제출 #1251234

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1251234		aryan12	선물 (IOI25_souvenirs)	C++20	4 / 100	1086 ms	412 KiB

souvenirs

#include "souvenirs.h"
#include <utility>
#include <vector>
#include <bits/stdc++.h>
using namespace std;

/*
P[i + 2] * 2 <= P[i + 1] + P[i + 2] <= P[i] <= 2 * P[i + 1]
So for a given value, P[i] (highest) cannot be < val / 3
*/

void buy_souvenirs(int N, long long P0) {
  vector<int> cnt(N, 0);
  vector<long long> lowest_price_for_which_highest(N + 1, -1);
  lowest_price_for_which_highest[0] = P0;
  iota(cnt.begin(), cnt.end(), 0);
  long long cur_val = P0 - 1;
  std::pair<std::vector<int>, long long> res = transaction(cur_val);
  lowest_price_for_which_highest[1] = cur_val - res.second;
  for(int x: res.first) {
    cnt[x] -= 1;
  }
  bool omk = false, cntt = true;
  while(true) {
    int sum = accumulate(cnt.begin(), cnt.end(), 0LL);
    if(sum == 0) break;
    if(!cntt) {
      for(int i = N - 1; i >= 0; i--) {
        if(lowest_price_for_which_highest[i] == -1) {
          lowest_price_for_which_highest[i] = lowest_price_for_which_highest[i + 2] * 2;
          cur_val = lowest_price_for_which_highest[i];
          omk = true;
          cntt = true;
          break;
        }
      }
    }
    else if(res.first.size() == 1) {
      lowest_price_for_which_highest[res.first[0]] = cur_val - res.second;
      lowest_price_for_which_highest[res.first[0] + 1] = cur_val - res.second - 1;

      for(int i = N - 1; i >= 0; i--) {
        if(lowest_price_for_which_highest[i] == -1) {
          lowest_price_for_which_highest[i] = lowest_price_for_which_highest[i + 2] * 2;
          cur_val = lowest_price_for_which_highest[i];
          omk = true;
          cntt = true;
          break;
        }
      }
    }
    else if(res.first.size() >= 3) {
      cur_val = (cur_val - res.second) / 3;
    }
    else {
      cur_val = (cur_val - res.second) / 2;
    }
    if(!omk || cntt) {
      res = transaction(cur_val);
      for(int x: res.first) {
        cnt[x] -= 1;
      }
      lowest_price_for_which_highest[res.first[0]] = cur_val - res.second;
      if(omk) cntt = false;
    }
  }
  // std::pair<std::vector<int>, long long> res = transaction(3);
  return;
}

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