atcoder_official's blog

By atcoder_official, history, 2 months ago, In English

We will hold AtCoder Beginner Contest 451.

We are looking forward to your participation!

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7 weeks ago, hide # |
Rev. 2  
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Rank 1(first comment)

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啊a~

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Vote: I like it -6 Vote: I do not like it

help~~~

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How to do D, wasted 1 hour trying to observe any sort of pattern

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Rev. 2  
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I tried Bs + digit dp to count number <=x I can make using power of 2 as string, Dp started over counting and i spent whole contest debugging

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Wasted way too much time on a randomized solution for E, didn't have enough time to solve G despite looking easy :( I'm guessing there was something obvious I missed

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    7 weeks ago, hide # ^ |
     
    Vote: I like it +3 Vote: I do not like it

    for E , i connected nodes with minimum distance first like a minimum spanning tree, and then did a dfs to generate distance matrix and checked whether it matched with given one.

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I spent 40 minutes for D, and the same time for E+F.

I tried to build a DP, but it turned out to be a brute force.

Really confused with my behavior

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I used mt19937 to solve E,no hacking!!

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For D, even if you don't delete any of the multiple numbers, there are only 1404832.

And you see in the sample, 1099898-->819264512, close to 1e6-->1e9.

According to our feeling, the numbers rarely be the same.

Only something like "1"+"2"+"8" and "128" would be like that, but these are few.

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E = 472D

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yo they just banned my acc for no reason (i write 3 question a, b, c lol) i dont understand :(

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G's editorial english translation is missing the definition $$$W'_e = W_e \oplus A_u \oplus A_v$$$ in the spanning tree $$$T$$$

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I think F is a very good educational problem for learning disjoint-set-union which maintains extra information. If you have not yet learnt it, I strongly recommend try this problem.

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6 weeks ago, hide # |
 
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has anyone solved E without resorting to tree construction?

i don't understand why finding a $$$k$$$ for every $$$i \lt j \space$$$ s.t. $$$\space A_{ij} = A_{1i} + A_{1j} - 2A_{1k} \space $$$ and $$$ \space A_{ij} = A_{ki} + A_{kj}$$$ isn't enough