Given an array, find a product of all array elements.
Examples :
Input : ar[] = {1, 2, 3, 4, 5}
Output : 120
Product of array elements is 1 x 2
x 3 x 4 x 5 = 120.
Input : ar[] = {1, 6, 3}
Output : 18
C
#include <stdio.h>
int product(int ar[], int n)
{
int result = 1;
for (int i = 0; i < n; i++)
result = result * ar[i];
return result;
}
int main()
{
int ar[] = { 1, 2, 3, 4, 5 };
int n = sizeof(ar) / sizeof(ar[0]);
printf("%d", product(a, n));
return 0;
}
|
Java
class GFG{
static int product(int ar[], int n)
{
int result = 1;
for (int i = 0; i < n; i++)
result = result * ar[i];
return result;
}
public static void main(String[] args)
{
int ar[] = { 1, 2, 3, 4, 5 };
int n = ar.length;
System.out.printf("%d", product(ar, n));
}
}
|
Python3
def product(ar, n):
result = 1
for i in range(0, n):
result = result * ar[i]
return result
ar = [ 1, 2, 3, 4, 5 ]
n = len(ar)
print(product(ar, n))
|
C#
using System;
class GFG {
static int product(int []ar, int n)
{
int result = 1;
for (int i = 0; i < n; i++)
result = result * ar[i];
return result;
}
public static void Main()
{
int []ar = { 1, 2, 3, 4, 5 };
int n = ar.Length;
Console.WriteLine(product(ar, n));
}
}
|
PHP
<?php
function product($ar, $n)
{
$result = 1;
for ($i = 0; $i < $n; $i++)
$result = $result * $ar[$i];
return $result;
}
$ar = array( 1, 2, 3, 4, 5 );
$n = count($ar);
print((int)product($ar, $n));
?>
|
Javascript
<script>
function product(ar,n)
{
let result = 1;
for (let i = 0; i < n; i++)
result = result * ar[i];
return result;
}
let ar = [ 1, 2, 3, 4, 5 ];
let n = ar.length;
document.write(parseInt(product(ar, n)));
</script>
|
Output :
120
The above code may cause overflow. Therefore, it is always desired to compute product under modulo. The reason for its working is the simple distributive property of modulo.
( a * b) % c = ( ( a % c ) * ( b % c ) ) % c
Below is a program to find and print the product of all the number in this array of Modulo (10^9 +7).
C
#include <stdio.h>
const int MOD = 1000000007;
int product(int ar[], int n)
{
int result = 1;
for (int i = 0; i < n; i++)
result = (result * ar[i]) % MOD;
return result;
}
int main()
{
int ar[] = { 1, 2, 3, 4, 5 };
int n = sizeof(ar) / sizeof(ar[0]);
printf("%d", product(ar, n));
return 0;
}
|
Java
class GFG {
static final int MOD = 1000000007;
static int product(int ar[], int n)
{
int result = 1;
for (int i = 0; i < n; i++)
result = (result * ar[i]) % MOD;
return result;
}
public static void main(String[] args)
{
int ar[] = { 1, 2, 3, 4, 5 };
int n = ar.length;
System.out.printf("%d", product(ar, n));
}
}
|
Python3
MOD = 1000000007
def product(ar, n):
result = 1
for i in range(0, n):
result = (result * ar[i]) % MOD
return result
ar = [1, 2, 3, 4, 5]
n = len(ar)
print(product(ar, n))
|
C#
using System;
class GFG {
static int MOD = 1000000007;
static int product(int []ar, int n)
{
int result = 1;
for (int i = 0; i < n; i++)
result = (result * ar[i]) % MOD;
return result;
}
public static void Main()
{
int []ar = { 1, 2, 3, 4, 5 };
int n = ar.Length;
Console.WriteLine(product(ar, n));
}
}
|
PHP
<?php
function product($ar, $n)
{
$result = 1;
for ($i = 0; $i < $n; $i++)
$result = ($result *
$ar[$i]) % 1000000007;
return $result;
}
$ar = array( 1, 2, 3, 4, 5 );
$n = count($ar);
print(product($ar, $n));
?>
|
Javascript
<script>
let MOD = 1000000007;
function product(ar, n)
{
let result = 1;
for (let i = 0; i < n; i++)
result = (result * ar[i]) % MOD;
return result;
}
let ar = [ 1, 2, 3, 4, 5 ];
let n = ar.length;
document.write(product(ar, n));
</script>
|
Output :
120
This article is contributed by Shivani Baghel. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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