HDU 5901 Count primes

 

[传送门]

 

存个模板...

 

#include <bits/stdc++.h>

#define MAXN 100
#define MAXM 10001
#define MAXP 40000
#define MAX 400000
#define clr(ar) meset(ar, 0, sizeof(ar))
#define chkbit(ar, i) (((ar[(i) >> 6]) & (1 << (((i) >> 1) & 31))))
#define setbit(ar, i) (((ar[(i) >> 6]) |= (1 << (((i) >> 1) & 31))))
#define isprime(x) (( (x) && ((x) & 1) && (!chkbit(ar, (x)))) || ((x) == 2))
#define ll long long

namespace pcf {
    ll dp[MAXN][MAXM];
    unsigned int ar[(MAX >> 6) + 5];
    int len, primes[MAXP], counter[MAX];
    void Sieve() {
        setbit(ar, 0), setbit(ar, 1);
        for (int i = 3; i * i < MAX; i += 2) {
            if (!chkbit(ar, i)) {
                int k = i << 1;
                for (int j = i * i; j < MAX; j += k) setbit(ar, j);
            }
        }
        for (int i = 1; i < MAX; i++) {
            counter[i] = counter[i - 1];
            if (isprime(i)) primes[len++] = i, counter[i]++;
        }
    }
    void init() {
        Sieve();
        for (int n = 0; n < MAXN; n++)
            for (int m = 0; m < MAXM; m++)
                if (!n) dp[n][m] = m;
                else dp[n][m] = dp[n - 1][m] - dp[n - 1][m / primes[n - 1]];
    }
    ll phi(ll m, int n) {
        if (n == 0) return m;
        if (primes[n - 1] >= m) return 1;
        if (m < MAXM && n < MAXN) return dp[n][m];
        return phi(m, n - 1) - phi(m / primes[n - 1], n - 1);
    }
    ll Lehmer(ll m) {
        if (m < MAX) return counter[m];
        ll res = 0;
        int s = sqrt(m + 0.5), y = cbrt(m + 0.5), c = y;
        int a = counter[y]; res = phi(m, a) + a - 1;
        for (int i = a; primes[i] <= s; i++)
            res = res - Lehmer(m / primes[i]) + Lehmer(primes[i]) - 1;
        return res;
    }
}

int main() {
    pcf::init();
    ll n;
    while (~scanf("%lld", &n)) {
        printf("%lld\n", pcf::Lehmer(n));
    }
    return 0;
}