1. Abridged Problem Statement  
Given an “irregular” chessboard with n rows, where row i has b_i contiguous cells in columns 1 through b_i. Count the number of ways to place exactly k rooks so that no two attack each other (no two share a row or a column).  

2. Detailed Editorial  
We need to choose k distinct rows and k distinct columns, and place one rook in each chosen row–column pair, subject to the constraint that column ≤ b_i for the rook in row i. Equivalently, if we process rows one by one and keep track of how many rooks we have placed so far, we need to know how many “new” columns are free in the current row.

Let b = [b₁, b₂, …, bₙ]. First sort b in non-decreasing order. Define a DP array dp[j] = number of ways, after processing some prefix of rows, to have placed exactly j rooks so far, using j distinct columns. Initially dp[0] = 1, dp[j>0] = 0.

When we consider a new row with length v, suppose we have already placed j–1 rooks in previous rows, occupying j–1 distinct columns. In this row, if we want to increase the total to j rooks, we must pick one of the columns 1…v that hasn’t been used yet. There are (v – (j–1)) such free columns. Thus the transition is

  new_dp[j] += old_dp[j–1] × (v – (j–1)).

We update dp in descending order of j (from k down to 1) to avoid overwriting states we still need in this iteration. After processing all n rows, dp[k] is the answer.

Time complexity: O(n·k). Space: O(k).

3. C++ Solution with Detailed Comments  
```cpp
#include <bits/stdc++.h>
#include <boost/multiprecision/cpp_int.hpp>
using namespace std;
// Use a big–integer type to hold arbitrarily large answers
using BigInt = boost::multiprecision::cpp_int;

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

    int n, k;
    // Read number of rows n and number of rooks k
    cin >> n >> k;

    vector<int> b(n);
    // Read the length of each row
    for(int i = 0; i < n; i++){
        cin >> b[i];
    }

    // Sort row lengths in non-decreasing order
    // This ensures that when we do the DP transitions,
    // (v - (j-1)) is computed in a stable order
    sort(b.begin(), b.end());

    // dp[j] = number of ways to place exactly j rooks after processing some rows
    vector<BigInt> dp(k+1);
    dp[0] = 1; // Zero rooks can always be placed in exactly one way

    // Process each row one by one
    for(int idx = 0; idx < n; idx++){
        int v = b[idx];  // current row has v columns

        // We must update dp[j] based on previous dp[j-1].
        // Iterate j downward so we don't reuse updated dp[j-1] from this same row.
        for(int j = k; j >= 1; j--){
            // If we want to place the j-th rook in this row,
            // we must choose a column among the v columns not yet occupied.
            // There have been (j-1) columns already used,
            // so there are (v - (j-1)) choices.
            BigInt waysToPickColumn = v - (j - 1);
            if(waysToPickColumn > 0){
                dp[j] += dp[j-1] * waysToPickColumn;
            }
            // If waysToPickColumn <= 0, we add nothing (no valid placements).
        }
    }

    // The answer is the number of ways to have placed exactly k rooks
    cout << dp[k] << "\n";
    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
def main():
    # Read n (number of rows) and k (number of rooks to place)
    n, k = map(int, input().split())
    # Read row lengths b[0..n-1]
    b = list(map(int, input().split()))

    # Sort row lengths so that transitions never see a “smaller” v after a larger one
    b.sort()

    # dp[j] = number of ways to place exactly j rooks so far
    dp = [0] * (k + 1)
    dp[0] = 1  # Base case: one way to place 0 rooks

    # Process each row length v in ascending order
    for v in b:
        # Update dp in reverse to avoid overwriting dp[j-1] needed in this iteration
        for j in range(k, 0, -1):
            # If we've already placed j-1 rooks, they've occupied j-1 columns.
            # In this row, we can choose any of the v total columns
            # except those j-1 already taken.
            free_columns = v - (j - 1)
            if free_columns > 0:
                dp[j] += dp[j - 1] * free_columns
            # If free_columns <= 0, no new placement is possible for this j.
    # Print the final count for exactly k rooks
    print(dp[k])

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

5. Compressed Editorial  
Sort the row lengths. Maintain a 1D DP array dp[j]: the number of ways to have placed j rooks so far, using j distinct columns. When adding a row of length v, you can place the j-th rook there in (v – (j–1)) ways (new columns available), so update  
  dp[j] += dp[j–1] * (v – (j–1)),  
in descending j. The final dp[k] is the answer.