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[6000], isnot[50010], mu[50010], ptot;
 
void init( int n ) {
    mu[1] = 1;
    for( int i=2; i<=n; i++ ) {
        if( !isnot[i] ) {
            prm[++ptot] = i;
            mu[i] = -1;
        }
        for( int j=1; j<=ptot && i*prm[j]<=n; j++ ) {
            isnot[i*prm[j]] = true;
            if( i%prm[j]==0 ) {
                mu[i*prm[j]] = 0;
                break;
            }
            mu[i*prm[j]] = -mu[i];
        }
    }
    for( int i=1; i<=n; i++ )
        mu[i] += mu[i-1];
}
dnt calc( int n, int m, int k ) {
    dnt rt = 0;
    if( n>m ) swap(n,m);
    n/=k;
    m/=k;
    for( int d=1; d<=n; d++ ) {
        int dd=min(n/(n/d),m/(m/d));
        rt += (dnt)(n/d)*(m/d)*(mu[dd]-mu[d-1]);
        d=dd;
    }
    return rt;
}
int main() {
    init(50000);
    int T;
    scanf( "%d", &T );
    while( T-- ) {
        int n, m, k;
        scanf( "%d%d%d", &n, &m, &k );
        printf( "%lld
", calc(n,m,k) );
    }
}