PROBLEM LINK:
Author:Egor Bobyk
Tester:Istvan Nagy
Editorialist:Xellos
DIFFICULTY:
SIMPLE
PREREQUISITES:
Greatest common divisor.
PROBLEM:
We have a permutation $\texttt{P}[0..N-1]$ with $\texttt{P}[x]=M+x$ for $x < N-M$ and $\texttt{P}[x]=x-(N-M)$ for $x \ge N-M$. Start at $x=0$ and keep moving from $x$ to $\texttt{P}[x]$. When visiting some $x$ for the second time, stop. How many different $x$-s have been visited?
QUICK EXPLANATION:
The number of robots with cakes is $N/gcd(N,M)$, which can be found using Euclid's GCD algorithm.
EXPLANATION:
For permutations, it's a well-known fact that if we start at $x=0$, we'll stop at $x=0$ as well - a permutation is made of cycles.
We can just simulate the first subtask. For the second subtask, this is too slow, but we can reduce our problem to a completely different one. The trick is to see that in each step, we're sent from $x$ to $x+M$ and if we see that $x \ge N$ afterwards (it really happens iff $x \ge N-M$ before this step), then we subtract $N$ to get it back into the range $[0,N)$. Imagine what'd happen if we wrote the permutation on paper and connected its ends - we're always just moving $M$ places to the right on the tape. More formally, after moving to the next robot $j$ times, we have $x=(jM)\% N$.
When will we visit $x=0$ again, then? It happens for the smallest positive $j$ such that $N\,|\,jM$; which is known to be $N/gcd(N,M)$ (think why). That's the answer to the question "how many robots will have a cake at the end?"; if it's $N$ - if $N$ and $M$ are coprime - we should print "YES".
The greatest common divisor can be computed using Euclid's algorithm in $O(\log N)$, which is the time complexity per test case. Memory: $O(1)$.
AUTHOR'S AND TESTER'S SOLUTIONS:
The author's solution can be found here.
The tester's solution can be found here.
The editorialist's solution can be found here.
