Producto de todos los elementos de un arreglo divisible por un número dado K

Dada una array que contiene N elementos y un número K. La tarea es encontrar el producto de todos los elementos de la array que son divisibles por K.

Ejemplos :  

Input : arr[] = {15, 16, 10, 9, 6, 7, 17}
        K = 3
Output : 810

Input : arr[] = {5, 3, 6, 8, 4, 1, 2, 9}
        K = 2
Output : 384 

La idea es recorrer la array y verificar los elementos uno por uno. Si un elemento es divisible por K, multiplique el valor de ese elemento con el producto hasta el momento y continúe este proceso hasta que se alcance el final de la array.
A continuación se muestra la implementación del enfoque anterior:  

// C++ program to find Product of all the elements
// in an array divisible by a given number K
 
#include <iostream>
using namespace std;
 
// Function to find Product of all the elements
// in an array divisible by a given number K
int findProduct(int arr[], int n, int k)
{
    int prod = 1;
 
    // Traverse the array
    for (int i = 0; i < n; i++) {
 
        // If current element is divisible by k
        // multiply with product so far
        if (arr[i] % k == 0) {
            prod *= arr[i];
        }
    }
 
    // Return calculated product
    return prod;
}
 
// Driver code
int main()
{
    int arr[] = { 15, 16, 10, 9, 6, 7, 17 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
 
    cout << findProduct(arr, n, k);
 
    return 0;
}

// C program to find Product of all the elements
// in an array divisible by a given number K
#include <stdio.h>
 
// Function to find Product of all the elements
// in an array divisible by a given number K
int findProduct(int arr[], int n, int k)
{
    int prod = 1;
 
    // Traverse the array
    for (int i = 0; i < n; i++) {
 
        // If current element is divisible by k
        // multiply with product so far
        if (arr[i] % k == 0) {
            prod *= arr[i];
        }
    }
 
    // Return calculated product
    return prod;
}
 
// Driver code
int main()
{
    int arr[] = { 15, 16, 10, 9, 6, 7, 17 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
    printf("%d",findProduct(arr, n, k));
 
    return 0;
}
 
// This code is contributed by kothavvsaakash.

// Java program to find Product of all the elements
// in an array divisible by a given number K
 
import java.io.*;
 
class GFG {
 
// Function to find Product of all the elements
// in an array divisible by a given number K
static int findProduct(int arr[], int n, int k)
{
    int prod = 1;
 
    // Traverse the array
    for (int i = 0; i < n; i++) {
 
        // If current element is divisible by k
        // multiply with product so far
        if (arr[i] % k == 0) {
            prod *= arr[i];
        }
    }
 
    // Return calculated product
    return prod;
}
 
// Driver code
    public static void main (String[] args) {
        int arr[] = { 15, 16, 10, 9, 6, 7, 17 };
    int n = arr.length;
    int k = 3;
 
    System.out.println(findProduct(arr, n, k));
    }
}
 
 
// This code is contributed by inder_verma..

# Python3 program to find Product of all
# the elements in an array divisible by
# a given number K
 
# Function to find Product of all the elements
# in an array divisible by a given number K
def findProduct(arr, n, k):
 
    prod = 1
 
    # Traverse the array
    for i in range(n):
 
        # If current element is divisible
        # by k, multiply with product so far
        if (arr[i] % k == 0):
            prod *= arr[i]
 
    # Return calculated product
    return prod
 
# Driver code
if __name__ == "__main__":
 
    arr= [15, 16, 10, 9, 6, 7, 17 ]
    n = len(arr)
    k = 3
 
    print (findProduct(arr, n, k))
 
# This code is contributed by ita_c

// C# program to find Product of all
// the elements in an array divisible
// by a given number K
using System;
 
class GFG
{
 
// Function to find Product of all
// the elements in an array divisible
// by a given number K
static int findProduct(int []arr, int n, int k)
{
    int prod = 1;
 
    // Traverse the array
    for (int i = 0; i < n; i++)
    {
 
        // If current element is divisible
        // by k multiply with product so far
        if (arr[i] % k == 0)
        {
            prod *= arr[i];
        }
    }
 
    // Return calculated product
    return prod;
}
 
// Driver code
public static void Main()
{
    int []arr = { 15, 16, 10, 9, 6, 7, 17 };
    int n = arr.Length;
    int k = 3;
     
    Console.WriteLine(findProduct(arr, n, k));
}
}
 
// This code is contributed by inder_verma

<?php
// PHP program to find Product of
// all the elements in an array
// divisible by a given number K
 
// Function to find Product of
// all the elements in an array
// divisible by a given number K
function findProduct(&$arr, $n, $k)
{
    $prod = 1;
 
    // Traverse the array
    for ($i = 0; $i < $n; $i++)
    {
 
        // If current element is divisible 
        // by k multiply with product so far
        if ($arr[$i] % $k == 0)
        {
            $prod *= $arr[$i];
        }
    }
 
    // Return calculated product
    return $prod;
}
 
// Driver code
$arr = array(15, 16, 10, 9, 6, 7, 17 );
$n = sizeof($arr);
$k = 3;
 
echo (findProduct($arr, $n, $k));
 
// This code is contributed
// by Shivi_Aggarwal
?>

<script>
// Function to find Product of all the elements
// in an array divisible by a given number K
function findProduct( arr, n,  k)
{
    var prod = 1;
 
    // Traverse the array
    for (var i = 0; i < n; i++) {
 
        // If current element is divisible by k
        // multiply with product so far
        if (arr[i] % k == 0) {
            prod *= arr[i];
        }
    }
 
    // Return calculated product
    return prod;
}
 
var arr = [15, 16, 10, 9, 6, 7, 17 ];
     
 
    document.write(findProduct(arr, 7, 3));
 
 
 
</script>

810

Complejidad de tiempo : O(N), donde N es el número de elementos en la array.
Espacio Auxiliar: O(1)