HDU 4497 GCD and LCM

I have written a similar one before. This should be more difficult than that.

(X/gcd (x, y, z) * (y/gcd (x, y, z) * (z * gcd (x, y, z) = lcm (x, y, z)/gcd (x, y, z );

Unique decomposition of LCM (x, y, z)/gcd (x, y, z;

For each PI ^ AI, X/gcd (x, y, z), y/gcd (x, y, z), Z/gcd (x, y, z) there must be one to take 0, one to take AI, and the other to any;

The repeat principle is removed;

(AI + 1) * (AI + 1) * (AI + 1)-2 * ai * Ai + (ai-1) * (ai-1) * (ai-1 );

PS: When AI is set to 1, it is 6;

 

 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 }

 

HDU 4497 GCD and LCM