既然a跟b的最小公倍数是y，那就是说a跟b都是y的约数
这样我们可以先把y的所有约数找到然后直接两重循环枚举约数检验是否满足条件即可

#include <iostream>
#include <cstdio>
#include <vector>

using namespace std;

typedef long long LL;

vector<LL> v;

LL gcd(int a, int b) {return b ? gcd(b, a % b) : a;}

int main()
{
    int x, y, res = 0;
    scanf("%d%d", &x, &y);
    for (int i = 1; i <= y; i ++ ) if (y % i == 0) v.push_back(i);
    for (int i = 0; i < v.size(); i ++ )
    {
        for (int j = 0; j < v.size(); j ++ )
        {
            int g = gcd(v[i], v[j]);
            if (g == x && v[i] * v[j] / g == y) res ++ ;
        }
    }
    cout << res << endl;
    return 0;
}