Count primes （Meissel-Lehmer算法）

Easy question! Calculate how many primes between [1…n]!
Input
Each line contain one integer n(1 <= n <= 1e11).Process to end of file.
Output
For each case, output the number of primes in interval [1…n]
Sample Input
2
3
10
Sample Output
1
2
4

[分析]
模板,有必要记录一下。

[代码]

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <vector>#include <cmath>#define MAXN 100    // pre-calc max n for phi(m, n)#define MAXM 10010 // pre-calc max m for phi(m, n)#define MAXP 40000 // max primes counter#define MAX 400010    // max prime#define setbit(ar, i) (((ar[(i) >> 6]) |= (1 << (((i) >> 1) & 31))))#define chkbit(ar, i) (((ar[(i) >> 6]) & (1 << (((i) >> 1) & 31))))#define isprime(x) (( (x) && ((x)&1) && (!chkbit(ar, (x)))) || ((x) == 2))using namespace std;    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;    }int main(){    init();    long long int n;    while(scanf("%lld",&n)!=EOF)    {        printf("%lld\n",Lehmer(n));    }    return 0;}