clever_person @ 2024-10-08 21:06:32
#include <bits/stdc++.h>
using namespace std;
int main () {
int x, y; cin >> x >> y;
long long z = 1ll * x * y, ans = 0, k = y / x;
for (int i = 1; i <= k; ++i) {
if (k % i != 0) continue;
long long gcd = __gcd(1ll * i, k / i);
if (gcd != 1) continue;
ans++;
}
cout << ans << endl;
}
by pi3_141592653589 @ 2024-10-09 21:40:39
特判y不为x的倍数的情况。
by pi3_141592653589 @ 2024-10-09 21:40:58
@clever_person