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Codeforces Round #530 — Editorial
By 300iq, 20 months ago, translation, In English
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20 months ago, # |
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"A fat fish can't dangerously eat a fish with smaller weight. Indeed, even if all the smaller fishes eat each other, the resulting fish would be too small. We can conclude that the danger is not greater k−t." How can you conclude that? All fat fish may as well take part only in fights with bigger eels, case in which the argument you have doesn't imply a "lost fight". They as well might fight with each other (you have around t/2 non-dangerous fights from here, and from then on, there's no argument as to why those newly created eels will not take part only in dangerous fights). Could you give a more formal proof for that observation?

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    20 months ago, # ^ |
      Vote: I like it +18 Vote: I do not like it

    Let's take fat fish, say it is k-th in sorted order. Let's give each fish its mark: it will be 1 for (k - 1)-th fish, 2 for each fish smaller than (k - 1)-th, and 0 for all bigger (including our fat fish). The result of the fight will be fish with mark min(x, y) where x and y are marks of fishes in the fight. In the end we will get fish with mark 0, so it is easy to see that there will be a fight between fishes with marks 0 and 1 at some moment. I claim that this fight is non-dangerous (any fish with mark 0 is more than twice bigger than any possible fish with marks 1 or 2). Also it is not hard to check that these fights are different for all fat fishes.

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20 months ago, # |
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Can someone explain their easy approach for div2 D with example ?

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    20 months ago, # ^ |
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    graph : https://imgur.com/a/Iac4Lm2

    Here we see,that s(1) = 1 ==> a(1) = 1.We can say,that a(3) = s(3) — a(1) and a(4) = s(4) — a(1). But can we reduce the answer?Yes.We can put on vertex 2 min(s(3) — a(1), s(4) — a(1)).And this value will embrace all children of vertex(2).So a(1) = 1, a(2) = 2, a(3) = 0, a(4) = 1. Ans = 4.

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      20 months ago, # ^ |
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      how a(4)=s(4)-a(3) ???? cant get this part

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        20 months ago, # ^ |
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        and also plz explain when will be the answer not possible

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          20 months ago, # ^ |
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          1)Fixed.2) If s(u) < s(v), where u is a child of the v

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20 months ago, # |
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can someone explain more clearly problem b div 2

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20 months ago, # |
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please provide proof of problem E- Nice Table.

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    20 months ago, # ^ |
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    If a row contains three characters, two neighbor rows of each row will be same. If a column contains three characters, two neighbor columns of each column will be same. Thus, in a good matrix, either each row contain at most two different characters, or each column contain at most two different characters.

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20 months ago, # |
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I have implemented code of @Shtef for problem E in a well commented manner, for easy to understand. If someone don't finds original code easier, you may have a look at my version of same for better understanding. link=> https://ideone.com/Ypsk8P .

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20 months ago, # |
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someone pls explain implementation of div 2 F problem, i commented it here but was not able to understand. https://ideone.com/kZAgdk

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20 months ago, # |
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How to find amount of integer solutions of inequality with Euclidean algorithm div1 E ?

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19 months ago, # |
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My solution for div 2 F: Cookies gives WA verdict on test 6. I followed the same approach as given in editorial. Can someone kindly point out my mistake? Thanks in advance. 49932059

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15 months ago, # |
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IN problem Sum in the tree 10 1 1 2 2 2 3 7 7 7 1 2 -1 -1 3 2 -1 2 3 4 As per question for this test case solution should provide ans as 7 but most of solution is giving -1 as answer. Correct me if I am wrong.

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5 months ago, # |
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In Div-2 C, when the required character(req) is more than the characters given in the string,

Let, si = req — the characters given in the string(excluding '?'&'*')

If we keep all the non-mendatory characters and repeat the character preceding first occurance of a snowflake si times, then we get a string with the length of required size.

But this approach is giving me wrong answer. Can someone point out what's wrong?

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