HDU 5901 Count primes （大素数模板题）

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题意：求1-n有多少个素数，n可以1e11左右

复杂度大概O（n^(3/4)）

#include <bits/stdc++.h>  #define ll long long  using namespace std;  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("%I64d",&n)!=EOF){          init();          cout<<f[1]<<endl;      }      return 0;  }

复杂度O（n^(2/3)）

//Meisell-Lehmer  //G++ 218ms 43252k  #include<cstdio>  #include<cmath>  using namespace std;  #define LL long long  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(~scanf("%lld",&n))      {          printf("%lld\n",lehmer_pi(n));      }      return 0;  }

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