找到最大因素的程序

Question

So I wrote a code for finding the largest factor for my learning procces on Project Euler and it worked fine and I got the right answer. Tho I saw diffrent solutions and they were "longer" and looked more complicated. So my question is, is there anything my code is no taking care of for lets say different number or is there a better way to do this task? I am trying to learn different ways of solving problems to get better at programming so I wonder if I should have done that code in a different way.

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace prime_factor
{
  class Program
  {
    static void Main(string[] args)
    {
        long n = 600851475143;
        int largest = 0;
        for (int i = 1; i<=n; i++)
        {
            if (n%i == 0)
            {
                Console.WriteLine(i);
                n /= i;
                largest = i;

            }
        }
        Console.WriteLine(largest);
        Console.ReadLine();
    }
  }
}

Answer 1

If you divide n by f every time you find a factor, you will automatically find only prime factors. Here's pseudocode that you can translate to your favorite language:

function factors(n)
    f := 2; fs := []
    while f * f <= n
        if n % f == 0
            fs := fs ++ f
            n := n / f
        else
            f := f + 1
    fs := fs ++ n
    return fs

Here fs is a linked list of factors, and ++ is the list-append operation. You can see a Python implementation here .

There are better ways to find factors, but this is sufficient for the Project Euler problem that you are trying to solve. If you are interested in programming with prime numbers, you might enjoy this essay at my blog.

Answer 2

I like it - you get around checking if i is prime, which is the most complicated part! If it were not prime, you would have already divided out it's composite factors.

However, I think there are cases where a non-prime i can sneak in - where multiples of prime factors divide your n . Try n = 36 (since 36 == 2^2 * 3^2). I think your algorithm returns 6.

You just need to add a loop to see if i can divide n multiple times and I think it will fix this issue.