getting prime factorial of an integer

[Sheilani]

Can anyone please help me with the code of prime factorization. Example input:135
The problem is I want to be an output of (3^3) (5^1) instead of 3,3,3,5.
here's the code:

#include <iostream>
#include <cstdlib>
#include <cmath>
using namespace std;

void get_divisors(int n);

int main() 
{
    int n = 0;

    cout << "Enter a number:";
    cin >> n;
    get_divisors(n);
    cout << endl; 
}

void get_divisors(int n)
 {
     int i;
     double sqrt_of_n = sqrt(n);

     for (i = 2; i <= sqrt_of_n; i++)
         if (n % i == 0) 
		 {
            cout << i << ", "; 
            get_divisors(n / i);
            return; 
                 }


     cout << n;
 }

[Nepomuk]

The reason you are only getting a list of factors is that you calculate one by one without remembering your last result. There are a few possible solutions to this but the one I'd suggest (as it needs the least changes in your code) is passing two further arguments to your get_divisors function: The last factor found and the number of times you've found it so far. Then do the output depending on those values.

[Sheilani]

thank for your reply,but i dont know how to do it.

[Nepomuk]

OK, here are a few hints:

• The signature of the changed function will be something like
  

  void get_divisors(int n, int lastFactor, int power)

• The initial call should be something like

  get_divisors(n, 2, 0)

• You can reduce the number of calls in your for-loop given the new parameters