大数的最大质数

Question

I did a Project Euler question - here

I am at the end of my wits. I am getting wrong solution. I am perplexed as to "why" particularly this code is not working.

Relevant - this

This was my generated answer.

long long x; // The max prime factor
long long n; // The number to be factored
while (n % 2 == 0)
{
    n /= 2;
}
x = 2;
while (n % 3 == 0)
{
    n /= 3;
}
x = 3;
for (int i = 5; i <= sqrt(n); i += 2)
{
    while (n % i == 0)
    {
        n /= i;
    }
    x = i;
}
std::cout << x;

n = 600851475143 answer = 6857>! I am getting x = 1471

Answer 1

At least this for loop

for (int i = 5; i <= sqrt(n); i += 2)
{
    while (n % i == 0)
    {
        n /= i;
    }
    x = i;
}

is wrong, The variable x gets the value of the last i that is less than or equal to sqrt( n ) .

Consider for example n equal to 22. After dividing it by 2 you will get n equal to 11.

After this code snippet

while (n % 3 == 0)
{
    n /= 3;
}
x = 3;

x will be equal to 3 and the for loop will be skipped due to its condition that evaluates to false.

The code can look for example the following way

    long long int n = 600851475143;

    long long int prime_factor = 0;

    if (n % 2 == 0)
    {
        prime_factor = 2;

        while (n % 2 == 0 ) n /= 2;
    }

    for (long long int i = 3; i <= n / i; i += 2)
    {
        if (n % i == 0)
        {
            prime_factor = i;
            while (n % i == 0) n /= i;
        }
    }

    if (n != 1) prime_factor = n;

    std::cout << "prime factor = " << prime_factor << '\n';

Also you should use the unsigned type unsigned long long int instead of the signed type long long int . Otherwise you will need to write code that will take into account the sign of the source number.

You could write a separate function as shown in the demonstration program below

#include <iostream>

unsigned long long max_prime_factor( unsigned long long int n )
{
    unsigned long long int prime_factor = 0;

    if (not ( n < 2 ))
    {
        if (n % 2 == 0)
        {
            prime_factor = 2;

            while (n % 2 == 0) n /= 2;
        }

        for (unsigned long long int i = 3; i <= n / i; i += 2)
        {
            if (n % i == 0)
            {
                prime_factor = i;
                while (n % i == 0) n /= i;
            }
        }
    }

    return n < 2 ? prime_factor : n;
}

int main()
{
    for (unsigned int i = 0; i < 20; i++)
    {
        std::cout << i << " -> " << max_prime_factor( i ) << '\n';
    }
}

The program output is

0 -> 0
1 -> 0
2 -> 2
3 -> 3
4 -> 2
5 -> 5
6 -> 3
7 -> 7
8 -> 2
9 -> 3
10 -> 5
11 -> 11
12 -> 3
13 -> 13
14 -> 7
15 -> 5
16 -> 2
17 -> 17
18 -> 3
19 -> 19

Answer 2

int main() {
  long long x = 0;            // The max prime factor
  long long n = 600851475143; // The number to be factored
  long long a = sqrt(n);
  while (n % 2 == 0) {
    n /= 2;
  }
  x = 2;
  while (n % 3 == 0) {
    n /= 3;
  }
  x = 3;
  for (int i = 5; i <= a; i += 2) {
    while (n % i == 0) {
      n /= i;
      x = i;
    }
  }
  
  std::cout << x << std::endl;
}

Here it works, the sqrt(n) stoped the for loop too soon so i put it in a long long, and the x = i; was in the wrong place