bzoj 2301

[n/x] = y

xx = [n/y]

则xx是最大的被除数，使得商向下取证不变。

/**************************************************************
    Problem: 2301
    User: idy002
    Language: C++
    Result: Accepted
    Time:10136 ms
    Memory:1880 kb
****************************************************************/
 
#include <cstdio>
#include <iostream>
using namespace std;
 
int a, b, c, d, k;
int prm[6000], isnot[50010], mu[50010], smu[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++ )
        smu[i] = smu[i-1]+mu[i];
}
int calc( int n, int m ) {
    n /= k;
    m /= k;
    if( n>m ) swap(n,m);
    int rt = 0;
    for( int i=1; i<=n; i++ ) {
        int ii = min( m/(m/i), n/(n/i) );
        rt += (smu[ii]-smu[i-1])*(m/i)*(n/i);
        i = ii;
    }
    return rt;
}
int main() {
    int T;
    scanf( "%d", &T );
    init( 50000 );
    while( T-- ) {
        scanf( "%d%d%d%d%d", &a, &b, &c, &d, &k );
        printf( "%d
", calc(b,d)-calc(a-1,d)-calc(b,c-1)+calc(a-1,c-1) );
    }
}