bzoj2301(莫比乌斯反演)

//
//  main.cpp
//  bzoj2301
//
//  Created by New_Life on 16/7/6.
//  Copyright © 2016年 chenhuan001. All rights reserved.
//

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define N 100100

//－－莫比乌斯反演函数－－//
//说明：利用线性素数筛选顺便求了个mu
//复杂度:O(n)
int mu[N];

void mobus()
{
    bool mark[N];
    int prime[N];
    int pcnt=0;
    memset(mark,0,sizeof(mark));
    mu[1] = 1;
    for(int i=2;i<N;i++)
    {
        if(mark[i] == 0)
        {
            prime[pcnt++] = i;
            mu[i] = -1;
        }
        for(int j=0;j<pcnt && i*prime[j]<N;j++)
        {
            int tmp = i*prime[j];
            mark[tmp] = 1;
            if( i%prime[j] == 0 )
            {
                mu[tmp] = 0;
                break;
            }
            
            mu[tmp] = mu[i]*-1;
        }
    }
}

int sum[N];


long long gaobili(int b,int d)
{
    if(b<=0||d<=0) return 0;
    int m = min(b,d);
    long long ans = 0;
    while(m>=1)
    {
        int tb = b/( b/m +1 )+1;
        int td = d/( d/m +1 )+1;
        //前进的最大位置
        int tm = max(tb,td);
        ans += (long long)(sum[m]-sum[tm-1])*(b/m)*(d/m);
        m = tm-1;
    }
    return ans;
}

int main() {
    mobus();
    for(int i=1;i<N;i++)
        sum[i] += sum[i-1]+mu[i];
    int T;
    cin>>T;
    while(T--)
    {
        int a,b,c,d,k;
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
        //a/=k; b/=k; c/=k; d/=k;//这里很关键
        //搞bd
        long long ans = gaobili(b/k, d/k)-gaobili((a-1)/k, d/k)-gaobili(b/k, (c-1)/k)+gaobili((a-1)/k, (c-1)/k);
        printf("%lld\n",ans);
    }
    return 0;
}
/*
2
2 5 1 5 1
1 5 1 5 2

 */