product 和 sum 相等的子数组的数量

给定一个包含 n 个数字的数组。我们需要计算乘积和元素总和相等的子数组的数量

例子：

Input  : arr[] = {1, 3, 2}
Output : 4
The subarrays are :
[0, 0] sum = 1, product = 1,
[1, 1] sum = 3, product = 3,
[2, 2] sum = 2, product = 2 and 
[0, 2] sum = 1+3+2=6, product = 1*3*2 = 6

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

这个想法很简单，我们检查每个子数组的乘积和其元素的总和是否相等。如果是，则将计数器变量增加 1

C++

// C++ program to count subarrays with
// same sum and product.
#include
using namespace std;
 
// returns required number of subarrays
int numOfsubarrays(int arr[] , int n)
{
    int count = 0; // Initialize result
 
    // checking each subarray
    for (int i=0; i

---

Java

// Java program to count subarrays with
// same sum and product.
 
class GFG
{
    // returns required number of subarrays
    static int numOfsubarrays(int arr[] , int n)
    {
        int count = 0; // Initialize result
      
        // checking each subarray
        for (int i=0; i

---

Python3

# python program to
# count subarrays with
# same sum and product.
 
# returns required
# number of subarrays
def numOfsubarrays(arr,n):
 
    count = 0 # Initialize result
  
    # checking each subarray
    for i in range(n):
     
        product = arr[i]
        sum = arr[i]
        for j in range(i+1,n):
         
            # checking if product is equal
            # to sum or not
            if (product==sum):
                count+=1
  
            product *= arr[j]
            sum += arr[j]
         
  
        if (product==sum):
            count+=1
     
    return count
 
# Driver code
 
arr = [1,3,2]
n =len(arr)
print(numOfsubarrays(arr , n))
 
# This code is contributed
# by Anant Agarwal.

C#

// C# program to count subarrays
// with same sum and product.
using System;
class GFG {
     
    // returns required number
    // of subarrays
    static int numOfsubarrays(int []arr ,
                              int n)
    {
         
        // Initialize result
        int count = 0;
     
        // checking each subarray
        for (int i = 0; i < n; i++)
        {
            int product = arr[i];
            int sum = arr[i];
            for (int j = i + 1; j < n; j++)
            {
                 
                // checking if product is
                // equal to sum or not
                if (product == sum)
                    count++;
     
                product *= arr[j];
                sum += arr[j];
            }
     
            if (product == sum)
                count++;
        }
        return count;
    }
     
    // Driver Code
    public static void Main()
    {
        int []arr = {1,3,2};
        int n = arr.Length;
        Console.Write(numOfsubarrays(arr , n));
    }
}
 
// This code is contributed by Nitin Mittal.

Javascript

 
// Javascript program to count subarrays
// with same sum and product.
 
// Returns required number
// of subarrays
function numOfsubarrays(arr, n)
{
     
    // Initialize result
    let count = 0;
   
    // Checking each subarray
    for(let i = 0; i < n; i++)
    {
        let product = arr[i];
        let sum = arr[i];
         
        for(let j = i + 1; j < n; j++)
        {
             
            // Checking if product is
            // equal to sum or not
            if (product == sum)
                count++;
   
            product *= arr[j];
            sum += arr[j];
        }
   
        if (product == sum)
            count++;
    }
    return count;
}
 
// Driver code
let arr = [ 1, 3, 2 ];
let n = arr.length;
 
document.write(numOfsubarrays(arr, n));
 
// This code is contributed by decode2207
 

输出：

4

时间复杂度：O(n 2 )