제출 #1250194

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1250194		s4dz	3개의 봉우리 (IOI25_triples)	C++20	0 / 100	27 ms	15684 KiB

triples

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
//sub4:
std::vector<int> construct_range(int M, int K)
{
    return{};
}
/*long long count_triples(vector<int> H)
{
    int n = H.size();
    ll ans = 0;
    for (int k = 0; k < n; k++)
    {
        int c = H[k];
        int i = k - c;
        if (i < 0 || i >= n) continue;

        int a = H[i];
        if (a > c - a) continue;

        int need = c - a;
        int j1   = i + a;
        int j2   = k - a;
        if (i < j1 && j1 < k && H[j1] == need) ans++;
        if (j2 != j1 && i < j2 && j2 < k && H[j2] == need) ans++;
    }
    return ans;
}*/

ll count_triples(vector<int> H) 
{
    int N = H.size();
    ll ans = 0;
    for(int i = 0; i < N; i++){
        int k = i + H[i];
        if(k >= N) continue;
        {
            int j = i + H[k];
            if(j > i && j < k && H[j] == k - j) ans++;
        }
        {
            int j = k - H[k];
            if(j > i && j < k && H[j] == j - i) ans++;
        }
    }
    for(int k = 0; k < N; k++)
    {
        int i = k - H[k];
        if(i < 0) continue;
        {
            int j = i + H[i];
            if(j > i && j < k && H[j] == k - j) ans++;
        }
        {
            int j = k - H[i];
            if(j > i && j < k && H[j] == j - i) ans++;
        }
    }
    for(int i = 0; i < N; i++)
    {
        int j = i + H[i];
        if(j >= N) continue;
        int k = i + H[j];
        if(k < N && H[k] == k - j) ans++;
    }
    int B = int(sqrt(N));
    vector<vector<int>> groups(2*N + 1);
    for(int i = 0; i < N; i++)
    {
        int key = H[i] - i + N;
        groups[key].push_back(i);
    }
    for(int key = 0; key <= 2*N; key++)
    {
        auto &v = groups[key];
        if(v.empty()) continue;
        int C = key - N;
        if((int)v.size() <= B)
        {
            for(int a = 0; a < (int)v.size(); a++)
            {
                for(int b = a+1; b < (int)v.size(); b++)
                {
                    int i = v[a], j = v[b];
                    int k = i + j + C;
                    if(k < N && j < k && H[k] == j - i)
                        ans++;
                }
            }
        } 
        else 
        {
            for(int k = 0; k < N; k++)
            {
                int sum = k - C;      
                int diff = H[k];     
                if(((sum + diff) & 1) != 0) continue;
                int j = (sum + diff) >> 1;
                int i = j - diff;
                if(0 <= i && i < j && j < k 
                   && H[i]-i == C 
                   && H[j]-j == C)
                {
                    ans++;
                }
            }
        }
    }

    return ans;
}

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