hdu5901 Count primes (计算1-1e11内有多少素数)

题意：计算1 ~ n内有多少素数(n < 1e11)

题解：刚开始心想分段打表，结果跑了好久，没出结果，放弃

           结果是Meisell-Lehmer，从来没听过，做为模板吧

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

//计算1-n内有多少素数
//复杂度O(n*(2/3))
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#pragma comment(linker,"/STACK:102400000,102400000")

using namespace std;
#define   MAX           100005
#define   MAXN          1000005
#define   maxnode       15
#define   sigma_size    30
#define   lson          l,m,rt<<1
#define   rson          m+1,r,rt<<1|1
#define   lrt           rt<<1
#define   rrt           rt<<1|1
#define   middle        int m=(r+l)>>1
#define   LL            long long
#define   ull           unsigned long long
#define   mem(x,v)      memset(x,v,sizeof(x))
#define   lowbit(x)     (x&-x)
#define   pii           pair<int,int>
#define   bits(a)       __builtin_popcount(a)
#define   mk            make_pair
#define   limit         10000

//const int    prime = 999983;
const int    INF   = 0x3f3f3f3f;
const LL     INFF  = 0x3f3f;
//const double pi    = acos(-1.0);
const double inf   = 1e18;
const double eps   = 1e-4;
const LL    mod    = 1e9+7;
const ull    mx    = 133333331;

/*****************************************************/
inline void RI(int &x) {
char c;
while((c=getchar())<'0' || c>'9');
x=c-'0';
while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
}
/*****************************************************/

const int N = 5e6 + 2;
bool np
;
int prime
, pi
;

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(cin >> n) {
cout << lehmer_pi(n) << endl;
}
return 0;
}