Let ~A_k~ be a set of angles ~\{a_1, a_2, \dots, a_n\}~ that Mirko can construct using at most ~k~ additions or subtractions of the initial angles. From ~A_k~ we obtain ~A_{k+1}~ by copying ~A_k~ into ~A_{k+1}~ and then for each initial angle adding it to all angles in ~A_k~ inserting the resulting angle in ~A_{k+1}~, if it is not already in. It can be easily seen that ~A_k~ can contain at most ~360~ angles, because there are only ~360~ possible values. Because the smallest number of angles we can add to ~A_{k+1}~ is ~1~, because if we did not add any new angles we could terminate our algorithm, we will perform at most ~360~ steps at which point ~A_{360}~ will contain all angles Mirko can construct. Of course ~A_0 = \{0\}~. This can easily translate into an efficient algorithm for this task. First we precalculate ~A_{360}~ and then for each angle Slavko gives, simply check if it is contained in ~A_{360}~.