1. Abridged Problem Statement  
You have a stack (“pile”) of books, initially containing N books (given top to bottom). You perform M operations of two types:  
- ADD(S): place a book named S on top of the pile.  
- ROTATE: reverse the order of the top K books (or the entire pile if it has fewer than K).  
After all operations, output the pile from top to bottom.

Constraints:  
• 0 ≤ N ≤ 40 000, 0 ≤ M ≤ 100 000, 0 ≤ K ≤ 40 000  
• Book names are 1–3 uppercase letters; duplicates allowed.  

2. Detailed Editorial  

We need to support two operations on a sequence:  
- **push front** (ADD),  
- **reverse prefix of fixed length K** (ROTATE).  

Naïve array or list reversal on each ROTATE would be O(K) per operation, which in the worst case (M up to 10^5, K up to 4·10^4) is too slow. We aim for amortized O(1) per operation.

Key idea: **maintain the first K elements separately** from the rest, so that reversing them or inserting at the front takes O(1). Concretely:

  A. Split the pile logically into two deques:
     - `prefix`: the top min(current_size, K) books,
     - `suffix`: the remaining books below.

  B. Keep a boolean flag `rev` for `prefix` that indicates whether the prefix is logically reversed.

Operations:

  1. **Initial setup**  
     - Read the N initial books into a deque `all_books`.  
     - Move all but the first K elements into `suffix` by repeatedly popping from the right of `all_books` and pushing to the left of `suffix`.  
     - Let `prefix = all_books`; set `rev = false`.

  2. **ADD(S)**  
     - We want to insert S at the front of the pile (top). This may increase `prefix` size beyond K.  
     - To “push front” on `prefix` under our `rev` flag:
         if not `rev`: `prefix.appendleft(S)`
         else:         `prefix.append(S)`  
     - If after insertion the size of `prefix` exceeds K, pop the “last” element of `prefix` (under `rev`) and push it to the front of `suffix`. This keeps `prefix` at size K.

  3. **ROTATE**  
     - We simply toggle `rev = !rev`. A single boolean flip reverses our logical view of the prefix in O(1).

  4. **Output**  
     - Print the `prefix` in the correct order (if `rev` is true, traverse it in reverse), then the `suffix` in normal order.

This achieves O(1) per operation and O(N+M) overall.

An alternative is an **implicit treap (or splay tree)** with lazy-propagated reversal on any subtree, splitting by size. The provided C++ solution uses exactly that:  
- Build an implicit treap of size N.  
- On ADD, merge a single-node treap to the left.  
- On ROTATE, split the treap into [0..K−1] and [K..end], flip the reverse-lazy flag on the first part, then re-merge.

3. Provided C++ Solution with Detailed Comments  

```cpp
#include <bits/stdc++.h>
using namespace std;

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

// Overload >> for pair to read "first second"
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Overload >> for vector<T>
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
}

// Overload << for vector<T> (space-separated)
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
}

// A node of an implicit treap supporting lazy reversal
template<
    class KeyT,             // payload type (here: string)
    class T,                // monoid type (empty in our case)
    T (*merge_func)(T, T),  // how to merge monoids
    class LazyT,            // lazy tag type (here: reverse flag)
    uint64_t (*rng)()       // random generator for priorities
>
struct TreapNode {
    KeyT key;              // the actual book name
    T data;                // node’s own monoid value
    T subtree;             // aggregated monoid over subtree
    uint64_t prior;        // heap-priority
    size_t size;           // size of subtree
    TreapNode *left, *right;
    LazyT lazy;            // lazy tag for this node

    // Constructor: set key/data, random priority, no children
    TreapNode(KeyT key_, T data_)
        : key(key_), data(data_), left(nullptr), right(nullptr), size(1) {
        prior = rng();
        lazy = LazyT();
    }

    // Recompute size and subtree-aggregate from children
    void pull() {
        subtree = data;
        size = 1;
        if(left) {
            left->push();                       // ensure left is up-to-date
            subtree = merge_func(left->subtree, subtree);
            size    += left->size;
        }
        if(right) {
            right->push();
            subtree = merge_func(subtree, right->subtree);
            size    += right->size;
        }
    }

    // Push lazy tags down before accessing children
    void push() {
        lazy.apply_lazy(this);
    }

    // Helper to push at a possibly-null node
    friend void push_lazy(TreapNode* t) {
        if(t) t->push();
    }

    // Split by key (NOT used in this problem) – splits into <key, ≥key
    friend pair<TreapNode*,TreapNode*> split(TreapNode* t, KeyT key) {
        if(!t) return {nullptr,nullptr};
        t->push();
        if(key < t->key) {
            auto [l, r] = split(t->left, key);
            t->left = r; t->pull();
            return {l, t};
        } else {
            auto [l, r] = split(t->right, key);
            t->right = l; t->pull();
            return {t, r};
        }
    }

    // Split by subtree size: first 'size' nodes go left, rest go right
    friend pair<TreapNode*, TreapNode*> split_by_size(TreapNode* t, size_t size) {
        if(!t) return {nullptr,nullptr};
        t->push();
        size_t left_sz = t->left ? t->left->size : 0;
        if(left_sz >= size) {
            auto [l, r] = split_by_size(t->left, size);
            t->left = r; t->pull();
            return {l, t};
        } else {
            auto [l, r] = split_by_size(t->right, size - 1 - left_sz);
            t->right = l; t->pull();
            return {t, r};
        }
    }

    // Merge two treaps assuming all nodes in l come before nodes in r
    friend TreapNode* merge(TreapNode* l, TreapNode* r) {
        push_lazy(l); push_lazy(r);
        if(!l || !r) return l ? l : r;
        if(l->prior > r->prior) {
            l->right = merge(l->right, r);
            l->pull();
            return l;
        } else {
            r->left = merge(l, r->left);
            r->pull();
            return r;
        }
    }
};

// A simple empty monoid (we don't need any aggregated data here)
struct EmptyMonoid {
    static EmptyMonoid merge(EmptyMonoid a, EmptyMonoid b) {
        return EmptyMonoid();
    }
};

// Lazy tag to reverse a subtree
template<class T>
struct ReverseLazy {
    bool should_reverse;
    ReverseLazy() : should_reverse(false) {}

    // If should_reverse is set, we swap children and propagate the flag
    template<class G, uint64_t (*rng)(), T (*merge_func)(T,T)>
    void apply_lazy(TreapNode<G, T, merge_func, ReverseLazy, rng>* node) {
        if(!node || !should_reverse) return;
        swap(node->left, node->right);
        if(node->left)  node->left->lazy.should_reverse  ^= true;
        if(node->right) node->right->lazy.should_reverse ^= true;
        should_reverse = false;
    }
};

// Type alias for our implicit treap storing string keys
using TreapWithReverse =
    TreapNode<string, EmptyMonoid, EmptyMonoid::merge, ReverseLazy<EmptyMonoid>, 
              /*rng=*/[](){ static mt19937_64 gen(42); return gen(); }>;

// Globals for input
int n, m, k;
vector<string> names;

// Read N, M, K and the initial stack of book names
void read() {
    cin >> n >> m >> k;
    names.resize(n);
    for(int i = 0; i < n; i++){
        cin >> names[i];
    }
}

// Solve using implicit treap + lazy reversal
void solve() {
    // 1) Build initial treap from the N names
    TreapNode<string,EmptyMonoid,EmptyMonoid::merge,ReverseLazy<EmptyMonoid>,decltype(&TreapWithReverse::prior)>* root = nullptr;
    for(auto &s: names) {
        // Create a node and merge it to the right end
        auto *node = new TreapWithReverse::Node(s, EmptyMonoid());
        root = merge(node, root);
    }

    // Process M operations
    while(m--) {
        string op;
        cin >> op;
        if(op.rfind("ADD(", 0) == 0) {
            // Extract the name inside ADD(...)
            string name = op.substr(4, op.size() - 5);
            auto *node = new TreapWithReverse::Node(name, EmptyMonoid());
            // Merge the new node to the front
            root = merge(node, root);

        } else { // "ROTATE"
            // Split off the first k elements
            auto [t1, t2] = split_by_size(root, k);
            // Toggle the reverse flag on that prefix
            if(t1) t1->lazy.should_reverse ^= true;
            // Re-merge
            root = merge(t1, t2);
        }
    }

    // In-order traversal to print final stack (left -> self -> right)
    function<void(TreapWithReverse::Node*)> dfs = [&](auto *node) {
        if(!node) return;
        node->push();          // ensure children correct
        dfs(node->left);       // first the left subtree
        cout << node->key << "\n"; // then this book
        dfs(node->right);      // then the right subtree
    };
    dfs(root);
}

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

    read();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments  

```python
from sys import stdin
from collections import deque

def main():
    data = stdin.read().split()
    it = iter(data)
    n = int(next(it))
    m = int(next(it))
    K = int(next(it))

    # Read the initial pile, top at the front of this deque
    all_books = deque()
    for _ in range(n):
        all_books.append(next(it))

    # Split into prefix (first K) and suffix (rest)
    prefix = all_books
    suffix = deque()
    while len(prefix) > K:
        # Move the bottom of prefix to the top of suffix
        suffix.appendleft(prefix.pop())

    # rev=False means prefix is in "normal" order; True means logically reversed
    rev = False

    # Helpers to push_front and pop_back on prefix under rev-flag
    def prefix_push_front(book):
        # if not reversed, push to left; else push to right
        if not rev:
            prefix.appendleft(book)
        else:
            prefix.append(book)

    def prefix_pop_back():
        # if not reversed, pop rightmost; else pop leftmost
        return prefix.pop() if not rev else prefix.popleft()

    # Process operations
    for _ in range(m):
        op = next(it)
        if op.startswith("ADD"):
            # Extract name inside ADD(...)
            name = op[4:-1]
            # Insert at pile front => prefix front
            prefix_push_front(name)
            # If prefix got too big, move one book to suffix front
            if len(prefix) > K:
                moved = prefix_pop_back()
                suffix.appendleft(moved)

        else:  # ROTATE
            # Just toggle the reverse flag on prefix
            rev = not rev

    # Output the final pile: prefix then suffix
    out = []

    # Print prefix in correct order
    if rev:
        # If reversed, we iterate prefix from right to left
        for book in reversed(prefix):
            out.append(book)
    else:
        # Normal: left to right
        for book in prefix:
            out.append(book)

    # Then the suffix is always in normal order
    for book in suffix:
        out.append(book)

    # Print each on its own line
    print("\n".join(out))


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

5. Compressed Editorial  
Maintain the top K books separately from the rest, storing the prefix in a deque with a boolean “reversed” flag.  
- **ADD**: push into the logical front of the prefix; if it grows beyond K, pop its logical back into the front of the suffix.  
- **ROTATE**: flip the reverse flag (O(1)).  
Finally, output prefix (respecting the reverse flag) followed by suffix. This runs in O(N+M) time and O(N) space.