47 lines
1.1 KiB
C++
47 lines
1.1 KiB
C++
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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/**
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测试数据：
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3 60
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答案：4
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*/
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//最大公约数
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LL gcd(LL x, LL y) {
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return y ? gcd(y, x % y) : x;
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}
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//最小公倍数
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int lcm(int x, int y) {
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return y / gcd(x, y) * x; //注意顺序，防止乘法爆int
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}
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int cnt;
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LL x; //最大公约数
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LL y; //最小公倍数
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int main() {
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cin >> x >> y;
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//理论依据：gcd(p,q)*lcm(p,q)=p*q
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//枚举最小公倍数y的约数,这个约数可能是p,也可能是y/p
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for (LL k = 1; k * k <= y; k++)
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if (y % k == 0) {
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//可能小因子是p，也可能大因子是p,但大小因子别一样，那样会算重了。
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LL p = k;
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LL q = x * y / k; //q是靠计算出来的。q=x*y/p
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if (gcd(p, q) == x) cnt++;
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//p=y/k,q=x*y/(y/k)=k*x 两组不能重复，就是 y/k!=k
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if (k == y / k)continue;
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//p是大因子的可能性
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p = y / k;
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q = k * x;
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if (gcd(p, q) == x) cnt++;
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}
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cout << cnt << endl;
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return 0;
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}
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