(Analysis by Dhruv Rohatgi )

This problem can be solved by dynamic programming. Let $dp[i]$ be the optimal sum of skill levels achievable by the first $i$ cows, assuming that the last team ends at cow $i$. If we knew that $k$ was the size of the last team (among the first $i$ cows), then the optimal sum of skill levels would be

$$dp[i-k] + \max_{0 \leq j < k} s_{i-j}.$$
But the size of the last team could be anywhere from $1$ to $K$. So we must try all possible sizes, and set $dp[i]$ to the best sum of skills achieved.

This immediately yields an $O(NK^2)$ algorithm, since there are $N$ states, $K$ transitions, and each transition takes $O(K)$ time to compute. However, we can speed up the transitions by maintaining

$$\max_{0 \leq j < k} s_{i-j}$$
as $k$ is increased. Updating the maximum takes $O(1)$ time, so each transition is now $O(1)$, for an overall time complexity of $O(NK)$. This is fast enough.

#include <iostream>
#include <algorithm>
using namespace std;
int N,K;
int A[10000];
int dp[10000];
int main()
	cin >> N >> K;
	for(int i=0;i<N;i++)
		cin >> A[i];
	dp[0] = A[0];
	for(int i=1;i<N;i++)
		int mx = A[i];
		for(int j=i;j>=0 && i+1-j <= K; j--)
			mx = max(mx, A[j]);
			if(j==0) dp[i] = max(dp[i],mx*(i+1-j));
			else dp[i] = max(dp[i],dp[j-1] + mx*(i+1-j));
	cout << dp[N-1] << '\n';