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.
#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++) {
if (n % i == 0) {
bool isPrime = true ;
for ( int j = 2; j <= i / 2; j++) {
if (i % j == 0) {
isPrime = false ;
break ;
}
}
if (isPrime) {
product = product * i;
}
}
}
return product;
}
int main()
{
int n = 44;
cout << productPrimeFactors(n);
return 0;
}
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Method 2 (Efficient)
The idea is based on Efficient program to print all prime factors of a given number
#include <bits/stdc++.h>
using namespace std;
long long int productPrimeFactors( int n)
{
long long int product = 1;
if (n % 2 == 0) {
product *= 2;
while (n % 2 == 0)
n = n / 2;
}
for ( int i = 3; i <= sqrt (n); i = i + 2) {
if (n % i == 0) {
product = product * i;
while (n % i == 0)
n = n / i;
}
}
if (n > 2)
product = product * n;
return product;
}
int main()
{
int n = 44;
cout << productPrimeFactors(n);
return 0;
}
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Please refer complete article on Product of unique prime factors of a number for more details!
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