hdu 1695 GCD(莫比乌斯反演)

设f(d)为gcd(x,y) ==k, 0<x<n , 0<y<m的个数

设F(d)为gcd(x,y)|d的个数

那么利用莫比乌斯反演公式直接求解即可，sqrt(n)的复杂度，面对三千组数据还是够的

注意long long 的使用，作为一个水笔，我竟然int了一发，wa的死死的

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#define MAX 150007

using namespace std;

typedef long long LL;

LL t,a,b,c,d,k;
int prime[MAX];
int cnt;
int mu[MAX];
int vis[MAX];

void init ( )
{
    memset ( vis , 0 , sizeof ( vis ) );
    cnt = 0;
    mu[1] = 1;
    for ( int i = 2 ; i < MAX ; i++ )
    {
        if ( !vis[i] )
        {
            prime[cnt++] = i;
            mu[i] = -1;
        }
        for ( int j = 0 ; j < cnt && i*prime[j] < MAX ; j++ )
        {
            vis[i*prime[j]] = 1;
            if ( i%prime[j] ) mu[i*prime[j]] = -mu[i];
            else
            {
                mu[i*prime[j]] = 0;
                break;
            }
        }
    }
}

LL get ( LL n , LL m , LL k )
{
    if ( n > m ) swap ( n , m );
    return (n/k)*(n/k+1)/2 + ((m-(n/k)*k)/k)*(n/k); 
}


int main ( )
{
    scanf ( "%lld" , &t );
    init ( );
    int cc = 1;
    while ( t-- )
    {
        scanf ( "%lld%lld%lld%lld%lld" , &a , &b , &c , &d , &k );
        LL ans = 0;
        int lim = min ( b , d );
        printf ( "Case %d: " , cc++ );
        if ( k == 0 )
        {
            puts ( "0" );
            continue;
        }
        for ( int i = k ; i<= lim ; i += k )
            ans += (LL)(mu[i/k])*(LL)(get( b , d , i));
        printf ( "%I64d\n" , ans );
    }
}