Submission #1087749

#	Time	Username	Problem	Language	Result	Execution time	Memory
1087749		juicy	Abracadabra (CEOI22_abracadabra)	C++17	100 / 100	447 ms	41544 KiB

Main

#include <bits/stdc++.h>

using namespace std;

#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) 42
#endif

const int N = 2e5 + 5, Q = 1e6 + 5;

int n, q;
int a[N], nxt[N], pos[N], s[4 * N], res[Q];

void upd(int i, int x, int id = 1, int l = 1, int r = n) {
  if (l == r) {
    s[id] = x;
    return;
  }
  int m = (l + r) / 2;
  if (i <= m) {
    upd(i, x, id * 2, l, m);
  } else {
    upd(i, x, id * 2 + 1, m + 1, r);
  }
  s[id] = s[id * 2] + s[id * 2 + 1];
}

array<int, 2> walk(int k) {
  int id = 1, l = 1, r = n;
  while (l ^ r) {
    int m = (l + r) / 2;
    id *= 2;
    if (s[id] >= k) {
      r = m;
    } else {
      l = m + 1;
      k -= s[id];
      id += 1;
    }
  }
  return {k, l};
}

void add(int i, int j) {
  while (i < j) {
    nxt[i] = min(j, nxt[i]);
    upd(a[i], nxt[i] - i);
    i = nxt[i];
  }
}

int main() {
  ios::sync_with_stdio(false); cin.tie(nullptr);

  cin >> n >> q;
  vector<int> st;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    pos[a[i]] = i;
    while (st.size() && a[st.back()] < a[i]) {
      nxt[st.back()] = i;
      st.pop_back();
    }
    st.push_back(i);
  }
  while (st.size()) {
    nxt[st.back()] = n + 1;
    st.pop_back();
  }
  vector<array<int, 3>> queries;
  for (int i = 1; i <= q; ++i) {
    int t, p; cin >> t >> p;
    queries.push_back({min(t, n), p, i});
  }
  sort(queries.begin(), queries.end());
  add(1, n + 1);
  bool c = 1;
  int m = n / 2;
  for (int i = 0, j = 0; i <= n; ++i) {
    while (j < q && queries[j][0] == i) {
      auto [t, p, id] = queries[j++];
      auto [l, x] = walk(p);
      res[id] = a[pos[x] + l - 1];
    }
    if (!c) {
      continue;
    }
    auto [l, x] = walk(m);
    if (l == nxt[pos[x]] - pos[x]) {
      c = 0;
      continue;
    }
    add(pos[x] + l, nxt[pos[x]]);
    nxt[pos[x]] = pos[x] + l;
    upd(x, l);
  }
  for (int i = 1; i <= q; ++i) {
    cout << res[i] << "\n";
  }
  return 0;
}

• Subtask 1: 10 / 10
• Subtask 2: 40 / 40
• Subtask 3: 25 / 25
• Subtask 4: 25 / 25