## 1) Abridged problem statement

You must typeset an article of **N words** on a page of fixed width **W** characters. Word *i* has length **Li** and words are placed in order using this greedy rule:

- Each word occupies **Li consecutive free cells** in a row.
- Between consecutive words in the same row there must be **at least 1 empty cell** (a "space").
- Among all valid placements consistent with previous words, choose the **earliest** possible position (topmost row, then leftmost column).

Before placing text, you must place one rectangular illustration of size **R rows × C columns** anywhere; its cells become occupied.

A row is "used" if after placing illustration and text it contains at least one occupied cell.
For each test case, find the **minimum possible number of used rows** over all illustration placements.

---

## 2) Detailed editorial (explaining the solution)

### Key viewpoint: we only care about "what word starts each row"
Without any illustration, greedy text layout is deterministic: it walks through words, maintaining a cursor (row, col). With an illustration, the layout is still greedy/deterministic, but now each row has a blocked interval (the illustration's columns) for **R consecutive rows**.

The illustration placement can be described by:
- its **top row** (somewhere; unbounded downward, but only relative to where text ends matters),
- its **left column** `col` (0…W−C),
- and therefore, in each affected row, there are two available free segments:
  - left segment width = `gap_left = col`
  - right segment width = `gap_right = W - col - C`

If a word doesn't fit entirely in a free segment (respecting spaces between words), the greedy algorithm moves forward until it can place it, potentially skipping to the next row.

The trick is to evaluate, for each column `col`, how the presence of R "split rows" changes which word index is reached after passing those R rows.

---

### Precompute: "how many words fit starting at v in a segment of width s"
Define a function:

`fit(v, s) = index of first word NOT placed after filling one row segment of usable width s, starting from word v`

So if `fit(v, s) = t`, it means words `v..t-1` fit in that segment.

We need this fast for all `v` and all `s <= W`, repeatedly.

The code builds:

`len_next[v][s]` for all `v in [0..N]`, `s in [0..W]`

in **O(N·W)** using a sliding process per starting word `v`. Then:

- If `v==N` or `s < L[v]`, return `v` (nothing fits)
- else return `len_next[v][s]`

This avoids an O(log N) binary search per query.

---

### Model one illustration row as a "jump" function next(v)
In a row that contains the illustration, usable cells are split into:
- left segment of width `gap_left`
- then blocked `C`
- then right segment of width `gap_right`

Greedy placement within that row will:
1) Place as many words as possible in left segment: `v1 = fit(v, gap_left)`
2) Then continue in right segment: `v2 = fit(v1, gap_right)`

So for that fixed column placement, one illustrated row maps:

`next(v) = fit(fit(v, gap_left), gap_right)`

Passing through **R** illustrated rows means applying this mapping R times:

`next^R(v)`

We need this for many starting `v` (the word at the beginning of each row).

---

### Get row starts in the "no illustration" layout
The solution computes `row_start[j]`: the word index that begins row `j` in normal layout (no illustration).

This is done by simulating the plain greedy layout with `advance(...)`, which updates the (row, col) where each word starts, respecting spaces and line breaks.

Let `num_lines` be the last used row index in the no-illustration layout (0-based in code); the baseline answer is then roughly `num_lines + 1` (rows count), but the code stores variants and takes mins.

---

### Fast computation of next^R(v): path compression memoization
Naively, applying `next` R times per starting row is O(R) each, too slow.

Binary lifting would be O(log R) but still heavier with tight 0.25s.

Instead, the code uses a *memoized recursion with compression*:

For each `v`, store:
- `comp[v].result`: where you end up after `comp[v].steps` applications
- `comp[v].steps`: how many steps that stored result corresponds to

When asked for `(v, steps)`:
- If already computed with some `comp[v].steps`, reuse it and recurse only for remaining steps.
- Otherwise compute one `next(v)` and recurse.

Because many queries share subpaths (successive rows' starts are close in word index), this "compresses" chains aggressively in practice.

This is created per column via `make_compress(gap_left, gap_right)`.

---

### Evaluate all columns
For each column `col` (0..W−C):
- Build the compressed `next^k` solver for this `col`.
- For each row `j` in the base layout:
  - Start word is `row_start[j]`
  - After R illustrated rows: `end = next^R(row_start[j])`
  - This yields a candidate "state" for dynamic combination with the rest of layout, stored in `best[end]`.

Interpretation of stored `best[x]`:
- It keeps the best (smallest) coordinate `(row, col)` representing the earliest position just before placing word `x` in the final combined layout.
- For ends reached after inserting the illustration block, we store a candidate row count like `j + R - (end==N)` (minor adjustment: if the text ends exactly at N, the last illustrated row might not need to be counted as used by text).

---

### Final pass: lay out the remaining words normally
Once `best` has initial states from trying all illustration columns, the code "propagates" them word-by-word:

For each `j` from 0 to N−1:
- `best[j+1] = min(best[j+1], advance(j+1, best[j]))`

This is just greedy placement continuing from the best known position for word `j`, producing a candidate position for word `j+1`.

Finally, `best[N].first + 1` is the number of used rows (row index + 1) in the best arrangement. Output the minimum.

---

### Complexity
Let `W ≤ (small, per statement)`, `sum N` bounded.

Per test:
- Precompute `len_next`: **O(N·W)**
- For each `col` in `0..W-C`:
  - compression structure reset O(N)
  - process each base row once; compression makes jumps fast in practice

Overall it is designed to fit very tight time limits with low constants.

---

## 3) 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, w, r, c;
vector<int> lens;

void read() {
    cin >> n >> w >> r >> c;
    lens.resize(n + 1);
    for(int i = 0; i < n; i++) {
        cin >> lens[i];
    }
    lens[n] = 0;
}

void solve() {
    // Given the constraints, we could afford a solution that is O(N*(R+W+C)),
    // or something of this sort. Naively, we could have tried all possible
    // locations for the upper left corner of the image in O((N + R) * C), and
    // then in O(N) checked how many rows this would actually take. To speed
    // this up, we can actually do something very similar - in stead of choosing
    // the full location, let's just choose the column in O(C). Now we can try
    // all possible options for the image starting below the row that ends with
    // word number i. More precisely, we can compute something like next[pos],
    // which corresponds to having our current row ending on word pos equaling
    // the number of the last word after one more row, but with an image here
    // (at the already fixed column). This next[.] can be computed easily in
    // O(log N) with binary search or O(1) with sliding window. Afterwards it's
    // enough to just try every possible position, and then apply
    // next[next[...]] from each position R times, to see where we would end up
    // after R rows. The candidate for the answer is then pref_rows[pos] + R +
    // suff_rows[next^R[pos]+1]. The slowest part here is computing next^R[pos],
    // as we can't afford a O(R) loop. However, we can do binary lifting and do
    // this in O(log R) instead. Overall, this solution would potentially be
    // still slow as it's in O(W * N * (log N + log R)).
    //
    // To speed this up further, we make two key improvements. First, we
    // precompute a table len_next[v][length] for every word v and every
    // available length, storing how many words fit starting from v with that
    // much space, using a sliding window in O(N * W) total, giving O(1) per
    // query instead of O(log N). Second, we replace binary lifting for
    // computing next^R[pos] with path compression similar to union-find. For
    // a fixed column, we need next^R[v] for each starting row's first word
    // v. We process rows j = 0, 1, ..., and for each, follow the chain
    // storing at each visited node both the final result and the number of
    // steps it compressed. When a later query passes through an already-
    // visited node v, it jumps directly to v's stored result and only extends
    // the chain by the remaining steps. Crucially, later rows always reach
    // shared nodes with more remaining steps (since the starting word is
    // closer to v, fewer steps were consumed reaching it), so the chain only
    // ever extends forward. This in practice is very quick and could maybe
    // be shown to be O(N) but I didn't get a clean argument after thinking for
    // a bit.
    //
    // Another option is to think of the tree where the parent of some position
    // is next[pos]. We want to find the K-th parent of each position, which 
    // can actually be done with a stack and a DFS: whenever we enter a node, 
    // we add it to the stack, while when we exit we pop it. It's always enough 
    // to just get the element at position depth(pos)-K, which we can do with 
    // a DFS in O(N).

    vector<vector<int>> len_next(n + 1, vector<int>(w + 1, n));
    for(int v = 0; v <= n; v++) {
        int nxt = min(n, v + 1);
        len_next[v][lens[v]] = nxt;
        for(int k = lens[v] + 1, gap = 0; k <= w - c; k++) {
            gap++;
            if(nxt < n && gap >= 1 + lens[nxt]) {
                nxt++;
                gap = 0;
            }
            len_next[v][k] = nxt;
        }
    }

    auto fit = [&](int v, int space) -> int {
        if(v == n || space < lens[v]) {
            return v;
        }
        return len_next[v][space];
    };

    auto advance = [&](int v, pair<int, int> prev) -> pair<int, int> {
        int col = prev.second + lens[v - 1] + (v < n ? 1 : 0);
        int row = prev.first;
        if(col + lens[v] > w) {
            col = 0;
            row++;
        }
        return {row, col};
    };

    vector<int> row_start(n + 1, 0);
    pair<int, int> cur = {0, 0};
    int num_lines = 0;
    for(int j = 1; j <= n; j++) {
        cur = advance(j, cur);
        if(cur.second == 0) {
            row_start[cur.first] = j;
        }
        if(j == n - 1) {
            num_lines = cur.first;
        }
    }

    struct Comp {
        int result = 0, steps = 0;
    };
    vector<Comp> comp(n + 1);

    auto make_compress = [&](int gap_left, int gap_right) {
        fill(comp.begin(), comp.end(), Comp{});
        auto impl = [&, gap_left,
                     gap_right](auto&& self, int v, int steps) -> int {
            if(v == n || steps == 0) {
                return v;
            }
            if(comp[v].steps) {
                if(comp[v].result != v) {
                    comp[v].result =
                        self(self, comp[v].result, steps - comp[v].steps);
                }
            } else {
                int nxt = fit(fit(v, gap_left), gap_right);
                comp[v].result = (nxt != v) ? self(self, nxt, steps - 1) : nxt;
            }
            comp[v].steps = steps;
            return comp[v].result;
        };
        return [impl](int v, int steps) { return impl(impl, v, steps); };
    };

    vector<pair<int, int>> best(n + 1, {n + r + 1, 0});

    int ans = num_lines + r + 1;
    for(int col = 0; col + c <= w; col++) {
        auto path_compress = make_compress(col, w - col - c);
        for(int j = 0; j <= num_lines; j++) {
            int end = path_compress(row_start[j], r);
            best[end] = min(best[end], {j + r - (end == n), 0});
        }
    }

    for(int j = 0; j < n; j++) {
        best[j + 1] = min(best[j + 1], advance(j + 1, best[j]));
    }

    ans = min(ans, best[n].first + 1);
    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();
        solve();
    }

    return 0;
}
```

---

## 4) Python solution (same approach, with detailed comments)

```python
import sys
sys.setrecursionlimit(1_000_000)

def solve_case(n, w, r, c, lens_input):
    # lens[0..n-1] are real words; lens[n] is sentinel 0
    lens = lens_input[:] + [0]

    # Precompute len_next[v][space] for v in [0..n], space in [0..w]
    # Only spaces up to (w - c) are needed for illustrated segments, but w is small enough.
    len_next = [[n] * (w + 1) for _ in range(n + 1)]
    max_seg = w - c

    for v in range(n + 1):
        nxt = min(n, v + 1)
        lv = lens[v]
        if lv <= w:
            len_next[v][lv] = nxt

        gap = 0
        # Fill for k = lv+1 .. max_seg
        for k in range(lv + 1, max_seg + 1):
            gap += 1
            if nxt < n and gap >= 1 + lens[nxt]:
                nxt += 1
                gap = 0
            len_next[v][k] = nxt

    def fit(v, space):
        """Return first word index not fitting in a contiguous segment of length space."""
        if v == n or space < lens[v]:
            return v
        return len_next[v][space]

    def advance(v, prev):
        """
        Normal-layout greedy placement transition.
        prev = (row, col_start_of_prev_word)
        returns (row, col_start_of_word_v)
        Words are 1-indexed in this transition, matching the C++ code style.
        """
        row, col_prev = prev
        # Move to after previous word and one mandatory space (unless v == n sentinel-ish)
        col = col_prev + lens[v - 1] + (1 if v < n else 0)
        if col + lens[v] > w:
            return (row + 1, 0)
        return (row, col)

    # Compute row_start for the base layout (no illustration)
    row_start = [0] * (n + 1)
    cur = (0, 0)
    num_lines = 0
    for j in range(1, n + 1):
        cur = advance(j, cur)
        if cur[1] == 0:
            row_start[cur[0]] = j
        if j == n - 1:
            num_lines = cur[0]

    # best[i] = best known (row, col) before placing word i
    INF = n + r + 10
    best = [(INF, 0)] * (n + 1)

    # Per-column path compression structure
    class Comp:
        __slots__ = ("result", "steps")
        def __init__(self):
            self.result = 0
            self.steps = 0

    comp = [Comp() for _ in range(n + 1)]

    def make_compress(gap_left, gap_right):
        # Reset memo for this column
        for i in range(n + 1):
            comp[i].result = 0
            comp[i].steps = 0

        def jump(v, steps):
            """Compute next^steps(v) with memoized compression."""
            if v == n or steps == 0:
                return v

            if comp[v].steps != 0:
                # Extend from stored result if needed
                if comp[v].result != v:
                    comp[v].result = jump(comp[v].result, steps - comp[v].steps)
            else:
                nxt = fit(fit(v, gap_left), gap_right)
                comp[v].result = jump(nxt, steps - 1) if nxt != v else nxt

            comp[v].steps = steps
            return comp[v].result

        return jump

    # Try all illustration columns
    for col in range(0, w - c + 1):
        jump = make_compress(col, w - col - c)
        for j in range(0, num_lines + 1):
            end = jump(row_start[j], r)
            candidate = (j + r - (1 if end == n else 0), 0)
            if candidate < best[end]:
                best[end] = candidate

    # Propagate remaining words with normal greedy layout
    for j in range(0, n):
        cand = advance(j + 1, best[j])
        if cand < best[j + 1]:
            best[j + 1] = cand

    return best[n][0] + 1  # row index + 1 => number of used rows


def main():
    data = list(map(int, sys.stdin.buffer.read().split()))
    t = data[0]
    idx = 1
    out = []
    for _ in range(t):
        n, w, r, c = data[idx:idx+4]
        idx += 4
        lens = data[idx:idx+n]
        idx += n
        out.append(str(solve_case(n, w, r, c, lens)))
    sys.stdout.write("\n".join(out))

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

---

## 5) Compressed editorial

- Precompute `fit(v, s)`: after placing as many words as possible starting from word `v` into a contiguous free segment of width `s`, return the next word index. Build a table `len_next[v][s]` in `O(NW)` using a sliding gap counter.
- For a fixed illustration column `col`, each affected row has two free segments: `gap_left = col`, `gap_right = W-col-C`. One illustrated row maps `v -> next(v) = fit(fit(v, gap_left), gap_right)`.
- If illustration spans `R` rows, we need `next^R(v)` for many `v` (row starts). Compute it fast using memoized path compression (store for each `v` the result after some number of steps).
- Compute `row_start[j]` from the normal layout.
- For each `col`, for each base row `j`, compute `end = next^R(row_start[j])` and update best known state for reaching `end`.
- Finally, propagate `best` through normal greedy placement for the remaining words and output the minimum used rows.
