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>

  1 //STATUS:C++_AC_2052MS_2052KB

  2 #include <functional>

  3 #include <algorithm>

  4 #include <iostream>

  5 //#include <ext/rope>

  6 #include <fstream>

  7 #include <sstream>

  8 #include <iomanip>

  9 #include <numeric>

 10 #include <cstring>

 11 #include <cassert>

 12 #include <cstdio>

 13 #include <string>

 14 #include <vector>

 15 #include <bitset>

 16 #include <queue>

 17 #include <stack>

 18 #include <cmath>

 19 #include <ctime>

 20 #include <list>

 21 #include <set>

 22 #include <map>

 23 using namespace std;

 24 //#pragma comment(linker,"/STACK:102400000,102400000")

 25 //using namespace __gnu_cxx;

 26 //define

 27 #define pii pair<int,int>

 28 #define mem(a,b) memset(a,b,sizeof(a))

 29 #define lson l,mid,rt<<1

 30 #define rson mid+1,r,rt<<1|1

 31 #define PI acos(-1.0)

 32 //typedef

 33 typedef long long LL;

 34 typedef unsigned long long ULL;

 35 //const

 36 const int N=50010;

 37 const int INF=0x3f3f3f3f;

 38 const int MOD=100000,STA=8000010;

 39 const LL LNF=1LL<<60;

 40 const double EPS=1e-8;

 41 const double OO=1e15;

 42 const int dx[4]={-1,0,1,0};

 43 const int dy[4]={0,1,0,-1};

 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

 45 //Daily Use ...

 46 inline int sign(double x){return (x>EPS)-(x<-EPS);}

 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}

 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}

 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}

 50 template<class T> inline T Min(T a,T b){return a<b?a:b;}

 51 template<class T> inline T Max(T a,T b){return a>b?a:b;}

 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}

 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}

 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}

 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}

 56 //End

 57 

 58 int T,n,a,b,c,d,k;

 59 int isprime[N],mu[N],prime[N],sum[N];

 60 int cnt;

 61 void Mobius(int n)

 62 {

 63     int i,j;

 64     //Init phi[N],prime[N],全局变量初始为0

 65     cnt=0;mu[1]=1;

 66     for(i=2;i<=n;i++){

 67         if(!isprime[i]){

 68             prime[cnt++]=i;  //prime[i]=1;为素数表

 69             mu[i]=-1;

 70         }

 71         for(j=0;j<cnt && i*prime[j]<=n;j++){

 72             isprime[i*prime[j]]=1;

 73             if(i%prime[j])

 74                 mu[i*prime[j]]=-mu[i];

 75             else {mu[i*prime[j]]=0;break;}

 76         }

 77     }

 78 }

 79 

 80 LL solve(int n,int m)

 81 {

 82     int i,j,la;

 83     LL ret=0;

 84     if(n>m)swap(n,m);

 85     for(i=1,la=0;i<=n;i=la+1){

 86         la=Min(n/(n/i),m/(m/i));

 87         ret+=(LL)(sum[la]-sum[i-1])*(n/i)*(m/i);

 88     }

 89     return ret;

 90 }

 91 

 92 int main(){

 93  //   freopen("in.txt","r",stdin);

 94     int i,j;

 95     LL ans;

 96     Mobius(50000);

 97     for(i=1;i<50000;i++)sum[i]=sum[i-1]+mu[i];

 98     scanf("%d",&T);

 99     while(T--)

100     {

101         scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);

102         ans=solve(b/k,d/k)-solve((a-1)/k,d/k)

103             -solve((c-1)/k,b/k)+solve((a-1)/k,(c-1)/k);

104         printf("%lld\n",ans);

105     }

106     return 0;

107 }