제출 #1249996

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1249996		Jakub_Wozniak	Triple Peaks (IOI25_triples)	C++20	42.07 / 100	2004 ms	2632 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
int ile = 0;

vector<int> construct_range(int M, int K)
{

  if(M == 0)return {};
  if(M == 1)return {1};
  if(M == 2)return {1,1};
  if(M == 3)return {1,2,1};
  if(M == 4)return {1,2,2,1};
  if(M == 5)return {1,1,2,1,1};
  if(M == 6)return {1,2,1,1,2,1};
  if(M == 7)return {1,2,1,2,1,2,1};
  if(M == 8)return {1,1,2,1,1,2,1,1};

  int S = 5;
  if(M > 100)S = 7;
  
  int R = M -((S-1) + S*((M-S+1)/S));
  M = ((S-1) + S*((M-S+1)/S));
  vector <int> temp = construct_range((M-S+1)/S,K);
  vector <int> r;
  
  for(int i = 0 ; i < S ; i++)
  {
    if(i%2 == 0)for(auto p : temp)r.push_back(p);
    else for(auto p : temp)r.push_back((M-S+1)/S+1-p);
    if(i != S-1)r.push_back((M-S+1)/S+1);
  }

  temp = construct_range(R , K);
  for(auto p : temp)r.push_back(p);

  return r;
}

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: 4 / 5
• Subtask 8: 1.235 / 5
• Subtask 9: 0.6431 / 5
• Subtask 10: 0.2642 / 5
• Subtask 11: 0.7755 / 5
• Subtask 12: 0.1564 / 5