三种素数筛法汇总

　　本地测试结果是，第三种方法速度最快，而且可以直接判断给定的数是否素数。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>

using namespace std;

const int N = 111111111;
int prm[6666666], pn, np[N >> 6];
bool mp[N];

void getPrm1() {
    pn = 0;
    prm[pn++] = 2;
    for (int i = 3; i < N; i++, i++) {
        if (!(np[i >> 6] & (1 << (i >> 1 & 31)))) {
            prm[pn++] = i;
            for (int j = (i << 1) + i; j < N; j += (i << 1)) {
                np[j >> 6] |= (1 << (j >> 1 & 31));
            }
        }
    }
//    for (int i = 0; i < 20; i++) {
//        cout << i << " : " << prm[i] << endl;
//    }
    cout << pn << endl;
}

void getPrm2() {
    pn = 0;
    int tmp;
    mp[0] = mp[1] = true;
    prm[pn++] = 2;
    for (int i = 3; i < N; i++, i++) {
        if (!mp[i]) prm[pn++] = i;
        for (int j = 0; j < pn; j++) {
            tmp = i * prm[j];
            if (tmp >= N) break;
            mp[tmp] = true;
            if (i % prm[j] == 0) break;
        }
    }
//    for (int i = 0; i < 20; i++) {
//        cout << i << " : " << prm[i] << endl;
//    }
    cout << pn << endl;
}

const int N2 = N >> 1;
const int Nsqrt = (int) sqrt((double) N) >> 1;
char ax[(N >> 4) + 2];

void sieve() {
    memset(ax, 255, sizeof(ax));
    ax[0] = 0xfe;
    for (int i = 1; i < Nsqrt; i++) {
        if (ax[i >> 3] & (1 << (i & 7))) {
            for (int j = (i << 1) + i + 1; j < N2; j += i << 1 | 1) {
                ax[j >> 3] &= ~(1 << (j & 7));
            }
        }
    }
}

inline bool isprime(int n) { return (n == 2) || ((n & 1) && (1 << (n >> 1 & 7) & ax[n >> 4]));}

void getPrm3() {
    pn = 0;
    sieve();
    prm[pn++] = 2;
    for (int i = 3; i < N; i++, i++) {
        if (isprime(i)) prm[pn++] = i;
    }
//    for (int i = 0; i < 20; i++) {
//        cout << i << " : " << prm[i] << endl;
//    }
    cout << pn << endl;
}

int main() {
    getPrm3();
    return 0;
}

——written by Lyon