Yes, it will be rated.

Thank you for your feedback, we have moved our contest by 1 hour.

If I happen to get red during this contest then maybe it will not feel like an achievement :(

Sorry, del

Currently, the only office where the software development takes place is our HQ in Redwood City.

Most probably yes. I work at Rocket Fuel and I was told about a week ago that we're planning to hire interns for summer.

We do offer internships but we do not sponsor visas for internships so if you are not in the US you will be responsible for providing your own work authorization.

If you are not planning to apply to Rocket Fuel, you do not need a CV. Please, wait some time for regular registration for the round.

"Contest Overview" section of the announcement has already answered the question.

Maybe it's "To be decided"?

Usually "to be done".

"to be defined" i guess

To be determined wiki

"To be declared" :P

To be Ferdinand de Lesseps.

Yes, it will be rated.

+

Are you sure that the file is non-empty?

http://bit.ly/1F57t42

Each problem will be worth a fixed amount of points and you'll be able to see these points. And these points are our estimation of the difficulty.

Штрафное время будет учитываться отдельно по каждой подзадаче.

did you read the first f ing comment?

Are you sure you've read the first comment and the first reply to it? If so, why doesn't it answer your question?

http://codeforces.me/blog/entry/16140#comment-210444

You can use this in case there are some not-so-big test in the large subproblem to verify correctness. However, once your code is too slow, you will have no idea whether it is because of big N or the solution itself. Anyway, it's not the big deal if you already have a brute force solution. I would just submit it for small tests and use the small tests to double check my solution for large test (I guess this works since problems are in ACM style).

Yes, our company will sponsor visas for full-time job offers.

We did the same scoring system last year and it wasn't bad, so I assume that it is completely fine to use it again this year.

I bet it will

It did

Each subproblem can be seen as single problem. Yes, you need to make a separate submission for each subproblem.

As you expected, tourist got the iphone6~

Because it is ACM style, so i guess they will count.

The most simple way I know is to add him to friends and open Friends Only list.

Add them as a friend (by click star next to their name in profile). Then in registration click the friends tab and you can see all of the people you have selected as friends that have registered.

You can add him as your friend and then check it. nic11 and poikniok were faster ..

It didn't sound that great when we had that queue :D

Since it's ACM-ICPC format, there's no need to put people into rooms (because there are no hacks) and thus there's no reason to stop accepting registrations after the contest starts (except that the late participants lose the time missed).

My task A (which submitted 10 minutes ago) haven't been judged. :(

Or we could have a tougher first problem to begin the contest with.

here's my guess:

1) Since it's ACM-ICPC format, it needs to run all the system tests instead of pretests.

2) The number of participants is simply a lot.

I have decided to go to the bed anyway :D

Прошу прощения. В этот раз изнутри совсем другая ситуация получилась. Просто снаружи все выглядит похоже. В данном случае на пределе стал работать головной nginx, а сервисы Codeforces нормально себя вели. Запас в requests-per-seconds есть не менее двух раз, тут оказалось дело в количестве concurrent sessions скорее + возможно в том, что он еще и статику отдает (хотя там все кэшируется).

Такое понятно как тестировать и понятно что с этим делать, я думаю, что этот issue будет скоро закрыт.

Тяжело побивать рекорды — к сожалению и такие неожиданности случаются. Но я вам гарантирую, что работа ведется ежедневная, в том числе и над подобными issues.

Еще раз приношу извинения. Сам переживаю сильно.

I request the organisers to please make this contest unrated being full of unexpected delays and problems.

Thank you sir.So is the round unrated?

Amazing! I have seen that at the same time there are at least 50 submissions judged! (Because there are only 50 submissions in one page...) How fast!

I hardcoded B1 . saw all the permutations but couldn't figure out any pattern :(

The same thing happened to me for quite a while on B :P

The real pattern is amazingly simple: start with an empty permutation, now put 1 in either end of the permutation, then put 2 in either end of the remaining blank space, then 3, ...

Is it permutation B2? Because I had that moment too :)

A pattern was floating in front of my minds. But could not code it. -_-

This feeling is probably the worst.

Всегда (по крайней мере пока что) мне хватало того, что матожидание некого числа — это сумма его во всех исходах, поделенная на их количество.

Пользовался только тем, что Мат. ожидание = значение * вероятность

Для C и G1 достаточно было знать определение мат.ожидания. Ну и влоб его насчитывать.

F was max flow problem.

• Obviously, we can decide the gender of the "other" one
• Binary search the answer.
• Construct graph with 6 layers of vertices:

1. Source
2. Males
3. Empty cells
4. Empty cells again
5. Females
6. Sink

• Add edges such that, 1 pair of (male — female) correspond to flow with value 1 from source to sink

"System" tests (at least for D1) were weak, too...

My brute-force solution passed... even though it stated there is solution for the following input:

3 3
1 2 LEFT
1 3 RIGHT
2 3 RIGHT

(which is clearly impossible as 3 would have to be both in the left and right subtree of 1).

same here lol, the statement of task D is really confusing...

yea, thought every node suppose to have 0/2 child... forgot the definition...

Can you post link, not full code there?

The first problem — K may take a value of 4, but the second index dp [1000] [4] can not. Second — do comparison "dp [x] [k]! = 0.0" is unsafe. Read about comparsion of doubles. Finally, you didn't calc inv[0] — inversion in start permutation.

My freaking English better than this solution, sorry.. Why dont write easy dfs without any states and get 3 points?

For each amount of money that the company who won has to pay, you calculate the prop by using bitmask.

Let fixed the amount of money x. 1. If the winning company bid x unit. Then there has to be at least another company bid the same x. 2. If the winning company didn't bid x unit, the company had to bid higher, and there are also at least one company bid the same amount x.

Here is my sol: Code

For each possible value v, compute the probability of: 1) [1 company bidding more than v] * [k companies bidding exactly v] * [n-k-1 companies bidding less than v] 2) [p companies bidding exactly v] * [n-p companies bidding less than v] (where p >= 2)

The probability that the value to be paid is v is equal to 1) + 2).

In order to compute 1) you can iterate over all companies (to choose the one that bids the maximum amount) and inside that iterate over all subsets of companies (ignoring the ones that contain maximum bidder) to pick the ones who are bidding exactly v.

To compute 2) you simple have to iterate over all subsets of companies that contain at least two companies.

The only primitives you need to compute 1) and 2) are p_bid_more_than(company, value), p_bid_less_than(company, value), p_bid_exactly(company, value) which can be computed in O(1) given L and R for each company.

You're a mean minion :P

I suppose smth to get rid of solutions O(n!), just huge amount of permutations.

probably something related to long long int. I got WA on #20 and it went away after I changed "m" to long long int.

Yes, I got WA 20 first and then fixed. The problem for was long long literal type. I did a bit shift 1 << length while it is supposed to be 1LL << length.

We inquired about this with our manufacturer. Unfortunately, the whole batch has to be either in normal or tall sizes so we'll stick with normal since this is the preference for most people.

Могло.

The possibility will be stable.... I tried to find the matrix A for dp(n-1)*A=dp(n).... But failed.

Yes.