hdu 5901 Count primes (meisell-Lehmer)

Count primes

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2625    Accepted Submission(s): 1337

Problem Description

Easy question! Calculate how many primes between [1...n]!

 

Input

Each line contain one integer n(1 <= n <= 1e11).Process to end of file.

 

Output

For each case, output the number of primes in interval [1...n]

 

Sample Input
2
3
10
 

Sample Output
1
2
4

help

C/C++：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 using namespace std;
 typedef long long ll;
 const int N=5e6+;
 bool np[N];
 int prime[N],pi[N];
 int getprime()
 {
     int cnt=;
     np[]=np[]=true;
     pi[]=pi[]=;
     for(int i=; i<N; ++i)
     {
         if(!np[i])
             prime[++cnt]=i;
         pi[i]=cnt;
         for(int j=; j<=cnt&&i*prime[j]<N; ++j)
         {
             np[i*prime[j]]=true;
             if(i%prime[j]==) break;
         }
     }
     return cnt;
 }
 const int M=;
 const int PM=******;
 int phi[PM+][M+],sz[M+];
 void init()
 {
     getprime();
     sz[]=;
     for(int i=; i<=PM; ++i)
         phi[i][]=i;
     for(int i=; i<=M; ++i)
     {
         sz[i]=prime[i]*sz[i-];
         for(int j=; j<=PM; ++j)
             phi[j][i]=phi[j][i-]-phi[j/prime[i]][i-];
     }
 }
 int sqrt2(ll x)
 {
     ll r=(ll)sqrt(x-0.1);
     while(r*r<=x) ++r;
     return int(r-);
 }
 int sqrt3(ll x)
 {
     ll r=(ll)cbrt(x-0.1);
     while(r*r*r<=x) ++r;
     return int(r-);
 }
 ll getphi(ll x,int s)
 {
     if(s==)
         return x;
     if(s<=M)
         return phi[x%sz[s]][s]+(x/sz[s])*phi[sz[s]][s];
     if(x<=prime[s]*prime[s])
         return pi[x]-s+;
     if(x<=prime[s]*prime[s]*prime[s]&&x<N)
     {
         int s2x=pi[sqrt2(x)];
         ll ans=pi[x]-(s2x+s-)*(s2x-s+)/;
         for(int i=s+; i<=s2x; ++i)
             ans+=pi[x/prime[i]];
         return ans;
     }
     return getphi(x,s-)-getphi(x/prime[s],s-);
 }
 ll getpi(ll x)
 {
     if(x<N) return pi[x];
     ll ans=getphi(x,pi[sqrt3(x)])+pi[sqrt3(x)]-;
     for(int i=pi[sqrt3(x)]+,ed=pi[sqrt2(x)]; i<=ed; ++i)
         ans-=getpi(x/prime[i])-i+;
     return ans;
 }
 ll lehmer_pi(ll x)
 {
     if(x<N) return pi[x];
     int a=(int)lehmer_pi(sqrt2(sqrt2(x)));
     int b=(int)lehmer_pi(sqrt2(x));
     int c=(int)lehmer_pi(sqrt3(x));
     ll sum=getphi(x,a)+ll(b+a-)*(b-a+)/;
     for(int i=a+; i<=b; i++)
     {
         ll w=x/prime[i];
         sum-=lehmer_pi(w);
         if(i>c)
             continue;
         ll lim=lehmer_pi(sqrt2(w));
         for(int j=i; j<=lim; j++)
             sum-=lehmer_pi(w/prime[j])-(j-);
     }
     return sum;
 }
 int main()
 {
     init();
     ll n;
     while(cin>>n)
         cout<<lehmer_pi(n)<<endl;
     return ;
 }