HDU 5901 Count primes （求1e11内素数个数）

死亡之题……

比赛打表打了3小时，想的是分段打表一些边界数据，类似这题：把1e9分成5000个20W大小的区间，打表出区间边界（1，200000，400000...）的答案。这样最坏查找最多也是20W-1

结局是三小时也没打出来，在第四小时的时候发现就算表打出来复杂度也在1e11左右，亲妈爆炸！

最最后的结局是，论文题：Wiki 或者 Lehmer快速求1e11以内质数个数

可惜！差这一题晋级。

ps：直接用cf665F 的代码交会TLE，实际上数组不用开那么大，预处理前

【代码】

#include 

#define MAXN 100
#define MAXM 10001
#define MAXP 40000
#define MAX 400000
#define clr(ar) memset(ar, 0, sizeof(ar))
#define read() freopen("lol.txt", "r", stdin)
#define dbg(x) cout << #x << " = " << x << endl
#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))

using namespace std;

namespace pcf{
    long long dp[MAXN][MAXM];
    unsigned int ar[(MAX >> 6) + 5] = {0};
    int len = 0, primes[MAXP], counter[MAX];

    void Sieve(){
        setbit(ar, 0), setbit(ar, 1);
        for (int i = 3; (i * i) < MAX; i++, i++){
            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]];
            }
        }
    }

    long long phi(long long 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);
    }

    long long Lehmer(long long m){
        if (m < MAX) return counter[m];

        long long w, res = 0;
        int i, a, s, c, x, y;
        s = sqrt(0.9 + m), y = c = cbrt(0.9 + m);
        a = counter[y], res = phi(m, a) + a - 1;
        for (i = a; primes[i] <= s; i++) res = res - Lehmer(m / primes[i]) + Lehmer(primes[i]) - 1;
        return res;
    }
}

long long solve(long long n){
    int i, j, k, l;
    long long x, y, res = 0;

    for (i = 0; i < pcf::len; i++){
        x = pcf::primes[i], y = n / x;
        if ((x * x) > n) break;
        res += (pcf::Lehmer(y) - pcf::Lehmer(x));
    }

    for (i = 0; i < pcf::len; i++){
        x = pcf::primes[i];
        if ((x * x * x) > n) break;
        res++;
    }

    return res;
}

int main(){
    pcf::init();
    long long n, res;

    while (scanf("%lld", &n) != EOF){
        //res = solve(n);
        printf("%lld\n",pcf::Lehmer(n));
        //printf("%lld\n", res);
    }
    return 0;
}