包含2的幂的数组，其XOR和元素之和等于X

给定整数X。该任务是查找并返回包含2的幂的数组，并且该数组的xor为X。

例子：

Input: X = 20
Output: 16 4

Input: X = 15
Output: 1 2 4 8

方法：答案在于数字X的二进制表示形式。

由于以2的幂表示，因此只有一个置位。如果存在两个2的不同幂，则xor将是两个数字的加法。

同样，如果将使用整个数组的xor，则它应等于X ，这将是该数字的二进制表示形式。

由于2的每个幂都有一个不同的置位位，因此xor和数组元素的总和将相同。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
  
// Function to return the required array
vector getArray(int n)
{
    vector ans;
  
    // Store the power of 2
    long p2 = 1;
  
    // while n is greater than 0
    while (n > 0) {
          
        // if there is 1 in binary
        // representation
        if (n & 1)
            ans.push_back(p2);
  
        // Divide n by 2
        // Multiply p2 by 2
        n >>= 1;
        p2 *= 2;
    }
  
    return ans;
}
  
// Driver code
int main()
{
    long n = 15;
  
    // Get the answer
    vector ans = getArray(n);
  
    // Printing the array
    for(int i : ans)
        cout << i << " ";
  
    return 0;
}

Java

// Java implementation implementation
// of the above approach
import java.util.*;
class GFG 
{
  
// Function to return the required array
static Vector getArray(int n)
{
    Vector ans = new Vector();
  
    // Store the power of 2
    long p2 = 1;
  
    // while n is greater than 0
    while (n > 0)
    {
          
        // if there is 1 in binary
        // representation
        if (n % 2 == 1)
            ans.add(p2);
  
        // Divide n by 2
        // Multiply p2 by 2
        n >>= 1;
        p2 *= 2;
    }
    return ans;
}
  
// Driver code
public static void main(String[] args) 
{
    int n = 15;
  
    // Get the answer
    Vector ans = getArray(n);
  
    // Printing the array
    for(Long i : ans)
        System.out.print(i + " ");
}
}
  
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the above approach 
  
# Function to return the required array 
def getArray(n) : 
  
    ans = []; 
  
    # Store the power of 2 
    p2 = 1; 
  
    # while n is greater than 0 
    while (n > 0) :
          
        # if there is 1 in binary 
        # representation 
        if (n & 1) :
            ans.append(p2); 
  
        # Divide n by 2 
        # Multiply p2 by 2 
        n >>= 1; 
        p2 *= 2; 
  
    return ans; 
  
# Driver code 
if __name__ == "__main__" : 
  
    n = 15; 
  
    # Get the answer 
    ans = getArray(n); 
  
    # Printing the array 
    for i in ans :
        print(i, end = " "); 
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
  
class GFG 
{
  
// Function to return the required array
static List getArray(int n)
{
    List ans = new List();
  
    // Store the power of 2
    long p2 = 1;
  
    // while n is greater than 0
    while (n > 0)
    {
          
        // if there is 1 in binary
        // representation
        if (n % 2 == 1)
            ans.Add(p2);
  
        // Divide n by 2
        // Multiply p2 by 2
        n >>= 1;
        p2 *= 2;
    }
    return ans;
}
  
// Driver code
public static void Main(String[] args) 
{
    int n = 15;
  
    // Get the answer
    List ans = getArray(n);
  
    // Printing the array
    foreach(long i in ans)
        Console.Write(i + " ");
}
}
  
// This code is contributed by Princi Singh

输出：

1 2 4 8