1. Abridged Problem Statement  
Given an n×n chessboard (1≤n≤10) and an integer k (0≤k≤n²), count the number of ways to place k identical queens so that no two attack each other (i.e., no two share a row, column, or diagonal).

2. Detailed Editorial  

We need to count all configurations of k queens on an n×n board so that no two are in line vertically, horizontally, or along either diagonal. A classic approach is backtracking (depth-first search) that proceeds row by row:

 1. We maintain a structure (here, a 2D array `attacked[row][col]`) that records how many queens attack a given square. A square is safe if its attacked count is zero.
 2. We write a recursive function `brute(row, k_left)` which returns the number of ways to place `k_left` queens on rows `row..n−1`.  
    - Prune immediately if `k_left > n − row` (not enough rows remain).
    - If `row == n`, we have considered all rows: return 1 if `k_left == 0`, else 0.
    - First, consider _not_ placing a queen on this row at all: `ans = brute(row+1, k_left)`.
    - Then, for each column `col` in `[0..n-1]`, if `(row,col)` is safe (`attacked[row][col]==0`) and `k_left>0`, we:
        a. Mark the new queen by incrementing `attacked` on its row, column, and both diagonals (`change_queen`).
        b. Recurse: `ans += brute(row+1, k_left-1)`.
        c. Unmark (subtract) to restore state.
 3. The function `change_queen(row, col, delta)` walks outwards in all eight directions from `(row,col)`, adding `delta` (±1) to `attacked`. We subtract an extra `3*delta` at the end to correct for the overcount at `(row,col)` itself (since the four “directional” loops each revisit it).
 4. The total number of valid placements is `brute(0, k)` with an initially zeroed `attacked` grid.
  
Time complexity: In the worst case (small k), the branching factor is O(n) per row, depth ≤ n, yielding something like O(nⁿ) in the absolute worst—but n≤10 makes this feasible, and pruning by `k_left` further reduces branches.

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, k;

void change_queen(int row, int col, vector<vector<int>>& attacked, int delta) {
    for(int i = 0; i < n; i++) {
        attacked[row][i] += delta;
        attacked[i][col] += delta;
        if(row + i < n && col + i < n) {
            attacked[row + i][col + i] += delta;
        }

        if(row + i < n && col - i >= 0) {
            attacked[row + i][col - i] += delta;
        }

        if(row - i >= 0 && col + i < n) {
            attacked[row - i][col + i] += delta;
        }

        if(row - i >= 0 && col - i >= 0) {
            attacked[row - i][col - i] += delta;
        }
    }

    attacked[row][col] -= 3 * delta;
}

int brute(int row, int k, vector<vector<int>>& attacked) {
    if(k > n - row) {
        return 0;
    }

    if(row == n) {
        return 1;
    }

    int ans = brute(row + 1, k, attacked);
    for(int col = 0; col < n; col++) {
        if(k && !attacked[row][col]) {
            change_queen(row, col, attacked, 1);
            ans += brute(row + 1, k - 1, attacked);
            change_queen(row, col, attacked, -1);
        }
    }

    return ans;
}

void read() { cin >> n >> k; }

void solve() {
    // Backtracking row by row, placing at most one queen per row. attacked[r][c]
    // counts how many already-placed queens attack cell (r, c) along a row,
    // column or diagonal, so a cell is free iff its counter is zero; placing or
    // removing a queen bumps every cell on its four lines by +/-1 (the queen's
    // own cell gets corrected for being counted on all four lines). At each row
    // we either skip it or try every free column, decrementing the remaining
    // queen budget k. The pruning k > n - row cuts branches that cannot fit the
    // remaining queens, and reaching row == n with the placements done counts as
    // one valid arrangement.

    vector<vector<int>> attacked(n, vector<int>(n, 0));
    cout << brute(0, k, attacked) << '\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 count_configurations(n, k):
    """
    Count ways to place k queens on an n×n board without attacks,
    proceeding row by row.
    """

    # Use bitmasks to track attacked columns and diagonals:
    # `cols`  : columns under attack
    # `d1`    : "main" diagonals under attack (row+col)
    # `d2`    : "anti" diagonals under attack (row-col + (n-1))
    ALL = (1 << n) - 1  # mask of n one-bits

    from functools import lru_cache

    @lru_cache(None)
    def dfs(row, placed, cols, d1, d2):
        # placed = number of queens already placed
        # row = next row index to consider
        if placed == k:
            # All required queens placed: any remaining rows can be empty
            return 1
        if row == n:
            # No more rows
            return 0
        # Prune: if even placing one queen per remaining row can't reach k
        if placed + (n - row) < k:
            return 0

        ways = 0
        # 1) Skip placing on this row
        ways += dfs(row+1, placed, cols, d1<<1, d2>>1)

        # 2) Place exactly one queen in any safe column on this row
        # safe positions = those columns not in cols, and not attacked on diagonals
        attacked = cols | d1 | d2
        free_positions = (~attacked) & ALL
        while free_positions:
            # pick lowest set bit
            bit = free_positions & -free_positions
            free_positions -= bit
            # Set that bit in cols;
            # shift diagonals for next row
            ways += dfs(
                row+1,
                placed+1,
                cols | bit,
                (d1 | bit) << 1,
                (d2 | bit) >> 1
            )
        return ways

    return dfs(0, 0, 0, 0, 0)

# Example usage:
if __name__ == "__main__":
    n, k = map(int, input().split())
    print(count_configurations(n, k))
```

Explanation of the Python approach:  
- We encode the board state in three bitmasks—columns, main diagonals, and anti-diagonals.  
- At each row, we can either skip placing a queen (so just update the diagonals' masks by shifting), or place one queen in any free column (bit not set in any of the three masks).  
- We recurse until we have placed k queens or run out of rows, using memoization (`lru_cache`) to avoid recomputation.

5. Compressed Editorial  

We use a depth-first search over rows, keeping track of attacked squares. At each row, we either skip placing a queen or place one in any safe column (attacked count zero). We prune when too few rows remain to fit the remaining queens. The complexity is acceptable for n≤10. A bitmask variant replaces the 2D attacked array with three integer masks—columns, main diagonals, and anti-diagonals—and shifts them as we move down the rows, with memoization to accelerate overlapping subproblems.