結果

問題 No.3075 Mex Recurrence Formula
ユーザー 👑 hos.lyric
提出日時 2025-03-28 23:17:15
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 36 ms / 2,000 ms
コード長 3,845 bytes
コンパイル時間 1,377 ms
コンパイル使用メモリ 115,336 KB
実行使用メモリ 8,472 KB
最終ジャッジ日時 2025-03-28 23:17:23
合計ジャッジ時間 4,041 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge4
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 * 46
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp: In function ‘int main()’:
main.cpp:131:12: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
  131 |       scanf("%d", &A[i]);
      |       ~~~~~^~~~~~~~~~~~~

ソースコード

diff #
プレゼンテーションモードにする

#include <cassert>
#include <cmath>
#include <cstdint>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <bitset>
#include <chrono>
#include <complex>
#include <deque>
#include <functional>
#include <iostream>
#include <limits>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
using namespace std;
using Int = long long;
template <class T1, class T2> ostream &operator<<(ostream &os, const pair<T1, T2> &a) { return os << "(" << a.first << ", " << a.second << ")"; };
template <class T> ostream &operator<<(ostream &os, const vector<T> &as) { const int sz = as.size(); os << "["; for (int i = 0; i < sz; ++i) { if (i 
    >= 256) { os << ", ..."; break; } if (i > 0) { os << ", "; } os << as[i]; } return os << "]"; }
template <class T> void pv(T a, T b) { for (T i = a; i != b; ++i) cerr << *i << " "; cerr << endl; }
template <class T> bool chmin(T &t, const T &f) { if (t > f) { t = f; return true; } return false; }
template <class T> bool chmax(T &t, const T &f) { if (t < f) { t = f; return true; } return false; }
#define COLOR(s) ("\x1b[" s "m")
struct Set {
  // max{ceil(log_64(n)), 1}
  int log64N, n;
  vector<unsigned long long> a[6];
  explicit Set(int n_ = 0) : n(n_) {
    assert(n >= 0);
    int m = n ? n : 1;
    for (int d = 0; ; ++d) {
      m = (m + 63) >> 6;
      a[d].assign(m, 0);
      if (m == 1) {
        log64N = d + 1;
        break;
      }
    }
  }
  bool empty() const {
    return !a[log64N - 1][0];
  }
  bool contains(int x) const {
    return (a[0][x >> 6] >> (x & 63)) & 1;
  }
  void insert(int x) {
    for (int d = 0; d < log64N; ++d) {
      const int q = x >> 6, r = x & 63;
      a[d][q] |= 1ULL << r;
      x = q;
    }
  }
  void erase(int x) {
    for (int d = 0; d < log64N; ++d) {
      const int q = x >> 6, r = x & 63;
      if ((a[d][q] &= ~(1ULL << r))) break;
      x = q;
    }
  }
  // max s.t. <= x (or -1)
  int prev(int x) const {
    if (x > n - 1) x = n - 1;
    for (int d = 0; d <= log64N; ++d) {
      if (x < 0) break;
      const int q = x >> 6, r = x & 63;
      const unsigned long long lower = a[d][q] << (63 - r);
      if (lower) {
        x -= __builtin_clzll(lower);
        for (int e = d; --e >= 0; ) x = x << 6 | (63 - __builtin_clzll(a[e][x]));
        return x;
      }
      x = q - 1;
    }
    return -1;
  }
  // min s.t. >= x (or n)
  int next(int x) const {
    if (x < 0) x = 0;
    for (int d = 0; d < log64N; ++d) {
      const int q = x >> 6, r = x & 63;
      if (static_cast<unsigned>(q) >= a[d].size()) break;
      const unsigned long long upper = a[d][q] >> r;
      if (upper) {
        x += __builtin_ctzll(upper);
        for (int e = d; --e >= 0; ) x = x << 6 | __builtin_ctzll(a[e][x]);
        return x;
      }
      x = q + 1;
    }
    return n;
  }
};
int N;
Int X;
vector<int> A;
vector<int> freq;
Set off;
void add(int a) {
  if (a <= N) if (!freq[a]++) off.erase(a);
}
void rem(int a) {
  if (a <= N) if (!--freq[a]) off.insert(a);
}
int mex() {
  return off.next(0);
}
int main() {
  for (; ~scanf("%d%lld", &N, &X); ) {
    --X;
    A.resize(N);
    for (int i = 0; i < N; ++i) {
      scanf("%d", &A[i]);
    }
    
    freq.assign(N + 1, 0);
    off = Set(N + 1);
    for (int a = 0; a <= N; ++a) off.insert(a);
    for (int i = 0; i < N; ++i) {
      add(A[i]);
    }
    for (int i = N; ; ++i) {
      const int res = mex();
      A.push_back(res);
      rem(A[i - N]);
      add(A[i]);
      if (A[i] == N) break;
    }
    
    int ans;
    if (X < (int)A.size()) {
      ans = A[X];
    } else {
      const int I = (int)A.size() - (N + 1);
      ans = A[I + (X - I) % (N + 1)];
    }
    printf("%d\n", ans);
  }
  return 0;
}
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