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HDU 5901 Count primes (求1e11内素数个数)

题目链接：点我~~

题意：求1e11内素数个数。

思路：Meisell-Lehmer的n^(2/3)方法，参考cf665F。

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long LL;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const int mod=1e9+7;
const double eps=1e-8;
const int inf=0x3f3f3f3f;
const double pi=acos(-1.0);
//LL powmod(LL a,LL b) {LL res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
const int MAXN=1000010;
#define MAX 1000010
#define sb(ar, i) (((ar[(i) >> 6]) |= (1 << (((i) >> 1) & 31))))
#define cb(ar, i) (((ar[(i) >> 6]) & (1 << (((i) >> 1) & 31))))
#define isprime(x) (( (x) && ((x)&1) && (!cb(ar, (x)))) || ((x) == 2))

LL dp[91][50010];
unsigned int ar[(10000010 >> 6) + 5] = {0};
int len = 0, pri[666666], cnt[1000010];

void solve()
{
    sb(ar, 0), sb(ar, 1);
    for (int i = 3; (i * i) < MAXN; i++, i++)
    {
        if (!cb(ar, i))
        {
            int k = i << 1;
            for (int j = (i * i); j < MAXN; j += k)
            {
                sb(ar, j);
            }
        }
    }
    for (int i = 1; i < MAXN; i++)
    {
        cnt[i] = cnt[i - 1];
        if (isprime(i)) pri[len++] = i, cnt[i]++;
    }
}

void init()
{
    solve();
    for (int n = 0; n < 90; n++)
    {
        for (int m = 0; m < 50010; m++)
        {
            if (!n) dp[n][m] = m;
            else dp[n][m] = dp[n - 1][m] - dp[n - 1][m / pri[n - 1]];
        }
    }
}

LL TT(LL m, int n)
{
    if (n == 0) return m;
    if (pri[n - 1] >= m) return 1;
    if (m < 50010  && n < 90) return dp[n][m];
    return TT(m, n - 1) - TT(m / pri[n - 1], n - 1);
}

LL Lehmer(LL m)
{
    if (m < MAX)
        return cnt[m];
    LL w, res = 0;
    int i, a, s, c, x, y;
    s = sqrt(0.9 + m), y = c = cbrt(0.9 + m);
    a = cnt[y], res = TT(m, a) + a - 1;
    for (i = a; pri[i] <= s; i++) res = res - Lehmer(m / pri[i]) + Lehmer(pri[i]) - 1;
    return res;
}


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