~~但能得30分~~
by The_Collector @ 2023-12-20 13:04:59
得保证a*b%i==0:
还有应该是lcm(i,j)==b&&_gcd(i,j)==a
by oldhorse1989 @ 2023-12-20 17:03:49
@[_thorns_](/user/1208314)
gcd 和 lcm 写反了
```cpp
#include<bits/stdc++.h>
#define int long long
using namespace std;
int a,b,ldd=0;
int lcm(int x,int y){
return x*y/__gcd(x,y);
}
signed main(){
cin>>a>>b;
for(int i=a;i<=b;i++){
int j=a*b/i;
if(__gcd(i,j)==a&&lcm(i,j)==b)
{
ldd++;
}
}
cout<<ldd;
return 0;
}
```
by Dream_Creator @ 2024-01-20 15:23:33