给定数组的所有可能的非空子集的值的总和

给定N个整数的数组arr [] ，任务是找到给定数组的所有可能的非空子集的值的总和。
例子：

Input: arr[] = {2, 3} 
Output: 10 
All non-empty subsets are {2}, {3} and {2, 3} 
Total sum = 2 + 3 + 2 + 3 = 10
Input: arr[] = {2, 1, 5, 6} 
Output: 112
 

方法：可以观察到，当从所有可能的子集中添加所有元素时，原始数组的每个元素都会出现2 ^{(N – 1)}次。这意味着最终答案中任何元素arr [i]的贡献将为arr [i] * 2 ^{(N – 1)} 。因此，所需的答案将是(arr [0] + arr [1] + arr [2] +…+ arr [N – 1])* 2 ^{(N – 1)} 。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the required sum
int sum(int arr[], int n)
{
 
    // Find the sum of the array elements
    int sum = 0;
    for (int i = 0; i < n; i++) {
        sum += arr[i];
    }
 
    // Every element appears 2^(n-1) times
    sum = sum * pow(2, n - 1);
    return sum;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 1, 5, 6 };
    int n = sizeof(arr) / sizeof(int);
 
    cout << sum(arr, n);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
    // Function to return the required sum
    static int sum(int arr[], int n)
    {
     
        // Find the sum of the array elements
        int sum = 0;
        for (int i = 0; i < n; i++)
        {
            sum += arr[i];
        }
     
        // Every element appears 2^(n-1) times
        sum = sum * (int)Math.pow(2, n - 1);
        return sum;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int arr[] = { 2, 1, 5, 6 };
        int n = arr.length;
        System.out.println(sum(arr, n));
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
 
# Function to return the required sum
def sum( arr, n):
 
    # Find the sum of the array elements
    sum = 0
    for i in arr :
        sum += i
     
    # Every element appears 2^(n-1) times
    sum = sum * pow(2, n - 1)
    return sum
 
# Driver code
arr = [ 2, 1, 5, 6 ]
n = len(arr)
 
print(sum(arr, n))
 
# This code is contributed by Arnab Kundu

C#

// C# implementation of the approach
using System;
class GFG
{
 
    // Function to return the required sum
    static int sum(int[] arr, int n)
    {
     
        // Find the sum of the array elements
        int sum = 0;
        for (int i = 0; i < n; i++)
        {
            sum += arr[i];
        }
     
        // Every element appears 2^(n-1) times
        sum = sum * (int)Math.Pow(2, n - 1);
        return sum;
    }
     
    // Driver code
    public static void Main ()
    {
        int[] arr = { 2, 1, 5, 6 };
        int n = arr.Length;
        Console.WriteLine(sum(arr, n));
    }
}
 
// This code is contributed by CodeMech

Javascript

// javascript implementation of the approach
 
// Function to return the required sum
function sum(arr, n)
{
 
// Find the sum of the array elements
var sum = 0;
for (i = 0; i < n; i++)
{
sum += arr[i];
}
 
// Every element appears 2^(n-1) times
sum = sum * parseInt(Math.pow(2, n - 1));
return sum;
}
 
// Driver code
 
var arr = [ 2, 1, 5, 6 ];
var n = arr.length;
document.write(sum(arr, n));
 
// This code is contributed by Amit Katiyar
 

输出：

112