We will hold AtCoder Regular Contest 203 (Div. 2).
- Contest URL: https://atcoder.jp/contests/arc203
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20250803T2100&p1=248
- Duration: 120 minutes
- Number of Tasks: 5
- Writer: vwxyz
- Tester: Nyaan, maspy
- Rated range: 1200 ~ 2399
The point values will be 400-500-600-800-800.
We are looking forward to your participation!
D is unexpectedly easy.
My friend asked me what is needed for D. After thinking for a while, I told him "a few arrays and variables".
C is harder than the usual Div.2 Cs. Unfortunately I made some foolish mistakes in B so I got a low rank :(
ad-hoc round.
Screencast with commentary
In this editorial, $$$A_i=\lfloor \frac{M}{2} \rfloor(i \lt 1)$$$?
I think it may be
(i>1)?(Sorry for my poor English)
Problem B is good, but unfortunately I spent too much time on the case where there is only one single 1 in both arrays.
How to simulate these operations?
Find subarray $$$[1,0]$$$ (or $$$[0,1]$$$) and subarray $$$[1]$$$, then swap them. Number of inversions increases, so the process is finite
Thanks. But I realized today that it is not necessary. We can always make the array look like 11...0... when there are more than 1 1's are present. So, I don't need to simulate it.
Damn! 10 downvotes. How did I offend so many people by asking for an implementation?
about the editorial of C,why can we only replace these H-V-H movements with V’ but not a H'?I mean,of course I know they lead to different results,but what's the reason behind it that we must replace right-up-right by a down(not a right)?
upd:get the idea.i stupid
Typo in A's editorial
it should be ceil(M/2) + (N-1)*floor(M/2)
I am not sure why you are downvoted. You are correct. There are typos in the editorial.