Divisors problem

[catawar2]

So I am a bit confused about how to find the prime divisors of a given number.
I figured I could start with 2 and go all the way up to sqrt(number).But now here is the problem, for big numbers is seems to work(or it doesn't, i can't be sure),but I am certain that is doesn't work for small numbers.
Ex:The divisors of 42 are:1 2 3 6 7 14 21 42 , but sqrt(42)=6.4 so it doesn't get to 7.
The code would look like this:

#include <fstream>
#include <math.h>
using namespace std;
int main()
{
    ifstream f ("data.in");
    ofstream g ("data.out");
    int n,i,j;
    bool prime=true;
    f>>n;
    for(i=2;i<=sqrt(n);i++)
    {prime=true;
     if(n%i==0)
     {
        for(j=2;j<=sqrt(i);j++)
        {
            if(i%j==0) {prime=false; break;}
        }
        if(prime==true) g<<i<<" ";
     }
    }
}

Where should I stop?Or is there a better way to find them?Can someone please tell me the fastest way to do it?

[zwA4jE8b]

you could use ceiling(sqrt(number)) that way it rounds up

or add .5 to it

[hamsterman]

Your method is wrong. You need to go to sqrt(n) when you're looking for all divisors. Then if n % i == 0 you find two divisors: i and n/i. So if i can go up to 6 it will find 42/6 = 7.

Consider number 26 = 2*13. 13 is prime, but it is nowhere near sqrt(26) ~ 5.

What you need to do, is go from 2 to n and every time you find n % i == 0 ,do n = n/i.
Note that a single prime divisor may go several times.

[catawar2]

Ok, I get it,thx for the help.

[catawar2]

Ok, I get it,thx for the help.