BZOJ 2301

分段優化即可。繼續學習中。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int MAXN = 100000;
bool check[MAXN+10];
int prime[MAXN+10];
int mu[MAXN+10];
void Moblus()
{
    memset(check,false,sizeof(check));
    mu[1] = 1;
    int tot = 0;
    for(int i = 2; i <= MAXN; i++)
    {
        if( !check[i] )
        {
            prime[tot++] = i;
            mu[i] = -1;
        }
        for(int j = 0; j < tot; 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];
            }
        }
    }
}
int sum[MAXN+10];
//找[1,n],[1,m]內互質的數的對數
long long solve(int n,int m)
{
    long long ans = 0;
    if(n > m)swap(n,m);
    for(int i = 1, la = 0; i <= n; i = la+1)
    {
        la = min(n/(n/i),m/(m/i));
        ans += (long long)(sum[la] - sum[i-1])*(n/i)*(m/i);
    }
    return ans;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    Moblus();
    sum[0] = 0;
    for(int i = 1;i <= MAXN;i++)
        sum[i] = sum[i-1] + mu[i];
    int a,b,c,d,k;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
        long long ans = solve(b/k,d/k) - solve((a-1)/k,d/k) - solve(b/k,(c-1)/k) + solve((a-1)/k,(c-1)/k);
        printf("%lld\n",ans);
    }
    return 0;
}