It is open for everyone.

They registered twice (probably because they registered on their own, then another team registered them). Just unregister one of which and it should be fine.

It won't prevent them from joining the contest either way, but last I heard, if they submit, a random entity out of the two will get counted for submission.

By 5 minutes??

Yes, we are aware of it. We think that the last 5 minutes of a 5 hours mirror and/or the first 5 minutes of google code jam round 1b are not super important and it's acceptable if they overlap.

You are absolutely right, this contest is completely useless for you.

Rule 5. If you want to participate "competitively" in the mirror, refrain from opening the scoreboard so that you don't obtain any information on the difficulty of the problems.

feeder1 needs to go to https://codeforces.net/contestRegistrants/1662/friends/true and unregister the team first. Just hit the red X. Then you're good to go

M is very straightforward, just put maxR 'R' and maxW 'W'. how to solve D? My thought is remove all occurence of AA, BB, CC and replace AB with BA, CB with BC, then check if the two string is same. However it fails. What did I miss?

K:

Note that due to continuity and such arguments, it is not hard to verify that the sum of lengths can be made the same for all three triangles when you reach the minimum $$$r$$$. Let the given points be $$$A, B, C$$$. Construct points $$$A'_i, B'_i, C'_i$$$ with $$$i \in \{1, 2\}$$$ such that $$$BCA'_i$$$ etc. are all equilateral. Let's call a triangle good if the minimum sum is achieved at a vertex, and bad otherwise. If we choose an optimal residence point $$$D$$$, then there are at least two triangles among $$$DBC, DCA, DAB$$$ which are both good or both bad, since a triangle is good iff one angle of the triangle is $$$> 120^\circ$$$. In the case they are good, say $$$DAB$$$ and $$$DCA$$$ are good, we just need to ternary search for the best point on the corresponding perpendicular bisector of $$$BC$$$ (do this for all permutations to find the answer in this case). In the case they are bad, say $$$DAB$$$ and $$$DCA$$$ are bad, note that we have $$$DB' = DC'$$$ where $$$B', C'$$$ are on the opposite sides of $$$CA, AB$$$ as $$$B$$$ and $$$C$$$. So we should ternary search for the answer on the perpendicular bisector of $$$B'C'$$$ (and do it for $$$A'B'$$$ and $$$C'A'$$$ as well).

Note that whenever we said that we need to do a ternary search, there are some more constraints to be satisfied. We note that the problem is to find a global minimum, and since the function is convex, we should just evaluate the answer for each case naively while ternary searching (without paying attention to such constraints). For instance, rather than using $$$DA'$$$, we can use $$$\max(DA', DA' ')$$$.

Here it is! https://codeforces.net/blog/entry/102042

Should be fixed now.

Yes, in the next few days the swerc.eu website will be updated with a copy of the scoreboard that includes team's photos.

Thank you :)