Bzoj-2301 [HAOI2011]Problem b 容斥原理,Mobius反演,分块

　　题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=2301

　　题意：多次询问，求有多少对数满足 gcd(x,y)=k, a<=x<=b, c<=y<=d。

　　对于有下界的区间，容易想到用容斥原理做。然后如果直接用Mobius反演定理做，那么每次询问的复杂度是O(n/k)，如果k=1的话，那么总体就是O(n^2)的复杂度了，会TLE。这样用到了分快优化，注意到 n/i ,在连续的k区间内存在，n/i=n/(i+k)，因此能用分块优化。由于n/d，最多有2*sqrt(n)不相同的数，因此每次询问复杂度2*sqrt(n)+2*sqrt(m)..

　　详细内容推荐看：<POI XIV Stage.1 Queries Zap>

//STATUS:C++_AC_2052MS_2052KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef long long LL;
typedef unsigned long long ULL;
//const
const int N=50010;
const int INF=0x3f3f3f3f;
const int MOD=100000,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

int T,n,a,b,c,d,k;
int isprime[N],mu[N],prime[N],sum[N];
int cnt;
void Mobius(int n)
{
    int i,j;
    //Init phi[N],prime[N],全局变量初始为0
    cnt=0;mu[1]=1;
    for(i=2;i<=n;i++){
        if(!isprime[i]){
            prime[cnt++]=i;  //prime[i]=1;为素数表
            mu[i]=-1;
        }
        for(j=0;j<cnt && i*prime[j]<=n;j++){
            isprime[i*prime[j]]=1;
            if(i%prime[j])
                mu[i*prime[j]]=-mu[i];
            else {mu[i*prime[j]]=0;break;}
        }
    }
}

LL solve(int n,int m)
{
    int i,j,la;
    LL ret=0;
    if(n>m)swap(n,m);
    for(i=1,la=0;i<=n;i=la+1){
        la=Min(n/(n/i),m/(m/i));
        ret+=(LL)(sum[la]-sum[i-1])*(n/i)*(m/i);
    }
    return ret;
}

int main(){
 //   freopen("in.txt","r",stdin);
    int i,j;
    LL ans;
    Mobius(50000);
    for(i=1;i<50000;i++)sum[i]=sum[i-1]+mu[i];
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
        ans=solve(b/k,d/k)-solve((a-1)/k,d/k)
            -solve((c-1)/k,b/k)+solve((a-1)/k,(c-1)/k);
        printf("%lld
",ans);
    }
    return 0;
}