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Each of the 
~C~ size-
~3~ items requires its very own knapsack. This amounts to ~C~ knapsacks.
Each of the 
~B~ size-
~2~ items cannot fit into a knapsack with any other size-~2~ items, so this amounts to another ~B~ knapsacks. However, each of those knapsacks also has space for a size-
~1~ item, so we should go ahead and dispose of 
~\min(A, B)~ size-~1~ items along the way.
Finally, the remaining size-~1~ items (of which there are 
~\max(0, A-B)~) should be packed into knapsacks ~3~ at a time, with one potentially non-full knapsack remaining at the end. This amounts to another 
~\lceil \frac{\max(0, A-B)}{3} \rceil~ knapsacks (or, equivalently, 
~\lfloor \frac{\max(0, A-B)+2}{3} \rfloor~ knapsacks).
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