2到n之间的所有素数：
算法1：

#include <iostream>
#include <cmath>
using namespace std;

bool isPrime(int num) {
    if (num <= 1) return false;
    for (int i = 2; i <= sqrt(num); i++)
        if (num % i == 0) return false;
    return true;
}

int main() {
    int n;
    cout << "Enter the value of n: ";
    cin >> n;

    for (int num = 2; num <= n; num++)
        if (isPrime(num))
            cout << num << " ";

    return 0;
}

算法2：

#include <iostream>
#include <vector>
using namespace std;

void sieveOfEratosthenes(int n) {
    // 初始化bool数组，所有元素默认为true
    vector<bool> prime(n+1, true);

    for (int p = 2; p*p <= n; p++) {
        // 如果prime[p]没有被改变，那么它一定是一个素数
        if (prime[p] == true) {
            // 更新所有p的倍数为非素数
            for (int i = p*p; i <= n; i += p)
                prime[i] = false;
        }
    }

    // 输出所有素数
    for (int p = 2; p <= n; p++)
        if (prime[p])
            cout << p << " ";
}

int main() {
    int n;
    cout << "Enter the value of n: ";
    cin >> n;

    cout << "Following are the prime numbers smaller than or equal to " << n << endl;
    sieveOfEratosthenes(n);
    return 0;
}

算法2简短版本：

#include <iostream>
#include <vector>
using namespace std;

int main() {
    int n;
    cout << "Enter the value of n: ";
    cin >> n;
    vector<bool> prime(n+1, true);

    for (int p = 2; p*p <= n; p++) {
        if (prime[p]) {
            for (int i = p*p; i <= n; i += p)
                prime[i] = false;
        }
    }

    for (int p = 2; p <= n; p++)
        if (prime[p]) cout << p << " ";
    return 0;
}

稍微不一样的算法2：

#include <iostream>
#include <cstring>
using namespace std;

int main() {
    int n;
    cin >> n;
    bool* prime = new bool[n+1];
    memset(prime, true, n+1);
    for (int p = 2; p*p <= n; p++)
        if (prime[p])
            for (int i = p*p; i <= n; i += p)
                prime[i] = false;
    for (int p = 2; p <= n; p++)
        if (prime[p]) cout << p << " ";
    delete[] prime;
    return 0;
}