HDU 4497 GCD and LCM

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 using namespace std;
 5 
 6 int sign[100000];
 7 int pri[100000];
 8 int tot;
 9 
10 void getpri (){
11     memset (sign,0,sizeof sign);
12     for (int i=2;i<100000;i++)
13         if (!sign[i])
14             for (int j=i*2;j<100000;j+=i)
15                 sign[j]=1;
16     tot=0;//cout<<"error";
17     for (int i=2;i<100000;i++)
18         if (!sign[i])
19             pri[tot++]=i;
20 }
21 
22 int mm[100000];
23 
24 int getpow (int n){
25     return n*n*n;
26 }
27 
28 void solved (int s){
29     int num=0;
30     int ans=1;
31     int x=s;
32     memset (mm,0,sizeof mm);
33     for (int i=0;i<tot;i++){//cout<<x<<endl;
34         if (x%pri[i]==0){
35             while (x%pri[i]==0){
36                 mm[i]++;
37                 x/=pri[i];
38             }
39             ans*=getpow(mm[i]+1)-2*getpow(mm[i])+getpow(mm[i]-1);//cout<<mm[i]<<" "<<pri[i]<<endl;
40             num++;
41         }
42     }
43     if (x!=1)
44         ans*=6;
45     cout<<ans<<endl;
46     
47 }
48 
49 int main (){
50     getpri();
51     int t;
52     int g,l;
53     cin>>t;
54     while (t--){
55         cin>>g>>l;
56         if (l%g!=0){
57             cout<<"0"<<endl;
58             continue ;
59         }
60         solved (l/g);
61         
62     }
63     return 0;
64 }