数组的所有子序列的总和

给定一个包含 n 个整数的数组。求一个数组所有可能子序列的总和。

例子：

Input : arr[] = { 1, 2 }
Output : 6
All possible subsequences are {}, {1}, {2} 
and { 1, 2 }

Input : arr[] = { 1, 2, 3 }
Output : 24

我们已经在下面的帖子中讨论了两种不同的解决方案。
所有子数组的总和 |设置 1

在这篇文章中讨论了一个不同的解决方案。让我们仔细看看这个问题并尝试找到一个模式

Let a[] = { 1, 2, 3 }

All subsequences are {}, {1}, {2}, {3}, {1, 2}, 
                     {1, 3}, {2, 3}, {1, 2, 3}

So sum of subsequences are 0 + 1 + 2 + 3 + 3 + 
                             4 + 5 + 6 = 24

Here we can observe that in sum every elements 
occurs 4 times. Or in general every element 
will occur 2^(n-1) times. And we can also 
observe that sum of array elements is 6. So
final result will be 6*4.

一般来说，我们可以通过将数组的所有元素相加乘以 2 ^(n-1)来找到所有子序列的总和，其中 n 是数组中的元素数。

C++

// CPP program to find sum of
// all subarrays of array
#include 
using namespace std;
  
// To find sum of all subsequences
int findSum(int arr[], int n)
{
    // Sum all array elements
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
  
    // Result is sum * 2^(n-1)
    return sum * (1 << (n - 1));
}
  
// Driver program to test findSum()
int main()
{
    int arr[] = { 1, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findSum(arr, n);
    return 0;
}

Java

// Java program to find sum of
// all subarrays of array
  
public class Main {
    // To find sum of all subsequences
    static int findSum(int arr[], int n)
    {
        // Sum all array elements
        int sum = 0;
        for (int i = 0; i < n; i++)
            sum += arr[i];
  
        // Result is sum * 2^(n-1)
        return sum * (1 << (n - 1));
    }
  
    // Driver program to test findSum()
    public static void main(String[] args)
    {
        int arr[] = { 1, 2 };
        int n = arr.length;
        System.out.print(findSum(arr, n));
    }
}

Python

# Python program to find sum of
# all subarrays of array
  
# To find sum of all subsequences
def findSum(arr, n):
      
    # Sum all array elements
    sum = 0
    for i in range(n):
        sum += arr[i]
   
    # Result is sum * 2^(n-1)
    return sum * (1 << (n - 1))
   
# Driver program to test findSum()
arr = [1, 2]
n = len(arr)
print findSum(arr, n)
  
# This code is submitted by Sachin Bisht

C#

// C# program to find sum of 
// all subarrays of array 
using System;
  
class GFG
{
  
// To find sum of all subsequences
static int findSum(int []arr, int n) 
{ 
    // Sum all array elements 
    int sum = 0; 
    for (int i = 0; i < n; i++) 
        sum += arr[i]; 
  
    // Result is sum * 2^(n-1) 
    return sum * (1 << (n - 1)); 
} 
  
// Driver Code
static public void Main ()
{
    int []arr = { 1, 2 }; 
    int n = arr.Length; 
    Console.WriteLine(findSum(arr, n)); 
} 
} 
  
// This code is contributed by ajit

PHP

Javascript

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