# 算法：如何计算乘积和总和相等的子数组的个数？

2021年3月28日15:46:22 发表评论 464 次浏览

## 本文概述

``````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``````

## C ++

``````// C++ program to count subarrays with
// same sum and product.
#include<bits/stdc++.h>
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<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 function
int main()
{
int arr[] = {1, 3, 2};
int n = sizeof (arr)/ sizeof (arr[0]);
cout << numOfsubarrays(arr , n);
return 0;
}``````

## 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<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 function
public static void main(String args[])
{
int arr[] = { 1 , 3 , 2 };
int n = arr.length;
System.out.println(numOfsubarrays(arr , n));
}
}``````

## 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.``````

## 的PHP

``````<?php
// PHP program to count subarrays
// with same sum and product.

// function returns required
// number of subarrays
function numOfsubarrays( \$arr , \$n )
{
// Initialize result
\$count = 0;

// checking each subarray
for ( \$i = 0; \$i < \$n ; \$i ++)
{
\$product = \$arr [ \$i ];
\$sum = \$arr [ \$i ];
for ( \$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
\$arr = array (1, 3, 2);
\$n = sizeof( \$arr );
echo (numOfsubarrays( \$arr , \$n ));

// This code is contributed by Ajit.
?>``````

``4``