给定一个集合,求所有子集的异或的异或。

问题是找到所有子集的XOR的XOR。i、 如果集合为{1,2,3},则为e。所有子集都是:[{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}]。求每个子集的异或,然后求每个子集结果的异或。 我们强烈建议您尽量减少浏览器,并先自己尝试。 如果我们正确地迈出第一步(也是唯一一步),这是一个非常简单的问题。当n>1时,XOR始终为0,当n为1时,设置[0]。

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C++

// C++ program to find XOR of XOR's of all subsets
#include <bits/stdc++.h>
using namespace std;
// Returns XOR of all XOR's of given subset
int findXOR( int Set[], int n)
{
// XOR is 1 only when n is 1, else 0
if (n == 1)
return Set[0];
else
return 0;
}
// Driver program
int main()
{
int Set[] = {1, 2, 3};
int n = sizeof (Set)/ sizeof (Set[0]);
cout << "XOR of XOR's of all subsets is "
<< findXOR(Set, n);
return 0;
}


JAVA

// Java program to find XOR of
// XOR's of all subsets
import java.util.*;
class GFG {
// Returns XOR of all XOR's of given subset
static int findXOR( int Set[], int n) {
// XOR is 1 only when n is 1, else 0
if (n == 1 )
return Set[ 0 ];
else
return 0 ;
}
// Driver code
public static void main(String arg[])
{
int Set[] = { 1 , 2 , 3 };
int n = Set.length;
System.out.print( "XOR of XOR's of all subsets is " +
findXOR(Set, n));
}
}
// This code is contributed by Anant Agarwal.


Python3

# Python program to find
# XOR of XOR's of all subsets
# Returns XOR of all
# XOR's of given subset
def findXOR( Set , n):
# XOR is 1 only when
# n is 1, else 0
if (n = = 1 ):
return Set [ 0 ]
else :
return 0
# Driver code
Set = [ 1 , 2 , 3 ]
n = len ( Set )
print ( "XOR of XOR's of all subsets is " ,
findXOR( Set , n));
# This code is contributed
# by Anant Agarwal.


C#

// C# program to find XOR of
// XOR's of all subsets
using System;
class GFG {
// Returns XOR of all
// XOR's of given subset
static int findXOR( int []Set, int n)
{
// XOR is 1 only when n
// is 1, else 0
if (n == 1)
return Set[0];
else
return 0;
}
// Driver code
public static void Main()
{
int []Set = {1, 2, 3};
int n = Set.Length;
Console.Write( "XOR of XOR's of all subsets is " +
findXOR(Set, n));
}
}
// This code is contributed by nitin mittal


PHP

<?php
// PHP program to find XOR
// of XOR's of all subsets
// Returns XOR of all
// XOR's of given subset
function findXOR( $Set , $n )
{
// XOR is 1 only when
// n is 1, else 0
if ( $n == 1)
return $Set [0];
else
return 0;
}
// Driver Code
$Set = array (1, 2, 3);
$n = count ( $Set );
echo "XOR of XOR's of all subsets is "
, findXOR( $Set , $n );
// This code is contributed by anuj_67.
?>


Javascript

<script>
// JavaScript program to find XOR of XOR's of all subsets
// Returns XOR of all XOR's of given subset
function findXOR(Set, n)
{
// XOR is 1 only when n is 1, else 0
if (n == 1)
return Set[0];
else
return 0;
}
// Driver program
let Set = [1, 2, 3];
let n = Set.length;
document.write( "XOR of XOR's of all subsets is "
+ findXOR(Set, n));
// This code is contributed by Surbhi Tyagi
</script>


输出:

XOR of XOR's of all subsets is 0

时间复杂性: O(1)

辅助空间: O(1)

相关问题: 所有可能子集的XOR之和 这是怎么回事? 逻辑很简单。让我们考虑NTH元素,它可以包含在剩余(N-1)元素的所有子集中。(n-1)元素的子集数等于2 (n-1) 当n>1时总是偶数。因此,在XOR结果中,每个元素包含偶数次,任何数字偶数出现的XOR为0。 本文由 埃克塔·戈尔 。如果您发现任何不正确的地方,或者您想分享有关上述主题的更多信息,请发表评论

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