Winner is a pupil ;)

The only real black handle is MikeMirzayanov :)

What is Codeforces Platinum?

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"The round will be hosted by rules of educational rounds (extended ACM-ICPC)"

Use custom invocation on cases like that, and then debug.

+1 the main website was broken and I wasnot able to open it until the contest was over.

I can't tell you how to solve it in English, but you can check my solution, maybe it will help you. https://mirror.codeforces.com/contest/1272/submission/66702493

Maintain 2 arrays, Increasing and Decreasing. Increasing[i] will store maximum increasing subarray till ith element, similarly Decreasing[i]. For each element check if (i+1)th element is greater than (i-1)th element, If true, then we can delete this element and updated ans=max(ans, Inc[i-1] + Dec[i+1]).

do dp. dp1[i] = maximum length of increasing contigous subarray having the element of i from 0 to n. dp2[i] = maximum length of increasing contigous subarray having the element of i from n — 1 to i. Find out max length of subarray without removing element.Initially, this is the ans. And then, for every i from 1 to n — 2, check if(dp1[i — 1] + dp2[i + 1] > ans && v[i — 1] < v[i + 1]), if yes, then ans = dp1[i — 1] + dp2[i + 1]. That's it.

Let pre[i] be the maximum possible length of strictly increasing contiguous subarray of array a from index 0 to index i

Let suf[i] be the maximum possible length of strictly increasing contiguous subarray of array a from index n — 1 to index i

Loop from i = 1 to n — 2, if a[i — 1] < a[i + 1], then ans = max(ans, pre[i — 1] + suf[i + 1]).

very similar to this problem: PIBRO

This AtCoder Problem is also an interesting one that you can solve using the above technique:

GCD On Blackboard https://atcoder.jp/contests/abc125/tasks/abc125_c

I've made a string of 0 and 1 of length N-1, where 1 means $$$a_i \lt a_{i+1}$$$.
Then applied a regex / 1+ 0 1+ /x, and checked elements near matched 0 if they increase after one deletion (i.e. if exists $$$j$$$ that $$$a_j \lt a_{j+2}$$$, where $$$j$$$ is near 0). Perl code 66779749.

No, it is dp.

Simple DP in $$$O(n^3)$$$. Notice that the answer's length cannot exceed $$$400$$$ and do dynamic programming. $$$dp_{i, j, bal} = $$$ minimum total length of the string with the first $$$i$$$ characters processed in the first string, $$$j$$$ first characters processed in the second string and the current balance is $$$bal$$$. Transitions can be made in a straight-forward way (nested loops) but you should do them carefully or by bfs/recursion.

Actually, because problem limits were low, I just used dynamic programming.

My state was (current unincluded character from S, current unincluded character from T, current bracket prefix). For some reason that I do not quite understand, the bracket prefix will not go above 200 and thus my state is 200^3=8000000 which is good.

My transition is "add open bracket/add close bracket" depending on whether they are possible operations. I store which transition is better so that I can backtrack later to construct the bracket sequence.

It passed the tests, so I think it's probably correct, but that was one of the ugliest case-check DP I have ever coded. Very unlikely to be official solution.

Multi-source bfs on the transposed graph.

Take 2 additional nodes, and reverse the graph...Think about it afterwards.

Form a graph from a[i]-i to i and a[i]+i to i.(a[i] -i >=1 , a[i]+i<=n) traverse the adjacency list and insert points on the other side of the edge in the queue if parity of both points are diffrent. (these points will have minimum distance which is 1). perform bfs.

Check this comment.

There are no empty strings in every test and empty strings are not allowed to input at all. Seems like it would be better to mark this in the problem statement more clearly.

The main site did not open for almost 2hr.I thought Round will not be rated but since 2200 users solved d, i think it will be rated.

Yeah mine also, code forces stopped working after solving A .

Not a valid reason because m1 website was up for the entire contest.

this is not at all a valid reason for contest to be unrated .I also face the same problem , but then i shifted to m1 .

Why you did not continue with any of the other options the management has given ( m1 , m2 , m3 websites ) ?

Still after being hacked ?? :))

Think of a dp with transitions similar to LCS and more state parameters.
You can find my submission here. :)

I think A can be solved on much easier way.

just calculate sum of current distance from given input. Let say it d. Then max(0,d-4) is answer.

Movement for friend who is in the middle doesn't make any change. Moving left one right will reduce distance by 2 similar for right one. You can reduce at most 4. Finally check reducing making it negative or not.

Hint: Try the move in a rectangular path.

If you go one step right then you also need one step left to come back ,similarly for up and down ...another case is if you can't move up of down then you can't move left and right more than 1 ..hope its enough for this problem

After moving to position 10, you can make 1 more move to position 6 (10 — 4 = 6), position 6 has value 5 which is odd. So that is why the answer is 3.