bzoj 1101

其实这个用的是Mobius反演的第二种形式

F(d) = (n div d) * (m div d)

f(d) = [ gcd(i,j)=d ] (i in [1,a], j in [1,b])

 /**************************************************************
     Problem: 1101
     User: idy002
     Language: C++
     Result: Accepted
     Time:6764 ms
     Memory:1688 kb
 ****************************************************************/

 #include <cstdio>
 #include <iostream>
 using namespace std;

 typedef long long dnt;

 int prm[], isnot[], mu[], ptot;

 void init( int n ) {
     mu[] = ;
     for( int i=; i<=n; i++ ) {
         if( !isnot[i] ) {
             prm[++ptot] = i;
             mu[i] = -;
         }
         for( int j=; j<=ptot && i*prm[j]<=n; j++ ) {
             isnot[i*prm[j]] = true;
             if( i%prm[j]== ) {
                 mu[i*prm[j]] = ;
                 break;
             }
             mu[i*prm[j]] = -mu[i];
         }
     }
     for( int i=; i<=n; i++ )
         mu[i] += mu[i-];
 }
 dnt calc( int n, int m, int k ) {
     dnt rt = ;
     if( n>m ) swap(n,m);
     n/=k;
     m/=k;
     for( int d=; d<=n; d++ ) {
         int dd=min(n/(n/d),m/(m/d));
         rt += (dnt)(n/d)*(m/d)*(mu[dd]-mu[d-]);
         d=dd;
     }
     return rt;
 }
 int main() {
     init();
     int T;
     scanf( "%d", &T );
     while( T-- ) {
         int n, m, k;
         scanf( "%d%d%d", &n, &m, &k );
         printf( "%lld\n", calc(n,m,k) );
     }
 }