Problem:
Score : 400 points

Problem Statement

Given are positive integers A and B.

Let us choose some number of positive common divisors of A and B.

Here, any two of the chosen divisors must be coprime.

At most, how many divisors can we choose?

Definition of common divisor

An integer d is said to be a common divisor of integers x and y when d divides both x and y.

Definition of being coprime

Integers x and y are said to be coprime when x and y have no positive common divisors other than 1.

Definition of dividing

An integer x is said to divide another integer y when there exists an integer \alpha such that y = \alpha x.

Constraints

All values in input are integers.

1 \leq A, B \leq 10^{12}

Input

Input is given from Standard Input in the following format:

A B

Output

Print the maximum number of divisors that can be chosen to satisfy the condition.

Sample Input 1

12 18

Sample Output 1

3

12 and 18 have the following positive common divisors: 1, 2, 3, and 6.

1 and 2 are coprime, 2 and 3 are coprime, and 3 and 1 are coprime, so we can choose 1, 2, and 3, which achieve the maximum result.

Sample Input 2

420 660

Sample Output 2

4

Sample Input 3

1 2019

Sample Output 3

1

1 and 2019 have no positive common divisors other than 1.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

/* Plan
 * ----
 * Run through half of the smaller number to get al the common multiples, for each number check if its a divisor to both numbers
 * then check the corresponding larger multiple to see if it also works.
 *
 * The next step is to check how many are all coprimes of each other. We could use a loop
 */

int main() {
    int a, b;
    vector<long int> multiples;

    cin >> a >> b;
    
    multiples.push_back(1);
    for (int i = 2; i < (min(a,b) / 2) + 1; i++)
    {
        if (a % i == 0 && b % i == 0 && find(multiples.begin(), multiples.end(), i) == multiples.end())
        {
            multiples.push_back(i);
            int ci = a / i;
            if (b % ci == 0 && find(multiples.begin(), multiples.end(), i) == multiples.end()) 
            {
                multiples.push_back(ci);
            }
        }
    }
    
    // Start at one because 1 will always be coprime, and index 1 because 0 would be 1
    int c = 1;
    // for (int i = multiples.size() - 1; i > 0; i--)
    // {
        // bool is_coprime = true;
        // for (int j = i - 1; j > 0; j--)
        // {
            // if (multiples[i] % multiples[j] == 0)
            // {
                // is_coprime = false;
                // break;
            // }
        // }

        // if (is_coprime)
        // {
            // c++;
        // }
    // }
    
    for (int i = 1; i < multiples.size(); i++)
    {
        bool is_coprime = true;
        
        for (int j = 1; j < i; j++)
        {
            for (long int k = 2; k < min(multiples[i], multiples[j]); k++)
            {
                if (multiples[j] % k == 0 && multiples[i] % k == 0)
                {
                    is_coprime = false;
                    break;
                }
            }

        }
            
        if (is_coprime)
        {
            c++;
        }
    }


    cout << c;
}