1) According to the results of the competition, top 5 students, who will take part in WCC, will receive prizes. 2) Rated, I hope =)

if it is not rated then why it is div2 only?

Участники из первого дивизиона могут участвовать в соревновании вне конкурса. В частности, школьники из 1 дивизиона могут участвовать в CF #289 вне конкурса, то есть для них раунд будет нерейтинговым, но их результаты будут учитываться при отборе на ЗКШ наравне с див.2 участниками.

Yes, of course

If you look closely, there are English translations below each Russian statement.

Да.

В группе пишут http://vk.com/wall-32943877_275

Да и на сайте висит: тык

Откройте список зарегистрировавшихся, найдите себя и нажмите на крестик

are you??

"we'll be rewarded on problem count and submission times with 20 mins penalty for wrong submissions" is correct. I can't promise anything about tasks.

Full feedback.

for div2 yes

Yes, of course

TODO: update comments before sending.

No freeze

Are you sure that people in here can understand what you say?

I think, every language supported by Codeforces, will be available.

yeah he did it! Congratulations! =D

i'll wait for dreamoon_love_AA vs sorry_dreamoon again :)

Why do you think so? I really liked problems, especially F, even though I have no idea how to solve it.

That problem C -_-

same problem

I don't know the test itself but what's your soln for this input?

2 300 300

Try 10 38 38 38 38 38 38 38 38 38 38

answer: 29999 38999 39899 39989 39998 47999 48899 48989 48998 49799

А я под конец так отчаялся, что удалил часть кода в решении по С и заслал, что привело к прорыву с WA6 до TL21.

you can have a look at my solution .. i did it so simple you even can't believe .

Let call a_i = 1 if s_i is a vowel else a_i = 0.

sum_i = a₁ + a₂ + ... + a_i

Number of vowels in all sub-strings with length x is:

P_i = (sum_x — sum₀) + (sum_x + 1 — sum₁) + ...

= (sum_x + ... + sum_n) — (sum₀ + ... + sum_n - x)

The result is sum of P_i / i

there can be something like this http://pastebin.com/E2pCyWmu

I didn't solve it either but I think it's dp too. We look at each letter independently and if it's vowel we add to answer sum of some harmonic serieses (we can calculate it using this value for last letter and array of prefix sums of harmonic series)

UPD: now I see that I was wrong and it can be solved easier another way.

Consider the vowel in the string at the position i. It contributes to the answer only addends of the form , where k ≥ i and j ≤ i, because it is only contained in the substrings of the form s_i..j, j ≤ i ≤ k. After what, we can sum this expression for all i, j, k such that s_i is a vowel and j ≤ i ≤ k. Plugging this into WolframAlpha, we can get the final formula for the answer: , where H_i = 1 + 1 / 2 + ... + 1 / i is the i-th harmonic number.

At the each step remember a last output value and a sum of digits in it. Then add 1 to last value and try to get the smallest integer which has the desirable sum of digits. To do it basically set less significant digits to 9 if you need to make sum of digits larger or set them to 0 (and +1 to digits before them) if you need to make it lesser.

Use bignum arithmetic.

I have seen two messages (English and Russian). Intelligence is not sleeping. Check your profile setting, maybe.)

Magic 9648904

It took me half an hour to push formula...And I solved this problem. answer=(1/2 + 1/3 + ... + 1/n)(a[1]+a[n])+2*(1/3 + ... + 1/n)(a[2]+a[n-1])+...+(n-1)(1/n)(a[n-1]+a[2])+a[1]+a[2]+a[3]+...+a[n] if a[i] = I or E or A or O or U or Y then a[i] = 1 n means length of string And it's my code: http://codeforces.me/contest/509/submission/9654437 :-)

I also wrote bad and long solution for E and I got accepted :-)

Динамика dp[l][r] -- сколько существует деревьев, таких, что корень в вершине с номером l и при dfs-е из условия получается последовательность b[l], ..., b[r].

Тогда дерево имеет вид l, (дерево с корнем в l + 1, состоящее из вершин l + 1, ..., r₁), (дерево с корнем в r₁ + 1 из вершин r₁ + 1, ..., r₂), ..., (дерево с корнем в r_k + 1 из вершин r_k + 1, ..., r), причём r_i < r_i + 1.

Это можно посчитать ещё одной динамикой dp₂[l + 1][r] -- сколько способов разбить отрезок на деревья указанным способом.

same here site page for submit will not open..

It is always laggy for me too while I am in college but works fine when I am at home . You could go to IPC or library for faster speed.