Gegeben sei ein Array arr[] aus N positiven ganzen Zahlen. Die Aufgabe besteht darin, ein Programm zu schreiben, um das Produkt aller Primzahlen des gegebenen Arrays zu finden.
Beispiele : 
 

Eingabe : arr[] = {1, 3, 4, 5, 7} 
Ausgabe : 105 
Es gibt drei Primzahlen, 3, 5 und 7, deren Produkt = 105 ist.
Eingabe : arr[] = {1, 2, 3, 4, 5, 6, 7} 
Ausgabe : 210 
 

Naiver Ansatz: Eine einfache Lösung besteht darin, das Array zu durchlaufen und für jedes Element zu prüfen, ob es eine Primzahl ist oder nicht, und gleichzeitig das Produkt des Primzahlelements zu berechnen.
Effizienter Ansatz: Erzeuge alle Primzahlen bis zum maximalen Element des Arrays mit dem Sieb von Eratosthenes und speichere sie in einem Hash. Durchqueren Sie nun das Array und finden Sie das Produkt der Elemente, die Primzahlen sind, mit dem Sieb.
Unten ist die Implementierung des obigen Ansatzes: 
 

// 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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