The upcoming contest is brought to you by me. My name is Arthur, I'm a student of Chelyabinsk lyceum 31.

I am glad to thank Gerald for contest preparation management, Delinur for translation of statements, MikeMirzayanov for a really excellent system, dolphinigle for test-solving, proof-reading and much valuable advice, fdoer, grey_wind, Skird and alger95 for help in problem development.

There is nothing more to say yet. As you can see, contest will be held in two divisions separately. The scoring system will be standard (not dynamic), but scores for tasks may be unusual (see update of this post).

There won't be volume stories. I hope you will like the contest and its problemset and everything will be OK.

**UPD:** Scores for tasks:

Division 1: 500-1000-2000-2000-2500

Division 2: 500-1000-1500-2000-3000

**UPD2:**

Congrats to winners =)

Div1:

Div2:

**UPD3** some part of English editoral posted. You will find the remaining part of the editorial later

**UPD4** English editoral was completed.

How can I register as virtual contestant?

Now you can only register as real contesant. There is no way to register out of competition.

Wow, dolphinigle as the tester? I would expect clear problem statements and a stable, rated round then :)

I agree. He always explains everything clearly in all of his problem statements. :)

the contest will be interesting if the scores for tasks are unusual. And I hope the contest be rated.

And the score distribution is... :-?

is ... published

Division 1: 500-1000-2000-2000-2500

Division 2: 500-1000-1500-2000-3000

it is writen now.

Can anyone give me a formula that can count new rating? Thanks very much. :-)

The formula is not public.

So what is the observation that solves B?

Eliminate any sequence that contains two consecutive XOR operations.

It's WA :) Sometimes you need to make only XOR operations without adding to get better result.

tourist's test:

Input: 1 2 1 10 20 -100 1

Output: -1200

Answer: -1000

Yes, because of that you assume that you can't do consecutive XOR operations, BUT ALSO NOTE: after that you should also assume that the length of operations can be u, u — 2, u — 4, ... u — 2k

While adding new operation if (u-operations_done%2==0) you can actualize result (later operations will be only xors).

Hmm I still got a time limit even after implementing something like that.

You need to check sequences shorter than

uas well, right?EDIT: Aha you already answered :) thanks.

Yes, please see the message above. Regarding to time limit, all possible sequences are about 2mln, and for each of them you need to do 30 operations, so 4s time limit should be ok.

Where/How do I report cheaters?

Check out this 2 guys codes for problem A:

http://codeforces.com/contest/193/submission/1756318

http://codeforces.com/contest/193/submission/1756337

Its the same code, they are both from the same country and they submitted at the same time..

You'd better contact either Gerald, or author of the contest, or MikeMirzayanov.

Their rating graphs since mid-April 2012 don't look vastly different either. Or their avatars, for that matter. Why would someone submit exactly the same stuff from two accounts anyway?

Maybe it is one user that has another account for testing and when he gets accepted he submits the same code with his real account. Either way it should be penalized by admins. Cheaters are never welcome.

Sorry, That accounts is mine, I always use 2 accounts to join codeforces beta round, but I am not a cheater, I don't know that, and I will use only one account to submit on next contest. Sorry, because I don't know.

Can't help wondering what could be the reason for always using two accounts, if not cheating?

Its only 8 minutes after the end of the contest and the system test for DIV-2 has completed 77% ... Blazing fast :)

Nice problemset. Hard, but fun :)

Thank you =)

Yes the contes was verry good and interesting and the there was no problems with the tasks.

Hi guys, I solved problem for the first time (in 2 participation) and my rating decreased as for the time I do not solve any problem. Is it normal ? Thanks for the contest,

Rating is change depend on your ranking, not from your submission :)

Thanks, i just do not noticed that I effectively less ranking than the first time one.

Really enjoyable and challenging contest. I could've done much better, but I still had lots of fun doing it and it's all that matters. Congratulations to the organizer!

I like the problem E in Div 2.

Any idea how to solve this equation?

I thought that queue is too slow for this, but only "trick" (I do not know if it's working — if it is enough) is the one from statement — we can combine 6 combination to get [ 4; 4; 4; 4; 4; 4 ]...

Although i don't think this equation related to Ediv1, Use some of method like Gauss-Jordan Elimination to solve it. After did elimination you will get 1 free variable, Bruteforce on that variable is enough to pass Ediv2 problem

you are correct, it's E in div2 (C in div1), thanks for hint ;-)

Great contest ... analytical problem set and clear statement ... really enjoyble .... and thtz the spirit of CodeForces ... :)

a really great contest. thanks for bringing such a great time to us. of course the problems were hard. :D

I hope there will be an editorial for this round too.

thanks again.

can u explain me how to solve the problem 194B? thanks so much :)

We take the square's fields as a linear sequence of 4*n fields. To simplify our calculations, we'll mark the fields starting from 0 (the bottom left corner).. all the way to 4*n — 1. In every move we will traverse n+1 fields, thus landing on the field marked as k * (n+1) mod 4*n, where k is the amount of moves we made. Now it is obvious that the solution to the problem is the smallest positive k for which k * (n+1) gives a remainder 0 when dividing with 4*n, plus 1 (because we have to count the initial cross too).

Hope this helped.

thanks very much :)

Hey skyocean1996 I post an explanation in my blog here

thanks. Your post's very clearly

So where is the English editoral?

It's hard for me to write it, but you can see editoral for the first 3 div2 problems. Today I hope, you will see editoral of all div2 problems.

I`m 204th coder in ranking, but my profile shows I got 206th.

Two cheaters were deleted from the standings. And places wasn't updated. I'm not sure, but it looks like truth.

Why does it say "Statement is not available." when I try to access a problem statement from this contest?

Fixed now. Thanks!

what time these rounds start in US?