Please explain this part of the editorial. Imagine what'd happen if we wrote the permutation on paper and connected its ends  we're always just moving MM places to the right on the tape. More formally, after moving to the next robot jj times, we have x=(jM)%Nx=(jM)%N. When will we visit x=0x=0 again, then? It happens for the smallest positive jj such that NjMNjM; which is known to be N/gcd(N,M)N/gcd(N,M) (think why). That's the answer to the question "how many robots will have a cake at the end?"; if it's NN  if NN and MM are coprime  we should print "YES". asked 26 Jan '16, 17:24
