Endagorion's blog

Technocup 2019 — Elimination Round 4 + Codeforces Round 528: editorial
By Endagorion, history, 6 years ago, translation, In English
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6 years ago, # |
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Thanks for fast editorial

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6 years ago, # |
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I want to ask questions about the solution F:the 2n polynomials,is it should be n2 polynomials?

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    6 years ago, # ^ |
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    2n since the inclusion-exclusion formula is used in the proof. But it is only the proof that the main function is a polynomial, we don't need to actually compute all these 2n summands.

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6 years ago, # |
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Out of curiosity, may I ask problemsetters about their views to criticism on Div1C? There seems to be quite negative feedback to this problem. Do you think it is reasonable to appear in CF rounds? Do you think people's criticism make sense? Thanks.

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    6 years ago, # ^ |
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    I am not a setter of this problem, but I considered it to be OK when it was proposed. Perhaps it would be better if we didn't ask for a certificate.

    But I still don't quite understand why it received such an amount of hate. There are problems which at first seem like handling a lot of corner cases without applying any specific idea, but if contestant thinks carefully before starting to write code, then he can reduce the length of this solution and the number of cases he needs to handle. I understand how this problem can be seen as a bad one if someone picks the first solution idea that comes to his mind and starts implementing it right away, but is it a good strategy on contests?

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6 years ago, # |
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can anyone please explain the following lines from "Minimum Diameter Tree" for dummies: "Note that the contribution to the sum on the right side of the inequality of the weight of each edge will be at least l−1, because any edge lies on ≥l−1 paths between the leaves of the tree. So, ∑1≤i<j≤ldistaiaj≥(l−1)⋅∑e∈Eweighte=(l−1)⋅s"

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5 years ago, # |
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could not understand problem 1085-B

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    4 years ago, # ^ |
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    As $$$(x\ mod\ k)*(x\ div\ k)=n$$$, both $$$x\ mod\ k$$$ and $$$x\ div\ k$$$ should be the divisors of $$$n$$$.

    As $$$k \leq 10^3$$$, which is very little, that means $$$x\ mod\ k$$$ is also less than $$$10^3$$$.

    Therefore, we can try all possibilities of $$$x\ mod\ k$$$ by assigning all values from $$$0$$$ to $$$k-1$$$ that are divisors of $$$n$$$ to it.

    Let $$$p$$$ be the assigned value of $$$x\ mod\ k$$$. How can we recover $$$x$$$?

    As we know right away that $$$x\ div\ k = \frac{n}{p}$$$, $$$x$$$ is almost going to be $$$k\cdot \frac{n}{p}$$$. However, if it ends that way, $$$x\ mod\ k$$$ would be $$$0$$$ which might be contradiction as we assigned $$$x\ mod\ k$$$ to be $$$p$$$.

    BUT $$$0 \leq p < k$$$. So if we add $$$p$$$ to that $$$x$$$, the $$$x\ div\ k$$$ wouldn't change and $$$x\ mod\ k$$$ is now $$$p$$$, the problem is solved.

    In conclusion, $$$x$$$ must be $$$k\cdot \frac{n}{p}+p$$$ if $$$x\ mod\ k=p$$$

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4 years ago, # |
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I am really confused of the definition of $$$dp2[i][k]$$$ in Div1 E.

"Calculate the following $$$dp$$$: $$$dp2[n][k]$$$ — the number of permutations of length $$$n$$$ of elements $$$1,2,…,n,n+1,n+2,…,2n−k$$$ such that $$$p_i \neq i$$$"

According to the definition, how can $$$dp2[n][0]$$$ is just only $$$n!$$$ ?

Because the number of ways just from using elements from $$${n+1,n+2,...,2n}$$$ only is already $$$n!$$$ If we take $$$1,2,3,...,n$$$ into account, it is even more.

Did you mean "the number of permutations of length $$$n$$$ of elements $$$2n,2n-1,2n-2,...,n+1,n,...,n+1-k$$$ such that $$$p_i \neq i$$$" ?

Even so, I still can't really understand the usage of it when we are trying to build the suffix of the current row ;_;

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