给定数组中所有有序对积的总和

给定一个大小为N的数组arr[] ，任务是找到可以从给定数组元素生成的有序对的所有乘积的总和。
例子：

Input: arr[] ={1, 2, 3}
Output: 36
Explanation:All possible pairs are {(1, 1), {1, 2}, {1, 3}, {2, 1}, {2, 2}, {2, 3}, {3, 1}, {3, 2}, {3, 3}}. Therefore, the sum of product of all pairs is 36.

Input : arr[]={3, 4, 1, 2, 5}
Output: 225 

朴素的方法：最简单的方法是遍历给定数组中的所有可能对，计算所有对产品的乘积之和。

时间复杂度： O(N ² )
辅助空间： O(1)

高效的方法：上述方法可以基于以下观察进行优化：

arr[]={a₁, a₂, a₃, a₄ ….., a_n-1, a_n}
Now, sum of product of all  possible pairs is =
{
(a_{1 *} a₁) + (a_{1 *} a₂) + (a_{1 *} a₃) + …..+ (a_{1 *} a_n-1) + (a₁, a_n) +
(a_{2 *} a₁) + (a_{2 *} a₂) + (a_{2 *} a₃) + ….. + (a_{2 *} a_n-1) + (a_{2 *} a_n) +
(a_{3 *} a₁) + (a_{3 *} a₂) + (a_{3 *} a₃) + ….. + (a_{3 *} a_n-1) + (a₃, a_n) +
…………………………………………………………………………..
(a_{n, *} a₁) + (a_{n *} a₂), +(a_{n *} a₃) + ….. + (a_{n *} a_n-1) + (a_n, a_n)
}

={
(a₁ + a₂+  a₃ + …..+ a_n-1 +  a_n ) _* (a₁ + a₂+  a₃ + …..+ a_n-1 +  a_n ) 
}
=(a₁ + a₂+  a₃ + …..+ a_n-1 +  a_n )²

请按照以下步骤解决问题：

1. 初始化变量， res=0存放数组元素的总和
2. 遍历数组arr[]并将数组的每个元素添加到res 。
3. 最后，打印res的平方作为所需答案。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to calculate the
// sum of all pair-products
int sumOfProd(int arr[], int N)
{
    // Stores sum of array
    int sum = 0;
 
    for (int i = 0; i < N; i++) {
 
        // Update sum of the array
        sum += arr[i];
    }
 
    return sum * sum;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 3, 1, 5, 4 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << sumOfProd(arr, N);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to calculate the
// sum of all pair-products
static int sumOfProd(int arr[], int N)
{
     
    // Stores sum of array
    int sum = 0;
 
    for(int i = 0; i < N; i++)
    {
         
        // Update sum of the array
        sum += arr[i];
    }
    return sum * sum;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 3, 1, 5, 4 };
    int N = arr.length;
     
    System.out.print(sumOfProd(arr, N));
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program to implement
# the above approach 
 
# Function to calculate the
# sum of all pair-products
def sumOfProd(arr, N):
     
    # Stores sum of array
    sum = 0
 
    for i in range(N):
         
        # Update sum of the array
        sum += arr[i]
 
    return sum * sum
 
# Driver Code
if __name__ == '__main__':
     
    arr = [ 2, 3, 1, 5, 4 ]
    N = len(arr)
     
    print(sumOfProd(arr, N))
 
# This code is contributed by SURENDRA_GANGWAR

C#

// C# program to implement 
// the above approach  
using System;
 
class GFG{
     
// Function to calculate the 
// sum of all pair-products 
static int sumOfProd(int[] arr, int N) 
{ 
     
    // Stores sum of array 
    int sum = 0; 
   
    for(int i = 0; i < N; i++)
    {
         
        // Update sum of the array 
        sum += arr[i]; 
    } 
    return sum * sum; 
}  
 
// Driver code
static void Main()
{
    int[] arr = { 2, 3, 1, 5, 4 }; 
    int N = arr.Length; 
     
    Console.WriteLine(sumOfProd(arr, N));
}
}
 
// This code is contributed by divyeshrabadiya07

Javascript

 
// JavaScript program for the above approach
 
// Function to calculate the
// sum of all pair-products
function sumOfProd(arr, N)
{
      
    // Stores sum of array
    let sum = 0;
  
    for(let i = 0; i < N; i++)
    {
          
        // Update sum of the array
        sum += arr[i];
    }
    return sum * sum;
}
  
// Driver Code
 
       let arr = [ 2, 3, 1, 5, 4 ];
    let N = arr.length;
      
    document.write(sumOfProd(arr, N));
 
// This code is contributed by souravghosh0416.

输出：

225

时间复杂度： O(N)

辅助空间： O(1)

如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live