PikMike's blog

By PikMike, history, 8 months ago, translation, In English,
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7 months ago, # |
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I'll try to explain my idea on F, because I couldn't get the editorial (I'm pretty sure they aren't same solution) and wanted to write mine.

My solution: http://codeforces.com/contest/946/submission/36022552

Let's get any subsequence. For every occurence of s in that subsequence of F(x), we'll increase the answer by 1. Instead, for every possible occurence we can increase the answer by number of subsequences that have that substring in it. Let's consider a subsequence i1, i2...., in (indices of F(x)) such that if we concatenate the characters at such index, we would obtain s. It means that in every following subsequence we should update the answer by 1.

  • For any j that is equal to some ik, it must contain it.
  • For any j that isn't equal to some ik but i1 < j < in, it can't contain it.
  • For any j < i1 or j > in, it can contain it, so we have 2 possibilities.

So, for every set of indices (called i) we can increase the answer by 2i1 - 1 + n - in and we can obtain the answer.

Let's create a function f(l, r, x) which is equal to answer when we are search for s[l..r] in F(x). (Not exactly but we'll come to that later.) F(x) is equal to F(x - 1)F(x - 2) so, that substring could be in F(x - 1), F(x - 2) or divided in both. If it's in F(x - 1), it means that we should calculate the number of ways and get the answer. Here's the key point, if r equals to n it means that we won't get anything from right so we should multiply it by 2|(F(x - 2)|.

Since 2x * 2y = 2x + y in the end we will get the each occurence multiplied by 2n - in. When we will handle the l = 1 cases, similarly, we will get the answer.

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    7 months ago, # ^ |
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    I've seen your code. There's "ret = add(ret, mul(f(l, k, x — 1), f(k + 1, r, x — 2)));". Why can they be simply multiplied? I think that f(l,k,x-1) may contain some subsequence like ...l,t... , and so to f(k+1,r,x-2). ...l,t... concatenate ...t+1,r... is ...l,t......t+1,r..., but not ...l,t,t+1,r... . How are those .... between l,t and t+1,r considered?

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      7 months ago, # ^ |
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      OK I got it. "which is equal to answer when we are search for s[l..r] in F(x)", This value depends on whether l==1 or r==n. If l==1 and r==n, it represent ...l,r... ; If l==1 and r<n , it represent ...l,r ; If l>1 and r==n, it represent l,r... ; If l>1 and r<n , it represent l,r .

      Thank you!

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        4 months ago, # ^ |
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        Can you explain me better? I don't understand you.

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    7 months ago, # ^ |
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    Wow, it is so clear!

    Thanks a lot, I understand the problem!

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7 months ago, # |
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I'm most probably missing something obvious, but why would greedy not work on Problem 946D ?

Like In O(mn) time, create an n*2 array greedy[day][i] where if there's still atleast one 1 in the ith string, then let greedy[i][0] denotes the number of hours that would reduce if you skip the first class in day i, and let greedy[i][1] denote the number of hours that would reduce if you skip the last class in day i (these two have same values if on that day, we have only one class), and if day i has no class, then set greedy[i][0] = greedy[i][1] =  - ∞

Then initialize an heap with those n*2 elements in time. Then at each step, I pick the largest element from the heap in time, say it's greedy[j][1], and I change the last 1 in day i to 0, and update greedy[j][0] and greedy[j][1] in the heap. I stop when I have picked k elements.

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    7 months ago, # ^ |
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    The greedy approach fails (for example) whenever there are equal options at the beggining and at the end but some side is better somehow.

    Imagine: k = 3 Case 1 : 1110000111111111111 Case 2 : 1111111111110000111

    Removing the first 1 reduces the time in 1 on both sides, but only one of them leads to the best solution (greedy approach won't know which). That's why you should do dp.

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7 months ago, # |
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I have a question about 946D. The problem statement was not clear to me. 2 6 2 010010 111100 For above example, I thought answer is 3. But, Passed code's answer is 4. It means skipped lessons is based on day, not week.

So, I felt very difficult to this problem^^ Thanks

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7 months ago, # |
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Can anyone plz explain the idea of this code http://codeforces.com/contest/946/submission/36016439 ? It's Problem F. I can't grasp the DP transition in this code clearly. As far as I studied, I guess it maybe use some math to simplify the transition. Can anyone help plz?

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    7 months ago, # ^ |
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    I appreciate this simple transition very much, so I want to make clear of it.

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    7 months ago, # ^ |
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    OK I got the idea. It's just a way by including the information of 2^length into the dp arrays. It's a very clever method, and I appreciate this very much!

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7 months ago, # |
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Can anyone explain why on local machine I get correct answer but on server not? http://codeforces.com/contest/946/submission/36021632

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7 months ago, # |
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Problem G can be solved with similar method as in the editorial, but without using segment tree. Let's calc dp[cnt][len] — minimum element ending sequence with cnt bad elements with length len. Calculating is similar to finding longest increasing sequence : for 1 bad element we use binary search for x-i+1, for 0 x-i.

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7 months ago, # |
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can anyone explain me solution E?

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7 months ago, # |
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Can someone explain in Problem B why using '%' gets AC and subtraction gets TLE?!

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    7 months ago, # ^ |
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    a = 1018 , b = 1, you subtracted 1018/2 times

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    7 months ago, # ^ |
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    Imagine test 10^18 1, you make 5*10^17 subtractions but just one division. You can find proof of Euclid's algorithm with division complexity (it's O(log n)) and apply it here.

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7 months ago, # |
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Does n^3 solution work where n = 500. Wont it be of order of 10^8, I left the D question for the same reason. I used to think order 10^7 is borderline for time limit of 2 seconds!!

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    7 months ago, # ^ |
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    10^7 is like the limit for python or some other slow lang. You can assume it to be about 2-4 * 10^8 for C++.

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      7 months ago, # ^ |
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      Ohh, alright thanks.
      By the way thats for 2 seconds, so if time limit is one second then i should expect 10^7 operations right ?

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7 months ago, # |
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Problem G -
I didn't use rollback. I have used only 1 segment tree with range query and point updates.

Solution

Brief Idea -
LNDS — Longest Non-decreasing Subsequence
A[i] = A[i] — i for every 'i'
For every index 'i', we want the length of LNDS ending at 'i-1', say E[i-1], and the length of LNDS from [i+1, n] whose 1st element is greater than A[i-1], say S[i+1]. We can easily calculate S[i+1] using segment tree. Calculation of E[i-1] is trivial.

Final answer would minimum of n — 1 — E[i-1] — S[i+1].

Complexity — O(N*logN)

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7 months ago, # |
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Can someone explain me the solution of C written in the editorial ?

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    7 months ago, # ^ |
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    Have a char variable C which is intitially equal to 'a'.You need to check if there exists a subsequence of the English Alphabet in the given string. So iterate through the string fully, for each iteration check if the current character of string is not greater than C and increment C. So in the end, if your C value is not equal to 'z', then it means you don't have a subsequence of the English Alphabet. Hope this helps :)

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      6 months ago, # ^ |
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      For question C, I understand how this: aacceeggiikkmmooqqssuuwwyy becomes: abcdefghijklmnopqrstuvwxyz

      However, I do not understand why the check is only if the current character is not greater than variable C.

      I do not understand why this: aada becomes: abdc

      Why does the final character become c from a? The impression I got from the question is that a character can only move to the next character in the alphabet (a -> b, f -> g etc). Given this it seems that the final character could only become b.

      If that makes sense, anyone got any ideas what i've missed?

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7 months ago, # |
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Does hacking in educational rounds give you points?

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7 months ago, # |
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nice editorial

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7 months ago, # |
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Can anyone explain why I (and some other people) get WA 16 in D problem? My code is here: 36048111. I coded the solution as in the editorial.

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7 months ago, # |
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It shows tutorial not available

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7 months ago, # |
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For problem E,i think the DFS algorithm is much easier.You can see my code. :)

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7 months ago, # |
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For problem G, there's a very concise code. http://codeforces.com/contest/946/submission/36058677

Using DP to manipulate the LIS arrays, I find there IS an optimal structure for them.

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7 months ago, # |
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What does 'length(cur)' mean in Problem D?

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7 months ago, # |
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For problem D: I tried solving it using a priority queue ( pq ). It's top is the day that if I skip a lesson in it, it will result in the maximum time saving...

Say we got these strings in the pq:

a: 100011011 --> saves 4 units of time if I skip a lesson

b: 100101111 --> saves 3 units of time if I skip a lesson

c: 010000111 --> saves 5 units of time if I skip a lesson

The pq top is c as it will result it will help to maximize time saving, hence minimizing the time spent at school..

What's wrong with this approach??

You can check my submissions..

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    7 months ago, # ^ |
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    And in case of a tie like this:

    1110111 --> saves 1 unit of time if I skip a lesson

    1111011 --> saves 1 unit of time if I skip a lesson

    It'll choose the second string as it has a 0 nearer to 1 so in the second skipped lesson the string becomes: 1111000 but if it chooses the first string, it will become: 1110100 hence less time saving. That's how my pq works.. what's wrong ?

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      7 months ago, # ^ |
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      Got it. This will break the pq sorting criteria..

      1100011
      0110110
      
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7 months ago, # |
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Can someone elaborate D some more? How should I calc this mn[i][j]? What should i put in mn[i][j] if I can't achieve j lessons in i-th day?

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    7 months ago, # ^ |
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    losmi247, my solution uses an array that I call opt[i][j], wich is the minimun amount of hours you have to spend in the i-th day if you kick off exactly j classes that day. That is the mn[i][j]. This can be done greedly by taking the best prefix and suffix of lessons of that day such len(prefix)+len(suffix)==j. Then you can do dp[i][j] = min(dp[i][j], dp[i-1][j-k] + opt[i][k]);

    PD: if the number of lessons you going to skip the i-th day is greater or equal than the number of lessons you have to attend that day opt[i][j] should be 0

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      7 months ago, # ^ |
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      I can't thank you enough for this, thank you very much

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7 months ago, # |
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can anyone please help me in understanding 946D timetable really i will appreciate the effort. I have no idea what is going on..

please please from scratch

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7 months ago, # |
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Please explain me problem F. Considering test 1:- 2 4 11 Judge answer is 14. But, I thought of problem as creating fibonacci string of 4 that is F(4) which is equal to 10110, then finding cost that is number of time string s that is 11 (entered by user) contained in F(4).

So, my answer comes out to be 1(int) as 11(string) is contained only one time is F(4) i.e. 10110(string).

Please, someone explain me. This is driving me crazy for days.

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    7 months ago, # ^ |
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    You misunderstand the problem description.

    "calculate the sum of costs of all subsequences of the string F(x)"

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      7 months ago, # ^ |
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      Yeah, I see . But can you explain me using 1st test case. I'll appreciate your efforts.

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        7 months ago, # ^ |
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        All subsequences of "10110" include 2^5=32 strings: 1, 0, 1, 1, 0, 10, 11, 11, 10...

        Maybe you need to check the definition of subsequence in wikipedia.

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7 months ago, # |
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Can someone explain why my solution for problem c is getting TLE[submission:36266766]

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    7 months ago, # ^ |
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    Got selected I use C# method to convert char array to string. C# is taking more time then usual dont know why

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      5 months ago, # ^ |
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      ans += s[i].ToString() give |s|^2, use StringBuilder instead

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4 months ago, # |
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Can someone point me to problems similar to Problem D : Timetable ? I usually get stuck at such problems which involve preprocessing before beginning DP.

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3 months ago, # |
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Problem F:

I wonder if my solution works ? If not, please explain and give example(s). Thanks in advance :D

For each step i, I store 2 strings pre and suf that is the prefix and suffix of F[i]. Each string's length is smaller than the length of the given string s. With this, at each step I can string match to find new occurences that appears due to the concatenation of F[i] and F[i — 1]. In addition, I notice that F[i]'s pre = F[i — 1]'s pre and F[i]'s suf = F[i — 2]'s suf.