【HDU5901】Count primes(大素数模板）

来源：互联网发布：软件开发有什么软件编辑：程序博客网时间：2024/06/05 16:15
记录一个菜逼的成长。。
纯记录，膜拜大神。
http://blog.csdn.net/xuanandting/article/details/52577406
    //Meisell-Lehmer      #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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