[hdu5901 Count primes]Meisell-Lehmer求质数个数PI(X)

题目链接：[hdu5901 Count primes]
题目描述：求区间[1,N]的质数的个数(1≤N≤1011)。
解题思路：套一个Meisell-Lehmer的模版。很强大~

#include<bits/stdc++.h>using namespace std;typedef long long LL;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(cin >> n) {        cout << lehmer_pi(n) << endl;    }    return 0;}