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://mirror.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...., i_n (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 i_k, it must contain it.
• For any j that isn't equal to some i_k but i₁ < j < i_n, it can't contain it.
• For any j < i₁ or j > i_n, it can contain it, so we have 2 possibilities.

So, for every set of indices (called i) we can increase the answer by 2^{i₁ - 1 + n - i_n} 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 2^x * 2^y = 2^x + y in the end we will get the each occurence multiplied by 2^n - i_n. When we will handle the l = 1 cases, similarly, we will get the answer.

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.

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

Can anyone plz explain the idea of this code http://mirror.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?

Can anyone explain why on local machine I get correct answer but on server not? http://mirror.codeforces.com/contest/946/submission/36021632

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.

Can someone explain in Problem B why using '%' gets AC and subtraction gets TLE?!

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!!

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)

Does hacking in educational rounds give you points?

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.

It shows tutorial not available

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

For problem G, there's a very concise code. http://mirror.codeforces.com/contest/946/submission/36058677

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

What does 'length(cur)' mean in Problem D?

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?

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.

Can someone explain why my solution for problem c is getting TLE[submission:36266766]

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

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.

ОбЬяснение задачи D немного неправильное

Hey awoo ,could you please elaborate the explanation of the following line of 946D

dp[i + 1][j + lengthcur] with dp[i][j] + mn[i][lengthcur]. Thanks,in advance.

Can E be solved using binary search, by finding the least significant position which can be reduced by 1, so that its possible to construct a maximal number which is a palindrome ?