Rmin's blog

By Rmin, 8 years ago, In English,

The contest of www.hsin.hr/coci is started at 5:55 PM at the time of the codeforces.

 
 
 
 
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8 years ago, # |
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the contest was end and i solved 3.5 problems.
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    8 years ago, # ^ |
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    I've done the contest very bad and my score is 136..
    What's your score???
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8 years ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

English statement of the second problem was so hard to understand...

BTW, who can help me with the last one? It seems me very interesting, but I still have no idea how to solve it.
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8 years ago, # |
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When can I see the results of all programmers?
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8 years ago, # |
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Could anyone tell how to solve the sixth problem?
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    8 years ago, # ^ |
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    Can you tell me - is my idea for the fifth problem right? I did Euler's composition of the given tree and then it was like update(l,r,x) and get(x) problem wich is easly solved with Fenwick's tree. I got 0 points for it and still can't understand why..) what do you think of my idea? 
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      8 years ago, # ^ |
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      My approach is almost the same. Here is my AC code.
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        8 years ago, # ^ |
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        Oh....! I had just the same, but your line 29 was in the "while" cycle after dfs calling. I changed it and got AC. I still wonder why didn't it work?! Anyway, the last value of timer will be puted as tout for curent vertex. I am stunned with it...
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    8 years ago, # ^ |
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    Я, пожалуй, по-русски отвечу.
    Попытаемся понять, какое условие на время d, в которое мы можем пускать на производство машину сложности β, если машина сложности α была запущена на производство в момент времени ноль.

    Необходимо, чтобы по каждому человеку i время когда он допилит первую машину было не больше, чем время, когда он приступит ко второй машине. Иными словами, α Si ≤ β Ti + d, где Si - сумма производительностей всех людей до i - ого, а Ti - до (i-1)-ого (слева стоит, собственно, одно время, а справа - другое).
    Иными словами, , где F(x) - это линейная функция, не зависящая от α и β. Ну а максимум из кучи линейных функций в точке ищется известным образом: построим в предподсчёте пересечение полуплоскостей, ими задаваемых, и бинпоиском будем искать отрезок, на который попадает . Как-то так.
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      8 years ago, # ^ |
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      Please don't write in Russian in English blog
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8 years ago, # |
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One common mistake in second task was when N = 0. There should be an empty file as an answer, but depending on realisation it could give runtime error or wrong output.
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8 years ago, # |
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Can anyone explain me, what are we supposed to do it second problem? I have no idea, what do they do?