rng_58's blog

By rng_58, history, 8 years ago, In English

AtCoder Grand Contest 005 will be held on Saturday (time). The writer is yosupo.

Contest Link

Contest Announcement

Let's discuss problems after the contest.

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8 years ago, # |
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Never mind. I was wrong.

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8 years ago, # |
  Vote: I like it -15 Vote: I do not like it

It clashes with the ICPC Preparatory Series contest on Codechef. :(

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8 years ago, # |
Rev. 3   Vote: I like it +23 Vote: I do not like it

I am really interested why all in top 10 are reds except semiexp whose color is green?Is it a bug?

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    8 years ago, # ^ |
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    Actually his color is #92D050. If you reach 3200, you can choose your own color (tourist and Petr haven't decided their colors though).

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8 years ago, # |
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I hate FFT :( the modulo is friendly though

Anyway, nice problems!

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8 years ago, # |
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The best thing about AtCoder is trying to read the Japanese editorial using Google Translate.

Edit: My bad, I see the English version now.

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    8 years ago, # ^ |
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    There's English editorial below Japanese editorial.

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      8 years ago, # ^ |
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      It would be much better to provide editorials in English and Japanese in separate file. I don't like seeing Japanese editorial at the top. Can something be done in this regards? Anyways, Thanks for great contests and well explained editorials.

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8 years ago, # |
  Vote: I like it +16 Vote: I do not like it

Can someone explain the solution for D better? I'm lost at the Inclusion-Exclusion part (where did the sum come from)

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8 years ago, # |
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Can someone please explain the editorial of D in a different way? I'm having some difficulty understanding it. How is that formula derived? And how to find M(k) in O(n^2) using DP?

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7 years ago, # |
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Sorry for necroposting, but how to solve D in time?

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    7 years ago, # ^ |
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    I think all you do is: notice that the only part of the editorial that needs O(N2) time is computing the Mk. You can compute all Mk for a single path of length l by straightforward combinatorics / dp.

    Let Pl(x) be the polynomial where the coefficient of xk is the number of matchings of size k on a path of size l. Notice that in the graph, all paths are gonna be length l or l + 1 for some l (). To find the overall Mk, this is equivalent to a convolution so you can just compute Pl(x) and Pl + 1(x) for the appropriate l and then raise them to the correct power and multiply. This can all be done by FFT in ,

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      7 years ago, # ^ |
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      How do we compute Pl(x) in time?

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        7 years ago, # ^ |
        Rev. 3   Vote: I like it +8 Vote: I do not like it

        Say you have a path of length l. You want to pick k of the edges so that you don't pick any two adjacent edges. This is just a binomial coefficient, something like or something.