"Tryam" to everyone!
I am an authors of the today's contest. First of all I want to thank Artem Rakhov [RAD] (thanks to him, you'll understand my problems) and Maria Belova [Delinur] (those who know my English level will understand a huge of my thanks). Also thanks to everyone staff for the wonderful of the codeforces system. And also thanks to Sergei Tarasov [Seryi] and Andrey Tkachenko [Tkach1024] for ideas generating and testing.
Good luck and have fun.
Today, a breakdown of score will be slightly different from the standard - for the second division is 500-1000-2000-2000-2500 and 1000-1000-1500-2000-2500 for the first.
And one more, but certainly pleasant news - register ends exactly at the start of the round.
At 19:35 Moscow time reported that in pretests for 123B - Squares have errors (a special thank to him). Indeed, a special case was not considered (a = 1, b! = 1 and symmetric to it). We have made decision to exclude tests covered by this case. There were a lot of such tests in pretests. Test generator has been fixed, and all tests were changed. After that all submits have been rejudged. As a result some submits with WA verdict could get the AC and vice versa.
What we have now.
We believe that the round should be rating. But if the situation with the problem 123B - Squares is strongly influenced to you, you can send an appeal with the evidence of this in the form of a personal message to RAD no later than 11pm on November 4. We can either do this round unrated for you or remove excess submits.
At the end of the system testing rating will be updated. If you have no appeals, it will be your final rating.
Sorry for the inconvenience.
In which language this word "Tryam" belong to i think it's meaning something like "Hello" ,i tried to use google translator but it doesn't know it :)
There is a very popular Soviet cartoon named "Tryam! Hello!" :)
Seems to be music onomatopoeia, like [trɐm]-[pam]-[pam], [pam]-[pam]-[pam], etc.
:)
looks like people who had a better contest than they usually do, is saying to make it rated(not everyone). But isn't it justified to make the a contest unrated when you change the problem statement? You never know how that has affected somebody.
Note : Making it rated/unrated is not going to change my rating massively, maybe that's why I'm neutral :P
http://en.wikipedia.org/wiki/Karma
In the context Petr used it, it's basically the same idea as saying "bad luck in the future". In other words, it could potentailly be bad luck to criticize a problem writer. That bad luck, or "bad karma" may affect you negatively in the future in the form of poor contest performances, or other such things.
It's the idea that whatever you do comes back to you in the future. Do good, it will come back to you, do bad, it also will come back to you. So it's better to do good, if you can :).
In the problem statement you can see this phrase "The only line contains the initial string s, consisting of small Latin letters"
quite on the contrary definition
Egor's earlier point ("Essential change of problem statement and pretests") holds for div 2 as well because that same problem he referred to was shared across divisions.
So div 2 people could have been affected by that same issue, so it wouldn't make sense to rate the contest for div 2 and not rate it for div 1.
I do remember seeing that some early submissions were made for D, but I can't say for sure if they were made before or after the announcement.
So if nobody was affected, then I think your point is valid.
it will be rated, it's written in russian version of the post
"Мы считаем, что раунд должен быть рейтинговым. Но если ситуация с задачей 123B - Клетки сильно на вас повлияла, вы можете подать апелляцию с доказательствами этого в виде личного сообщению RAD’у до 23:00 4 ноября. Мы можем либо сделать этот раунд нерейтинговым для вас, либо убрать лишние посылки."
It means(shortly):
"If you want round to be unrated, and problem 125B-Squares made big changes in your points, you should apeal to RAD until that time."
Sorry for bad english
If someone wants this round to be rated, let him have it. There were problems, statement and tests were changed so this round should be unrated. Let's consider Egor's solution for problem B. He has the last submission skipped, which is accepted. Instead his earlier submission is taken into account, which fails. It's a joke, unrate this round.
Sorry, but I just can't imagine. Statement was changed 35 minutes after the contest start. Tests were also changed. Status of some solutions were changed (someone was allowed to hack earlier than he should, someone lost time, someone made wrong hack). The judging is ridiculous (look above) and you still want to make this round rated, with option to write some strange appeals and made changes by hand - unbelievable.
But we are violet and should be very carefull! If you do not test, you take a big risk
Don't worry. I submitted A+B+E. Only E survived the tests. It looks funny in the scoreboard. After that A and B, each with one line more, got accepted :)
Nevermind
I reduced Problem A (Prime Permutation) to the problem ``given two sets , determine whether there exists a partition of A such that .'' Actually Problem A was a very special version of the latter problem in which only one element of B is greater than 1, which I didn't realize during the contest :-(.
The question is how to solve the latter problem in general. During the contest I thought it could be solved by DP and bipartite matching, but I couldn't finish coding.
http://ideone.com/Go65C judge is giving WA on 2nd test case ?
I don't understand why making this contest rated would make it any better? I think it should be unrated without any discussion (I didn't participate, so I don't have personal motives).
Suppose the entrance and exit are fixed. For a given edge x, what is the expected number of times that the DFS will traverse it before reaching the exit? For edges that lie on the path from entrance to exit, it's 1, because we will always traverse them exactly once. What about, say, another edge from entrance - what is the possible cases for it? Can we traverse it 0 times? once? twice? more? What are the probabilities of those cases?
could anyone tell me how to solve problem D?