Who won that challenge?

Consider just solving it for dist0 with as many edges possible. Now, take the intersection of these two solutions and see if it satisfies the original constraints.

First rule out the obvious non solutions (d0[0] ≠ 0, d1[1] ≠ 0 and d0[1] ≠ d1[0]).

Consider a pair of vertices (u, v). If there is an edge between them, then |d0[u] - d0[v]| ≤ 1 and |d1[u] - d1[v]| ≤ 1. Although this is not an equivalent condition, one possible solution is to include all edges where the above conditions satisfy. Do a Floyd-Warshall in the end to check if the solution is valid.

EDIT: typo.

A harder version was here: http://cf16-exhibition.contest.atcoder.jp/tasks/codefestival_2016_ex_a, and there is an editorial for this problem here (problem A, english version is at the end).

Please send an email to [email protected]. "unused code rule" is handled manually.

I did. And also screwed up B. Yay, 0 points :)

It takes me one attempt! :(

I did it like this: let's find all reachable pairs for A, B <= 50 using straightforward dynamic programming. If c, s <= 50, we're done. Otherwise, let's try to subtract each reachable pair from c and s and check if the gcd of the remaining numbers is not 1 for at least one such pair.

A good thing about this solution is that it's impossible to miss a corner case as there're no corner cases (unlike in a solution that uses case analysis).

What do you mean? I had an idea but didn't have time to complete all its details. What I tried to do is to consider how an array b[i] = (a[i] ^ a[i + 1]) changes after a move.