2301: [HAOI2011]Problem b ( 分块+莫比乌斯反演+容斥）

/*
* @Author: LyuC
* @Date:   2017-10-08 16:54:59
* @Last Modified by:   LyuC
* @Last Modified time: 2017-10-08 21:06:44
*/
/*
 直接处理会超时，对于6/6=1 6/5=1 6/4=1这样的实际和已合并同类项一次
    计算出来，能节约不少时间
*/
#include <bits/stdc++.h>

#define LL long long 
#define MAXN 50005
using namespace std;

int t;
int a, b, c, d, k;
int sum [ MAXN ];
bool check[MAXN];
int mu[MAXN];
int prime[MAXN];

void mobi(){
     memset(check,false,sizeof check);
     mu[1]=1;
     int tol=0;
     for(int i=2;i<MAXN;i++){
         if(!check[i]){
             prime[tol++]=i;
             mu[i]=-1;
         }
         for(int j=0;j<tol;j++){
             if(i*prime[j]>MAXN) break;
             check[i*prime[j]]=true;
             if(i%prime[j]==0){
                 mu[i*prime[j]]=0;
                 break;
             }else{
                 mu[i*prime[j]]=-mu[i];
             }
         }
     }
}


inline int Count (int a, int b) {
    int s=0;
    if (a > b) {
        swap (a, b);
    }
    for (int i = 1, last = 0; i <= a; i = last + 1) {
        last = min( a / (a / i), b / (b / i) );
        s += (sum [ last ] - sum [ i - 1 ]) * (a / i) * (b / i);
    }
    return s;
}

inline void init () {
    sum [ 0 ] = 0;
    for (int i = 1;i < MAXN; i ++) {
        sum [ i ] = sum[ i - 1 ] + mu [ i ];
    }
}

int main () {
    // freopen ("in.txt", "r", stdin);
    mobi ();
    init ();
    scanf ("%d", &t);
    while (t -- ) {
        scanf ("%d%d%d%d%d", &a, &b, &c, &d, &k);
        int res = Count ( b / k, d / k ) - Count ( ( a - 1 ) / k, d / k ) - Count ( b / k, (c - 1) / k ) + Count ( ( a - 1 ) / k, ( c -1 ) / k );
        printf ( "%d\n", res );
    }
    return 0;
}