## 1. Abridged problem statement

Given an `N × N` grid (`1 ≤ N ≤ 20`) containing a set `Q` of marked cells (`1`), decide whether `Q` can be divided into two disjoint subsets `A` and `B` such that one subset can be transformed into the other by:

- rotating by `0`, `90`, `180`, or `270` degrees,
- then shifting by some vector.

If possible, print `YES` and output one valid subset `A` as an `N × N` binary grid. Otherwise print `NO`.

---

## 2. Detailed editorial

We need split the marked cells `Q` into two parts `A` and `B` such that:

```text
B = f(A)
```

where `f` is one of the allowed transformations: a rotation by a multiple of `90°`, followed by a translation.

Since `A` and `B` form a partition of `Q`, they must have equal size. Therefore, if `|Q|` is odd, the answer is immediately `NO`.

### Representing transformations

Use grid coordinates `(r, c)`.

The four rotations around the origin can be represented as:

```text
0°:   (r, c)
90°:  (c, -r)
180°: (-r, -c)
270°: (-c, r)
```

After rotation, we add a translation vector `(dr, dc)`.

So every transformation has the form:

```text
f(cell) = rotate(cell) + shift
```

### Reducing the number of transformations

There are infinitely many possible shifts, but we only need to try finitely many.

Let `c0` be any marked cell, for example the first one in row-major order.

Assume `c0 ∈ A`. Then `f(c0)` must be some other marked cell in `Q`.

So, for each rotation, we can try all shifts that map `c0` to another marked cell.

What if in the true solution `c0 ∈ B` instead? Then using the inverse transformation swaps the roles of `A` and `B`, so we still find a valid solution.

Thus, we try at most:

```text
4 * |Q|
```

candidate transformations.

Since `N ≤ 20`, `|Q| ≤ 400`, this is easily small enough.

### Checking one fixed transformation

For a fixed transformation `f`, we need know whether there exists a subset `A` such that:

```text
Q = A ∪ f(A)
A ∩ f(A) = ∅
```

Build a directed graph on cells of `Q`:

- each cell `x` points to `f(x)` if `f(x)` is also in `Q`.

Because `f` is bijective on the infinite grid, every cell has:

- out-degree at most `1`,
- in-degree at most `1`.

Therefore, connected components are only:

- paths,
- cycles.

Along such a component, cells must alternate between `A` and `B`.

For a path:

```text
v1 -> v2 -> v3 -> ... -> vk
```

- `v1` has no predecessor, so it cannot be in `B`; it must be in `A`.
- `vk` has no successor, so it cannot be in `A`; it must be in `B`.

Thus the path length must be even.

For a cycle, alternating is possible iff the cycle length is even.

So for each component, we color cells alternately `0/1`. If a component has odd length, this transformation fails.

If all components have even length, all cells colored `0` form one valid subset `A`.

### Complexity

Let `M = |Q| ≤ 400`.

There are `O(M)` transformations to try, and each check takes `O(N²)` or `O(M)` time.

Total complexity:

```text
O(M * N²)
```

which is tiny for `N ≤ 20`.

---

## 3. Provided C++ solution with detailed comments

```cpp
#include <bits/stdc++.h> // Includes almost all standard C++ headers.

using namespace std; // Allows using std:: names without prefix.

// Output operator for pairs, useful for debugging.
// Not essential for this problem.
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Input operator for pairs.
// Not essential for this problem.
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Input operator for vectors.
// Reads all elements of the vector from input.
// This is used to read the grid strings.
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

// Output operator for vectors.
// Not essential for this problem.
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n;              // Size of the grid.
vector<string> g;   // The input grid.

// Reads input.
void read() {
    cin >> n;              // Read grid size.
    g.assign(n, "");       // Create n empty strings.
    cin >> g;              // Read n grid rows using overloaded vector input.
}

void solve() {
    // Store all marked cells of Q.
    vector<pair<int, int>> cells;

    // Scan the grid.
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
            // A cell belongs to Q iff it contains '1'.
            if(g[i][j] == '1') {
                cells.push_back({i, j});
            }
        }
    }

    // Number of marked cells.
    int total = cells.size();

    // Empty set can be split into two empty sets.
    if(total == 0) {
        cout << "YES" << '\n';

        // Output an empty subset.
        for(int i = 0; i < n; i++) {
            cout << string(n, '0') << '\n';
        }

        return;
    }

    // The two parts must have equal size.
    // Therefore the total number of cells must be even.
    if(total % 2 != 0) {
        cout << "NO" << '\n';
        return;
    }

    // Rotate point (r, c) by k * 90 degrees around the origin.
    // k = 0: identity
    // k = 1: 90 degrees
    // k = 2: 180 degrees
    // k = 3: 270 degrees
    auto rotate_k = [](int r, int c, int k) {
        // Apply 90-degree rotation k times.
        for(int i = 0; i < k; i++) {
            // 90-degree rotation: (r, c) -> (c, -r).
            int nr = c, nc = -r;

            // Update coordinates.
            r = nr;
            c = nc;
        }

        // Return rotated coordinates.
        return make_pair(r, c);
    };

    // Checks whether cell (r, c) is inside the grid and belongs to Q.
    auto in_q = [&](int r, int c) {
        return 0 <= r && r < n && 0 <= c && c < n && g[r][c] == '1';
    };

    // Choose an anchor cell.
    // If a valid partition exists, either this cell is in A,
    // or it is in B and the inverse transformation will handle it.
    auto [c0r, c0c] = cells[0];

    // Tries one fixed transformation:
    // rotate by k * 90 degrees, then shift by (dr, dc).
    auto try_transform = [&](int k, int dr, int dc) -> bool {
        // Transformation f.
        auto f = [&](int r, int c) {
            // First rotate.
            auto [nr, nc] = rotate_k(r, c, k);

            // Then translate.
            return make_pair(nr + dr, nc + dc);
        };

        // Inverse transformation f^{-1}.
        auto f_inv = [&](int r, int c) {
            // First undo the translation.
            r -= dr;
            c -= dc;

            // Then undo the rotation.
            // The inverse of rotation k is rotation (4 - k) % 4.
            auto [nr, nc] = rotate_k(r, c, (4 - k) % 4);

            return make_pair(nr, nc);
        };

        // color[r][c]:
        // -1 = unvisited
        //  0 = chosen subset A
        //  1 = other subset B
        vector<vector<int>> color(n, vector<int>(n, -1));

        // Visit every cell of Q.
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                // Ignore non-Q cells and already processed cells.
                if(g[i][j] != '1' || color[i][j] != -1) {
                    continue;
                }

                // We need to find the beginning of this path component.
                // If the component is a cycle, this loop eventually detects it.
                int sr = i, sc = j;

                while(true) {
                    // Look at predecessor under f.
                    auto [pr, pc] = f_inv(sr, sc);

                    // If predecessor is not in Q, we found path start.
                    // If predecessor is the original cell, we detected a cycle.
                    if(!in_q(pr, pc) || (pr == i && pc == j)) {
                        break;
                    }

                    // Move one step backward.
                    sr = pr;
                    sc = pc;
                }

                // Alternately color the component starting from (sr, sc).
                int col = 0;      // Start with color 0, meaning subset A.
                int length = 0;   // Number of cells in this component.

                int cr = sr, cc = sc;

                // Traverse forward until leaving Q or returning to a colored cell.
                while(color[cr][cc] == -1) {
                    // Assign current color.
                    color[cr][cc] = col;

                    // Alternate color for next cell.
                    col ^= 1;

                    // Increase component length.
                    length++;

                    // Move forward by f.
                    auto [nr, nc] = f(cr, cc);

                    // If next cell is not in Q, this path ends.
                    if(!in_q(nr, nc)) {
                        break;
                    }

                    // Continue from next cell.
                    cr = nr;
                    cc = nc;
                }

                // Odd-length path or cycle cannot be split alternately.
                if(length % 2 != 0) {
                    return false;
                }
            }
        }

        // If all components have even length, we found a valid split.
        cout << "YES" << '\n';

        // Output all cells colored 0 as the first subset.
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                // Print 1 only for cells in Q and in subset A.
                if(g[i][j] == '1' && color[i][j] == 0) {
                    cout << '1';
                } else {
                    cout << '0';
                }
            }

            cout << '\n';
        }

        return true;
    };

    // Try all four rotations.
    for(int k = 0; k < 4; k++) {
        // Rotate the anchor cell.
        auto [rr, rc] = rotate_k(c0r, c0c, k);

        // Try mapping the anchor cell to every other marked cell.
        for(auto [tr, tc]: cells) {
            // The anchor cannot map to itself in a valid disjoint partition.
            if(tr == c0r && tc == c0c) {
                continue;
            }

            // Compute the shift needed to send rotated c0 to target cell.
            int dr = tr - rr;
            int dc = tc - rc;

            // If this transformation works, we are done.
            if(try_transform(k, dr, dc)) {
                return;
            }
        }
    }

    // No transformation worked.
    cout << "NO" << '\n';
}

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

    // Only one test case.
    int T = 1;

    // Process test case.
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    return 0;
}
```

---

## 4. Python solution with detailed comments

```python
import sys


def rotate_k(r, c, k):
    """
    Rotate point (r, c) by k * 90 degrees around the origin.

    k = 0: (r, c)
    k = 1: (c, -r)
    k = 2: (-r, -c)
    k = 3: (-c, r)
    """
    for _ in range(k):
        r, c = c, -r
    return r, c


def solve():
    data = sys.stdin.read().strip().split()

    # Read N.
    n = int(data[0])

    # Read grid.
    grid = data[1:1 + n]

    # Collect all cells belonging to Q.
    cells = []
    for r in range(n):
        for c in range(n):
            if grid[r][c] == '1':
                cells.append((r, c))

    total = len(cells)

    # Empty set is trivially splittable.
    if total == 0:
        print("YES")
        for _ in range(n):
            print("0" * n)
        return

    # Two congruent parts must have equal size.
    if total % 2 == 1:
        print("NO")
        return

    # Check whether a cell is inside Q.
    def in_q(r, c):
        return 0 <= r < n and 0 <= c < n and grid[r][c] == '1'

    # Use the first marked cell as anchor.
    c0r, c0c = cells[0]

    def try_transform(k, dr, dc):
        """
        Try transformation:
            f(x) = rotate_k(x, k) + (dr, dc)

        Return the resulting subset grid if valid, otherwise None.
        """

        def f(r, c):
            # Apply rotation.
            nr, nc = rotate_k(r, c, k)

            # Apply translation.
            return nr + dr, nc + dc

        def f_inv(r, c):
            # Undo translation.
            r -= dr
            c -= dc

            # Undo rotation.
            return rotate_k(r, c, (4 - k) % 4)

        # color[r][c]:
        # -1 means unvisited,
        #  0 means chosen subset A,
        #  1 means transformed subset B.
        color = [[-1] * n for _ in range(n)]

        # Process every marked cell.
        for i in range(n):
            for j in range(n):
                # Skip cells not in Q or already visited.
                if grid[i][j] != '1' or color[i][j] != -1:
                    continue

                # Find start of the component.
                sr, sc = i, j

                while True:
                    pr, pc = f_inv(sr, sc)

                    # If predecessor is outside Q, this is path start.
                    # If predecessor is the original cell, this is a cycle.
                    if not in_q(pr, pc) or (pr == i and pc == j):
                        break

                    sr, sc = pr, pc

                # Traverse forward and color alternately.
                cr, cc = sr, sc
                current_color = 0
                length = 0

                while color[cr][cc] == -1:
                    # Assign color.
                    color[cr][cc] = current_color

                    # Alternate between 0 and 1.
                    current_color ^= 1

                    # Count component size.
                    length += 1

                    # Move to next cell under f.
                    nr, nc = f(cr, cc)

                    # Stop if next cell is outside Q.
                    if not in_q(nr, nc):
                        break

                    cr, cc = nr, nc

                # Odd component cannot be alternately split.
                if length % 2 == 1:
                    return None

        # Build output subset: cells colored 0.
        answer = []
        for r in range(n):
            row = []
            for c in range(n):
                if grid[r][c] == '1' and color[r][c] == 0:
                    row.append('1')
                else:
                    row.append('0')
            answer.append(''.join(row))

        return answer

    # Try all four rotations.
    for k in range(4):
        # Anchor after rotation.
        rr, rc = rotate_k(c0r, c0c, k)

        # Try mapping anchor to every other marked cell.
        for tr, tc in cells:
            if tr == c0r and tc == c0c:
                continue

            # Translation needed.
            dr = tr - rr
            dc = tc - rc

            result = try_transform(k, dr, dc)

            # If valid, print answer.
            if result is not None:
                print("YES")
                print("\n".join(result))
                return

    # Nothing worked.
    print("NO")


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

---

## 5. Compressed editorial

Let `Q` be the set of marked cells. If `|Q|` is odd, answer is `NO`.

Try all transformations of the form:

```text
f(x) = rotation(x) + shift
```

There are four rotations. To limit shifts, pick one marked anchor cell `c0`. In a valid solution, either `c0` belongs to the first part, or the inverse transformation handles the case. Therefore, for each rotation, try shifts that map `c0` to every other marked cell.

For a fixed transformation `f`, create directed edges:

```text
x -> f(x)
```

whenever both cells are in `Q`.

Since `f` is bijective, every vertex has in-degree and out-degree at most one, so components are paths or cycles. A valid partition exists exactly when every component has even length, because cells must alternate between the two subsets.

If all components are even, color them alternately and output one color class. If no transformation works, output `NO`.

Complexity is small: at most `4|Q|` transformations, each checked in `O(N²)`, with `N ≤ 20`.