1. Abridged Problem Statement  
Given an n×n chessboard and a number k, count the number of ways to place k bishops so that no two attack each other. Bishops move along diagonals; two bishops conflict if they share a diagonal. Output the total count. Constraints: 1≤n≤10, 0≤k≤n².

2. Detailed Editorial  

Overview  
– Two bishops never attack each other if placed on different-colored squares (a standard chessboard coloring). Thus we can split the board into its “black” and “white” cells and count independently how many ways to place i bishops on black cells and k−i on white cells, then sum over i.  

Counting on one color  
– Every cell is uniquely identified by two diagonals: main diagonal d1 = r+c (0 to 2n−2) and anti-diagonal d2 = r−c+(n−1) (0 to 2n−2). Two bishops attack if they share either d1 or d2.  
– We list all cells of one color, then backtrack: for each cell we choose either to skip it or place a bishop there if both its diagonals are unused. We maintain two boolean arrays tracking used d1 and d2. Whenever we have considered all cells, we record in ways[color][placed]++. We do this for black and white separately.  
– Finally, the answer is  
  ans = Σ_{i=0..k} waysBlack[i] * waysWhite[k−i].

Complexity  
– Black and white each have roughly ⌈n²/2⌉ cells. The backtracking explores 2^{cells} branches but prunes heavily by checking diagonal usage and the limit k. For n≤10 this runs within time limits.  

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;
vector<pair<int, int>> black_cells, white_cells;
vector<bool> used_diag1_black, used_diag2_black;
vector<bool> used_diag1_white, used_diag2_white;
vector<int64_t> ways_black, ways_white;

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

void backtrack_black(int idx, int placed) {
    if(idx == (int)black_cells.size()) {
        if(placed <= k) {
            ways_black[placed]++;
        }

        return;
    }

    backtrack_black(idx + 1, placed);

    auto [r, c] = black_cells[idx];
    int d1 = r + c;
    int d2 = r - c + (n - 1);
    if(!used_diag1_black[d1] && !used_diag2_black[d2]) {
        used_diag1_black[d1] = used_diag2_black[d2] = true;
        backtrack_black(idx + 1, placed + 1);
        used_diag1_black[d1] = used_diag2_black[d2] = false;
    }
}

void backtrack_white(int idx, int placed) {
    if(idx == (int)white_cells.size()) {
        if(placed <= k) {
            ways_white[placed]++;
        }

        return;
    }

    backtrack_white(idx + 1, placed);

    auto [r, c] = white_cells[idx];
    int d1 = r + c;
    int d2 = r - c + (n - 1);
    if(!used_diag1_white[d1] && !used_diag2_white[d2]) {
        used_diag1_white[d1] = used_diag2_white[d2] = true;
        backtrack_white(idx + 1, placed + 1);
        used_diag1_white[d1] = used_diag2_white[d2] = false;
    }
}

void solve() {
    // Bishops attack along diagonals. Rotating the board 45 degrees, the two
    // diagonal directions become rows and columns, and the board splits into
    // two independent sub-problems: cells of one colour (by (r+c) parity) never
    // share a diagonal with cells of the other colour. So we enumerate, for
    // each colour separately, how many ways there are to place exactly i
    // non-attacking bishops on that colour's cells, recording the counts in
    // ways_black[i] / ways_white[i] via backtracking that tracks which "/" and
    // "\" diagonals are already occupied. The total number of placements of k
    // bishops is then the convolution sum over i of
    // ways_black[i] * ways_white[k - i].

    black_cells.clear();
    white_cells.clear();
    for(int r = 0; r < n; r++) {
        for(int c = 0; c < n; c++) {
            if(((r + c) & 1) == 0) {
                black_cells.emplace_back(r, c);
            } else {
                white_cells.emplace_back(r, c);
            }
        }
    }

    used_diag1_black.assign(2 * n, false);
    used_diag2_black.assign(2 * n, false);
    used_diag1_white.assign(2 * n, false);
    used_diag2_white.assign(2 * n, false);

    ways_black.assign(k + 1, 0);
    ways_white.assign(k + 1, 0);

    backtrack_black(0, 0);
    backtrack_white(0, 0);

    int64_t ans = 0;
    for(int i = 0; i <= k; i++) {
        ans += ways_black[i] * ways_white[k - i];
    }

    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
import sys
sys.setrecursionlimit(10000)

def read_input():
    return map(int, sys.stdin.read().split())

def diag_lengths(n, color):
    """
    Returns a list diag_len[d] = number of squares of given color
    on main diagonal r+c = d, for d = 0..2n-2.
    color=0 for black ((r+c)%2==0), color=1 for white.
    """
    lens = [0] * (2*n - 1)
    for r in range(n):
        for c in range(n):
            if (r + c) % 2 == color:
                d = r + c
                lens[d] += 1
    return lens

def ways_on_color(n, k, color):
    """
    Compute ways[b] = number of ways to place b bishops on all
    squares of the given color, using DP over main diagonals.
    """
    lens = diag_lengths(n, color)
    # dp[j] = number of ways to place j bishops so far
    dp = [0] * (k + 1)
    dp[0] = 1  # zero bishops on zero diagonals

    # Process each diagonal in turn
    for cnt in lens:
        new_dp = dp[:]  # start by not placing any new bishop on this diag
        # Try placing t bishops on this diagonal (t is 1.. up to k)
        # But on each diag we can place at most one bishop: we do transitions
        for placed in range(1, k + 1):
            # If we want 'placed' total bishops, one of them on this diagonal,
            # we had placed-1 before. Available squares = cnt - (placed-1)
            free_spots = cnt - (placed - 1)
            if free_spots > 0:
                new_dp[placed] += dp[placed - 1] * free_spots
        dp = new_dp
    return dp  # dp[b] for b=0..k

def main():
    n, k = read_input()
    # Compute for black (color=0) and white (color=1)
    ways_black = ways_on_color(n, k, 0)
    ways_white = ways_on_color(n, k, 1)
    # Combine: sum_{i=0..k} ways_black[i] * ways_white[k-i]
    ans = sum(ways_black[i] * ways_white[k - i] for i in range(k + 1))
    print(ans)

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

5. Compressed Editorial  
- Split board into black/white squares; bishops on different colors never attack.  
- Independently count ways to place i bishops on black squares and k−i on white squares.  
- Use backtracking with diagonal-usage arrays (or DP over diagonals) to get counts.  
- Convolve the two color-specific counts to obtain the final answer.