如何对大数进行素因数分解？

Question

I'm trying to solve Project Euler third question but while my code works perfectly with small numbers when I try to use big number it doesn't give me any answer.

#include<iostream>

using std :: cout;
using std :: cin;
using std :: endl;

int main()
{
   long long int a = 0, bigPrime = 0, smallPrime = 2, prime = 0;
   cout << "Please enter a number...!" << endl;
   cin >> a;

   for(long long int i = 2 ; i < a ; i++)
   {
      for(long long int c = 2 ; c < i ; c++)
      {
         if(i % c != 0)
         {
            prime = i;
         }
         else
         {
            prime = 0;
            break;
         }
      }
      if(prime > 0 )
      {
         if(a % prime == 0)
         {
            bigPrime = prime;
         }
      }

  }


  cout << "The biggest prime is = " << bigPrime << endl;
  return 0;

}

That's my bad code :) i am using ubuntu linux and g++ what is wrong with my code and how can i improve it?

Answer 1

You can improve your program using one simple trick:

Every time you find a divisor d , divide your number by d .

That means that for every divisor found, your number gets smaller, making the remaining part easier to factor.

As a bonus, that means you don't need to be so careful about only using primes as divisors. Every time a divisor is found, it's the smallest divisor of the current number, and since it's the smallest divisor, it must be a prime. That saves a whole level of looping.

The factors are extracted in order from smallest to highest, so in the end what you have is the highest prime factor - the answer to this challenge.

This is not a fast algorithm, but 600851475143 is not a large number and this algorithm will factor it no problem.

For examle ( on ideone ):

for (long long int d = 2; d * d <= a; d++) {
    if (a % d == 0) {
        a /= d;
        d--; // this is to handle repeated factors
    }
}

I also used the old d * d <= a trick but you don't even need it here. It helps if the highest factor is high, and in this example it is not.

Answer 2

But the problem states that you just need to find the biggest prime of 600851475143, right? Why don't you just iterate from sqrt(600851475143) to 2 and return the first number that is a prime?

bool isPrime(uint64_t num)
{
  bool result = true;
  for(uint64_t i = 2; i < std::sqrt(num); ++i)
  {
    if(num % i == 0)
    {
      result = false;
      break;
    }
  }
  return result;
}

int main()
{
  uint64_t num = 600851475143;
  uint64_t i = std::sqrt(num);
  while (i > 1)
  {
    i--;
    if (num%i != 0) continue;
    if (isPrime(i))
    {
      break;
    }
  }
  std::cout << i << std::endl;
  return 0;
}

For sure it can be done faster, but it takes 10ms on my machine, so I guess it's not terrible.

Answer 3

#include <iostream>

using namespace std;

typedef long long ulong;

int main()
{
ulong num = 600851475143;
ulong div = num;
ulong p = 0;

ulong i = 2;
while (i * i <= div)
{
    if (div % i == 0)
    {
        div /= i;
        p = i;
    }
    else {
        i++;
    }
}

if (div > p) 
{ 
    p = div;
}

cout << p << endl;

return 0;
}