每个元素 x 后的数组元素之和本身 XORed x 次

给定一个整数数组，任务是在每个元素 x 与自身 x 次 XOR 之后计算所有数组元素的总和。例如，如果元素是 4，那么我们将这个数字与自身进行 XOR 4 次 Like:= 4^4^4^4
例子：

Input :  arr[] = { 1, 2, 3, 5 }
Output :  9 
explanation:  1 + 2^2 + 3^3^3 + 5^5^5^5^5 : 9
                     
Input :   arr[] ={ 5, 6, 7, 9 }
Output :  21

一个简单的解决方案是一个一个地选择每个数组元素，并根据其值与自身进行异或。最后将 XOR 值添加到结果中。
下面是上述想法的实现。

C++

// C++ program to compute sum of all element after
// doing Xor with itself ( element_time)
#include 
using namespace std;
 
// function return sum of all XOR element of array
int XorSum(int arr[], int n)
{
    // store result
    int result = 0;
 
    // Traverse array element and apply XOR
    // operation on it
    for (int i = 0; i < n; i++) {
         
        // XOR of current element with itself
        // according to value.
        int k = 0;
        for (int j = 1; j <= arr[i]; j++)
            k ^= arr[i];
 
        result += k;
    }
    return result;
}
// Driver program
int main()
{
    int arr[] = { 1, 2, 6, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << XorSum(arr, n) << endl;
    return 0;
}

Java

// Java program to compute sum of all
// element after doing Xor with itself
// ( element_time)
import java.io.*;
 
class GFG {
     
    // function return sum of all XOR
    // element of array
    static int XorSum(int arr[], int n)
    {
        // store result
        int result = 0;
     
        // Traverse array element and apply
        // XOR operation on it
        for (int i = 0; i < n; i++) {
             
            // XOR of current element with
            // itself according to value.
            int k = 0;
            for (int j = 1; j <= arr[i]; j++)
                k ^= arr[i];
     
            result += k;
        }
         
        return result;
    }
     
    // Driver program
    public static void main(String args[])
    {
        int arr[] = { 1, 2, 6, 3, 4, 5 };
        int n = arr.length;
        System.out.println(XorSum(arr, n));
    }
}
 
/*This code is contributed by Nikita Tiwari.*/

Python

# Python 3 program to compute sum of
# all element after doing Xor with
# itself ( element_time)
 
# function return sum of all XOR
# element of array
def XorSum(arr, n) :
     
    # store result
    result = 0
 
    # Traverse array element and
    # apply XOR operation on it
    for i in range(0, n) :
         
        # XOR of current element
        # with itself according to
        # value.
        k = 0
        for j in range(1, arr[i]+1) :
            k = k ^ arr[i]
 
        result = result + k
     
    return result
 
 
# Driver program
 
arr = [ 1, 2, 6, 3, 4, 5 ]
n = len(arr)
print(XorSum(arr, n))
 
 
# This code is contributed by Nikita Tiwari.

C#

// C# program to compute sum of all
// element after doing Xor with itself
// ( element_time)
using System;
 
class GFG {
     
    // function return sum of all XOR
    // element of array
    static int XorSum(int []arr, int n)
    {
        // store result
        int result = 0;
     
        // Traverse array element and apply
        // XOR operation on it
        for (int i = 0; i < n; i++) {
             
            // XOR of current element with
            // itself according to value.
            int k = 0;
            for (int j = 1; j <= arr[i]; j++)
                k ^= arr[i];
     
            result += k;
        }
         
        return result;
    }
     
    // Driver program
    public static void Main()
    {
        int []arr = { 1, 2, 6, 3, 4, 5 };
        int n = arr.Length;
        Console.WriteLine(XorSum(arr, n));
    }
}
 
// This code is contributed by vt_m.

Javascript

// Javascript program to compute sum of all element after
// doing Xor with itself ( element_time)
 
// function return sum of all XOR element of array
function XorSum(arr, n)
{
    // store result
    let result = 0;
 
    // Traverse array element and apply XOR
    // operation on it
    for (let i = 0; i < n; i++) {
         
        // XOR of current element with itself
        // according to value.
        let k = 0;
        for (let j = 1; j <= arr[i]; j++)
            k ^= arr[i];
 
        result += k;
    }
    return result;
}
// Driver program
    let arr = [ 1, 2, 6, 3, 4, 5 ];
    let n = arr.length;
    document.write(XorSum(arr, n));
 

输出：

9

时间复杂度：O(n*m)（这里 m 是数组中的最大元素）
辅助空间：O(1)
这个问题的有效解决方案是基于这样一个事实：如果我们对任何数字进行 XOR（偶数次）它会产生 0，如果我们进行 XOR 奇数次它会产生相同的数字。
例如

let number be  : 3  do XOR with itself 3 time
             3^3^3 = 3 
   let number be :  4 do XOR with itself 4 time 
             4^4^4^4 = 0 
  so if number is odd it's mean output is number 
  itself. Else zero  

以下是上述想法的实现：

C++

// C++ program to compute sum of all element after
// doing XOR with itself ( element_time)
#include 
using namespace std;
 
// function return sum of all XOR element of array
int XorSum(int arr[], int n)
{
    int result = 0;
    for (int i = 0; i < n; i++) {
 
        // if number is odd then add it to the
        // result else not
        if (arr[i] % 2 != 0)
            result += arr[i];
    }
 
    return result;
}
 
// Driver program
int main()
{
    int arr[] = { 1, 2, 6, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << XorSum(arr, n) << endl;
    return 0;
}

Java

// Java program to compute sum of
// all element after doing XOR
// with itself ( element_time)
class GFG {
     
// function return sum of all
// XOR element of array
static int XorSum(int arr[], int n) {
     
    int result = 0;
    for (int i = 0; i < n; i++) {
 
    // if number is odd then add it to the
    // result else not
    if (arr[i] % 2 != 0)
        result += arr[i];
    }
 
    return result;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = {1, 2, 6, 3, 4, 5};
    int n = arr.length;
    System.out.println(XorSum(arr, n));
}
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python program to compute
# sum of all element after
# doing XOR with itself
# ( element_time)
 
# function return sum of
# all XOR element of array
def XorSum(arr,n):
 
    result = 0
    for i in range(n):
  
        # if number is odd then add it to the
        # result else not
        if (arr[i] % 2 != 0):
            result += arr[i]
     
  
    return result
 
# Driver program
arr = [ 1, 2, 6, 3, 4, 5 ]
n = len(arr)
 
print(XorSum(arr, n))
 
# This code is contributed
# by Anant Agarwal.

C＃

// C# program to compute sum of
// all element after doing XOR
// with itself ( element_time)
using System;
 
class GFG {
     
    // function return sum of all
    // XOR element of array
    static int XorSum(int []arr, int n)
    {
         
        int result = 0;
        for (int i = 0; i < n; i++) {
     
        // if number is odd then add it to the
        // result else not
        if (arr[i] % 2 != 0)
            result += arr[i];
        }
     
        return result;
    }
     
    // Driver code
    public static void Main()
    {
        int []arr = {1, 2, 6, 3, 4, 5};
        int n = arr.Length;
        Console.WriteLine(XorSum(arr, n));
    }
}
 
// This code is contributed by vt_m.

Javascript

 
// Javascript program to compute
// sum of all element after
// doing XOR with itself ( element_time)
 
// function return sum of all XOR element of array
function XorSum(arr, n)
{
    let result = 0;
    for (let i = 0; i < n; i++) {
 
        // if number is odd then add it to the
        // result else not
        if (arr[i] % 2 != 0)
            result += arr[i];
    }
 
    return result;
}
 
// Driver program
    let arr = [ 1, 2, 6, 3, 4, 5 ];
    let n = arr.length;
    document.write(XorSum(arr, n));
 

输出：

9