editorial

#include <iostream>

using namespace std;

int main()
{
    ///Problem C
    for (int l = 0; l < n; l++) {
        int cur_sum = 0;
        for (int r = l; r < n; r++) {
            cur_sum += a[r];
            if (cur_sum == 0) {
                cout << l << ' ' << r << endl;
            }
        }
    }

    ///Problem D
    ///pref[i] = a[0] + a[1] + ... + a[i]
    ///pref[i] = pref[i - 1] + a[i]
    ///a[l] + a[l + 1] + ... + a[r] = pref[r] - pref[l - 1] = 0
    ///for r = 0..n
    /// if there is such l <= r so that pref[l - 1] = pref[r] ??

    map<int, bool> was;
    was[0] = 0;
    int cur_sum = 0;
    for (int r = 0; i < n; i++) {
        cur_sum += a[r];
        if (was[cur_sum]) {
            cout << was[cur_sum] << ' ' << r << endl;
            return 0;
        }
        was[cur_sum] = r;
    }

    ///Problems E-F
    ///1. calculate number of L's = x
    ///2. calculate number of R's among first x elements = y
    ///3. y is the answer
    ///[] [] [R] [] [R] [] ... [] [] []


    ///problem H
    ///Dijkstra

    ///problem I
    ///1. If the size of the clique (n / 2) => approx. (n/2)^2 edges
    ///2.1. DSU
    ///2.2. clique 2 (v1, v2, v3) => add c2, connect it with v1, v2, v3
    ///                                v1 with v2, v1 with v3
    return 0;
}