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C/C++ Program to find Product of unique prime factors of a number
  • Last Updated : 05 Dec, 2018

Given a number n, we need to find the product of all of its unique prime factors. Prime factors: It is basically a factor of the number that is a prime number itself.

Examples:

Input: num = 10
Output: Product is 10
Explanation:
Here, the input number is 10 having only 2 prime factors and they are 5 and 2.
And hence their product is 10.

Input : num = 25
Output: Product is 5
Explanation:
Here, for the input to be 25  we have only one unique prime factor i.e 5.
And hence the required product is 5.

Method 1 (Simple)
Using a loop from i = 2 to n and check if i is a factor of n then check if i is prime number itself if yes then store product in product variable and continue this process till i = n.




// C++ program to find product of
// unique prime factors of a number
#include <bits/stdc++.h>
using namespace std;
  
long long int productPrimeFactors(int n)
{
    long long int product = 1;
  
    for (int i = 2; i <= n; i++) {
  
        // Checking if 'i' is factor of num
        if (n % i == 0) {
  
            // Checking if 'i' is a Prime number
            bool isPrime = true;
            for (int j = 2; j <= i / 2; j++) {
                if (i % j == 0) {
                    isPrime = false;
                    break;
                }
            }
  
            // condition if 'i' is Prime number
            // as well as factor of num
            if (isPrime) {
                product = product * i;
            }
        }
    }
  
    return product;
}
  
// driver function
int main()
{
    int n = 44;
    cout << productPrimeFactors(n);
    return 0;
}
Output:
22

Method 2 (Efficient)
The idea is based on Efficient program to print all prime factors of a given number






// C++ program to find product of
// unique prime factors of a number
#include <bits/stdc++.h>
using namespace std;
  
// A function to print all prime factors of
// a given number n
long long int productPrimeFactors(int n)
{
    long long int product = 1;
  
    // Handle prime factor 2 explicitly so that
    // can optimally handle other prime factors.
    if (n % 2 == 0) {
        product *= 2;
        while (n % 2 == 0)
            n = n / 2;
    }
  
    // n must be odd at this point. So we can
    // skip one element (Note i = i +2)
    for (int i = 3; i <= sqrt(n); i = i + 2) {
        // While i divides n, print i and
        // divide n
        if (n % i == 0) {
            product = product * i;
            while (n % i == 0)
                n = n / i;
        }
    }
  
    // This condition is to handle the case when n
    // is a prime number greater than 2
    if (n > 2)
        product = product * n;
  
    return product;
}
  
// driver function
int main()
{
    int n = 44;
    cout << productPrimeFactors(n);
    return 0;
}
Output:
22

Please refer complete article on Product of unique prime factors of a number for more details!

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