For the same reason as MikeMirzayanov. He is a member of Codeforces team.

rip in pepperonis kursatbakis0's contribution.

He is the author of the most-downvote comment on Codeforces

You meant a lot of math with no programming right? Because that's two different things.

Wow, apparently other people share my ideas as well :o

As it turns out, all problems are math (with some programming)...

you know you are in a math oriented contest when memory of your first 3 solutions is 0 Kb.

Most of programming doesn't involve math at all: look what typical job postings say.

Это баян :D

Darn, I wonder how you would know..

Registration will be opened later (see on Contests tab).

by not getting downvotes

Strangely nobody has answered your question in details. To get contribution scores you can try to:

• write editorials for the round which do not have any.

• prepare a CF round or at least a gym contest. Announcement might bring a lot of contribution points, though it depends on the announcement.

• post some funny jokes on CF. This seems to be the most effective way but you need to stay on-topic while posting the jokes.

• maybe write some good programming related articles here? Didn't try this one though.

standard

Претесты проходило даже решение вроде cout << (r+g+b)/3;

Полное решение. КАК????

You find the maximum possible height (approximately ) and do dynamic programming: DP[h][r] = the number of possibilities to build the tower with height h and using r red blocks (and h * (h + 1) / 2 - r green blocks).

R and G are 2e5 so number of levels are at most 600 ! It can be solved like Knapsack problem !

Because you have participated in so many contest, right?

What was the hack test case ?

try to formulate a top-down DP solution that takes O((r+g)*h), this will give you MLE, then try to formulate the same DP bottom-up, this will require only O(r) in space and runtime of O((r+g)*h), if you see my code you will see both approaches

абсолютно то же самое. очень обидно

Скорее всего вы говорите о более-менее одном и том же необычном случае.

В раунде была пара сабмитов, в которых фигурировали строки вида:

int a = 0;
for (int i = 0; i < b; i++)
  a += x;

Где b имеет порядок 10⁹. То, что такое решение работает, и время исполнения — 15 мс не значит, что сервера Codeforces научились выполнять десятки миллиардов операций в секунду. Это значит, что шибко умный компилятор оптимизирует данный цикл в что-то наподобие:

int a = b * x;

Так что шансов почелленджить подобное решение немного =)

For example "0 0 3" to crack some of the solutions. TLE was also achievable on others by entering numbers close to the set variable limits.

1000000000 2000000000 2000000000

7 2 1 answer-3

case 0 0 0 0 0

it is easy to forget about 0 case, ant first time I had fogotten too/

Yeah, I'm one of them. It's so easy to miss the non-zero requirement. Wish they hadn't put this case in pretests for another gazillion of hacks ;d

LOL ! He's translating the same message into different languages !! what a guy :/

so overall complexity would be O(r.sqrt(r)) ? I didnt't realise that would pass, so i didn't code it up :\

that was exactly what i did, but received RTE ): no idea why.. EDIT: not exactly, i used a map to store answer, instead of optimizing space.

This should be O(10^8) which i think it should give TLE :D

However I am surprised because (10^8) gives MLE but don't give TLE :D

the first observation is false (and the reason I got RTE during contest). For r = g = 2 * 10⁵, H = 793. My upper limit for H was 700, using 800 should result in TLE instead of RTE :D

Всегда же был моноширинный.

Can you please explain the solution? I can't get it till now. Thanks

you are right

you forgot to sort t =)

I solved it using the same solution .. do you have a proof of correctness?

Hi can you please explain it? Thanks :)

assume t[0] <= t[1] <= t[2]. and sum = t[0] + t[1] + t[2].

best case would be sum / 3. because no matter what we do we will always be left with sum % 3 balloons. now when can we get sum / 3 tables and when we can not:

case 1: if 2 * (t[0] + t[1]) <= t[2] than it is obvious that we can not get more then t[0] + t[1] tables because at each table at least 1 balloon is subtracted from t[0] + t[1], and by taking 2 from t[2] and 1 from t[0] + t[1] at each turn we can get exactly t[0] + t[1] tables.

case 2: if 2 * (t[0] + t[1]) > t[2]. let's prove that we can get sum / 3 tables in such case. by tactic taking 2 from t[2] and one from max(t[0], t[1]), t[0] + t[1] will not become zero before t[2] becomes zero, which means that at some point t[2] will become at most max(t[0], t[1]). and at that point difference between t[2] and max(t[0], t[1]) will not be more than 2. no we know that max(t[0], t[1], t[2]) — second_max(t[0], t[1], t[2]) is not more then 2. we can use this as an invariant. tactic taking 2 from max and one from second_max does not change invariant. now suppose we have some choosing tactic, we are stuck when we have situation like this a 0 0 where a >= 0 or 1 1 0. if a <= 2 it means choosing was optimal because no matter what we do we will be left with sum % 3 ballons. and second finish(1, 1, 0) is also optimal by same reason. answer in both cases will be sum / 3. according to our invariant we will not be left with a 0 0 with a > 2 because if a > 2 then our invariant does not hold. so we now that our choosing was optimal and answer is sum / 3.

now what min(sum / 3, t[0] + t[1]) does is exactly checking if 2 * (t[0] + t[1]) <= t[2]. because if t[2] >= 2 * (t[0] + t[1]) (case 1) then sum / 3 = (t[0] + t[1] + t[2]) / 3 >= (t[0] + t[1] + 2 * (t[0] + t[1])) / 3 = t[0] + t[1]. so min(sum / 3, t[0] + t[1]) will be t[0] + t[1] (exactly answer to case 1).

if t[2] < 2 * (t[0] + t[1]) (case 2) then sum / 3 = (t[0] + t[1] + t[2]) / 3 <= (t[0] + t[1] + 2 * (t[0] + t[1])) / 3 = t[0] + t[1] so min(sum / 3, t[0] + t[1]) will be sum / 3 (exactly answer to case 2).

hope it was helpful :D

It's really weird case ! :/

http://codeforces.com/contest/478/submission/8255696

I've seen solutions skipped because of plagiarism before. Don't know about you.

Plagiarism. I refer to this 8255446.

I solved 3 problems and became purple and I am not surprised :D

Read the problem statement. It says: "A wavy number is such positive integer that for any digit of its decimal representation except for the first one and the last one following condition holds: the digit is either strictly larger than both its adjacent digits or strictly less than both its adjacent digits." For the first and last digits the adjacent digit if exits should also be strictly larger or strictly less than the digit. "strictly larger" or "strictly less" does imply that two adjacent digits should be different. Hence the number "11" cannot be a wavy number.

in A don't forget 0 case (0 0 0 0 0)!

In B your max looks OK, but... I think, it is good to use function like my, to make code easier (c++):

 long long pair(long long a){// pairs in command
	return (a*(a-1)/2);
}

Link to your solution is the 7digit number written on the left.

Open it and send the URL

if you wont i will send my solution. my solution is to short. i think you will uderstand

First of all take the maximum if it is >2*(sum of other two) then answer will be (sum of other two) because for every element taken of rest two take 2 elements of maximum. now suppose max<= 2*(sum of other two) then answer will be (sum of three)/3. you can observe this by take the balls greedily.

your rate is a good proof -_-