解法1:唯一分解定理
通过$gcd$和$lcm$对$x$的质因数个数的限制,算出每个质因数的能取的$min~max$个数
然后用乘法原理乘起来。
解法2(code↓):
考虑$lcm(x,b_{0})=b_{1}$
转化一下:$x*b_{0}=b_{1}*gcd(x,b_{0})$
$x=b_{1}/b_{0}*gcd(x,b_{0})$
枚举一下$b_{0}$的因数,筛一筛就完事了
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<cctype> 6 #define re register 7 using namespace std; 8 void read(int &x){ 9 char c=getchar();x=0; 10 while(!isdigit(c)) c=getchar(); 11 while(isdigit(c)) x=(x<<3)+(x<<1)+(c^48),c=getchar(); 12 } 13 int gcd(int a,int b){return b?gcd(b,a%b):a;} 14 int n,a0,a1,b0,b1,ans; 15 int main(){ 16 read(n); 17 for(re int i=1;i<=n;++i){ 18 read(a0);read(a1);read(b0);read(b1); 19 int m=sqrt(b0+0.5); ans=0; 20 for(re int j=1;j<=m;++j){//直接枚举约数 21 if(b0%j) continue; 22 int x=b1/b0*j; 23 ans+=(gcd(x,b0)==j&&gcd(x,a0)==a1); 24 if(j*j!=b0) 25 x=b1/j,ans+=(gcd(x,b0)==b0/j&&gcd(x,a0)==a1); 26 }printf("%d ",ans); 27 }return 0; 28 }