Product of all prime numbers in an Array in C++

Given an integer array arr[] with some elements, the task is to find the product of all the prime number of that numbers.

Prime number are those number which are either divided by 1 or the number itself, or a prime number is a number which is not divisible by any other number except 1 and the number itself. Like 1, 2, 3, 5, 7, 11, etc.

we have to find the solution for the given array −

Input −arr[] = { 11, 20, 31, 4, 5, 6, 70 }

Output − 1705

Explanation − Prime numbers in array are − 11, 31, 5 their product is 1705

Input − arr[] = { 1, 2, 3, 4, 5, 6, 7 }

Output − 210

Explanation − Prime numbers in array are − 1, 2, 3, 5, 7 their product is 210

Approach used below is as follows to solve the problem

• Take input array arr[].

• Loop every element and check if it is prime or not.

• Product all the present primes in an array.

• Return the product.

Algorithm

Start
In function int prodprimearr(int arr[], int n)
   Step 1→ Declare and initialize max_val as max_val *max_element(arr, arr + n)
   Step 2→ Declare vector<bool> isprime(max_val + 1, true)
   Step 3→ Set isprime[0] and isprime[1] as false
   Step 4→ Loop For p = 2 and p * p <= max_val and p++
      If isprime[p] == true then,
         Loop For i = p * 2 and i <= max_val and i += p
            Set isprime[i] as false
   Step 5→ Set prod as 1
   Step 6→ For i = 0 and i < n and i++
      If isprime[arr[i]]
         Set prod = prod * arr[i]
   Step 6→ Return prod
In function int main(int argc, char const *argv[])
   Step 1→ Declare and initilalize arr[] = { 11, 20, 31, 4, 5, 6, 70 }
   Step 2→ Declare and initialize n = sizeof(arr) / sizeof(arr[0])
   Step 3→ Print the results of prodprimearr(arr, n)
Stop

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
int prodprimearr(int arr[], int n){
   // To find the maximum value of an array
   int max_val = *max_element(arr, arr + n);
   // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
   // THAN OR EQUAL TO max_val
   vector<bool> isprime(max_val + 1, true);
   isprime[0] = false;
   isprime[1] = false;
   for (int p = 2; p * p <= max_val; p++) {
      // If isprime[p] is not changed, then
      // it is a prime
      if (isprime[p] == true) {
         // Update all multiples of p
         for (int i = p * 2; i <= max_val; i += p)
            isprime[i] = false;
      }
   }
   // Product all primes in arr[]
   int prod = 1;
   for (int i = 0; i < n; i++)
      if (isprime[arr[i]])
         prod *= arr[i];
      return prod;
}
int main(int argc, char const *argv[]){
   int arr[] = { 11, 20, 31, 4, 5, 6, 70 };
   int n = sizeof(arr) / sizeof(arr[0]);
   cout << prodprimearr(arr, n);
   return 0;
}

Output

If run the above code it will generate the following output −

1705