Submission #1074801

#	Time	Username	Problem	Language	Result	Execution time	Memory
1074801		fv3	Rarest Insects (IOI22_insects)	C++17	43.30 / 100	163 ms	712 KiB

insects

#include "insects.h"
#include <bits/stdc++.h>
 
using namespace std;
 
int min_cardinality(int N) 
{
  // Find the number of different insect types:
  // Maximum number of insects with 1 being maximum
  // Let that number be K
  // Find another set of K numbers with 2 being the maximum
  // Then 3 and 4 and so on untill it is no longer possible
 
  // The grader is not adaptive:
  vector<int> indicies(N);
  iota(indicies.begin(), indicies.end(), 0);
 
  mt19937 mt(time(0));
  shuffle(indicies.begin(), indicies.end(), mt);
 
  vector<int> in(N);
  vector<int> order;

  int k = 0;
  int insect_cnt = 0;
  for (int i = 0; i < N; i++)
  {
    move_inside(indicies[i]);
    if (press_button() > 1)
    {
      move_outside(indicies[i]);
    }
    else
    {
      k++;
      insect_cnt++;
      order.push_back(i);
      in[i] = 1;
    }
  }

  if (k < sqrt(N))
  {
    int res = 1;
    while (true)
    {
      int cnt = 0;
      for (int i = 0; i < N && cnt < k; i++)
      {
        if (in[i]) continue;
   
        move_inside(indicies[i]);
        if (press_button() > res + 1)
        {
          move_outside(indicies[i]);
        }
        else
        {
          cnt++;
          in[i] = 1;
        }
      }
   
      if (cnt < k)
        return res;
      res++;
    }
   
    return res;
  }

  int l = 1, r = N / k;
  while (l < r)
  {
    int c = (l + r + 1) / 2;
    while (insect_cnt > k * l)
    {
      move_outside(indicies[order.back()]);
      in[order.back()] = 0;
      order.pop_back();
      insect_cnt--;
    }

    for (int i = 0; i < N && insect_cnt < k * c; i++)
    {
      if (in[i]) continue;

      move_inside(indicies[i]);
      if (press_button() > c)
      {
        move_outside(indicies[i]);
      }
      else
      {
        insect_cnt++;
        order.push_back(i);
        in[i] = 1;
      }
    }
    
    if (insect_cnt >= k * c)
      l = c;
    else
      r = c - 1;
  }
 
  return l;
}

• Subtask 1: 10 / 10
• Subtask 2: 15 / 15
• Subtask 3: 18.3 / 75