# 从二进制字符串中删除一个元素，使XOR变为0的方法

2021年4月4日18:50:47 发表评论 847 次浏览

## 本文概述

``````Input : 10000
Output : 1
We only have 1 ways to

Input : 10011
Output : 3
There are 3 ways to make XOR 0. We
can remove any of the three 1's.

Input : 100011100
Output : 5
There are 5 ways to make XOR 0. We
can remove any of the give 0's``````

An有效的解决方案基于以下事实。如果1的计数是奇数, 那么我们必须删除1才能使计数0, 并且我们可以删除任何1。如果1的计数为偶数, 则XOR为0, 我们可以删除任何0, 而XOR仍为0。

## C ++

``````// C++ program to count number of ways to
// remove an element so that XOR of remaining
// string becomes 0.
#include <bits/stdc++.h>
using namespace std;

// Return number of ways in which XOR become ZERO
// by remove 1 element
int xorZero(string str)
{
int one_count = 0, zero_count = 0;
int n = str.length();

// Counting number of 0 and 1
for ( int i = 0; i < n; i++)
if (str[i] == '1' )
one_count++;
else
zero_count++;

// If count of ones is even
// then return count of zero
// else count of one
if (one_count % 2 == 0)
return zero_count;
return one_count;
}

// Driver Code
int main()
{
string str = "11111" ;
cout << xorZero(str) << endl;
return 0;
}``````

## Java

``````// Java program to count number of ways to
// remove an element so that XOR of remaining
// string becomes 0.
import java.util.*;

class CountWays
{
// Returns number of ways in which XOR become
// ZERO by remove 1 element
static int xorZero(String s)
{
int one_count = 0 , zero_count = 0 ;
char [] str=s.toCharArray();
int n = str.length;

// Counting number of 0 and 1
for ( int i = 0 ; i < n; i++)
if (str[i] == '1' )
one_count++;
else
zero_count++;

// If count of ones is even
// then return count of zero
// else count of one
if (one_count % 2 == 0 )
return zero_count;
return one_count;
}

// Driver Code to test above function
public static void main(String[] args)
{
String s = "11111" ;
System.out.println(xorZero(s));
}
}

// This code is contributed by Mr. Somesh Awasthi``````

## Python3

``````# Python 3 program to count number of
# ways to remove an element so that
# XOR of remaining string becomes 0.

# Return number of ways in which XOR
# become ZERO by remove 1 element
def xorZero( str ):
one_count = 0
zero_count = 0
n = len ( str )

# Counting number of 0 and 1
for i in range ( 0 , n, 1 ):
if ( str [i] = = '1' ):
one_count + = 1
else :
zero_count + = 1

# If count of ones is even
# then return count of zero
# else count of one
if (one_count % 2 = = 0 ):
return zero_count
return one_count

# Driver Code
if __name__ = = '__main__' :
str = "11111"
print (xorZero( str ))

# This code is contributed by
# Surendra_Gangwar``````

## C#

``````// C# program to count number
// of ways to remove an element
// so that XOR of remaining
// string becomes 0.
using System;

class GFG
{
// Returns number of ways
// in which XOR become
// ZERO by remove 1 element
static int xorZero( string s)
{
int one_count = 0, zero_count = 0;

int n = s.Length;

// Counting number of 0 and 1
for ( int i = 0; i < n; i++)
if (s[i] == '1' )
one_count++;
else
zero_count++;

// If count of ones is even
// then return count of zero
// else count of one
if (one_count % 2 == 0)
return zero_count;
return one_count;
}

// Driver Code
public static void Main()
{
string s = "11111" ;
Console.WriteLine(xorZero(s));
}
}

// This code is contributed by anuj_67.``````

## 的PHP

``````<?php
// PHP program to count number
// of ways to remove an element
// so that XOR of remaining
// string becomes 0.

// Return number of ways in
// which XOR become ZERO
// by remove 1 element

function xorZero( \$str )
{
\$one_count = 0; \$zero_count = 0;
\$n = strlen ( \$str );

// Counting number of 0 and 1
for ( \$i = 0; \$i < \$n ; \$i ++)
if ( \$str [ \$i ] == '1' )
\$one_count ++;
else
\$zero_count ++;

// If count of ones is even
// then return count of zero
// else count of one
if ( \$one_count % 2 == 0)
return \$zero_count ;
return \$one_count ;
}

// Driver Code
\$str = "11111" ;
echo xorZero( \$str ), "\n" ;

// This code is contributed by aj_36
?>``````

``5``