # ιε―Ήη§°ζδ½ε―η¨θ―₯η―θ
node[nl] = node[nr] = node[now] >> 1
label[nl] = label[nr] = label[now]
label[now] = -1
if l < mx: set_range(lx, mx, v, nl)
if r > mx: set_range(mx, rx, v, nr)
node[now] = node[nl] + node[nr]
def get_range(lx, rx, now):
if l <= lx and r >= rx: return node[now]
mx = (lx + rx) >> 1
nl = now << 1
nr = nl + 1
if label[now] != -1: return min(rx, r) - max(lx, l) if label[now] else 0
m1 = get_range(lx, mx, nl) if l < mx else 0
m2 = get_range(mx, rx, nr) if r > mx else 0
return m1 + m2
def match(lx, rx, now):
if lx >= n: return False
if now >= sn: return node[now] != s[lx]
if label[now] != -1: return any(v != label[now] for v in s[lx:rx])
mx = (rx + lx) >> 1
return match(lx, mx, now << 1) or match(mx, rx, now << 1 | 1)
for _ in range(N()):
n, q = RL()
s, t = [int(c) for c in S()], [int(c) for c in S()]
p = [tuple(RL()) for _ in range(q)]
sn = 1 << n.bit_length()
node = [0] * sn + t + [0] * (sn - n)
label = [-1] * (2 * sn)
for i in range(sn - 1, 0, -1): node[i] = node[i << 1] + node[i << 1 | 1]
f = True
for l, r in p[::-1]:
l -= 1
count = get_range(0, sn, 1) * 2
if count == r - l:
f = False
break
set_range(0, sn, 1 if count > r - l else 0, 1)
print('YES' if f and not match(0, sn, 1) else 'NO')
```
|
instruction
| 0
| 7,791
| 0
| 15,582
|
Yes
|
output
| 1
| 7,791
| 0
| 15,583
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nezzar has a binary string s of length n that he wants to share with his best friend, Nanako. Nanako will spend q days inspecting the binary string. At the same time, Nezzar wants to change the string s into string f during these q days, because it looks better.
It is known that Nanako loves consistency so much. On the i-th day, Nanako will inspect a segment of string s from position l_i to position r_i inclusive. If the segment contains both characters '0' and '1', Nanako becomes unhappy and throws away the string.
After this inspection, at the i-th night, Nezzar can secretly change strictly less than half of the characters in the segment from l_i to r_i inclusive, otherwise the change will be too obvious.
Now Nezzar wonders, if it is possible to avoid Nanako being unhappy and at the same time have the string become equal to the string f at the end of these q days and nights.
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β the number of test cases.
The first line of each test case contains two integers n,q (1 β€ n β€ 2 β
10^5, 0 β€ q β€ 2 β
10^5).
The second line of each test case contains a binary string s of length n.
The third line of each test case contains a binary string f of length n.
Then q lines follow, i-th of them contains two integers l_i,r_i (1 β€ l_i β€ r_i β€ n) β bounds of the segment, that Nanako will inspect on the i-th day.
It is guaranteed that the sum of n for all test cases doesn't exceed 2 β
10^5, and the sum of q for all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print "YES" on the single line if it is possible to avoid Nanako being unhappy and have the string f at the end of q days and nights. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
4
5 2
00000
00111
1 5
1 3
2 1
00
01
1 2
10 6
1111111111
0110001110
1 10
5 9
7 10
1 7
3 5
6 10
5 2
10000
11000
2 5
1 3
Output
YES
NO
YES
NO
Note
In the first test case, \underline{00000} β \underline{000}11 β 00111 is one of the possible sequences of string changes.
In the second test case, it can be shown that it is impossible to have the string f at the end.
Submitted Solution:
```
"""
import sys
readline = sys.stdin.readline
"""
import io,os
readline = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from operator import add
class Lazysegtree:
def __init__(self, A, fx, ex, fm, em, fa, initialize = True):
self.N = len(A)
self.N0 = 2**(self.N-1).bit_length()
self.fx = fx
self.ex = ex
self.em = em
self.lazy = [em]*(2*self.N0)
if initialize:
self.data = [self.ex]*self.N0 + A + [self.ex]*(self.N0 - self.N)
for i in range(self.N0-1, -1, -1):
self.data[i] = self.fx(self.data[2*i], self.data[2*i+1])
else:
self.data = [self.ex]*(2*self.N0)
def fa(self, ope, idx):
if ope is not None:
return ope
return self.data[idx]
def fm(self, ope1, ope2):
return ope1
def __repr__(self):
s = 'data'
l = 'lazy'
cnt = 1
for i in range(self.N0.bit_length()):
s = '\n'.join((s, ' '.join(map(str, self.data[cnt:cnt+(1<<i)]))))
l = '\n'.join((l, ' '.join(map(str, self.lazy[cnt:cnt+(1<<i)]))))
cnt += 1<<i
return '\n'.join((s, l))
def _ascend(self, k):
k = k >> 1
c = k.bit_length()
for j in range(c):
idx = k >> j
self.data[idx] = self.fx(self.data[2*idx], self.data[2*idx+1])
def _descend(self, k):
k = k >> 1
idx = 1
c = k.bit_length()
for j in range(1, c+1):
idx = k >> (c - j)
if self.lazy[idx] == self.em:
continue
self.data[2*idx] = self.fa(self.lazy[idx], 2*idx)
self.data[2*idx+1] = self.fa(self.lazy[idx], 2*idx+1)
self.lazy[2*idx] = self.fm(self.lazy[idx], self.lazy[2*idx])
self.lazy[2*idx+1] = self.fm(self.lazy[idx], self.lazy[2*idx+1])
self.lazy[idx] = self.em
def query(self, l, r):
L = l+self.N0
R = r+self.N0
self._descend(L//(L & -L))
self._descend(R//(R & -R)-1)
sl = self.ex
sr = self.ex
while L < R:
if R & 1:
R -= 1
sr = self.fx(self.data[R], sr)
if L & 1:
sl = self.fx(sl, self.data[L])
L += 1
L >>= 1
R >>= 1
return self.fx(sl, sr)
def operate(self, l, r, x):
L = l+self.N0
R = r+self.N0
Li = L//(L & -L)
Ri = R//(R & -R)
self._descend(Li)
self._descend(Ri-1)
while L < R :
if R & 1:
R -= 1
self.data[R] = self.fa(x, R)
self.lazy[R] = self.fm(x, self.lazy[R])
if L & 1:
self.data[L] = self.fa(x, L)
self.lazy[L] = self.fm(x, self.lazy[L])
L += 1
L >>= 1
R >>= 1
self._ascend(Li)
self._ascend(Ri-1)
def propall(self):
for idx in range(1, self.N0):
self.data[2*idx] = self.fa(self.lazy[idx], 2*idx)
self.data[2*idx+1] = self.fa(self.lazy[idx], 2*idx+1)
self.lazy[2*idx] = self.fm(self.lazy[idx], self.lazy[2*idx])
self.lazy[2*idx+1] = self.fm(self.lazy[idx], self.lazy[2*idx+1])
T = int(readline())
Ans = ['NO']*T
for qu in range(T):
N, Q = map(int, readline().split())
"""
S = list(map(int, readline().strip()))
T = list(map(int, readline().strip()))
"""
S = list(map(lambda x: int(x)-48, readline().strip()))
T = list(map(lambda x: int(x)-48, readline().strip()))
Qs = [tuple(map(int, readline().split())) for _ in range(Q)]
K = Lazysegtree(T, add, 0, None, None, None)
for l, r in Qs[::-1]:
l -= 1
leng = r-l
x = K.query(l, r)
if 2*x > leng:
K.operate(l, r, 0)
elif 2*x == leng:
break
else:
K.operate(l, r, 1)
else:
if [K.query(i, i+1) for i in range(N)] == S:
Ans[qu] = 'YES'
print('\n'.join(Ans))
```
|
instruction
| 0
| 7,792
| 0
| 15,584
|
No
|
output
| 1
| 7,792
| 0
| 15,585
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nezzar has a binary string s of length n that he wants to share with his best friend, Nanako. Nanako will spend q days inspecting the binary string. At the same time, Nezzar wants to change the string s into string f during these q days, because it looks better.
It is known that Nanako loves consistency so much. On the i-th day, Nanako will inspect a segment of string s from position l_i to position r_i inclusive. If the segment contains both characters '0' and '1', Nanako becomes unhappy and throws away the string.
After this inspection, at the i-th night, Nezzar can secretly change strictly less than half of the characters in the segment from l_i to r_i inclusive, otherwise the change will be too obvious.
Now Nezzar wonders, if it is possible to avoid Nanako being unhappy and at the same time have the string become equal to the string f at the end of these q days and nights.
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β the number of test cases.
The first line of each test case contains two integers n,q (1 β€ n β€ 2 β
10^5, 0 β€ q β€ 2 β
10^5).
The second line of each test case contains a binary string s of length n.
The third line of each test case contains a binary string f of length n.
Then q lines follow, i-th of them contains two integers l_i,r_i (1 β€ l_i β€ r_i β€ n) β bounds of the segment, that Nanako will inspect on the i-th day.
It is guaranteed that the sum of n for all test cases doesn't exceed 2 β
10^5, and the sum of q for all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print "YES" on the single line if it is possible to avoid Nanako being unhappy and have the string f at the end of q days and nights. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
4
5 2
00000
00111
1 5
1 3
2 1
00
01
1 2
10 6
1111111111
0110001110
1 10
5 9
7 10
1 7
3 5
6 10
5 2
10000
11000
2 5
1 3
Output
YES
NO
YES
NO
Note
In the first test case, \underline{00000} β \underline{000}11 β 00111 is one of the possible sequences of string changes.
In the second test case, it can be shown that it is impossible to have the string f at the end.
Submitted Solution:
```
from sys import stdin
def log2(x):
if x == 1:
return 1
if x > 1 and x <= 2:
return 2
return log2(x / 2) * 2
def findsum(qs, qe, seg, lz):
summ = 0
stack = [[0, 1, length]]
while stack:
pos, ss, se = stack.pop()
if lz[pos] != -1:
seg[pos] = lz[pos] * (se - ss + 1)
if ss != se:
lz[pos * 2 + 1] = lz[pos]
lz[pos * 2 + 2] = lz[pos]
lz[pos] = -1
if (qs > se) or (qe < ss):
continue
if (qs <= ss) and (qe >= se):
summ += seg[pos]
continue
if ss != se:
mid = (ss + se) // 2
stack.append([pos * 2 + 1, ss, mid])
stack.append([pos * 2 + 2, mid + 1, se])
return summ
def update(pos, ss, se, direc):
if lz[pos] != -1:
seg[pos] = (se - ss + 1) * lz[pos]
if (ss != se):
lz[pos * 2 + 1] = lz[pos]
lz[pos * 2 + 2] - lz[pos]
lz[pos] = -1
if (qs > se) or (qe < ss):
return
elif (qs <= ss) and (qe >= se):
seg[pos] = (se - ss + 1) * direc
if ss != se:
lz[pos * 2 + 1] = direc
lz[pos * 2 + 2] = direc
return
mid = (se + ss) // 2
update(pos * 2 + 1, ss, mid, direc)
update(pos * 2 + 2, mid + 1, se, direc)
seg[pos] = seg[pos * 2 + 1] + seg[pos * 2 + 2]
def clean(seg, lz):
stack = [[0, 1, length]]
while stack:
pos, ss, se = stack.pop()
if lz[pos] != -1:
if ss != se:
lz[pos * 2 + 1] = lz[pos]
lz[pos * 2 + 2] = lz[pos]
else:
seg[pos] = (se - ss + 1) * lz[pos]
if ss != se:
mid = (se + ss) // 2
stack.append([pos * 2 + 1, ss, mid])
stack.append([pos * 2 + 2, mid + 1, se])
t = int(stdin.readline())
for _ in range(t):
n, q = map(int,stdin.readline().split())
s = stdin.readline().strip()
f = stdin.readline().strip()
length = log2(n)
queries = [map(int, stdin.readline().split()) for quer in range(q)]
seg = [0] * (length - 1) + list(map(int, list(f))) + [0] * (length - n)
for pos in range(length - 2, -1, -1):
seg[pos] = seg[pos * 2 + 1] + seg[pos * 2 + 2]
lz = [-1] * (2 * length - 1)
flag = True
for qs, qe in queries[::-1]:
dis = qe - qs + 1
tong = findsum(qs, qe, seg, lz)
if tong * 2 == dis:
flag = False
break
else:
if tong * 2 > dis:
update(0, 1, length, 1)
else:
update(0, 1, length, 0)
#print(seg, lz)
if flag == False:
print('no')
else:
clean(seg, lz)
#print(seg, lz)
if seg[-length: -length + n] == [int(s[i]) for i in range(n)]:
print('yes')
else:
print('no')
```
|
instruction
| 0
| 7,793
| 0
| 15,586
|
No
|
output
| 1
| 7,793
| 0
| 15,587
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nezzar has a binary string s of length n that he wants to share with his best friend, Nanako. Nanako will spend q days inspecting the binary string. At the same time, Nezzar wants to change the string s into string f during these q days, because it looks better.
It is known that Nanako loves consistency so much. On the i-th day, Nanako will inspect a segment of string s from position l_i to position r_i inclusive. If the segment contains both characters '0' and '1', Nanako becomes unhappy and throws away the string.
After this inspection, at the i-th night, Nezzar can secretly change strictly less than half of the characters in the segment from l_i to r_i inclusive, otherwise the change will be too obvious.
Now Nezzar wonders, if it is possible to avoid Nanako being unhappy and at the same time have the string become equal to the string f at the end of these q days and nights.
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β the number of test cases.
The first line of each test case contains two integers n,q (1 β€ n β€ 2 β
10^5, 0 β€ q β€ 2 β
10^5).
The second line of each test case contains a binary string s of length n.
The third line of each test case contains a binary string f of length n.
Then q lines follow, i-th of them contains two integers l_i,r_i (1 β€ l_i β€ r_i β€ n) β bounds of the segment, that Nanako will inspect on the i-th day.
It is guaranteed that the sum of n for all test cases doesn't exceed 2 β
10^5, and the sum of q for all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print "YES" on the single line if it is possible to avoid Nanako being unhappy and have the string f at the end of q days and nights. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
4
5 2
00000
00111
1 5
1 3
2 1
00
01
1 2
10 6
1111111111
0110001110
1 10
5 9
7 10
1 7
3 5
6 10
5 2
10000
11000
2 5
1 3
Output
YES
NO
YES
NO
Note
In the first test case, \underline{00000} β \underline{000}11 β 00111 is one of the possible sequences of string changes.
In the second test case, it can be shown that it is impossible to have the string f at the end.
Submitted Solution:
```
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
import copy
import time
starttime = time.time()
mod = int(pow(10, 9) + 7)
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def L(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
try:
# sys.setrecursionlimit(int(pow(10,4)))
sys.stdin = open("input.txt", "r")
# sys.stdout = open("../output.txt", "w")
except:
pass
def pmat(A):
for ele in A:
print(*ele,end="\n")
# def seive():
# prime=[1 for i in range(10**6+1)]
# prime[0]=0
# prime[1]=0
# for i in range(10**6+1):
# if(prime[i]):
# for j in range(2*i,10**6+1,i):
# prime[j]=0
# return prime
import sys
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def indexes(L,R):
INDLIST=[]
R-=1
L>>=1
R>>=1
while L!=R:
if L>R:
INDLIST.append(L)
L>>=1
else:
INDLIST.append(R)
R>>=1
while L!=0:
INDLIST.append(L)
L>>=1
return INDLIST
def updates(l,r,x):
L=l+seg_el
R=r+seg_el
L//=(L & (-L))
R//=(R & (-R))
UPIND=indexes(L,R)
for ind in UPIND[::-1]:
if LAZY[ind]!=None:
update_lazy = LAZY[ind] *(1<<(seg_height - 1 - (ind.bit_length())))
LAZY[ind<<1]=LAZY[1+(ind<<1)]=LAZY[ind]
SEG[ind<<1]=SEG[1+(ind<<1)]=update_lazy
LAZY[ind]=None
while L!=R:
if L > R:
SEG[L]=x * (1<<(seg_height - (L.bit_length())))
LAZY[L]=x
L+=1
L//=(L & (-L))
else:
R-=1
SEG[R]=x * (1<<(seg_height - (R.bit_length())))
LAZY[R]=x
R//=(R & (-R))
for ind in UPIND:
SEG[ind]=SEG[ind<<1]+SEG[1+(ind<<1)]
def getvalues(l,r):
L=l+seg_el
R=r+seg_el
L//=(L & (-L))
R//=(R & (-R))
UPIND=indexes(L,R)
for ind in UPIND[::-1]:
if LAZY[ind]!=None:
update_lazy = LAZY[ind] *(1<<(seg_height - 1 - (ind.bit_length())))
LAZY[ind<<1]=LAZY[1+(ind<<1)]=LAZY[ind]
SEG[ind<<1]=SEG[1+(ind<<1)]=update_lazy
LAZY[ind]=None
ANS=0
while L!=R:
if L > R:
ANS+=SEG[L]
L+=1
L//=(L & (-L))
else:
R-=1
ANS+=SEG[R]
R//=(R & (-R))
return ANS
t=int(input())
for tests in range(t):
n,q=map(int,input().split())
S=input().strip()
F=input().strip()
Q=[tuple(map(int,input().split())) for i in range(q)]
seg_el=1<<(n.bit_length())
seg_height=1+n.bit_length()
SEG=[0]*(2*seg_el)
LAZY=[None]*(2*seg_el)
for i in range(n):
SEG[i+seg_el]=int(F[i])
for i in range(seg_el-1,0,-1):
SEG[i]=SEG[i*2]+SEG[i*2+1]
for l,r in Q[::-1]:
SUM=r-l+1
xx=getvalues(l-1,r)
if xx*2==SUM:
print("NO")
break
else:
if xx*2>SUM:
updates(l-1,r,1)
else:
updates(l-1,r,0)
else:
for i in range(n):
if getvalues(i,i+1)==int(S[i]):
True
else:
print("NO")
break
else:
print("YES")
endtime = time.time()
# print(f"Runtime of the program is {endtime - starttime}")
```
|
instruction
| 0
| 7,794
| 0
| 15,588
|
No
|
output
| 1
| 7,794
| 0
| 15,589
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nezzar has a binary string s of length n that he wants to share with his best friend, Nanako. Nanako will spend q days inspecting the binary string. At the same time, Nezzar wants to change the string s into string f during these q days, because it looks better.
It is known that Nanako loves consistency so much. On the i-th day, Nanako will inspect a segment of string s from position l_i to position r_i inclusive. If the segment contains both characters '0' and '1', Nanako becomes unhappy and throws away the string.
After this inspection, at the i-th night, Nezzar can secretly change strictly less than half of the characters in the segment from l_i to r_i inclusive, otherwise the change will be too obvious.
Now Nezzar wonders, if it is possible to avoid Nanako being unhappy and at the same time have the string become equal to the string f at the end of these q days and nights.
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β the number of test cases.
The first line of each test case contains two integers n,q (1 β€ n β€ 2 β
10^5, 0 β€ q β€ 2 β
10^5).
The second line of each test case contains a binary string s of length n.
The third line of each test case contains a binary string f of length n.
Then q lines follow, i-th of them contains two integers l_i,r_i (1 β€ l_i β€ r_i β€ n) β bounds of the segment, that Nanako will inspect on the i-th day.
It is guaranteed that the sum of n for all test cases doesn't exceed 2 β
10^5, and the sum of q for all test cases doesn't exceed 2 β
10^5.
Output
For each test case, print "YES" on the single line if it is possible to avoid Nanako being unhappy and have the string f at the end of q days and nights. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
4
5 2
00000
00111
1 5
1 3
2 1
00
01
1 2
10 6
1111111111
0110001110
1 10
5 9
7 10
1 7
3 5
6 10
5 2
10000
11000
2 5
1 3
Output
YES
NO
YES
NO
Note
In the first test case, \underline{00000} β \underline{000}11 β 00111 is one of the possible sequences of string changes.
In the second test case, it can be shown that it is impossible to have the string f at the end.
Submitted Solution:
```
import sys
input = sys.stdin.readline
def segfunc(x,y):
return x + y
class LazySegTree_RUQ:
def __init__(self, init_val, segfunc, ide_ele):
n = len(init_val)
self.segfunc = segfunc
self.ide_ele = ide_ele
self.num = 1 << (n - 1).bit_length()
self.tree = [ide_ele] * 2 * self.num
self.lazy = [None] * 2 * self.num
for i in range(n):
self.tree[self.num + i] = init_val[i]
for i in range(self.num - 1, 0, -1):
self.tree[i] = self.segfunc(self.tree[2 * i], self.tree[2 * i + 1])
def gindex(self, l, r):
l += self.num
r += self.num
lm = l >> (l & -l).bit_length()
rm = r >> (r & -r).bit_length()
while r > l:
if l <= lm:
yield l
if r <= rm:
yield r
r >>= 1
l >>= 1
while l:
yield l
l >>= 1
def propagates(self, *ids):
for i in reversed(ids):
v = self.lazy[i]
if v is None:
continue
self.lazy[i] = None
self.lazy[2 * i] = v
self.lazy[2 * i + 1] = v
self.tree[2 * i] = v
self.tree[2 * i + 1] = v
def update(self, l, r, x):
*ids, = self.gindex(l, r)
self.propagates(*ids)
l += self.num
r += self.num
while l < r:
if l & 1:
self.lazy[l] = x
self.tree[l] = x
l += 1
if r & 1:
self.lazy[r - 1] = x
self.tree[r - 1] = x
r >>= 1
l >>= 1
for i in ids:
self.tree[i] = self.segfunc(self.tree[2 * i], self.tree[2 * i + 1])
def query(self, l, r):
*ids, = self.gindex(l, r)
self.propagates(*ids)
res = self.ide_ele
l += self.num
r += self.num
while l < r:
if l & 1:
res = self.segfunc(res, self.tree[l])
l += 1
if r & 1:
res = self.segfunc(res, self.tree[r - 1])
l >>= 1
r >>= 1
return res
def main():
n, q = map(int, input().split())
S = list(map(int, input().strip()))
T = list(map(int, input().strip()))
lr = [list(map(int, input().split())) for _ in range(q)]
seg = LazySegTree_RUQ(T,segfunc,0)
for l, r in lr[::-1]:
c = seg.query(l - 1, r)
z = r - l + 1
if z == c * 2:
print("NO")
return
if c == z or c == 0:
continue
elif c * 2 < z:
seg.update(l - 1, r, 1)
else:
seg.update(l - 1, r, 0)
for i in range(n):
if seg.query(i, i + 1) != S[i]:
print("NO")
return
print("YES")
for _ in range(int(input())):
main()
```
|
instruction
| 0
| 7,795
| 0
| 15,590
|
No
|
output
| 1
| 7,795
| 0
| 15,591
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You will receive 5 points for solving this problem.
Manao has invented a new operation on strings that is called folding. Each fold happens between a pair of consecutive letters and places the second part of the string above first part, running in the opposite direction and aligned to the position of the fold. Using this operation, Manao converts the string into a structure that has one more level than there were fold operations performed. See the following examples for clarity.
We will denote the positions of folds with '|' characters. For example, the word "ABRACADABRA" written as "AB|RACA|DAB|RA" indicates that it has been folded three times: first, between the leftmost pair of 'B' and 'R' letters; second, between 'A' and 'D'; and third, between the rightmost pair of 'B' and 'R' letters. Here are several examples of folded strings:
"ABCDEF|GHIJK" | "A|BCDEFGHIJK" | "AB|RACA|DAB|RA" | "X|XXXXX|X|X|XXXXXX"
| | | XXXXXX
KJIHG | KJIHGFEDCB | AR | X
ABCDEF | A | DAB | X
| | ACAR | XXXXX
| | AB | X
One last example for "ABCD|EFGH|IJ|K":
K
IJ
HGFE
ABCD
Manao noticed that each folded string can be viewed as several piles of letters. For instance, in the previous example, there are four piles, which can be read as "AHI", "BGJK", "CF", and "DE" from bottom to top. Manao wonders what is the highest pile of identical letters he can build using fold operations on a given word. Note that the pile should not contain gaps and should start at the bottom level. For example, in the rightmost of the four examples above, none of the piles would be considered valid since each of them has gaps, starts above the bottom level, or both.
Input
The input will consist of one line containing a single string of n characters with 1 β€ n β€ 1000 and no spaces. All characters of the string will be uppercase letters.
This problem doesn't have subproblems. You will get 5 points for the correct submission.
Output
Print a single integer β the size of the largest pile composed of identical characters that can be seen in a valid result of folding operations on the given string.
Examples
Input
ABRACADABRA
Output
3
Input
ABBBCBDB
Output
3
Input
AB
Output
1
Note
Consider the first example. Manao can create a pile of three 'A's using the folding "AB|RACAD|ABRA", which results in the following structure:
ABRA
DACAR
AB
In the second example, Manao can create a pile of three 'B's using the following folding: "AB|BB|CBDB".
CBDB
BB
AB
Another way for Manao to create a pile of three 'B's with "ABBBCBDB" is the following folding: "AB|B|BCBDB".
BCBDB
B
AB
In the third example, there are no folds performed and the string is just written in one line.
|
instruction
| 0
| 7,900
| 0
| 15,800
|
Tags: brute force
Correct Solution:
```
t = {}
for i, c in enumerate(input()):
if c not in t: t[c] = (i, 1)
elif (t[c][0] - i) & 1: t[c] = (i, t[c][1] + 1)
print(max(b for a, b in t.values()))
```
|
output
| 1
| 7,900
| 0
| 15,801
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You will receive 5 points for solving this problem.
Manao has invented a new operation on strings that is called folding. Each fold happens between a pair of consecutive letters and places the second part of the string above first part, running in the opposite direction and aligned to the position of the fold. Using this operation, Manao converts the string into a structure that has one more level than there were fold operations performed. See the following examples for clarity.
We will denote the positions of folds with '|' characters. For example, the word "ABRACADABRA" written as "AB|RACA|DAB|RA" indicates that it has been folded three times: first, between the leftmost pair of 'B' and 'R' letters; second, between 'A' and 'D'; and third, between the rightmost pair of 'B' and 'R' letters. Here are several examples of folded strings:
"ABCDEF|GHIJK" | "A|BCDEFGHIJK" | "AB|RACA|DAB|RA" | "X|XXXXX|X|X|XXXXXX"
| | | XXXXXX
KJIHG | KJIHGFEDCB | AR | X
ABCDEF | A | DAB | X
| | ACAR | XXXXX
| | AB | X
One last example for "ABCD|EFGH|IJ|K":
K
IJ
HGFE
ABCD
Manao noticed that each folded string can be viewed as several piles of letters. For instance, in the previous example, there are four piles, which can be read as "AHI", "BGJK", "CF", and "DE" from bottom to top. Manao wonders what is the highest pile of identical letters he can build using fold operations on a given word. Note that the pile should not contain gaps and should start at the bottom level. For example, in the rightmost of the four examples above, none of the piles would be considered valid since each of them has gaps, starts above the bottom level, or both.
Input
The input will consist of one line containing a single string of n characters with 1 β€ n β€ 1000 and no spaces. All characters of the string will be uppercase letters.
This problem doesn't have subproblems. You will get 5 points for the correct submission.
Output
Print a single integer β the size of the largest pile composed of identical characters that can be seen in a valid result of folding operations on the given string.
Examples
Input
ABRACADABRA
Output
3
Input
ABBBCBDB
Output
3
Input
AB
Output
1
Note
Consider the first example. Manao can create a pile of three 'A's using the folding "AB|RACAD|ABRA", which results in the following structure:
ABRA
DACAR
AB
In the second example, Manao can create a pile of three 'B's using the following folding: "AB|BB|CBDB".
CBDB
BB
AB
Another way for Manao to create a pile of three 'B's with "ABBBCBDB" is the following folding: "AB|B|BCBDB".
BCBDB
B
AB
In the third example, there are no folds performed and the string is just written in one line.
|
instruction
| 0
| 7,901
| 0
| 15,802
|
Tags: brute force
Correct Solution:
```
t = {}
for i, c in enumerate(input()):
if c not in t: t[c] = (i, 1)
elif (t[c][0] - i) & 1: t[c] = (i, t[c][1] + 1)
print(max(b for a, b in t.values()))
# Made By Mostafa_Khaled
```
|
output
| 1
| 7,901
| 0
| 15,803
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp invented a new way to encode strings. Let's assume that we have string T, consisting of lowercase English letters. Let's choose several pairs of letters of the English alphabet in such a way that each letter occurs in at most one pair. Then let's replace each letter in T with its pair letter if there is a pair letter for it. For example, if you chose pairs (l, r), (p, q) and (a, o), then word "parallelogram" according to the given encoding principle transforms to word "qolorreraglom".
Polycarpus already has two strings, S and T. He suspects that string T was obtained after applying the given encoding method from some substring of string S. Find all positions mi in S (1 β€ mi β€ |S| - |T| + 1), such that T can be obtained fro substring SmiSmi + 1... Smi + |T| - 1 by applying the described encoding operation by using some set of pairs of English alphabet letters
Input
The first line of the input contains two integers, |S| and |T| (1 β€ |T| β€ |S| β€ 2Β·105) β the lengths of string S and string T, respectively.
The second and third line of the input contain strings S and T, respectively. Both strings consist only of lowercase English letters.
Output
Print number k β the number of suitable positions in string S.
In the next line print k integers m1, m2, ..., mk β the numbers of the suitable positions in the increasing order.
Examples
Input
11 5
abacabadaba
acaba
Output
3
1 3 7
Input
21 13
paraparallelogramgram
qolorreraglom
Output
1
5
Submitted Solution:
```
def kmp(s):
k = [0] * len(s)
j = 0
for i in range(1, len(s)):
j = k[i - 1]
while True:
if s[i] == s[j]:
k[i] = j + 1
break
if j == 0:
k[i] = 0
break
j = k[j - 1]
return k
def mns(s, c):
a = []
for i in range(len(s)):
if s[i] == c:
a.append(1)
else:
a.append(0)
return a
def check(pairs, alf):
for c in alf:
ind = 0
for p in pairs:
a, b = p
if a == c or b == c:
ind+=1
if ind!=1:
return False
return True
if __name__ == '__main__':
n, m = [int(e) for e in input().split()]
s = input()
s1 = input()
sets = []
for i in range(n):
sets.append(set())
alf = list(set(s+s1))
alf.sort()
for a in alf:
for b in alf:
q = mns(s, a)
q1 = mns(s1, b)
kmp_res = kmp(q1 + [-1] + q)
pos = []
for i in range(len(kmp_res)):
if kmp_res[i] == m:
sets[i - 2 * m].add((min(a,b), max(a,b)))
res = []
for i in range(n):
if check(sets[i], alf):
res.append(i)
print(len(res))
print(' '.join([str(e+1) for e in res]))
```
|
instruction
| 0
| 7,940
| 0
| 15,880
|
No
|
output
| 1
| 7,940
| 0
| 15,881
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This task is very simple. Given a string S of length n and q queries each query is on the format i j k which means sort the substring consisting of the characters from i to j in non-decreasing order if k = 1 or in non-increasing order if k = 0.
Output the final string after applying the queries.
Input
The first line will contain two integers n, q (1 β€ n β€ 105, 0 β€ q β€ 50 000), the length of the string and the number of queries respectively.
Next line contains a string S itself. It contains only lowercase English letters.
Next q lines will contain three integers each i, j, k (1 β€ i β€ j β€ n, <image>).
Output
Output one line, the string S after applying the queries.
Examples
Input
10 5
abacdabcda
7 10 0
5 8 1
1 4 0
3 6 0
7 10 1
Output
cbcaaaabdd
Input
10 1
agjucbvdfk
1 10 1
Output
abcdfgjkuv
Note
First sample test explanation:
<image>
<image>
<image>
<image>
<image>
Submitted Solution:
```
n,t=map(int,input().split())
s=input()
while(t):
t=t-1
st,end,k=map(int,input().split())
sub=s[st:end+1]
#print(sub)
l=list(sub)
#print(l)
if(k==0):
l.sort()
else:
l.sort(reverse=True)
r="".join(l)
#print(r)
res=s[0:st]+r+s[end+1:n]
print(res)
```
|
instruction
| 0
| 7,941
| 0
| 15,882
|
No
|
output
| 1
| 7,941
| 0
| 15,883
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This task is very simple. Given a string S of length n and q queries each query is on the format i j k which means sort the substring consisting of the characters from i to j in non-decreasing order if k = 1 or in non-increasing order if k = 0.
Output the final string after applying the queries.
Input
The first line will contain two integers n, q (1 β€ n β€ 105, 0 β€ q β€ 50 000), the length of the string and the number of queries respectively.
Next line contains a string S itself. It contains only lowercase English letters.
Next q lines will contain three integers each i, j, k (1 β€ i β€ j β€ n, <image>).
Output
Output one line, the string S after applying the queries.
Examples
Input
10 5
abacdabcda
7 10 0
5 8 1
1 4 0
3 6 0
7 10 1
Output
cbcaaaabdd
Input
10 1
agjucbvdfk
1 10 1
Output
abcdfgjkuv
Note
First sample test explanation:
<image>
<image>
<image>
<image>
<image>
Submitted Solution:
```
import textwrap
n,q=list(map(int,input().split()))
print(n,q)
stroka=textwrap.wrap(input(),1)
for x in range(q):
i,j,k=list(map(int,input().split()))
stroka[i-1:j]=sorted(stroka[i-1:j],reverse=abs(k-1))
print(''.join(stroka))
```
|
instruction
| 0
| 7,942
| 0
| 15,884
|
No
|
output
| 1
| 7,942
| 0
| 15,885
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This task is very simple. Given a string S of length n and q queries each query is on the format i j k which means sort the substring consisting of the characters from i to j in non-decreasing order if k = 1 or in non-increasing order if k = 0.
Output the final string after applying the queries.
Input
The first line will contain two integers n, q (1 β€ n β€ 105, 0 β€ q β€ 50 000), the length of the string and the number of queries respectively.
Next line contains a string S itself. It contains only lowercase English letters.
Next q lines will contain three integers each i, j, k (1 β€ i β€ j β€ n, <image>).
Output
Output one line, the string S after applying the queries.
Examples
Input
10 5
abacdabcda
7 10 0
5 8 1
1 4 0
3 6 0
7 10 1
Output
cbcaaaabdd
Input
10 1
agjucbvdfk
1 10 1
Output
abcdfgjkuv
Note
First sample test explanation:
<image>
<image>
<image>
<image>
<image>
Submitted Solution:
```
n, q = map(int, input().split(" "))
s = list(input())
for a in range(0, q):
i, j, k = map(int, input().split(" "))
s[i-1:j] = sorted(s[i-1:j], reverse=1-k)
print(s)
```
|
instruction
| 0
| 7,943
| 0
| 15,886
|
No
|
output
| 1
| 7,943
| 0
| 15,887
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This task is very simple. Given a string S of length n and q queries each query is on the format i j k which means sort the substring consisting of the characters from i to j in non-decreasing order if k = 1 or in non-increasing order if k = 0.
Output the final string after applying the queries.
Input
The first line will contain two integers n, q (1 β€ n β€ 105, 0 β€ q β€ 50 000), the length of the string and the number of queries respectively.
Next line contains a string S itself. It contains only lowercase English letters.
Next q lines will contain three integers each i, j, k (1 β€ i β€ j β€ n, <image>).
Output
Output one line, the string S after applying the queries.
Examples
Input
10 5
abacdabcda
7 10 0
5 8 1
1 4 0
3 6 0
7 10 1
Output
cbcaaaabdd
Input
10 1
agjucbvdfk
1 10 1
Output
abcdfgjkuv
Note
First sample test explanation:
<image>
<image>
<image>
<image>
<image>
Submitted Solution:
```
n,q=[int(x) for x in input().split()]
s=input()
L=[]
M=[]
for i in range(q):
i,j,k=[int(x) for x in input().split()]
if k==0:
try:
if i< min(L) and j in L:
s=s[:i-1]+"".join(sorted(s[i-1:min(L)-1],reverse=True))+s[min(L)-1:]
if i in M and min(L)-1 in M:
del M[i-1:min(L)-1]
L.extend(range(i,min(L)))
elif i in L and j>max(L):
s=s[:max(L)]+"".join(sorted(s[max(L):j],reverse=True))+s[j:]
if max(L)+1 in M and j in M:
del M[max(L):j]
L.extend(range(max(L)+1,j+1))
elif j not in L and i not in L:
s=s[:i-1]+"".join(sorted(s[i-1:j],reverse=True))+s[j:]
if i in M and j in M:
del M[i-1:j]
L.extend(range(i,j+1))
else:
pass
except ValueError:
s=s[:i-1]+"".join(sorted(s[i-1:j],reverse=True))+s[j:]
if i in M and j in M:
del M[i-1:j]
L.extend(range(i,j+1))
L.sort()
else:
try:
if i< min(M) and j in M:
s=s[:i-1]+"".join(sorted(s[i-1:min(M)-1]))+s[min(M)-1:]
if i in L and min(M)-1 in L:
del L[i-1:min(M)-1]
M.extend(range(i,min(M)))
elif i in L and j>max(L):
s=s[:max(L)]+"".join(sorted(s[max(M):j]))+s[j:]
if max(M)+1 in L and j in L:
del L[max(M):j]
M.extend(range(max(M)+1,j+1))
elif j not in M and i not in M:
s=s[:i-1]+"".join(sorted(s[i-1:j]))+s[j:]
if i in L and j in L:
del L[i-1:j]
M.extend(range(i,j+1))
else:
pass
except ValueError:
s=s[:i-1]+"".join(sorted(s[i-1:j]))+s[j:]
if i in L and j in L:
del L[i-1:j]
M.extend(range(i,j+1))
M.sort()
print(s)
```
|
instruction
| 0
| 7,944
| 0
| 15,888
|
No
|
output
| 1
| 7,944
| 0
| 15,889
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Marina loves strings of the same length and Vasya loves when there is a third string, different from them in exactly t characters. Help Vasya find at least one such string.
More formally, you are given two strings s1, s2 of length n and number t. Let's denote as f(a, b) the number of characters in which strings a and b are different. Then your task will be to find any string s3 of length n, such that f(s1, s3) = f(s2, s3) = t. If there is no such string, print - 1.
Input
The first line contains two integers n and t (1 β€ n β€ 105, 0 β€ t β€ n).
The second line contains string s1 of length n, consisting of lowercase English letters.
The third line contain string s2 of length n, consisting of lowercase English letters.
Output
Print a string of length n, differing from string s1 and from s2 in exactly t characters. Your string should consist only from lowercase English letters. If such string doesn't exist, print -1.
Examples
Input
3 2
abc
xyc
Output
ayd
Input
1 0
c
b
Output
-1
|
instruction
| 0
| 7,945
| 0
| 15,890
|
Tags: constructive algorithms, greedy, strings
Correct Solution:
```
n, k = map(int, input().split())
a = input()
b = input()
cnt = 0
for i in range(n):
if (a[i] == b[i]):
cnt += 1
dif = n - cnt
ok = n - (cnt + dif // 2)
if (k < ok):
print(-1)
else:
res, cur, req, i, nxt = {}, 0, n - k, 0, True
while (cur < req):
if (a[i] == b[i]):
res[i], cur = a[i], cur + 1
elif (nxt):
pos, nxt = i, not nxt
else:
res[pos], res[i], nxt, cur = a[pos], b[i], not nxt, cur + 1
i += 1
if (nxt == False):
if (('a' != a[pos]) and ('a' != b[pos])):
res[pos] = 'a'
elif (('b' != a[pos]) and ('b' != b[pos])):
res[pos] = 'b'
elif (('c' != a[pos]) and ('c' != b[pos])):
res[pos] = 'c'
else:
res[pos] = 'd'
while (i < n):
if (('a' != a[i]) and ('a' != b[i])):
res[i] = 'a'
elif (('b' != a[i]) and ('b' != b[i])):
res[i] = 'b'
elif (('c' != a[i]) and ('c' != b[i])):
res[i] = 'c'
else:
res[i] = 'd'
i += 1
for i in range(n):
print(res[i], end = '')
```
|
output
| 1
| 7,945
| 0
| 15,891
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Marina loves strings of the same length and Vasya loves when there is a third string, different from them in exactly t characters. Help Vasya find at least one such string.
More formally, you are given two strings s1, s2 of length n and number t. Let's denote as f(a, b) the number of characters in which strings a and b are different. Then your task will be to find any string s3 of length n, such that f(s1, s3) = f(s2, s3) = t. If there is no such string, print - 1.
Input
The first line contains two integers n and t (1 β€ n β€ 105, 0 β€ t β€ n).
The second line contains string s1 of length n, consisting of lowercase English letters.
The third line contain string s2 of length n, consisting of lowercase English letters.
Output
Print a string of length n, differing from string s1 and from s2 in exactly t characters. Your string should consist only from lowercase English letters. If such string doesn't exist, print -1.
Examples
Input
3 2
abc
xyc
Output
ayd
Input
1 0
c
b
Output
-1
|
instruction
| 0
| 7,946
| 0
| 15,892
|
Tags: constructive algorithms, greedy, strings
Correct Solution:
```
n, t = map(int, input().split())
s1 = input()
s2 = input()
same = n - t
ans = ["" for i in range(n)]
for i in range(n):
if(same==0):
break
if s1[i] == s2[i]:
same -= 1
ans[i] = s1[i]
same1 = same2 = same
for i in range(n):
if(ans[i]!=""):
continue
if same1 != 0 and same1>=same2:
ans[i] = s1[i]
same1 -= 1
elif same2 != 0 and same2>=same1:
ans[i] = s2[i]
same2 -= 1
elif same1 == 0 and same2 == 0:
if s1[i] != 'a' and s2[i] != 'a':
ans[i] = 'a'
elif s1[i] != 'b' and s2[i] != 'b':
ans[i] = 'b'
elif s1[i] != 'c' and s2[i] != 'c':
ans[i] = 'c'
if same1 > 0 or same2 > 0:
print(-1)
else:
print("".join(ans))
```
|
output
| 1
| 7,946
| 0
| 15,893
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Marina loves strings of the same length and Vasya loves when there is a third string, different from them in exactly t characters. Help Vasya find at least one such string.
More formally, you are given two strings s1, s2 of length n and number t. Let's denote as f(a, b) the number of characters in which strings a and b are different. Then your task will be to find any string s3 of length n, such that f(s1, s3) = f(s2, s3) = t. If there is no such string, print - 1.
Input
The first line contains two integers n and t (1 β€ n β€ 105, 0 β€ t β€ n).
The second line contains string s1 of length n, consisting of lowercase English letters.
The third line contain string s2 of length n, consisting of lowercase English letters.
Output
Print a string of length n, differing from string s1 and from s2 in exactly t characters. Your string should consist only from lowercase English letters. If such string doesn't exist, print -1.
Examples
Input
3 2
abc
xyc
Output
ayd
Input
1 0
c
b
Output
-1
|
instruction
| 0
| 7,947
| 0
| 15,894
|
Tags: constructive algorithms, greedy, strings
Correct Solution:
```
n,k=map(int,input().split())
a=input()
b=input()
c=[]
ans=0
for i in range(n):
if a[i]==b[i]:
ans+=1
si=n-k
d=['a','b','c']
if ans>=si:
for i in range(n):
if a[i]==b[i] and si>0:
c.append(a[i])
si-=1
else:
ee=[a[i],b[i]]
if 'a' not in ee:
c.append('a')
elif 'c' not in ee:
c.append('c')
else:
c.append('b')
print(''.join(c))
else:
re1=si-ans
re2=si-ans
for i in range(n):
if a[i]==b[i] :
c.append(a[i])
else:
if re1>0:
c.append(a[i])
re1-=1
elif re2>0:
c.append(b[i])
re2-=1
else:
ee=[a[i],b[i]]
if 'a' not in ee:
c.append('a')
elif 'c' not in ee:
c.append('c')
else:
c.append('b')
#print(re1,re2)
if re1+re2==0:
print(''.join(c))
else:
print(-1)
```
|
output
| 1
| 7,947
| 0
| 15,895
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Marina loves strings of the same length and Vasya loves when there is a third string, different from them in exactly t characters. Help Vasya find at least one such string.
More formally, you are given two strings s1, s2 of length n and number t. Let's denote as f(a, b) the number of characters in which strings a and b are different. Then your task will be to find any string s3 of length n, such that f(s1, s3) = f(s2, s3) = t. If there is no such string, print - 1.
Input
The first line contains two integers n and t (1 β€ n β€ 105, 0 β€ t β€ n).
The second line contains string s1 of length n, consisting of lowercase English letters.
The third line contain string s2 of length n, consisting of lowercase English letters.
Output
Print a string of length n, differing from string s1 and from s2 in exactly t characters. Your string should consist only from lowercase English letters. If such string doesn't exist, print -1.
Examples
Input
3 2
abc
xyc
Output
ayd
Input
1 0
c
b
Output
-1
|
instruction
| 0
| 7,948
| 0
| 15,896
|
Tags: constructive algorithms, greedy, strings
Correct Solution:
```
from sys import exit
def cdiff(c1, c2):
ch = 'a'
while ch in (c1, c2):
ch = chr(ord(ch) + 1)
return ch
n, t = map(int, input().split())
a = input()
b = input()
ra = 0
rb = 0
c = [-1] * n
for i in range(n):
c[i] = (a[i], b[i])[(ra + rb) % 2]
ra += c[i] != a[i]
rb += c[i] != b[i]
if rb < ra:
ra, rb = rb, ra
a, b = b, a
if ra < rb:
for i in range(n):
if c[i] != b[i] and c[i] == a[i]:
c[i] = cdiff(a[i], b[i])
ra += 1
break
if ra < rb:
print(-1)
exit(0)
if t < ra:
print(-1)
exit(0)
i = 0
lst = -1
while ra < t and i < n:
if a[i] == b[i]:
c[i] = cdiff(a[i], b[i])
ra += 1
rb += 1
elif c[i] not in (a[i], b[i]):
pass
elif lst == -1:
lst = i
else:
c[lst] = cdiff(a[lst], b[lst])
c[i] = cdiff(a[i], b[i])
lst = -1
ra += 1
rb += 1
i += 1
if ra != t:
print(-1)
exit(0)
print(*c, sep='')
```
|
output
| 1
| 7,948
| 0
| 15,897
|
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