题意

题目链接

Sol

充满套路的数学题。。

如果你学过莫比乌斯反演的话不难得到两个等式

\[gcd(\frac{x}{a_1}, \frac{a_0}{a_1}) = 1\]

\[gcd(\frac{b_1}{b_0}, \frac{b_1}{x}) = 1\]

然后枚举\(b_1\)的约数就做完了。。

// luogu-judger-enable-o2
// luogu-judger-enable-o2
#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int MAXN = 1e6;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int T, a0, a1, b0, b1, ans;
int gcd(int a, int b) {
    return b == 0 ? a : gcd(b, a % b);
}
void check(int x) {
    if(x % a1) return ;
    ans += (gcd(x / a1, a0 / a1) == 1 && gcd(b1 / b0, b1 / x) == 1);
}
int main() {
    T = read();
    while(T--) {
        a0 = read(), a1 = read(), b0 = read(), b1 = read(); ans = 0;
        for(int x = 1; x * x <= b1; x++) {
            if(b1 % x == 0) {
                check(x);
                if(b1 != x) check(b1 / x);
            }
        }
        cout << ans << endl;
    }
    return 0; 
}