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 $A_{360}$.