HDU 5901 Count primes（ Meisell-Lehmer算法模板 ）

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链接：****传送门

题意：计算 [ 1 , n ] 之间素数的个数，(1 <= n <= 1e11)

思路：Meisell-Lehmer算法是计算超大范围内素数个数的一种算法，原理并不明白，由于英语太渣看不懂WIKI上的原理，附WIKI链接：Here

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/*************************************************************************
    > File Name: hdu5901.cpp
    > Author:    WArobot 
    > Blog:      http://www.cnblogs.com/WArobot/ 
    > Created Time: 2017年05月23日 星期二 19时38分39秒
 ************************************************************************/

// Meisell-Lehmer算法，快速计算超大范围（1e11）内素数个数
// HDU 5901 数据范围1e11 319MS

#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
",lehmer_pi(n));
    }
    return 0;
}