数组中所有素数的乘积

给定一个包含 N 个正整数的数组 arr[]。任务是编写一个程序来找到给定数组的所有素数的乘积。
例子：

Input: arr[] = {1, 3, 4, 5, 7}
Output: 105
There are three primes, 3, 5 and 7 whose product = 105.
Input: arr[] = {1, 2, 3, 4, 5, 6, 7}
Output: 210

编程需要懂一点英语

朴素方法：一个简单的解决方案是遍历数组并不断检查每个元素是否为素数，同时计算素数元素的乘积。
有效方法：使用 Eratosthenes 的筛子生成直到数组最大元素的所有素数，并将它们存储在哈希中。现在遍历数组并使用筛子找到那些素数元素的乘积。
下面是上述方法的实现：

C++

// CPP program to find product of
// primes in given array.
#include 
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 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

// 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 prime = new Vector(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

# 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#

// 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 prime = new List(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

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