题意：

1——n（10^11）的素数个数

思路：

参考：http://blog.csdn.net/chaiwenjun000/article/details/52589457

第一个O（n^(3/4)）

/* ***********************************************
Author        :devil
************************************************ */
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <stack>
#include <map>
#include <string>
#include <cmath>
#include <stdlib.h>
#define inf 0x3f3f3f3f
#define LL long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define ou(a) printf("%d\n",a)
#define pb push_back
#define mkp make_pair
template<class T>inline void rd(T &x){char c=getchar();x=0;while(!isdigit(c))c=getchar();while(isdigit(c)){x=x*10+c-'0';c=getchar();}}
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
using namespace std;
const int mod=1e9+7;
const int N=310;
LL f[340000],g[340000],n;
void init()
{
    LL i,j,m;
    for(m=1;m*m<=n;m++) f[m]=n/m-1;
    for(i=1;i<=m;i++) g[i]=i-1;
    for(i=2;i<=m;i++)
    {
        if(g[i]==g[i-1]) continue;
        for(j=1;j<=min(m-1,n/i/i);j++)
        {
            if(i*j<m) f[j]-=f[i*j]-g[i-1];
            else f[j]-=g[n/i/j]-g[i-1];
        }
        for(j=m;j>=i*i;j--) g[j]-=g[j/i]-g[i-1];
    }
}
int main()
{
    while(~scanf("%lld",&n))
    {
        init();
        cout<<f[1]<<endl;
    }
    return 0;
}

第二个O（n^(2/3)）

/* ***********************************************
Author        :devil
************************************************ */
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <stack>
#include <map>
#include <string>
#include <cmath>
#include <stdlib.h>
#define inf 0x3f3f3f3f
#define LL long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define ou(a) printf("%d\n",a)
#define pb push_back
#define mkp make_pair
template<class T>inline void rd(T &x){char c=getchar();x=0;while(!isdigit(c))c=getchar();while(isdigit(c)){x=x*10+c-'0';c=getchar();}}
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
using namespace std;
const int mod=1e9+7;
const int N=5e6+2;
bool np[N];
int prime[N],pi[N];
int getprime()
{
    int cnt=0;
    np[0]=np[1]=1;
    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]]=1;
            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(~scanf("%lld",&n))
    {
        printf("%lld\n",lehmer_pi(n));
    }
    return 0;
}