2023번: 신기한 소수 ↗

/*
[2023: 신기한 소수](https://www.acmicpc.net/problem/2023)

Tier: Gold 5 
Category: math, number_theory, backtracking, primality_test
*/

#include <bits/stdc++.h>

using namespace std;

#define for1(s, e) for(int i = s; i < e; i++)
#define for1j(s, e) for(int j = s; j < e; j++)
#define forEach(k) for(auto i : k)
#define forEachj(k) for(auto j : k)
#define sz(vct) vct.size()
#define all(vct) vct.begin(), vct.end()
#define sortv(vct) sort(vct.begin(), vct.end())
#define uniq(vct) sort(all(vct));vct.erase(unique(all(vct)), vct.end())
#define fi first
#define se second
#define INF (1ll << 60ll)

typedef unsigned long long ull;
typedef long long ll;
typedef ll llint;
typedef unsigned int uint;
typedef unsigned long long int ull;
typedef ull ullint;

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
typedef pair<double, int> pdi;
typedef pair<string, string> pss;

typedef vector<int> iv1;
typedef vector<iv1> iv2;
typedef vector<ll> llv1;
typedef vector<llv1> llv2;

typedef vector<pii> piiv1;
typedef vector<piiv1> piiv2;
typedef vector<pll> pllv1;
typedef vector<pllv1> pllv2;
typedef vector<pdd> pddv1;
typedef vector<pddv1> pddv2;

const double EPS = 1e-8;
const double PI = acos(-1);

template<typename T>
T sq(T x) { return x * x; }

int sign(ll x) { return x < 0 ? -1 : x > 0 ? 1 : 0; }
int sign(int x) { return x < 0 ? -1 : x > 0 ? 1 : 0; }
int sign(double x) { return abs(x) < EPS ? 0 : x < 0 ? -1 : 1; }

ll N;

llv2 primes = {
  {},
  {2, 3, 5, 7}
};

bool is_prime(ll k) {
  for(int i = 2; i * i <= k; i++) {
    if (k % i == 0) return false;
  }
  return true;
}

void solve() {
  cin >> N;

  for(int i = 2; i <= N; i++) {
    primes.push_back({});

    for(ll prev : primes[i - 1]) {
      for(int j = 1; j <= 9; j += 2) {
        ll nxt = prev * 10 + j;

        if(is_prime(nxt)) {
          primes[i].push_back(nxt);
        }
      }
    }
  }

  for(auto p : primes[N]) {
    cout << p << "\n";
  }
}

int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);cout.tie(NULL);
  int tc = 1; // cin >> tc;
  while(tc--) solve();

}