题解 | #素数判定#素数筛法模板题，记住就行。

//埃氏筛法
//C++版代码
#include <iostream>
#include <vector>
using namespace std;
int main() {
    vector<bool> isPrime(1001, true);
    isPrime[0] = isPrime[1] = false;
    for (int i = 2; i * i <= 1000; i++) {
        if (isPrime[i]) {
            for (int j = i; j * i <= 1000; j++) isPrime[i * j] = false;
        }
    }
    int n;
    while (cin >> n) {
        if (isPrime[n]) cout << "yes" << endl;
        else cout << "no" << endl;
    }
    return 0;
}
//Java版代码
import java.util.Arrays;
import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        boolean[] isPrimes = new boolean[10001];
        Arrays.fill(isPrimes, 2, 10001, true);
        for (int i = 2; i * i <= 10000; i++) {
            if (isPrimes[i]) {
                for (int j = i; j * i <= 10000; j++) isPrimes[j * i] = false;
            }
        }
        while (sc.hasNextInt()) {
            int n = sc.nextInt();
            if (isPrimes[n]) System.out.println("yes");
            else System.out.println("no");
        }
    }
}
//欧拉筛法
//C++版代码
#include <iostream>
#include <vector>
using namespace std;
int main() {
    vector<int> primes;
    bool isPrime[10001];
    for (int i = 2; i <= 10000; i++) isPrime[i] = true;
    for (int i = 2; i <= 10000; i++) {
        if (isPrime[i]) primes.push_back(i);
        for (int j = 0; j < primes.size() && i * primes[j] <= 10000; j++) {
            isPrime[i * primes[j]] = false;
            if (i % primes[j] == 0) break;
        }
    }
    int n;
    while (cin >> n) {
        if (isPrime[n]) cout << "yes" << endl;
        else cout << "no" << endl;
    }
    return 0;
}
//Java版代码
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        List<Integer> primes = new ArrayList<>();
        boolean[] isPrimes = new boolean[10001];
        Arrays.fill(isPrimes, 2, 10001, true);
        for (int i = 2; i <= 10000; i++) {
            if (isPrimes[i]) primes.add(i);
            for (int j = 0; j < primes.size() && i * primes.get(j) <= 10000; j++) {
                isPrimes[i * primes.get(j)] = false;
                if (i % primes.get(j) == 0) break;
            }
        }
        while (sc.hasNextInt()) {
            int n = sc.nextInt();
            if (isPrimes[n]) System.out.println("yes");
            else System.out.println("no");
        }
    }
}