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 `f`.
  • 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 `f` by 1).
- Whenever `f` 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. Since the problem guarantees all buttons are distinct, at most one removal happens per check.
- If `f` equaled 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 `f` (current floor) and `tgt`. Initialize `tgt = a[0]`.
- The inner loop removes all pending entries that equal `f` before checking whether we have reached the target and before advancing. Because all buttons are distinct, at most one removal fires per visit.
- Move `f` one step toward `tgt` after handling any removal at the current position.
- 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
```cpp
#include <bits/stdc++.h>

using namespace std;

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;
vector<int> a;

void read() {
    cin >> n >> f;
    a.resize(n);
    cin >> a;
}

void solve() {
    // Direct simulation of the elevator. The pressed buttons stay in their
    // press order in the vector a, so the current destination tgt is always
    // a[0] (the earliest pressed among the still-highlighted buttons). We move
    // the current floor f one step at a time towards tgt; whenever some
    // highlighted button equals the current floor we record a stop there and
    // remove it. When we reach tgt we pick the next a[0] as the new
    // destination, stopping once no highlighted buttons remain.

    int tgt = a[0];
    vector<int> ans;
    while(!a.empty()) {
        bool added = false;
        while(true) {
            auto it = find(a.begin(), a.end(), f);
            if(it != a.end()) {
                if(!added) {
                    ans.push_back(f);
                }
                added = true;
                a.erase(it);
            } else {
                break;
            }
        }

        if(tgt == f) {
            if(a.empty()) {
                break;
            } else {
                tgt = a[0];
            }
        }

        if(tgt > f) {
            f++;
        } else {
            f--;
        }
    }

    cout << ans << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        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 `f` and current target `tgt` = first in list. While the list isn't empty, if `f` matches a pending floor, record and remove it; if it was the target, reset `tgt` to the new first element. Otherwise move `f` by ±1 toward `tgt`. Repeat until all are visited; output stops.
