結果

問題 No.233 めぐるはめぐる (3)
ユーザー fumiphysfumiphys
提出日時 2019-10-08 02:44:51
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 281 ms / 1,000 ms
コード長 3,600 bytes
コンパイル時間 1,977 ms
コンパイル使用メモリ 180,048 KB
実行使用メモリ 13,568 KB
最終ジャッジ日時 2024-10-15 16:15:24
合計ジャッジ時間 5,938 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge2
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 96 ms
13,056 KB
testcase_01 AC 51 ms
9,600 KB
testcase_02 AC 23 ms
6,528 KB
testcase_03 AC 61 ms
10,368 KB
testcase_04 AC 251 ms
13,440 KB
testcase_05 AC 234 ms
13,440 KB
testcase_06 AC 234 ms
13,568 KB
testcase_07 AC 276 ms
13,440 KB
testcase_08 AC 281 ms
13,440 KB
testcase_09 AC 2 ms
5,248 KB
testcase_10 AC 2 ms
5,248 KB
testcase_11 AC 2 ms
5,248 KB
testcase_12 AC 53 ms
9,728 KB
testcase_13 AC 108 ms
13,312 KB
権限があれば一括ダウンロードができます

ソースコード

diff # 

// includes
#include <bits/stdc++.h>
using namespace std;

// macros
#define pb emplace_back
#define mk make_pair
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define bit(n) (1LL<<(n))
// functions
template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}
template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}
template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}
template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << p.first << " " << p.second; return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
//  types
using ll = long long int;
using P = pair<int, int>;
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1000000007;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
// io
struct fast_io{
  fast_io(){ios_base::sync_with_stdio(false); cin.tie(0); cout << fixed << setprecision(20);}
} fast_io_;


int main(int argc, char const* argv[])
{
  int n; cin >> n;
  set<string> st;
  rep(i, n){
    string s; cin >> s;
    st.insert(s);
  }
  vector<char> a = {'i', 'a', 'a', 'e', 'u', 'u'};
  vector<char> b = {'n', 'b', 'm', 'g', 'r'};
  vector<int> e1(6), e2(5);
  rep(i, 6)e1[i] = i;
  rep(i, 5)e2[i] = i;
  do{
    do{
      rep(k, 6){
        string t = "";
        rep(i, 5){
          if(k == i)t += a[e1[0]];
          t += b[e2[i]];
          t += a[e1[i+1]];
        }
        if(k == 5)t += a[e1[0]];
        if(st.find(t) == st.end()){
          cout << t << endl;
          return 0;
        }
      }
    }while(next_permutation(all(e2)));
  }while(next_permutation(all(e1)));
  cout << "NO" << endl;
  return 0;
}
0