计算与N的差等于与N的XOR的数字

给定数字N。任务是计算x的所有可能值，以使n $\oplus$ x等于(Nx)，其中$\oplus$表示按位XOR运算。
例子：

Input: N = 3
Output: 4
The all possible values of x are respectively 0, 1, 2, 3.

Input: N = 6
Output: 4
The all possible values of x are respectively 0, 2, 4, 6.

方法：如果两个位的符号相反，则两个位的XOR值分别为1；如果两个位的符号相同，则XOR值为0。因此，根据XOR的性质，我们可以说n $\oplus$ x始终大于或等于nx。其值等于nx的唯一条件是x的位形成n的位的子集。因为如果在第i个位置x和n都设置了位，则xor之后的值将减小，减小的值将为$2^{i}$ ，其中i是从0开始的位置。
所以答案是第n个位的子集的总数为$2^{k}$ ，其中k是n中设置位的计数。
下面是上述方法的实现：

C++

#include 
using namespace std;
 
// function to Count all values of x
void count_values(int n)
{
    // Count set bits in n
    // by using stl function
    int set_bits = __builtin_popcount(n);
 
    // count all subset of set bits
    cout << pow(2, set_bits) << "\n";
}
 
// Driver code
int main()
{
 
    int n = 27;
    count_values(n);
 
    return 0;
}

Java

import java.util.*;
 
class Solution
{
//count number of set bits
static int __builtin_popcount(int n)
{
    //count variable
    int count=0;
     
    while(n>0)
    {
        //if the bit is 1
        if(n%2==1)
        count++;
         
        n=n/2;
    }
    return count;
}
     
// function to Count all values of x
static void count_values(int n)
{
    // Count set bits in n
    // by using stl function
    int set_bits = __builtin_popcount(n);
   
    // count all subset of set bits
    System.out.println((int)Math.pow(2, set_bits));
}
   
// Driver code
public static void main(String args[])
{
   
    int n = 27;
    count_values(n);
   
 
}
}
 
// This code is contributed
// by Arnab Kundu

Python 3

# Python3 program to implement
# above approach
 
# from math import pow method
from math import pow
 
# count number of set bits
def __builtin_popcount(n) :
 
    # count variable
    count = 0
 
    while n > 0 :
 
        # if the bit is 1
        if n % 2 == 1 :
            count += 1
 
        n = n//2
         
    return count
 
 
# function to Count all values of x
def count_values(n) :
 
    set_bits = __builtin_popcount(n)
 
    # count all subset of set bits
    print(int(pow(2, set_bits)))
 
 
# Driver code
if __name__ == "__main__" :
 
    n = 27
    count_values(n)
 
# This code is contributed by
# ANKITRAI1

C#

using System;
class GFG
{
// count number of set bits
static int __builtin_popcount(int n)
{
    // count variable
    int count = 0;
     
    while(n > 0)
    {
        //if the bit is 1
        if(n % 2 == 1)
        count++;
         
        n = n / 2;
    }
    return count;
}
     
// function to Count all values of x
static void count_values(int n)
{
    // Count set bits in n
    // by using stl function
    int set_bits = __builtin_popcount(n);
     
    // count all subset of set bits
    Console.Write((int)Math.Pow(2, set_bits));
}
     
// Driver code
public static void Main()
{
    int n = 27;
    count_values(n);
}
}
 
// This code is contributed by Smitha

PHP

 0)
    {
        //if the bit is 1
        if($n % 2 == 1)
            $count++;
         
        $n = $n / 2;
    }
    return $count;
}
     
// function to Count all values of x
function count_values($n)
{
    // Count set bits in n
    // by using stl function
    $set_bits = __builtin_popcount($n);
 
    // count all subset of set bits
    echo (int)pow(2, $set_bits);
}
 
// Driver code
$n = 27;
count_values($n);
 
// This code is contributed
// by Akanksha Rai(Abby_akku)
?>

Javascript

 
// count number of set bits
function __builtin_popcount(n)
{
    // count variable
    let count = 0;
     
    while(n > 0)
    {
        //if the bit is 1
        if(n % 2 == 1)
        count++;
         
        n = parseInt(n / 2);
    }
    return count;
}
 
// function to Count all values of x
function count_values(n)
{
    // Count set bits in n
    // by using stl function
    let set_bits = __builtin_popcount(n);
 
    // count all subset of set bits
    document.write(Math.pow(2, set_bits) + "<br>");
}
 
// Driver code
 
    let n = 27;
    count_values(n);
 

输出：

16

时间复杂度： O(k)，其中k是N中的设置位数。