所有可能子阵列的乘积之和

给定一个由N 个正整数组成的数组arr[] ，任务是找到所有可能子数组元素的乘积之和。

例子：

Input: arr[] = {1, 2, 3}
Output: 20
Explanation: Possible Subarrays are: {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 2, 3}.
Products of all the above subarrays are 1, 2, 3, 2, 6 and 6 respectively.
Sum of all products = 1 + 2 + 3 + 2 + 6 + 6 = 20.

Input: arr[] = {1, 2, 3, 4}
Output: 84
Explanation:
Possible Subarrays are: {1}, {2}, {3}, {4}, {1, 2}, {2, 3}, {3, 4}, {1, 2, 3}, {2, 3, 4}, {1, 2, 3, 4}. Products of all the above subarrays are 1, 2, 3, 4, 2, 6, 12, 6, 24 and 24.
Sum of all products = 1 + 2 + 3 + 4 + 2 + 6 + 12 + 6 + 24 + 24 = 84.

朴素方法：解决问题的最简单方法是生成所有可能的子数组，并计算每个子数组的所有元素的乘积并将其添加到最终总和中。

时间复杂度： O(N ² )
辅助空间： O(1)

高效方法：优化上述方法，其思想是将问题观察到某种模式中。假设我们有四个数字a 、 b 、 c和d 。我们可以将所有可能的子数组乘积写为：

a + b + c + ab + bc + cd + abc + bcd + abcd
   = (a + ab + abc + abcd) + (b + bc + bcd) + (c + cd) + d
   = a * (1+ b + bc + bcd) + (b + bc + bcd) + (c + cd) + d

Now, the value of underlined expression (b + bc + bcd) can be calculated once and use twice.
Again, (b+ bc + bcd) + (c + cd) = b * (1 + c + cd) + (c + cd)

In the same way, the expression (c + cd) can be used twice.
The latter part is the same as above.

请按照以下步骤解决问题：

• 迭代最后一个元素，并使用每个元素更新重复出现的表达式并进一步使用它。在这个过程中，相应地更新结果。
• 初始化ANS为0将保存最后一笔和资源为0，将保留以前的子阵列中的所有元素的产品价值的轨道。
• 从后面遍历数组，对于每个元素， arr[i]执行以下操作：
  • 通过arr[i]和(1 + res)的乘积增加 ans。
  • 将 res 更新为arr[i]*(1 + res) 。
• 完成上述步骤后，打印ans 中存储的所有子数组的乘积之和。  

下面是上述方法的实现：

C++14

// C++ program for the above approach
#include 
using namespace std;
 
// Function that finds the sum of
// products of all subarray of arr[]
void sumOfSubarrayProd(int arr[], int n)
{
     
    // Stores sum of all subarrays
    int ans = 0;
    int res = 0;
 
    // Iterate array from behind
    for(int i = n - 1; i >= 0; i--)
    {
        int incr = arr[i] * (1 + res);
 
        // Update the ans
        ans += incr;
 
        // Update the res
        res = incr;
    }
 
    // Print the final sum
    cout << (ans);
}
 
// Driver Code
int main()
{
     
    // Given array arr[]
    int arr[] = { 1, 2, 3 };
 
    // Size of array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    sumOfSubarrayProd(arr, N);
}
 
// This code is contributed by mohit kumar 29

Java

// Java program for the above approach
 
import java.io.*;
class GFG {
 
    // Function that finds the sum of
    // products of all subarray of arr[]
    static void sumOfSubarrayProd(int arr[],
                                  int n)
    {
        // Stores sum of all subarrays
        int ans = 0;
        int res = 0;
 
        // Iterate array from behind
        for (int i = n - 1; i >= 0; i--) {
            int incr = arr[i] * (1 + res);
 
            // Update the ans
            ans += incr;
 
            // Update the res
            res = incr;
        }
 
        // Print the final sum
        System.out.println(ans);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given array arr[]
        int arr[] = { 1, 2, 3 };
 
        // Size of array
        int N = arr.length;
 
        // Function Call
        sumOfSubarrayProd(arr, N);
    }
}

Python3

# Python3 program for the above approach
 
# Function that finds the sum of
# products of all subarray of arr[]
def sumOfSubarrayProd(arr, n):
     
    # Stores sum of all subarrays
    ans = 0
    res = 0
 
    # Iterate array from behind
    i = n - 1
    while(i >= 0):
        incr = arr[i] * (1 + res)
 
        # Update the ans
        ans += incr
 
        # Update the res
        res = incr
        i -= 1
 
    # Print the final sum
    print(ans)
 
# Driver Code
if __name__ == '__main__':
     
    # Given array arr[]
    arr = [ 1, 2, 3 ]
 
    # Size of array
    N = len(arr)
 
    # Function call
    sumOfSubarrayProd(arr, N)
     
# This code is contributed by ipg2016107

C#

// C# program for the
// above approach
using System;
 
// Function that finds
// the sum of products
// of all subarray of arr[]
class solution{
 
static void sumOfSubarrayProd(int []arr,
                              int n)
{   
  // Stores sum of all
  // subarrays
  int ans = 0;
  int res = 0;
 
  // Iterate array from behind
  for(int i = n - 1; i >= 0; i--)
  {
    int incr = arr[i] * (1 + res);
 
    // Update the ans
    ans += incr;
 
    // Update the res
    res = incr;
  }
 
  // Print the final sum
  Console.WriteLine(ans);
}
 
// Driver Code
public static void Main(String[] args)
{   
  // Given array arr[]
  int []arr = {1, 2, 3 };
 
  // Size of array
  int N = arr.Length;
  // Function call
  sumOfSubarrayProd(arr, N);
}
}
 
// This code is contributed by SURENDRA_GANGWAR

Javascript

// Javascript program to implement
// the above approach
 
// Function that finds the sum of
    // products of all subarray of arr[]
    function sumOfSubarrayProd(arr, n)
    {
        // Stores sum of all subarrays
        let ans = 0;
        let res = 0;
  
        // Iterate array from behind
        for (let i = n - 1; i >= 0; i--) {
            let incr = arr[i] * (1 + res);
  
            // Update the ans
            ans += incr;
  
            // Update the res
            res = incr;
        }
  
        // Prlet the final sum
        document.write(ans);
    }
 
    // Driver Code
     
     // Given array arr[]
        let arr = [ 1, 2, 3 ];
  
        // Size of array
        let N = arr.length;
  
        // Function Call
        sumOfSubarrayProd(arr, N);
      

输出：

20

时间复杂度： O(N)
辅助空间： O(1)

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