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=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //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=;mu[]=;
     for(i=;i<=n;i++){
         if(!isprime[i]){
             prime[cnt++]=i;  //prime[i]=1;为素数表
             mu[i]=-;
         }
         for(j=;j<cnt && i*prime[j]<=n;j++){
             isprime[i*prime[j]]=;
             if(i%prime[j])
                 mu[i*prime[j]]=-mu[i];
             else {mu[i*prime[j]]=;break;}
         }
     }
 }

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

 int main(){
  //   freopen("in.txt","r",stdin);
     int i,j;
     LL ans;
     Mobius();
     for(i=;i<;i++)sum[i]=sum[i-]+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-)/k,d/k)
             -solve((c-)/k,b/k)+solve((a-)/k,(c-)/k);
         printf("%lld\n",ans);
     }
     return ;
 }