How do I prime factorize large numbers? - c++

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;
}