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 }

 

 

 

posted @ 2013-08-20 02:16  zhsl  阅读(...)  评论(...编辑  收藏