hdu5901--数论（1到n的素数个数）模板题

模板 n最大1e11

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N=5e6+2;
bool np[N];
int prime[N],pi[N];
int getprime()
{
    int cnt=0;
    np[0]=np[1]=true;
    pi[0]=pi[1]=0;
    for(int i=2;i<N;++i)
    {
        if(!np[i])
            prime[++cnt]=i;
        pi[i]=cnt;
        for(int j=1;j<=cnt&&i*prime[j]<N;++j)
        {
            np[i*prime[j]]=true;
            if(i%prime[j]==0) break;
        }
    }
    return cnt;
}
const int M=7;
const int PM=2*3*5*7*11*13*17;
int phi[PM+1][M+1],sz[M+1];
void init()
{
    getprime();
    sz[0]=1;
    for(int i=0;i<=PM;++i)
        phi[i][0]=i;
    for(int i=1;i<=M;++i)
    {
        sz[i]=prime[i]*sz[i-1];
        for(int j=1;j<=PM;++j)
            phi[j][i]=phi[j][i-1]-phi[j/prime[i]][i-1];
    }
}
int sqrt2(ll x)
{
    ll r=(ll)sqrt(x-0.1);
    while(r*r<=x) ++r;
    return int(r-1);
}
int sqrt3(ll x)
{
    ll r=(ll)cbrt(x-0.1);
    while(r*r*r<=x) ++r;
    return int(r-1);
}
ll getphi(ll x,int s){
    if(s==0)
        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+1;
    if(x<=prime[s]*prime[s]*prime[s]&&x<N)
    {
        int s2x=pi[sqrt2(x)];
        ll ans=pi[x]-(s2x+s-2)*(s2x-s+1)/2;
        for(int i=s+1;i<=s2x;++i)
            ans+=pi[x/prime[i]];
        return ans;
    }
    return getphi(x,s-1)-getphi(x/prime[s],s-1);
}
ll getpi(ll x)
{
    if(x<N) return pi[x];
    ll ans=getphi(x,pi[sqrt3(x)])+pi[sqrt3(x)]-1;
    for(int i=pi[sqrt3(x)]+1,ed=pi[sqrt2(x)];i<=ed;++i)
        ans-=getpi(x/prime[i])-i+1;
    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-2)*(b-a+1)/2;
    for(int i=a+1;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-1);
    }
    return sum;
}
int main()
{
    init();
    ll n;
    while(cin>>n)
        cout<<lehmer_pi(n)<<endl;
    return 0;
}