5 replies to this topic

[Cruel Hand]

I need to write a program that finds the prime factors of a certain number. I have the code for finding the factors of a number, but I don't know how to single out the numbers and determine if they're prime. This is the code I have thus far:

public class PrimeFactor {

	public static void main(String[] args){

		double i = 400, factor = 0;

		for(double counter = 2; counter <= i; counter++){

			if(i % counter == 0){

				factor = i / counter;

				System.out.println(factor);

				}

			}

		}

	}

can anyone point me in the right direction for my next step?

Thanks :D

Edited by Cruel Hand, 26 November 2011 - 02:17 PM.
mis-spelled word

[fread]

If you working with large numbers this may not be very efficient. Best to look up sieve of eratosthenes. This is what i used:

public int isPrime(int x)
{ 
    if( x == 1 ) return 1; /* if you are considering one as prime */ 
    if( x == 2 ) return 1; 
    if( x %  2 == 0) return 0; 
    for(int i = 3; i * i <= x; i = i + 2) /* increment by two; only capture odd numbers
      if( x % i == 0) return 0;   
    return 1; 
}

[WingedPanther]

If you change the if to a while, you can divide i by counter so that you divide out all the prime factors as you find them, thus eliminating the chance of composites.

[Cruel Hand]

@Fread, idk what the # means in java

@wingedpanther, do you mean like this?

while(i % counter == 0){

				factor = i / counter;

				System.out.println(factor);

				}

or did I misunderstand you?

as you can both tell, my skills in java are novice at best haha

[fread]

Quote

@Fread, idk what the # means in java

That is just x mod 2. The percent sign prints out like that sometimes if there is no space after it.

[WingedPanther]

That's exactly what I meant.