朴素筛法

O(nlogn)

#include <iostream>

using namespace std;

const int N = 1e6 + 10;

int primes[N], cnt;
bool st[N];

void get_primes(int n)
{
    for (int i = 2; i <= n; i++)
    {
        if (!st[i])
        {
            primes[cnt++] = i;
        }

        for (int j = i + i; j <= n; j += i) st[j] = true;
    }
}

int main()
{
    int n;
    cin >> n;

    get_primes(n);

    cout << cnt << endl;

    for (int i = 0; i < cnt; i++) cout << primes[i] << ' ' << endl;

    return 0;
}

埃式筛法

O(n) O(nloglogn)

#include <iostream>

using namespace std;

const int N = 1e6 + 10;

int primes[N] ,cnt;
bool st[N];  // st[x] 存储x是否被筛掉

void get_primes(int n)
{
    for (int i = 2; i <= n; i++)
    {
        if (st[i]) continue;
        primes[cnt++] = i;
        for (int j = i; j <= n; j += i)
            st[j] = true;
    }
}

int main()
{
    int n;
    cin >> n;

    get_primes(n);

    cout << cnt << endl;

    return 0;
}

线性筛

核心：n只会被他的最小质因子筛掉

#include <iostream>

using namespace std;

const int N = 1e6 + 10;

int primes[N] ,cnt;
bool st[N];  // st[x] 存储x是否被筛掉

void get_primes(int n)
{
    for (int i = 2; i <= n; i++)
    {
        if (!st[i]) primes[cnt++] = i;
        for (int j = 0; primes[j] <= n / i; j++)
        {
            st[primes[j] * i] = true;
            if (i % primes[j] == 0) break;
        }
    }
}

int main()
{
    int n;
    cin >> n;

    get_primes(n);

    cout << cnt << endl;

    return 0;
}