Submission #1173802

#	Time	Username	Problem	Language	Result	Execution time	Memory
1173802		xyz__	Doktor (COCI17_doktor)	C++20	10 / 100	24 ms	2116 KiB

doktor

#include <iostream>
#include <vector>
using namespace std;
 
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n;
    cin >> n;
    // use 1-indexed vector for convenience.
    vector<int> p(n+1);
    for (int i = 1; i <= n; i++){
        cin >> p[i];
    }
    
    // Option 1: Look for a pair that gives +2 improvement.
    for (int i = 1; i <= n; i++){
        int j = p[i];
        // Check valid index and that it is a mutual pair and not fixed originally.
        if(i != j && j >= 1 && j <= n && p[j] == i) {
            // We want to output the contiguous subarray from min(i, j) to max(i, j).
            int l = min(i, j);
            int r = max(i, j);
            // Output the card numbers at positions l and r.
            cout << p[l] << " " << p[r] << "\n";
            return 0;
        }
    }
    
    // Option 2: Look for an adjacent pair that gives +1 improvement.
    for (int i = 1; i < n; i++){
        // Consider subarray [i, i+1]
        // After reversal, position i gets p[i+1] and becomes fixed if p[i+1]== i,
        // and position i+1 gets p[i] and becomes fixed if p[i]== i+1.
        if(p[i+1] == i && p[i] != i){ // ensure that the improvement is not cancelling an already fixed point
            cout << p[i] << " " << p[i+1] << "\n";
            return 0;
        }
        if(p[i] == i+1 && p[i+1] != i+1){
            cout << p[i] << " " << p[i+1] << "\n";
            return 0;
        }
    }
    
    // Option 3: If no improvement can be made, choose a single element reversal.
    // We can pick the first card.
    cout << p[1] << " " << p[1] << "\n";
    return 0;
}

• Subtask 1: 10 / 10
• Subtask 2: 0 / 10
• Subtask 3: 0 / 10
• Subtask 4: 0 / 10
• Subtask 5: 0 / 10
• Subtask 6: 0 / 10
• Subtask 7: 0 / 10
• Subtask 8: 0 / 10
• Subtask 9: 0 / 10
• Subtask 10: 0 / 10