結果

問題 No.2671 NUPC Decompressor
ユーザー 2qbingxuan
提出日時 2024-03-15 22:13:38
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 3 ms / 2,000 ms
コード長 3,742 bytes
コンパイル時間 3,424 ms
コンパイル使用メモリ 258,984 KB
実行使用メモリ 5,248 KB
最終ジャッジ日時 2024-09-30 01:33:42
合計ジャッジ時間 4,131 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 13
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 #
プレゼンテーションモードにする

// An AC a day keeps the doctor away.
#include <bits/stdc++.h>
using namespace std;
/*{{{*/
#define all(x) begin(x), end(x)
#ifdef CKISEKI
#include <experimental/iterator>
#define safe cerr<<__PRETTY_FUNCTION__<<" line "<<__LINE__<<" safe\n"
#define debug(a...) debug_(#a, a)
#define orange(a...) orange_(#a, a)
void debug_(auto s, auto ...a) {
  cerr << "\e[1;32m(" << s << ") = (";
  int f = 0;
  (..., (cerr << (f++ ? ", " : "") << a));
  cerr << ")\e[0m\n";
}
void orange_(auto s, auto L, auto R) {
  cerr << "\e[1;33m[ " << s << " ] = [ ";
  using namespace experimental;
  copy(L, R, make_ostream_joiner(cerr, ", "));
  cerr << " ]\e[0m\n";
}
#else
#define safe ((void)0)
#define debug(...) safe
#define orange(...) safe
#endif
template <typename T, T MOD> class Modular {
  public:
    constexpr Modular() : v() {}
    template <typename U> Modular(const U &u) { v = (0 <= u && u < MOD ? u : (u%MOD+MOD)%MOD); }
    template <typename U> explicit operator U() const { return U(v); }
    T operator()() const { return v; }
#define REFOP(type, expr...) Modular &operator type (const Modular &rhs) { return expr, *this; }
    REFOP(+=, v += rhs.v - MOD, v += MOD & (v >> width)) ; REFOP(-=, v -= rhs.v, v += MOD & (v >> width))
    // fits for MOD^2 <= 9e18
    REFOP(*=, v = static_cast<T>(1LL * v * rhs.v % MOD)) ; REFOP(/=, *this *= inverse(rhs.v))
#define VALOP(op) friend Modular operator op (Modular a, const Modular &b) { return a op##= b; }
    VALOP(+) ; VALOP(-) ; VALOP(*) ; VALOP(/)
    Modular operator-() const { return 0 - *this; }
    friend bool operator == (const Modular &lhs, const Modular &rhs) { return lhs.v == rhs.v; }
    friend bool operator != (const Modular &lhs, const Modular &rhs) { return lhs.v != rhs.v; }
    friend std::istream & operator>>(std::istream &I, Modular &m) { T x; I >> x, m = x; return I; }
    friend std::ostream & operator<<(std::ostream &O, const Modular &m) { return O << m.v; }
  private:
    constexpr static int width = sizeof(T) * 8 - 1;
    T v;
    static T inverse(T a) {
      // copy from tourist's template
      T u = 0, v = 1, m = MOD;
      while (a != 0) {
        T t = m / a;
        m -= t * a; std::swap(a, m);
        u -= t * v; std::swap(u, v);
      }
      assert(m == 1);
      return u;
    }
};
using lld = int64_t;
using llf = long double;
template <typename T> using max_heap = std::priority_queue<T,vector<T>,less<T> >;
template <typename T> using min_heap = std::priority_queue<T,vector<T>,greater<T> >;
template <typename V, typename T> int get_pos(const V &v, T x) { return lower_bound(all(v),x) - begin(v); }
template <typename V> void sort_uni(V &v) { sort(all(v)), v.erase(unique(all(v)),end(v)); }
template <typename T> bool chmin(T &x, const T &v) { return v < x ? (x=v, true) : false; }
template <typename T> bool chmax(T &x, const T &v) { return x < v ? (x=v, true) : false; }
constexpr inline lld cdiv(lld x, lld m) { return x/m + (x%m ? (x<0) ^ (m>0) : 0); } // ceiling divide
constexpr inline lld modpow(lld e,lld p,lld m) { lld r=1; for(e%=m;p;p>>=1,e=e*e%m) if(p&1) r=r*e%m; return r; }/*}}}*/
constexpr llf eps = 1e-9;
constexpr lld maxn = 200025, INF = 1e18, mod = 998244353, K = 14699, inf = 1e9;
using Mint = Modular<int, mod>;
Mint modpow(Mint e, uint64_t p) { Mint r = 1; while (p) (p&1) && (r *= e), e *= e, p >>= 1; return r; } // 0^0 = 1
const auto dummy = [] { return cin.tie(nullptr)->sync_with_stdio(false); }();
signed main() {
  vector<string> a;
  string s = "NUPC";
  for (int m = 0; m < (1 << 4); m++) {
    string t;
    for (int i = 0; i < 4; i++) {
      t += s[i];
      if (m >> i & 1) t = t + t;
    }
    a.push_back(t);
  }
  sort(all(a));
  debug(a.size());
  int k;
  cin >> k;
  cout << a[k-1] << '\n';
}
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