calculate the prime number lower and equal than n quickly.

Meisell-Lehmer.cpp

#include<bits/stdc++>
/Meisell-Lehmer
const int MAX_N = 5e6 + 2;
bool np[MAX_N];
int prime[MAX_N], pi[MAX_N];
int getprime()
{
    int cnt = 0;
    np[0] = np[1] = true;
    pi[0] = pi[1] = 0;
    for(int i = 2; i < MAX_N; ++i)
    {
        if(!np[i]) prime[++cnt] = i;
        pi[i] = cnt;
        for(int j = 1; j <= cnt && i * prime[j] < MAX_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 < MAX_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 < MAX_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 < MAX_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;
}