Producto de todos los números primos en un Array

Dada una array arr[] de N enteros positivos. La tarea es escribir un programa para encontrar el producto de todos los números primos de la array dada.
Ejemplos : 
 

Entrada : arr[] = {1, 3, 4, 5, 7} 
Salida : 105 
Hay tres primos, 3, 5 y 7 cuyo producto = 105.
Entrada : arr[] = {1, 2, 3, 4, 5, 6, 7} 
Salida : 210 
 

Enfoque ingenuo: una solución simple es atravesar la array y seguir verificando cada elemento si es primo o no y calcular el producto del elemento primo al mismo tiempo.
Enfoque eficiente: genere todos los números primos hasta el elemento máximo de la array utilizando el tamiz de Eratóstenes y guárdelos en un hash. Ahora recorra la array y encuentre el producto de aquellos elementos que son primos usando el tamiz.
A continuación se muestra la implementación del enfoque anterior: 
 

// CPP program to find product of
// primes in given array.
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the product of prime numbers
// in the given array
int primeProduct(int arr[], int n)
{
    // Find maximum value in the array
    int max_val = *max_element(arr, arr + n);
 
    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
    // THAN OR EQUAL TO max_val
    // Create a boolean array "prime[0..n]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    vector<bool> prime(max_val + 1, true);
 
    // Remaining part of SIEVE
    prime[0] = false;
    prime[1] = false;
    for (int p = 2; p * p <= max_val; p++) {
 
        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p] == true) {
 
            // Update all multiples of p
            for (int i = p * 2; i <= max_val; i += p)
                prime[i] = false;
        }
    }
 
    // Product all primes in arr[]
    int prod = 1;
    for (int i = 0; i < n; i++)
        if (prime[arr[i]])
            prod *= arr[i];
 
    return prod;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << primeProduct(arr, n);
 
    return 0;
}

// Java program to find product of
// primes in given array.
import java.util.*;
 
class GFG
{
 
// Function to find the product of prime numbers
// in the given array
static int primeProduct(int arr[], int n)
{
    // Find maximum value in the array
    int max_val = Arrays.stream(arr).max().getAsInt();
 
    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
    // THAN OR EQUAL TO max_val
    // Create a boolean array "prime[0..n]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    Vector<Boolean> prime = new Vector<Boolean>(max_val + 1);
    for(int i = 0; i < max_val + 1; i++)
        prime.add(i, Boolean.TRUE);
 
    // Remaining part of SIEVE
    prime.add(0, Boolean.FALSE);
    prime.add(1, Boolean.FALSE);
    for (int p = 2; p * p <= max_val; p++)
    {
 
        // If prime[p] is not changed, then
        // it is a prime
        if (prime.get(p) == true)
        {
 
            // Update all multiples of p
            for (int i = p * 2; i <= max_val; i += p)
                prime.add(i, Boolean.FALSE);
        }
    }
 
    // Product all primes in arr[]
    int prod = 1;
    for (int i = 0; i < n; i++)
        if (prime.get(arr[i]))
            prod *= arr[i];
 
    return prod;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int n = arr.length;
 
    System.out.print(primeProduct(arr, n));
}
}
 
// This code has been contributed by 29AjayKumar

# Python3 program to find product of
# primes in given array
import math as mt
 
# function to find the product of prime
# numbers in the given array
def primeProduct(arr, n):
     
    # find the maximum value in the array
    max_val = max(arr)
     
    # USE SIEVE TO FIND ALL PRIME NUMBERS
    # LESS THAN OR EQUAL TO max_val
    # Create a boolean array "prime[0..n]". A
    # value in prime[i] will finally be false
    # if i is Not a prime, else true.
    prime = [True for i in range(max_val + 1)]
     
    # remaining part of SIEVE
    prime[0] = False
    prime[1] = False
     
    for p in range(mt.ceil(mt.sqrt(max_val))):
         
        # Remaining part of SIEVE
         
        # if prime[p] is not changed,
        # than it is prime
        if prime[p]:
             
            # update all multiples of p
            for i in range(p * 2, max_val + 1, p):
                prime[i] = False
     
    # product all primes in arr[]
    prod = 1
     
    for i in range(n):
        if prime[arr[i]]:
            prod *= arr[i]
     
    return prod
 
# Driver code
arr = [1, 2, 3, 4, 5, 6, 7]
 
n = len(arr)
 
print(primeProduct(arr, n))
 
# This code is contributed
# by Mohit kumar 29
                

// C# program to find product of
// primes in given array.
using System;
using System.Linq;
using System.Collections.Generic;
 
class GFG
{
 
// Function to find the product of prime numbers
// in the given array
static int primeProduct(int []arr, int n)
{
    // Find maximum value in the array
    int max_val = arr.Max();
 
    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
    // THAN OR EQUAL TO max_val
    // Create a boolean array "prime[0..n]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    List<bool> prime = new List<bool>(max_val + 1);
    for(int i = 0; i < max_val + 1; i++)
        prime.Insert(i, true);
 
    // Remaining part of SIEVE
    prime.Insert(0, false);
    prime.Insert(1, false);
    for (int p = 2; p * p <= max_val; p++)
    {
 
        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p] == true)
        {
 
            // Update all multiples of p
            for (int i = p * 2; i <= max_val; i += p)
                prime.Insert(i, false);
        }
    }
 
    // Product all primes in arr[]
    int prod = 1;
    for (int i = 0; i < n; i++)
        if (prime[arr[i]])
            prod *= arr[i];
 
    return prod;
}
 
// Driver code
public static void Main()
{
    int []arr = { 1, 2, 3, 4, 5, 6, 7 };
    int n = arr.Length;
 
    Console.Write(primeProduct(arr, n));
}
}
 
/* This code contributed by PrinciRaj1992 */

<?php
// PHP program to find product of
// primes in given array.
 
// Function to find the product of
// prime numbers in the given array
function primeProduct($arr, $n)
{
    // Find maximum value in the array
    $max_val = max($arr);
 
    // USE SIEVE TO FIND ALL PRIME NUMBERS
    // LESS THAN OR EQUAL TO max_val
    // Create a boolean array "prime[0..n]".
    // A value in prime[i] will finally be false
    // if i is Not a prime, else true.
    $prime = array_fill(0, $max_val + 1, True);
     
    // Remaining part of SIEVE
    $prime[0] = false;
    $prime[1] = false;
    for ($p = 2; $p * $p <= $max_val; $p++)
    {
 
        // If prime[p] is not changed,
        // then it is a prime
        if ($prime[$p] == true)
        {
 
            // Update all multiples of p
            for ($i = $p * 2;
                 $i <= $max_val; $i += $p)
                $prime[$i]= false;
        }
    }
 
    // Product all primes in arr[]
    $prod = 1;
    for ($i = 0; $i < $n; $i++)
        if ($prime[$arr[$i]])
            $prod *= $arr[$i];
 
    return $prod;
}
 
// Driver code
$arr = array(1, 2, 3, 4, 5, 6, 7);
$n = sizeof($arr);
 
echo(primeProduct($arr, $n));
 
// This code contributed by Code_Mech
?>

<script>
// Javascript program to find product of
// primes in given array.
 
// Function to find the product of
// prime numbers in the given array
function primeProduct(arr, n)
{
    // Find maximum value in the array
    let max_val = arr.sort((a, b) => b - a)[0];
 
    // USE SIEVE TO FIND ALL PRIME NUMBERS
    // LESS THAN OR EQUAL TO max_val
    // Create a boolean array "prime[0..n]".
    // A value in prime[i] will finally be false
    // if i is Not a prime, else true.
    let prime = new Array(max_val + 1).fill(true);
     
    // Remaining part of SIEVE
    prime[0] = false;
    prime[1] = false;
    for (let p = 2; p * p <= max_val; p++)
    {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == true)
        {
 
            // Update all multiples of p
            for (let i = p * 2;
                i <= max_val; i += p)
                prime[i]= false;
        }
    }
 
    // Product all primes in arr[]
    let prod = 1;
    for (let i = 0; i < n; i++)
        if (prime[arr[i]])
            prod *= arr[i];
 
    return prod;
}
 
// Driver code
let arr = new Array(1, 2, 3, 4, 5, 6, 7);
let n = arr.length;
 
document.write(primeProduct(arr, n));
 
// This code contributed by _Saurabh_Jaiswal.
</script>

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