This problem can be solved with dynamic programming. Let ~DP[i][j]~ be the minimum number of treats required to satisfy the first ~i~ cows (cows ~1 \dots i~) while dividing them into ~j~ troughs. The base case is ~DP[0][0] = 0~, and the final answer will be ~DP[N][K+1]~, as we'll be installing ~K~ dividers to divide the ~N~ cows into ~K+1~ troughs.

From a given state ~(i, j)~, the only choice is regarding how many cows should be included in the next trough. The first cow in the trough will be ~i~, and the last will be ~k~ (for some ~k~ between ~i~ and ~N~, inclusive). We should consider each possible value of ~k~. For a given ~k~, there will be ~c = k-i+1~ cows in the next trough. The number of treats ~t~ required to satisfy each of those ~c~ cows can easily be computed by iterating over them (and comparing their ~C~ values to the cow count ~c~). We end up with the transition ~DP[k][j+1] = \min(DP[k][j+1], DP[i][j]+t)~.

A direct implementation of the approach described above would have a time complexity of ~\mathcal O(N^3 K)~ – we'll consider each of the ~\mathcal O(NK)~ states in turn (for ~i~ from ~0~ to ~N-1~ and from ~j~ from ~0~ to ~K~), consider each of the ~\mathcal O(N)~ possible transitions from it, and then compute each transition's ~t~ cost in another ~\mathcal O(N)~ time. This is too slow to earn full marks. However, the time complexity can be improved to ~\mathcal O(N^2(N+K))~ by essentially just swapping the nesting order of the ~j~ and ~k~ loops. Note that each transition's ~t~ cost depends only on ~i~ and ~k~, so there's no need to recompute it ~\mathcal O(K)~ times for all possible values of ~j~.