# Product of every K’th prime number in an array in C++

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Given an array arr[n] containing n prime numbers and k; the task is to find the product of every k’th prime number in an array.

Like, we have an array arr[] = {3, 5, 7, 11} and k = 2 so the prime number after every k i.e 5 and 11 we have to find their product which will be 5x11 = 55 and print the result as output.

What are prime numbers?

A prime number is a natural number which can’t be divided by any other number except 1 or the number itself. Some of the prime numbers are 2, 3, 5, 7, 11, 13 etc.

## Example

Input: arr[] = {3, 5, 7, 11, 13} k= 2
Output: 55
Explanation: every 2nd element of the array are 5 and 11; their product will be 55

Input: arr[] = {5, 7, 13, 23, 31} k = 3
Output: 13
Explanation: every 3rd element of an array is 13 so the output will be 13.

Approach we will be using to solve the above problem

• Take an input array of n elements and k, for finding product of every k’th element.
• Create a sieve for storing the prime numbers.
• Then we have to traverse the array and get the k’th element and multiply it with the product variable recursively for every k’th element.
• Print the product.

## Algorithm

Start
Step 1-> Define and initialize MAX 1000000
Step 2-> Define bool prime[MAX + 1]
Step 3-> In function createsieve()
Call memset(prime, true, sizeof(prime));
Set prime[1] = false
Set prime[0] = false
Loop For p = 2 and p * p <= MAX and p++
If prime[p] == true then,
For i = p * 2 and i <= MAX and i += p
Set prime[i] = false
Step 4-> void productOfKthPrimes(int arr[], int n, int k)
Set c = 0
Set product = 1
Loop For i = 0 and i < n and i++
If prime[arr[i]] then,
Increment c by 1
If c % k == 0 {
Set product = product * arr[i]
Set c = 0
Print the product
Step 5-> In function main()
Call function createsieve()
Set n = 5, k = 2
Set arr[n] = { 2, 3, 11, 13, 23 }
Call productOfKthPrimes(arr, n, k)
Stop

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
#define MAX 1000000
bool prime[MAX + 1];
void createsieve() {
memset(prime, true, sizeof(prime));
// 0 and 1 are not prime numbers
prime[1] = false;
prime[0] = false;
for (int p = 2; p * p <= MAX; p++) {
if (prime[p] == true) {
// finding all multiples of p
for (int i = p * 2; i <= MAX; i += p)
prime[i] = false;
}
}
}
void productOfKthPrimes(int arr[], int n, int k) {
// count the number of primes
int c = 0;
// find the product of the primes
long long int product = 1;
// traverse the array
for (int i = 0; i < n; i++) {
// if the number is a prime
if (prime[arr[i]]) {
c++;
if (c % k == 0) {
product *= arr[i];
c = 0;
}
}
}
cout << product << endl;
}
//main block
int main() {
// create the sieve
createsieve();
int n = 5, k = 2;
int arr[n] = { 2, 3, 11, 13, 23 };
productOfKthPrimes(arr, n, k);
return 0;
}

## Output

39
Updated on 23-Dec-2019 11:04:17