jQuery火箭图标返回顶部代码

题目
x和(a_0)的最大公约数是(a_1)
x和(b_0)的最小公倍数是(b_1)
(gcd(x,a_0)=a_1)
(gcd(x/a_1,a_0/a_1)=1) ( ag{1})
(lcm(x_1,b_0)=b_1)
(frac{x*b_0}{gcd(x,b_0)}=b_1)
(gcd(x,b_0)=frac{x*b_0}{b_1})
(gcd(x/frac{x * b_0}{b_1},b_0/frac{x*b_0}{b_1})=1)
化简可得
(gcd(frac{b_1}{b_0},frac{b_1}{x})=1) ( ag{2})
观察((1),(2))可得(x)既是(a_1)的倍数又是(b_1)的约数
枚举(b_1)的约数判断是否满足他是(a_1)的倍数并且满足((1),(2))即可
注：那个(1),(2)在最右边..........

Code：

#include <cmath>
#include <cstdio>
#include <iostream>
using namespace std;
int P, ans, a0, a1, b0, b1;

int read() {
	int s = 0, w = 1;
	char ch = getchar();
	while(!isdigit(ch)) {if(ch == '-') w = -1; ch = getchar();}
	while(isdigit(ch)) {s =s  * 10 + ch - '0'; ch = getchar();}
	return s * w;
}
int gcd(int x, int y) {
	return y == 0 ? x : gcd(y, x % y);
}
int main() {
	P = read();
	while(P--) {
		a0 = read(), a1 = read(), b0 = read(), b1 = read();
		ans = 0;
		int wz = sqrt(b1);
		for(int i = 1; i <= wz; i++) {
			if(!(b1 % i)) {
				if(!(i % a1) && gcd(i / a1, a0 / a1) == 1 && gcd(b1 / b0, b1 / i) == 1) ans++;
				int j = b1 / i;
				if (i != j) if(!(j % a1) && gcd(j / a1, a0 / a1) == 1 && gcd(b1 / b0, b1 / j) == 1) ans++;
			}
		}
		printf("%d
", ans);
	}
	return 0;
}