제출 #1249961

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1249961		Jakub_Wozniak	3개의 봉우리 (IOI25_triples)	C++20	39.89 / 100	2095 ms	2736 KiB

triples

#include "triples.h"

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<vector <int> ,ll> pvl;
#define st first
#define nd second

vector <int> rek(int N)
{
  if(N == 0)return {};
  if(N == 1)return {1};
  if(N == 2)return {1,1};
  if(N%3 != 2)
  {
    vector <int> temp = rek(N-1); temp.push_back(1);
    return temp;
  }
  vector <int > r;
  int s1 = N/3 ;
  int s2 = s1+N/3+1;
  vector <int> temp = rek((N/3));
  for(auto p : temp)r.push_back(p);
  r.push_back(N/3+1);
  for(auto p : temp)r.push_back(N/3+1-p);
  r.push_back(N/3+1);
  for(auto p : temp)r.push_back(p);

  return r;
}

vector<int> construct_range(int M, int K)
{
    int pot = 1;
    while(pot < M)pot *= 3;
    pot /= 3;
    vector <int> res = rek(pot);
    if(2*res.size() <= M)
    {
      vector <int> temp = res;
      for(auto p : temp)res.push_back(p);
    }
    return res;
}

long long count_triplesnd(vector<int> H) 
{
  int N = H.size();
  ll ile = 0;
  for(int i = N-1 ; i >= 0 ; i--)
  {
    int k = i-H[i];
    if(k < 0)continue;
    int j = k+H[k];
    if(j < 0 || j >= N);
    else if(H[k]+H[j] == H[i] && (i-j) == H[j] && k < j && j < i)ile++;
    j = i-H[k];
    if(k+H[k] == j)continue;
    if(j < 0 || j >= N);
    else if(H[k]+H[j] == H[i] && (j-k) == H[j] && k < j && j < i)ile++;
  }
  return ile;
}

long long count_triples(vector<int> H) 
{
  int N = H.size();
  int B = 0;
  bool czynd = 1;
  for(int i = 0; i < N ; i++)B = max(B , H[i]);

  for(int i = 1; i < N ; i++)
  {
    if(H[i-1] > H[i])czynd = 0;
  }

  if(czynd)return count_triplesnd(H);

  ll ile = 0;
  for(int i = 0 ; i < N ; i++)
  {
    for(int j = i+1 ; j < min(N,i+B+1) ; j++)
    {
      int d = j-i;
      if(H[i] == d || H[j] == d)
      {
        int R = 0;
        if(H[i]  == d)R = H[j];
        else R = H[i];
        int k = j+R;
        if(k >= 0 && k < N && k > j && H[k] == R+d)ile++;
        k = i+R;
        if(k >= 0 && k < N && k > j && R == H[k]+d)ile++;
      }
      else
      {
        int k = j+min(H[i] , H[j]);
        if(k >= 0 && k < N && H[k] == d && min(H[i],H[j])+d == max(H[i],H[j]))ile++;
      }
    }
  }
  return ile;
}

• Subtask 0: 0 / 0
• Subtask 1: 8 / 8
• Subtask 2: 6 / 6
• Subtask 3: 10 / 10
• Subtask 4: 11 / 11
• Subtask 5: 0 / 16
• Subtask 6: 0 / 19
• Subtask 7: 1.3333 / 5
• Subtask 8: 1.38 / 5
• Subtask 9: 0.7752 / 5
• Subtask 10: 0.4232 / 5
• Subtask 11: 0.5876 / 5
• Subtask 12: 0.3901 / 5