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 of balls to paint so that every contiguous segment of length M contains at least two of these positions. Equivalently, the gap between any two consecutive painted balls must be at most M, and the same must hold on both ends (the run before the first painted ball and the run after the last one).

A convenient DP formulation 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. The first real painted ball at index i has its "previous" at position −1, so the initial distance is i−(−1) = i+1. We only allow i < m−1 (so dp[i][i+1] = a[i]); otherwise the very first window of length M would contain at most one black.

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

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

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

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, m;
vector<int> a;

void read() {
    cin >> n >> m;
    a.resize(n);
    cin >> a;
}

void solve() {
    // We must paint balls so every window of M consecutive balls holds at
    // least two black ones; this is equivalent to forcing the gap between any
    // two consecutive black balls to be at most M (and the same on both ends).
    // dp[i][d] = minimum cost to paint ball i black given that the previous
    // black ball is exactly d positions earlier (d in 1..M). Transitions move
    // to the next black ball nxt = i + dist with the constraint prv_dist +
    // dist <= M (so the two windows straddling ball i still contain two black
    // balls). A state is a valid finishing state when the remaining tail
    // n - i together with prv_dist still fits within M, and the answer is the
    // minimum such dp value.

    vector<vector<int>> dp(n, vector<int>(m + 1, (int)1e9));
    for(int i = 0; i < m - 1; i++) {
        dp[i][i + 1] = a[i];
    }

    int ans = (int)1e9;
    for(int i = 0; i < n; i++) {
        for(int prv_dist = 1; prv_dist <= m; prv_dist++) {
            for(int dist = 1; dist <= m; dist++) {
                int nxt = i + dist;
                if(nxt < n && prv_dist + dist <= m) {
                    dp[nxt][dist] =
                        min(dp[nxt][dist], dp[i][prv_dist] + a[nxt]);
                }
            }

            int left_to_end = n - i;
            if(prv_dist + left_to_end <= m) {
                ans = min(ans, dp[i][prv_dist]);
            }
        }
    }

    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

```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 i in range(n):
        # Check tail: if last painted gap + (n - i) <= m, we can finish here.
        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:
        # from (i, prv_d), next at j = i + dist with dist in 1..m, requiring prv_d + dist <= m.
        for prv_d in range(1, m+1):
            base = dp[i][prv_d]
            if base >= INF:
                continue
            for dist in range(1, m - prv_d + 1):
                j = i + dist
                if j >= n:
                    break
                new_cost = base + 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[nxt][dist]` for nxt=i+dist when `prv_dist + dist ≤ M`, and track solutions that satisfy the end-gap constraint. The result is the minimal accumulated painting cost. Complexity O(N·M²), which suffices for N≤10⁴, M≤10².
