제출 #1328262

#제출 시각아이디문제언어결과실행 시간메모리
1328262adiyerTriple Peaks (IOI25_triples)C++20
41.21 / 100
2094 ms25260 KiB
#include <bits/stdc++.h>
#include "triples.h"

using namespace std;

typedef long long ll;

const int N = 200000;

ll ans;

int n;
int a[N], c[N];

void print(ll x, char sep = ' '){
    cout << x << sep;
}

bool check(int i, int j, int k){
    c[k - j]++, c[j - i]++, c[k - i]++, c[a[i]]--, c[a[j]]--, c[a[k]]--;
    if(!c[k - j] && !c[j - i] && !c[k - i]) return 1;
    c[k - j] = c[j - i] = c[k - i] = c[a[i]] = c[a[j]] = c[a[k]] = 0;
    return 0;
}

ll count_triples(vector < int > h){
    n = h.size(), ans = 0;
    map < vector < int >, bool > was;
    for(int i = 0; i < n; i++) a[i] = h[i];
    for(int i, k, j = 0; j < n; j++){
        i = j - a[j];
        if(i >= 0){
            k = i + a[i];
            if(k < n && j < k && check(i, j, k) && !was[{i, j, k}]) ans++, was[{i, j, k}] = 1;
            k = j + a[i];
            if(k < n && j < k && check(i, j, k) && !was[{i, j, k}]) ans++, was[{i, j, k}] = 1;
        }
        k = j + a[j];
        if(k < n){
            i = k - a[k];
            if(i >= 0 && i < j && check(i, j, k) && !was[{i, j, k}]) ans++, was[{i, j, k}] = 1;
            i = j - a[k];
            if(i >= 0 && i < j && check(i, j, k) && !was[{i, j, k}]) ans++, was[{i, j, k}] = 1;
         }
    }
    for(int j = 0; j < n; j++){
        for(int i = 0; i < j; i++){
            int k = i + a[j];
            if(a[j] == k - i && a[i] == j - i && a[k] == k - j){
                if(0 <= i && i < j && j < k && k < n && check(i, j, k)) ans++;
            }
        }
    }
    for(int j = 0; j < n; j++){
        for(int i = 0; i < j; i++){
            int k = i + a[j];
            if(a[j] == k - i && a[i] == k - j && a[k] == j - i && k - j != j - i){
                if(0 <= i && i < j && j < k && k < n && check(i, j, k)) ans++;
            }
        }
    }
    //1)
    // a[j] = k - i -> i + a[a[i] + i] = k
    // a[i] = j - i -> a[i] + i = j
    // a[k] = k - j -> a[k] - k = j
    //2)
    // a[j] = k - i -> a[j] = a[i] + j - i -> a[j] - j = a[i] - i
    // a[i] = k - j -> a[i] + j = k
    // a[k] = j - i -> a[k] = a[i] + a[j]
    // a[a[i] + j] = a[i] + a[j] && a[j] - j == a[i] - i
    return ans;
}

vector < int > construct_range(int n, int k){    
    vector < int > v;
    n /= 5;
    while(n--) v.push_back(1), v.push_back(1), v.push_back(2), v.push_back(2), v.push_back(3);
    return v;
}
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