1. Abridged Problem Statement  
Given N white balls in a row, each ball i costs Ci to paint black. You must paint some balls so that in every consecutive block of M balls there are at least two black balls. Compute the minimum total painting cost.

2. Detailed Editorial  

We need to select positions 1 ≤ p1 < p2 < … < pk ≤ N (balls to paint) so that every contiguous segment of length M contains at least two of these positions. Equivalently:
- The “gap” (number of balls) before the first painted ball plus the gap between the first two painted balls must not exceed M, so that the first window is valid.
- Between any two consecutive painted balls, say at pi and pj, and the next pj and pk, the sum of those two gaps must be ≤ M (this enforces two blacks in every sliding window straddling these).
- Similarly, the gap after the last painted ball to the end must be small enough so that windows touching the end also see two blacks.

A convenient DP formulation treats an imaginary “painted ball” at position 0, and tracks distances between consecutive painted balls. Define
  dp[i][d] = minimum cost to have painted ball i (0-based) as black, with the distance from the previous painted ball equal to d.
Here d runs from 1 to M. We initialize the first real painted ball at index i by imagining its “previous” was at position –1, so the initial distance is i–(–1) = i+1. We only allow those i with i+1 ≤ M–1 (otherwise the very first window of length M would contain at most one black).

Transition:
  From state (i, prv_d), we may paint a next ball at index j = i + dist, for dist from 1 to M, provided
      prv_d + dist ≤ M
  (so that no window of length M spanning those two blacks lacks a second black).
  Then
      dp[j][dist] = min(dp[j][dist], dp[i][prv_d] + cost[j]).

Finally, to ensure the tail end (after the last black) does not form a violating window, we look at any dp[i][prv_d] where prv_d + (N − i) ≤ M; among those we take the minimum cost.

Complexity: O(N·M²) transitions in the naive form. Since M ≤ 100 and N ≤ 10000, this runs in time in optimized C++.  

3. C++ Solution with Detailed Comments  

```cpp
#include <bits/stdc++.h>
using namespace std;

static const int INF = 1000000000;

int n, m;
vector<int> a;

// Read input: N, M, then costs a[0..n-1].
void read() {
    cin >> n >> m;
    a.resize(n);
    for (int &x : a) cin >> x;
}

void solve() {
    // dp[i][d] = minimum cost where ball i is painted,
    // and distance from the previous painted ball is exactly d (1 ≤ d ≤ M).
    vector<vector<int>> dp(n, vector<int>(m+1, INF));

    // Initialize: painting the first real black ball at index i (0-based).
    // Distance from the imaginary painted ball at -1 is (i - (-1)) = i+1.
    // We require i+1 ≤ M-1 so that the very first window of length M
    // contains at least two blacks (the imaginary one at -1 and the real one at i).
    for (int i = 0; i < n; i++) {
        int d = i + 1;
        if (d < m) {
            dp[i][d] = a[i];
        } else {
            break;  // once i+1 >= m, no further first positions are valid
        }
    }

    int answer = INF;

    // Main DP: for each painted state (i, prv_d),
    // try to paint a next ball at j = i + dist, dist from 1..M.
    // The constraint prv_d + dist ≤ M enforces every sliding window
    // between these two painted balls has two blacks.
    for (int i = 0; i < n; i++) {
        for (int prv_d = 1; prv_d <= m; prv_d++) {
            int cur_cost = dp[i][prv_d];
            if (cur_cost >= INF) continue;  // unreachable state

            // If from this state to the end we still satisfy the tail constraint,
            // we can finish here and update the answer.
            int tail_gap = n - i;  // gap from ball i to position n (one past end)
            if (prv_d + tail_gap <= m) {
                answer = min(answer, cur_cost);
            }

            // Otherwise, try to choose a next painted ball
            // at j = i + dist, dist = 1..M, with prv_d+dist ≤ M.
            for (int dist = 1; dist <= m; dist++) {
                int j = i + dist;
                if (j >= n) break;  // beyond the last ball
                if (prv_d + dist > m) break;  // would violate window constraint
                int new_cost = cur_cost + a[j];
                if (new_cost < dp[j][dist]) {
                    dp[j][dist] = new_cost;
                }
            }
        }
    }

    // Output the best achievable answer.
    cout << answer << "\n";
}

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

    read();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments  

```python
import sys
input = sys.stdin.readline

def main():
    INF = 10**18
    n, m = map(int, input().split())
    a = list(map(int, input().split()))
    
    # dp[i][d]: minimal cost if ball i is painted black,
    # and distance to the previous painted ball is exactly d (1..m).
    dp = [ [INF]*(m+1) for _ in range(n) ]
    
    # Initialize first painted ball at index i: distance from "virtual" at pos -1 is i+1.
    # We require i+1 < m so that the first window of size m sees at least two blacks.
    for i in range(n):
        d = i + 1
        if d < m:
            dp[i][d] = a[i]
        else:
            break

    answer = INF

    # For each i, we will:
    # 1) see if we can finish here (ensure tail constraint),
    # 2) transition to future painted balls.
    for i in range(n):
        # Precompute prefix minima of dp[i][1..m]:
        # prefix_min[d] = min(dp[i][1], ..., dp[i][d]).
        prefix_min = [INF] * (m+1)
        running = INF
        for d in range(1, m+1):
            if dp[i][d] < running:
                running = dp[i][d]
            prefix_min[d] = running

        # Check tail: if last painted gap + (n - i) ≤ m, we can finish.
        tail_gap = n - i
        for d in range(1, m+1):
            cost_here = dp[i][d]
            if cost_here >= INF: 
                continue
            if d + tail_gap <= m:
                answer = min(answer, cost_here)

        # Transition to next paint positions:
        # If current was at (i,prv_d), next at j = i + dist with dist in 1..m,
        # requiring prv_d + dist ≤ m => prv_d ≤ m - dist.
        for dist in range(1, m+1):
            j = i + dist
            if j >= n:
                break
            # minimal cost among dp[i][1..(m-dist)]
            lim = m - dist
            if lim <= 0:
                continue
            best_prev = prefix_min[lim]
            new_cost = best_prev + a[j]
            if new_cost < dp[j][dist]:
                dp[j][dist] = new_cost

    print(answer)

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

5. Compressed Editorial  

We model painting decisions by a DP on pairs (i, d), where i is the index of the last painted ball and d the distance back to the previous painted ball. Enforcing that any two consecutive painted balls (including a virtual one before the start and the tail end) span ≤ M keeps every block of M balls containing at least two painted ones. We initialize feasible first paints, iterate i from 0..N−1, update `dp[j][dist]` for j=i+dist when `prev_d + dist ≤ M`, and track solutions that satisfy the end‐gap constraint. The result is the minimal accumulated painting cost. Complexity O(N·M²) in the naive form, or O(N·M) via prefix minima, which suffices for N≤10⁴, M≤10².