 Solution

Use positive integral item weights and a nonnegative integral capacity. For the unbounded knapsack problem, define $K(c)$ to be the best value with total weight at most $c$. ThenThe zero candidate is the empty selection, so unused capacity is allowed and an empty set of fitting items causes no problem. Every nonempty feasible selection has some last item $i$; removing one copy leaves a feasible solution at capacity $c-w_i$. Its value is bounded by the corresponding candidate. Conversely, taking an optimal smaller selection and adding that item attains the candidate, proving the recurrence relation by optimal substructure. Positive weights make every dependency strictly smaller.

Fill capacities in increasing order. There are $W+1$ entries, each scanning $n$ items, so the dynamic programming takes $O(nW)$ time and $O(W)$ storage. A winning item pointer at each capacity reconstructs the chosen multiplicities. These are capacity-dependent, pseudopolynomial bounds. For zero-weight items, any positive value makes the repetition problem unbounded; nonpositive-value zero-weight items may simply be discarded before using the recurrence.