А если бы они не были земляками Вашими, то раунд не удался бы?

Hahaha so much hate for you

Why there is so difference two people with exactly same comments but one has 65 upvotes while other -16 downvotes !

he is just 8 rating away from becoming Grandmaster. let's hope that he can achieve this in the next round. :)
seriously, downvotes for wishing a participant good luck? seriously??!

Rating graph of randrew is also inspiring... :)
minimum rating is 1306 and maximum is 1889 ... just 11 rating away from being Master...

I hope that means we get a holiday from maths problems

i do too! [for geometry especially :D]

this is CF, not IMO

Good luck! :))

Wish you good luck :D I'm just hoping to get a positive rating this round.

Hope short problem statements too.

actually there's a shorter announcement here it was only first 4 lines when the author first posted it.

брат за брата :D

Welcome

No. For example, aropan was the author of Codeforces Beta Round 92 (Div. 1 Only) and Codeforces Beta Round 92 (Div. 2 Only) before, and tourist was one of the authors of Codeforces Round 127 (Div. 1) and Codeforces Round 127 (Div. 2). So it's at least the third Belarusian contest :)

mine is correct. F5?

I think the registrations close 5 minutes before contest :)

It's simple: the time shown corresponds to what's written above it. If it's "Registration's running", then the time shown is how long the registration will be running. If it's "Before start", then the time shown is how long until the contest starts.

Look carefully! Bottom right one says Registration is running

Видимо туда же, куда кусок логотипа.

magic

Она исчезла. (с) Mike

http://codeforces.net/blog/entry/12980#comment-176995

А вдруг на цфе рейтинг unsigned, и поэтому если уйти в минус можно выйти в топ?

мне вот просто интересно, на сколько он поднимется если хотя бы решит 3-4 задачи за контест?

I'm ready

O RLY???

Haha :)

Good problems, but contest was ruined by slow testing.

I saw 20 pretests in some problems. Which means no hacks and maybe 500 tests in final tests. This could take some time to finish.

Thanks. We don't thank Delinur because when I write the announce we didn't know who was the translator.

Seems rope doesn't suck 7400437, but how does it work?

I guess it was linear DP. dp[i] = max(dp[i-2]+(freq[i]*i),dp[i-1])

I too faced the same problem. :(

It should be dynamic programming, you should consider two circumstances of the number. Reserve it or throw it away. Suppose that c is the count of a precise number, we have: f_i, 0 = f_{i - 1, 1} + c[i] * i, f_i, 1 = max(f_i, 0, f_i, 1). Note that you should use long long in c++ or int64 in pascal.

Ok, but what is the mistake with my approach? http://ideone.com/MVT6SI

Простая динамика

dp[1] = cnt[1];
for (int i = 2; i < MAXN; ++i)
     dp[i] = max(dp[i - 2] + i * cnt[i], dp[i - 1]);
cout << dp[MAXN - 1] << endl;

cnt[i] — кол чисел i

invalid input after 10 minutes :(

Всю жизнь пишу массивы от i64, внезапно сегодня написал от инта, контесты утром для меня плохо складываются.

Who ?

The chain in the problem C could be maintained just by using a set. After merging two vertexes (a, b), the answer will be .

For the ones who have studied problem 1A — Dreaming in IOI 2013 the solution will be quite clear. Fortunately I have read that problem and solved it before :D

We have solution and it's much easier then yours.

One can divide an array into sqrt(n) blocks and maintain a deque for each block. Shift operation is rather simple: for fully covered blocks, it is just one push_front and one pop_back. Partially covered blocks can be handled naively.

To answer the second type of queries, one can maintain a count array for each block(count[i] — the number of occurrences of i within a block). For fully covered blocks the answer is just count[x]. Partially covered blocks can be handled naively again. Shift operation changes the count array in at most 2 positions per block, so it is possible to update it quickly.

The complexity is O((m + n) * sqrt(n)).

From what I could read in kutengine's solution: keep linked lists containing elements each, and remember the count of each integer for each list. Then you can do the rotating query by removing from the end of one list and prepending to the start of the next.

That's quite beautiful, actually. I'm always surprised by sqrt-decomposition brilliance :P

You try not use #include <bits/stdc++.h>

Try a different compiler (e.g. C++0x) -- it works fine on my computer.

I guess, D is based on game theory..

waiting for the editorial :)

I think we can use bor and then for each vertex set: 1) can we Win from this position 2) can we loose from this position After it see on root vertex and print answer. http://codeforces.net/contest/455/submission/7387998

trie + 2 tree traversals to check if it's always possible for first player to win and if it's always possible for first player to loose. then just check for each case:

• first player can choose if he wins or looses a single game -> "First"

• first player always wins a single game AND k%2 == 0 -> "Second"

• first player always wins a single game AND k%2 == 1 -> "First"

• first player always looses a single game-> "Second"

yes ,but one of them give me 100 pionts. others even passed,very angry...

How do you write int main?

16. Solution code obfuscation and creation of obstacles to reading and understanding the solution is prohibited. It is forbidden to use any special techniques designed to complicate the reading of the code, to understand its workflow.

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