Note that the following solution is incorrect: Instead of using binary search as the official solution does, a knapsack can be done. An array of talents can be kept, where $\texttt{talent}[i]$ is the minimum weight required to create a talent sum of $i$. To make sure each cow is used at most once and the order of the cows doesn't matter, the indices of $\texttt{talent}$ are iterated though from greatest to least, and the $\texttt{cows}$ are iterated though in the outer loop; the implementation should make this clear. This maximizes the ratio $\frac{t}{w}$ as desired. Then, the maximum ratio of $\frac{t_i}{w_i}$ can be easily found by going through $\texttt{talent}$ and ignoring any $w_i$ that is less than $W$.

My $\mathcal{O}(\text{Total talent } \cdot n)$ code is below:

However, even though this passes all of the test cases on USACO, there is a fatal flaw to this. We are finding the minimum weight required for each talent sum. What if the minimum weight for a talent sum is less than the weight threshold? Consider if the threshold is indeed attainable with a particular talent sum but the DP concluded it is not. The DP never considers a slightly higher weight, one that can pass the threshold, as a candidate for the answer. This means the DP fails on cases where the aforementioned scenario occurs and is optimal; one example is as follows:

4 16
11 2
17 2
12 2
4 4

The correct answer is 375; the above solution outputs 296.