1. Abridged Problem Statement  
You have an elevator starting at floor f. A sequence of distinct buttons e₁…eₙ (none equal to f) are pressed in quick succession and become “highlighted.” The elevator then visits all highlighted floors as follows:
- It targets the earliest-pressed highlighted floor a.
- It moves one floor at a time toward a. If it passes any highlighted floor b along the way, it stops there, unhighlights b, and records that stop.
- Upon reaching a, it stops, unhighlights a, and records that stop.
- It then repeats, targeting the next earliest-pressed remaining highlighted floor, until none remain.

Output the floors in the order the elevator stops.

2. Detailed Editorial  
We need to simulate the elevator’s movement under the rules given. Key points:
- The set of pending stops is exactly the list of pressed buttons, in the order pressed.
- At any moment, the elevator has:
  • A current floor `cur`.  
  • A current target floor `tgt`, which is always the earliest-pressed among the still-pending floors.
- The elevator moves one floor at a time toward `tgt` (incrementing or decrementing `cur` by 1).  
- Whenever `cur` matches one of the pending floors, the elevator stops there immediately, unhighlights that floor (removes it from the pending list), and appends it to the answer.  
- If that floor was also the current target, we select a new `tgt` as the first element in the (now-updated) pending list.  
- Repeat until the pending list is empty.

Implementation details:
- Store the pending floors in a vector in the original pressed order.
- Keep track of `cur` and `tgt`. Initialize `cur = f`, `tgt = pending[0]`.
- Loop while `pending` is not empty:
   1. Check if `cur` is in `pending`. If so, record `cur` in the answer and erase it from `pending`.
   2. If `pending` became empty, break.
   3. If the erased floor was equal to `tgt`, update `tgt = pending[0]`.
   4. Move `cur` one step toward `tgt` (i.e., if `cur < tgt`, do `cur++`; else `cur--`).
- Finally, print the answer sequence.

Time complexity: each step either removes a floor (≤ n removals) or moves one floor (≤ 100 total movement per removal), and searching/removing in a vector of size ≤ 100 is O(100). Overall worst-case O(n·max_floor·n) which is fine for n,f ≤ 100.

3. Provided C++ Solution with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// Overloads to read/write pairs and vectors conveniently
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for (auto& x : a) in >> x;
    return in;
}
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for (auto x : a) out << x << ' ';
    return out;
}

int n, f;           // n = number of button presses, f = starting floor
vector<int> a;      // list of pressed buttons (highlighted floors)

void read() {
    cin >> n >> f;     // read n and initial floor
    a.resize(n);
    cin >> a;          // read the pressed floors in order
}

void solve() {
    int cur = f;               // current floor
    int tgt = a[0];            // target = earliest-pressed pending floor
    vector<int> ans;           // to store the stops

    // Continue until we have unhighlighted all floors
    while (!a.empty()) {
        // If current floor is highlighted, stop here
        auto it = find(a.begin(), a.end(), cur);
        if (it != a.end()) {
            ans.push_back(cur);         // record stop
            bool wasTarget = (cur == tgt);
            a.erase(it);                // unhighlight this floor
            if (a.empty()) break;       // done if none remain
            if (wasTarget) tgt = a[0];  // update target if needed
        }
        // Move one floor toward tgt
        if (cur < tgt) cur++;
        else if (cur > tgt) cur--;
        // if cur == tgt, the next iteration will detect it and remove it
    }

    // Output the sequence of stops
    cout << ans << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    read();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
def main():
    import sys
    data = sys.stdin.read().split()
    n, cur = map(int, data[:2])
    pending = list(map(int, data[2:]))

    # tgt = earliest-pressed pending floor
    tgt = pending[0]
    ans = []

    # Loop until all pending floors are visited
    while pending:
        # If we are on a pending floor, visit it immediately
        if cur in pending:
            ans.append(cur)
            was_target = (cur == tgt)
            pending.remove(cur)
            if not pending:
                break
            if was_target:
                # next target is the new earliest-pressed
                tgt = pending[0]
        # Move one floor toward the target
        if cur < tgt:
            cur += 1
        elif cur > tgt:
            cur -= 1
        # if cur == tgt, the next loop iteration will handle removal

    # Print result
    print(' '.join(map(str, ans)))

if __name__ == '__main__':
    main()
```

5. Compressed Editorial  
Simulate the elevator: keep a list of outstanding floors in press order. Maintain current floor `cur` and current target `tgt` = first in list. While the list isn’t empty, if `cur` matches a pending floor, record and remove it; if it was the target, reset `tgt` to the new first element. Otherwise move `cur` by ±1 toward `tgt`. Repeat until all are visited; output stops.