Problem - D - Codeforces

题意：

思路：

这种题一定要从小数据入手，不然很有可能走歪思路

先考虑n = 1的情况，直接输出即可

然后是n = 2的情况，如果相加是质数，就输出2个，否则就输出一个

然后是n = 3的情况

看这个集合能不能是3个数

2 3 5不行

4 5 8不行

注意到三个数一定是奇 偶 奇，只要没有1，就一定不行

想到这就很简单了，分类讨论即可

Code：

#include <bits/stdc++.h>

#define int long long

using i64 = long long;

constexpr int N = 2e3 + 10;
constexpr int M = 2e6 + 10;
constexpr int P = 2600;
constexpr i64 Inf = 1e18;
constexpr int mod = 1e9 + 7;
constexpr double eps = 1e-6;

int n;
int len = 0;
int a[N];
int vis[M], prime[M];

void p_init(int n) {
    for (int i = 2; i <= n; i ++) {
        if (!vis[i]) prime[++len] = i;
        for (int j = 1; prime[j] <= n / i; j ++) {
            vis[i * prime[j]] = 1;
            if (i % prime[j] == 0) break;
        }
    }
}
void solve() {
    std::cin >> n;
    
    int cnt1 = 0;
    for (int i = 1; i <= n; i ++) {
        std::cin >> a[i];
        if (a[i] == 1) cnt1 ++;
    }

    std::sort(a + 1, a + 1 + n);

    if (cnt1) {
        for (int i = cnt1 + 1; i <= n; i ++) {
            if (!vis[a[i] + 1]) {
                std::cout << cnt1 + 1 << "\n";
                for (int i = 1; i <= cnt1; i ++) std::cout << 1 << " ";
                std::cout << a[i] << "\n";
                return;
            }
        }

        if (cnt1 >= 2) {
            std::cout << cnt1 << "\n";
            for (int i = 1; i <= cnt1; i ++) std::cout << 1 << " \n" [i == cnt1];
            return;
        }
    }

    for (int i = 1; i <= n; i ++) {
        for (int j = i + 1; j <= n; j ++) {
            if (!vis[a[i] + a[j]]) {
                std::cout << 2 << "\n";
                std::cout << a[i] << " " << a[j] << "\n";
                return;
            }
        }
    }

    std::cout << 1 << "\n" << a[1] << "\n";
}
signed main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    
    int t = 1;
    p_init(2e6);

    while (t--) {
        solve();
    }
    
    return 0;
}