uhhhh lucky you are , i got an exam tomorrow so i guess i won't be participating :/ Good luck everyone :) !

Do you have the keys of every round. I really want to learn

In my experience, there were times when there are some late changes like two problems were swapped or we concluded that a B problem should cost 1500. Since often there is much stuff to do for the authors also in the contest day, we usually decided the score distribution after everything was done.

If I was an author, I would never post the score distribution before the contest. As well as the first letter of title for each problem, number of samples, or, maybe size of each statement in bytes. I don't see why any of this useless (before the contest starts) information should be published — if you are interested in it, you can find it just after round begins, it is the same as asking author to announce topics of forthcoming problems.

Or are you planning your tactics on sheet of paper depending of scoring before the round?

No, your rating will stay the same.

If you are not sure to participate then you can always Unregister before closing registrations.

No, as long as you didn't make any submission.

and by the way you can unregister from here (click on the red X next to your handle)

it is starting at regular time (1930 MSK). this is just a guess, but it maybe due to daylight savings that it's one hour earlier in ur timezone.

Just keep in mind that not every problem is solvable in Python when the constraints are tailored for natively compiled languages, and you'll be fine ^^ .

Are you a vampire ? :-s

it will happen! just a bit later)

Wow, 52 unsuccessful hacking attempts. That's truly the result of the red coder :)

It's really harder than C, but can be solved using relatively easy binary search. T don't understand, why so many people did not solve this problem. Especially you, Xellos (no offense, just surprised).

B<C in my mind

i've used an interesting idea which is too complicated for div2B actualy.

It is obvious that allways b <= 2000 so lets define next[b][i] = after 2^i step what will be b then lets define add[b][i] = after 2^i step how many time a will decrease. So binary search on answer and check answer O(log MAX_ANSWER). Overall complexty is O(log MAX_ANSWER)^2 Also here is the proof of why we can use binary search :

let f(x) = after x steps what will be new a value

let g(x) = after x steps what will be new c value

let df(x) = f(x) - f(x - 1)

let dg(x) = g(x) - g(x - 1)

As you see this functions are monotonic and also dg(x) is allways less then df(x) .So they have only one intersection point.

it's not just you.

Not just you, definitely.

That was my worst participation ever. I'm glad I'm Div1 and have rating saved :D

Пусть dp[i][j][b] — количество деревьев с i вершинами, размером максимального паросочетания j и обязательно включенным корнем/или необязательно. Теперь перебираем сколько вершин кидаем в правое поддерево, в левое идут остальные. Если b = 1, то мы ставим ребро в корень, а в детях третьи параметры могут принимать значения (0, 0), (0, 1), (1, 0), если b = 0, то только (1, 1). В процессе перебора размера поддеревьев мы должны выбрать номер корню и отправить сколько-то вершин в поддерево — это choose(n — 1, i) * n. А потом еще и поделить на 2, чтобы не учитывать изоморфные деревья. Если ребенок один, то все понятно. Ну и в конце поделить на n, т.к. корень фиксирован.

the tree is a binary tree, one parent and two children at most.

let f[n][k] be the number of trees of n nodes, max matching k such that the root is required for the maximum matching (i.e there's no augmentation that unlatches the root and keep the matching size k).

let g[n][k] be the number of trees of n nodes, max matching k such that the root can be unmatched and keep the same matching size.

to calculate f, g we have to split the nodes we have into two subtrees and select the root of each subtree, and distribute the required matching (k) among the subtrees, you can force the node with the smallest index is always in the left subtree and recurse.

for a root to be required in the matching one of it's children must be not required in the matching.
for a root to be not required in the matching both children must be required in the matching.

based on these rules you can calculate f[n] using f[n-1], g[n-1], and calculate g[n] using f[n-1].

the answer is f[n][k]+g[n][k], watch out for the cases when you distribute the nodes and one of the subtrees is empty I had to handle those separately.

here's my AC sub 5723599, I got RTE during the contest because I didn't handle the case where k>n.

1 1 of course. perhaps that's the 7th pretest

1 1

1 1 of course

Advice: If you don't want hard problems, never register for "div2 only" contests ;)

That's why I used BFS instead.

But, nevertheless, hadn't enough time :(

You can use this Thread constructor.

i got MLE on test 35 :(

1 3 was the answer, not your output. the checker log says "output contains 0 elements". so it seems that your code produces no output at all for this input.

This is a common behavior in Codeforces Round (only div 2) !!!

Looks like 2011 is a lucky number... http://mirror.codeforces.com/contest/380/submission/5706257

How !!

When you insert a character into string, and the string's storage space is too small, the string will allocate a new array which is  ≈ 2 times bigger and copy the data into the new array. As a result, for each of the strings the reserved memory could be greater than 2000 bytes.

Try t[i].reserve(2000); for each i in the start, then memory will not be reallocated (and it is also slightly faster).

If it still fails, then this is due to dynamic memory (maybe the allocated blocks can't be packed so closely as in char t[2000][2000]).

EDIT: I realized that 2000 * 2000 is too small to hit the memory limit anyway. Perhaps there is something else wrong.

Вероятно, использование стека в рекурсии разное.