Endagorion's blog

By Endagorion, history, 2 months ago, In English,
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2 months ago, # |
Rev. 2   Vote: I like it +13 Vote: I do not like it

For C,is there a way to get the maximal solution? :D

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2 months ago, # |
  Vote: I like it +23 Vote: I do not like it

Can someone help me with understanding of problem E. When we fix T, we need to know whether exist ai's, such that Sum of ai = a and dpei, k, ai > T. For that, we find ai's, such that Sum of ai <= a and dpei, k, ai > T. So here we use, that dpei, k, x <= dpei, k, y if x <= y, but I believe it's wrong. I will describe the graph below.

5
1 2 1
1 3 1
1 4 1
2 5 1

Here dp(1->2), 2, 1 = 4, while dp(1->2), 2, 2 = 1.

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    2 months ago, # ^ |
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    I think we don't actually use dpei, k, x ≤ dpei, k, y, x ≤ y. We can leave the residual in our current node for example. Then we use dpei, x, k ≤ dpei, x + s, k + s, x ≤ y, s is some non negative integer, which is true. I hope i didn't misunderstand something.

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2 months ago, # |
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I read codes of problem D and can't understand them. In most of the codes, when string length exceed 1000 on concatenation, they consider only first and last 500 characters of the string and leave the rest. I don't understand why they are doing so. Shouldn't they try to keep the meeting point as that is the point where new strings will be formed. One such code doing this : http://codeforces.com/contest/868/submission/31066000

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    2 months ago, # ^ |
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    I deem what STD wanted to tell us is that the answer k will not exceed 9 and we may maintain arrays of distinct substrings,or bitset as we merge the two demanded strings.But in fact the solutions you mentioned seem a little different from the original solution,because these solution get ans from three strings(itself,left and right strings) for each query.Hmm...I have read similar codes from submissions so I put forward my idea here~

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2 months ago, # |
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Can someone please elaborate how the maximum answer is 9 in DIV 2 D?

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    2 months ago, # ^ |
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    The editorial assumes 10 is the answer and tries to see after all queries, if the number of unique substrings of length 10 can reach 2^10.

    At first it explains how many substrings of length 10 there are in the original strings. Assume input is one long string of length 100, maximum number of unique substrings of length 10 won't be more than 100.

    Now we need to know after each query, whats the maximum amount of new substrings of length 10 that can be added each time.

    When we concatenate two strings, the new substrings of length 10 must start in the first string and end in the second. If it started and ended in the first string, that means its not newly created (same with the second). The maximum amount of new substrings of length 10 for each query therefore is 9.

    So we started with 100 unique substrings of length ten, each query added 9.

    Maximum amount when done is 100 + 9 *100 which is less than 2 ^10. So 10 cant be a possible answer.

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      2 months ago, # ^ |
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      Why is the number of unique substrings after concatenating two strings is 100 +9?

      Lets say two strings are there S1 and S2 of length 100. Both of them will have not more than 100 unique substrings of length 10. It is also possible that the 100 unique subtrings of S1 are not present in S2. So when we concatenate them we get a total of 100 (from S1) + 100(from S2) + 9(from border of two strings) unique substrings.

      So if there are 100 such strings then the number of unique substrings can be: 100*100 + 9*100.

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        2 months ago, # ^ |
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        The input states that the total sum length of all strings in the beginning is 100. Then the maximum amount of unique length 10 substrings at start is 100 for all input.

        So the only way there can be 100 unique substrings in both S1 and S2 is if minimum 100 of those 200 were created new. If they were created new its already accounted for.

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      7 weeks ago, # ^ |
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      Nice explanation, really appreciable!!

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      7 weeks ago, # ^ |
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      Thanks for your explanation!

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2 months ago, # |
Rev. 4   Vote: I like it 0 Vote: I do not like it

For problem F, shouldn't j1 ≤ j2 since we must try partitioning at all x where x < p(j2) ≤ j2?

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2 months ago, # |
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In Problem G, What is meant by smoothest partition? According to wikipedia ,In mathematical analysis, the smoothness of a function is a property measured by the number of derivatives it has which are continuous. A smooth function is a function that has derivatives of all orders everywhere in its domain. If this definition is correct , then how it is related to the problem.

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2 months ago, # |
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In Problem G Endagorion How E0 = (E5 / 2) + 1 ? where Ei is be the expected number of days to find the treasure in spot i. Can you elaborate how you got this.

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7 weeks ago, # |
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On Div2C, the number of problems needed is at most 2, if k was bigger, lets say 6 or 8, would 2 be enough?

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7 weeks ago, # |
Rev. 4   Vote: I like it 0 Vote: I do not like it

How is optimisation part O(nlogn) in F?

Edit: Got it