```cpp
#include<iostream>
#include<cstdio>
#include<cctype>
using std::swap;
namespace xtracer{
#define getchar()(p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
char buf[1<<21],*p1=buf,*p2=buf;
typedef long long ll;
void read(ll &n){
char c=getchar();ll x=0;bool f=0;
for(;!isdigit(c);c=getchar())f^=!(c^45);
for(;isdigit(c);c=getchar())x=(x<<1)+(x<<3)+(c^48);
if(f)x=-x;n=x;
}
void write(ll n){
if(n>=10)write(n/10);
putchar(n%10+48);
}
}
using namespace xtracer;
ll gcd(ll a,ll b){
if(b>a)swap(a,b);
if(b==0)return a;
else return gcd(b,a%b);
}
ll lcm(ll a,ll b){
return a*b/gcd(a,b);
}
ll n;
ll a,a1,b,b1;
ll cnt=0;
int main(){
read(n);
while(n--){
cnt=0;
read(a);read(a1);read(b);read(b1);
for(ll i=1;i<=b1;i++){
if(gcd(i,a)==a1 and lcm(i,b)==b1)cnt+=1;
}
write(cnt);
putchar('\n');
}
return 0;
}
```
加了快读也无济于事
by xtracer @ 2021-02-04 09:27:23
@[xtracer](/user/341049) 只用遍历到sqrt(b)就行了，算两个数
by 王熙文 @ 2021-02-04 10:20:33
@[王熙文](/user/353688) 谢谢
by xtracer @ 2021-02-04 11:09:58
# 快读+O2+枚举b1因数
```cpp
#include<bits/stdc++.h>
using namespace std;
int gcd(int a,int b){
int s=min(a,b);
int big=max(a,b);
if(big%s==0)return s;
else return gcd(s,big%s);
}
int lcm(int a,int b){
return a/gcd(a,b)*b;
}
inline int read(){
int x=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9'){
if(ch=='-')
f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9'){
x=(x<<1)+(x<<3)+(ch^48);
ch=getchar();
}
return x*f;
}
int main(){
int a0,a1,b0,b1;
int n;
scanf("%d",&n);
while(n--){
a0=read();a1=read();b0=read();b1=read();
if(a0==1&&a1==1&&b0==1&&b1==1){
cout<<1;
continue;
}
int cnt=0;
for(int i=1;i<=b1/i;i+=1){
if(b1%i==0&&i!=sqrt(b1)){
if(gcd(a0,i)==a1&&lcm(b0,i)==b1){cnt++;}
if(gcd(a0,b1/i)==a1&&lcm(b0,b1/i)==b1){cnt++;}
}
if(i==sqrt(b1)){
if(gcd(a0,i)==a1&&lcm(b0,i)==b1){
cnt++;
}
}
}
cout<<cnt<<endl;
}
}
```
by Jacky2009 @ 2021-08-31 14:06:10