1) Abridged problem statement

- There are a boys and b girls. Initially, 0 boys are with the teacher and all b girls are with the teacher.
- The teacher took n notes. Just before note i:
  - Some number of boys can join (never leave afterward).
  - Some number of girls can leave (never return).
  - Then the teacher counts the group size; it must equal the i-th recorded number.
- Find any schedule (for each note i: boys_joined_i, girls_left_i) that produces the notes. If impossible, print "ERROR".


2) Detailed editorial

Model:
- Let cnt[i] be the group size recorded at note i (1-based).
- Insert an initial “note” cnt[0] = b to represent the starting state (0 boys, b girls).
- Let at step i (after applying that step’s moves and before counting):
  - B_i = number of boys with the teacher,
  - G_i = number of girls with the teacher.
- Constraints:
  - B_0 = 0, G_0 = b
  - B_i ≥ B_{i-1} (boys only join and stay)
  - G_i ≤ G_{i-1} (girls only leave)
  - 0 ≤ B_i ≤ a, 0 ≤ G_i ≤ b
  - B_i + G_i = cnt[i] for i = 0..n (with cnt[0] = b)

Transition:
- Suppose we are at position pos (0 ≤ pos < n) with B_pos boys. Then G_pos is determined:
  - G_pos = cnt[pos] − B_pos
  - If G_pos is outside [0, b], this state is invalid.
- For the next note, pos+1:
  - Let y = number of girls leaving at this step, where 0 ≤ y ≤ G_pos.
  - Then G_{pos+1} = G_pos − y.
  - The recorded size cnt[pos+1] fixes B_{pos+1}:
    B_{pos+1} = cnt[pos+1] − G_{pos+1}.
  - Validate:
    - B_{pos+1} ≥ B_pos (non-decreasing boys),
    - 0 ≤ B_{pos+1} ≤ a, 0 ≤ G_{pos+1} ≤ b.
  - If valid, we can move to state (pos+1, B_{pos+1}), and the step’s action is:
    boys_joined = B_{pos+1} − B_pos, girls_left = y.

Algorithm:
- Perform DP/graph search over states (pos, B_pos), pos ∈ [0..n], B_pos ∈ [0..a].
- Start from (0, 0).
- For each state, try all possible y (girls leaving), deduce B_{pos+1}, check validity, and mark reachable.
- Store parent pointers to reconstruct moves:
  par[pos+1][B_{pos+1}] = (B_pos, boys_joined, girls_left).
- After filling DP, if any state (n, B_n) is reachable, reconstruct the path backwards and output the sequence of moves for i = 1..n. Otherwise print "ERROR".

Correctness:
- Every valid schedule obeys the monotonicity of B and G and the sum constraint at each note; the DP enumerates exactly those transitions, so every valid schedule is representable in the DP graph.
- Conversely, any path found by the DP defines nonnegative join/leave counts that satisfy all constraints and produce the recorded sequence.

Complexity:
- States: (n+1) × (a+1) ≤ 101 × 101.
- From each state, we try up to G_pos ≤ b possibilities for y, so O(n × a × b) ≤ about 10^6 operations.
- Memory: O(n × a).


3) Provided C++ solution with detailed comments

#include <bits/stdc++.h>             // Pull in standard headers (GNU extension).
using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second; // Print pair as "first second".
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second; // Read pair from stream.
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;                      // Read all elements of a vector.
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';              // Print vector elements separated by spaces.
    }
    return out;
};

int a, b, n;                          // a = boys total, b = girls total, n = number of notes.
vector<int> cnt;                      // cnt[i] = recorded group sizes (we will prepend b).

void read() {
    cin >> a >> b;                    // Read the number of boys and girls.
    cin >> n;                         // Read the number of notes.
    cnt.resize(n);                    // Resize cnt to hold n notes.
    cin >> cnt;                       // Read all recorded sizes.
    cnt.insert(cnt.begin(), b);       // Prepend initial size: initially 0 boys, b girls -> size b.
}

void dfs(
    int pos,                          // Current position in the sequence (0..n-1). pos=0 means before 1st note.
    int cnt_boys,                     // Current number of boys with the teacher at this pos.
    vector<vector<bool>>& visited,    // visited[pos][boys] indicates reachable state.
    vector<vector<tuple<int, int, int>>>& par // Parent pointers for reconstruction:
                                              // par[pos][boys] = {prev_boys, boys_joined, girls_left} to reach this state.
) {

    int cnt_girls = cnt[pos] - cnt_boys; // Current number of girls with teacher is determined by total at pos.

    // If counts are invalid, stop exploring this path.
    if(cnt_girls < 0 || cnt_boys < 0 || cnt_boys > a || cnt_girls > b) {
        return;
    }

    if(pos == n) {                    // If we've processed all notes, no further transitions.
        return;
    }

    // Try all possibilities: how many girls leave before the next note is taken.
    for(int cnt_girls_leave = 0; cnt_girls_leave <= cnt_girls; cnt_girls_leave++) {
        int new_cnt_girls = cnt_girls - cnt_girls_leave;         // Girls remaining after leaving.
        int new_cnt_boys = cnt[pos + 1] - new_cnt_girls;         // Boys needed to match next recorded total.

        // Validate new counts and monotonicity of boys, and avoid revisiting states.
        if(new_cnt_boys >= cnt_boys && new_cnt_boys <= a &&
           !visited[pos + 1][new_cnt_boys]) {

            // Record how we reached (pos+1, new_cnt_boys):
            // from 'cnt_boys' boys, with 'new_cnt_boys - cnt_boys' joining,
            // and 'cnt_girls_leave' girls leaving.
            par[pos + 1][new_cnt_boys] = {
                cnt_boys, new_cnt_boys - cnt_boys, cnt_girls_leave
            };

            visited[pos + 1][new_cnt_boys] = true;               // Mark as reachable.
            dfs(pos + 1, new_cnt_boys, visited, par);            // Recurse to next position.
        }
    }
}

void solve() {
    // visited[pos][boys] indicates we can reach the (pos, boys) state.
    vector<vector<bool>> visited(n + 1, vector<bool>(a + 1, false));

    // par[pos][boys] stores the parent state and the move used to reach (pos, boys).
    // Tuple fields: (prev_boys, boys_joined, girls_left); initialize with invalid markers.
    vector<vector<tuple<int, int, int>>> par(
        n + 1, vector<tuple<int, int, int>>(a + 1, {-1, -1, -1})
    );

    // Start DFS from the initial state: pos=0, boys=0 (girls are implicitly cnt[0]-boys = b).
    dfs(0, 0, visited, par);

    // Among all possible numbers of boys at the end, find any reachable one.
    for(int cnt_boys = 0; cnt_boys <= a; cnt_boys++) {
        if(visited[n][cnt_boys]) {
            vector<pair<int, int>> ans;     // Store (boys_joined, girls_left) for each step.
            int pos = n, c = cnt_boys;      // Reconstruct backwards from (n, cnt_boys).
            while(pos > 0) {
                auto [prev_boys, move_boys, move_girls] = par[pos][c]; // Retrieve how we got here.
                ans.push_back({move_boys, move_girls});                // Append this step.
                c = prev_boys;                                         // Move to previous state.
                pos--;
            }

            reverse(ans.begin(), ans.end()); // Reverse to get steps 1..n in correct order.

            // Output each step: boys joined, girls left.
            for(auto [x, y]: ans) {
                cout << x << ' ' << y << '\n';
            }
            return;                           // Any valid solution is acceptable.
        }
    }
    cout << "ERROR" << endl;                  // No reachable ending state found.
}

int main() {
    ios_base::sync_with_stdio(false);         // Fast I/O.
    cin.tie(nullptr);

    int T = 1;                                // Single test case (can be extended).
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();                               // Read input.
        // cout << "Case #" << test << ": ";
        solve();                              // Solve and print the answer.
    }

    return 0;
}


4) Python solution (well-commented)

import sys

def solve():
    data = sys.stdin.read().strip().split()
    it = iter(data)
    a = int(next(it))  # total boys
    b = int(next(it))  # total girls
    n = int(next(it))  # number of notes
    rec = [int(next(it)) for _ in range(n)]  # recorded group sizes
    # Prepend the implicit initial "note": starting size is exactly b (0 boys, b girls).
    cnt = [b] + rec

    # dp[pos][boys] is True if we can reach after processing 'pos' notes (pos in [0..n])
    # having exactly 'boys' boys with the teacher.
    dp = [[False] * (a + 1) for _ in range(n + 1)]

    # par[pos][boys] = (prev_boys, boys_joined_this_step, girls_left_this_step)
    # to reconstruct the schedule. Only meaningful when dp[pos][boys] is True and pos > 0.
    par = [[None] * (a + 1) for _ in range(n + 1)]

    # Initial state: before any recorded note, we are at pos = 0 with 0 boys.
    # Girls are implicitly cnt[0] - 0 = b.
    dp[0][0] = True

    # Iterate over positions 0..n-1 and try to reach pos+1.
    for pos in range(n):
        for boys in range(a + 1):
            if not dp[pos][boys]:
                continue
            # Compute current girls; this must match the recorded size at pos.
            girls = cnt[pos] - boys
            if girls < 0 or girls > b:
                # This state cannot be valid; skip.
                continue

            # Try all possibilities for the number of girls leaving before next note.
            for girls_leave in range(girls + 1):
                new_girls = girls - girls_leave
                # Next recorded total fixes the new number of boys.
                new_boys = cnt[pos + 1] - new_girls

                # Validate non-decreasing boys and bounds.
                if new_boys < boys or new_boys < 0 or new_boys > a:
                    continue

                if not dp[pos + 1][new_boys]:
                    dp[pos + 1][new_boys] = True
                    par[pos + 1][new_boys] = (boys, new_boys - boys, girls_leave)

    # Find any reachable end state after n notes.
    end_boys = None
    for c in range(a + 1):
        if dp[n][c]:
            end_boys = c
            break

    if end_boys is None:
        print("ERROR")
        return

    # Reconstruct actions from par.
    actions = []
    pos = n
    boys = end_boys
    while pos > 0:
        prev = par[pos][boys]
        # prev must exist because dp[pos][boys] is True.
        prev_boys, joined, left = prev
        actions.append((joined, left))
        boys = prev_boys
        pos -= 1

    actions.reverse()
    for joined, left in actions:
        print(joined, left)

if __name__ == "__main__":
    solve()


5) Compressed editorial

- Insert an initial “note” cnt[0] = b to represent the starting state with 0 boys and b girls.
- DP over states (pos, B): after pos notes, B boys are with the teacher; girls are determined as G = cnt[pos] − B.
- From (pos, B), try all y in [0..G] girls leaving; then G' = G − y and B' = cnt[pos+1] − G'. Require B' ≥ B, 0 ≤ B' ≤ a, 0 ≤ G' ≤ b. Mark (pos+1, B') reachable and store parent (B, B'−B, y).
- If any (n, B_n) is reachable, backtrack to output per step: boys_joined and girls_left; else print ERROR.
- Complexity: O(n·a·b) time, O(n·a) memory.