Nth prime & numbers of primes （模板）

都是取的模板，这几天做的素数题挺多的，所以整理了放在这里，感觉有一天回用到的！

SPOJ:Nth Prime：     求第N个素数，N<1e9。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll; 
const int N=2000000,P=510510,Q=92160;
struct getnthprime
{   
    int prime[N+1],pi[N+1],e[P];
    void init(){
       for(int i=2;i<=N;i++) {
          if(!prime[i]) prime[++prime[0]]=i,pi[i]=pi[i-1]+1;
          else pi[i]=pi[i-1];
          for(int j=1;j<=prime[0]&&i<=N/prime[j];j++) {
              prime[i*prime[j]]=1;
              if(i%prime[j]==0) break;
          }
       }
       for(int i=0;i<P;i++) e[i]=i;
       for(int i=1;i<=7;i++) {
          for(int j=P-1;j>=0;j--)
            e[j]-=e[j/prime[i]];
       }
    }
    ll get_phi(ll m,int n) {
       if (n==7) return m/P*Q+e[m%P];
       if (m<prime[n]) return 1;
       if (m<=N&&m<=(ll)prime[n]*prime[n]*prime[n]) {
          ll ans=pi[m]-n+1;
          for(int i=n+1,l=pi[(int)sqrt(m+0.1)];i<=l;i++)
             ans+=pi[m/prime[i]]-i+1;
          return ans;
       }
       return get_phi(m,n-1)-get_phi(m/prime[n],n-1);
    }

    ll get_pi(ll m){
       if(m<=N) return pi[m];
       int n=pi[(int)cbrt(m-0.1)+1];
       ll ans=get_phi(m,n)+n-1;
       for(int i=n+1,l=pi[(int)sqrt(m+0.1)];i<=l;i++)
          ans-=get_pi(m/prime[i])-i+1;
       return ans;
   }

   bool f[1000010];
   ll get_pn(ll n) {
      if (n<=prime[0]) return prime[n];
      ll x=n*(log(n)+log(log(n))-1)+n*(log(log(n))-2)/log(n)-6*n/1000;
      ll y=n*(log(log(n)))*(log(log(n)))/log(n)/log(n);
      y=min(y,ll(3500000));
      ll l=x,r=x+y,flag = 0;
      for (int i=0;i<2;i++) {
         ll m=(l+r)>> 1;
         ll pm=get_pi(m);
         if(pm>=n) r=m,flag=0;
         else l=m+1,flag=pm;
      }
      ll count=flag?flag:get_pi(l-1);
      for(int i=1,li=pi[(int)sqrt(r+0.1)];i<=li;i++) {
         for(int j=((l-1)/prime[i]+1)*prime[i]-l;j<=r-l+1;
             j+=prime[i]){
             f[j]=true;
         } 
      }
      for(int i=0;i<=r-l+1;i++) {
         if(!f[i]){
             count++;
             if(count==n) return i+l;
         }
      }
      return -1;
   }
}NP;

int main() {
    NP.init();
    ll n; scanf("%lld",&n);
    cout<<NP.get_pn(n)<<endl;
    return 0;
}

HDU5901:Count primes:    求1到N有多少个素数。N<1e11。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=5e6+2,M=7,PM=2*3*5*7*11*13*17;
struct countprimes
{
    bool np[N],did[N];
    int prime[N],pi[N],phi[PM+1][M+1],sz[M+1];
    vector<ll>v;
    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;
    }
    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);//开三次方 
       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;
   }
}CP;
int main()
{
    CP.init();
    ll n,ans=0;
    while(~scanf("%lld",&n)){
        cout<<CP.lehmer_pi(n)<<endl;
    }
}