HDU-5901-Count primes（大素数模板）

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链接：http://acm.hdu.edu.cn/showproblem.php?pid=5901

题意：

求区间[1,N]的质数的个数(1≤N≤1011)

详见：https://en.wikipedia.org/wiki/Prime-counting_function

CODE：

1，O(n^(3/4))

#include <bits/stdc++.h>#define ll long longusing 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;}

2，o(n^(2/3))

#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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