1000亿以内的素数统计（Meisell-Lehmer算法）

绝地灬酷狼 2022-06-10 02:27 50阅读 0赞

今年沈阳有个素数计数的题，但是数据到了1e11,大概用Meisell-Lehmer能跑到2、3秒的样子，但是据说还有一种Deleglise Rivat的算法特别快，大概1e16差不多才3、4秒的样子，但是网上没有找到板子，所以就发一个Meisell-Lehmer的模板，1e11内的数差不多都能在5秒内处理完：

#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;
    }