We will hold AtCoder Regular Contest 151.
- Contest URL: https://atcoder.jp/contests/arc151
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20221016T2100&p1=248
- Duration: 120 minutes
- Number of Tasks: 6
- Writer: leaf1415
- Tester: maspy, tatyam
- Rated range: — 2799
The point values will be 300-500-600-700-700-1000.
We are looking forward to your participation!
Only hope I won't fall back to 1 Kyu. I climbed to 1 Dan yesterday.
Wrote https://atcoder.jp/contests/abc273/editorial/5063
Didn't take part in the contest.
It probably won't go away once you achieve Kyu or Dan.
really?I'll try it next time
Kyu and Dan are based on max rating, so they don't go away, but your rating still can.
Thanks
rp++!
rp++
Hope you all have a good time!
such no math, so thanks
Grundy Number is not needed in C.
But it's easier to solve this problem by using Grundy Number.
It's cool without it xD
Can you please explain?
First we flip all odd squares, then adjacent squares must be the same. For an empty interval sandwiched between two known squares, the parity of operation times is fixed. Let $$$x=a_1-1,y=n-a_m$$$, we have solved the case $$$x=y$$$ because we can ignore the left and right parts, there one can always imitate the other. For the case $$$|x-y|>1$$$ the first player wins. He can use one step to make $$$x=y$$$ and decide the parity of an interval. For the case $$$|x-y|=1$$$ the first wins if the parity(ignore left and right) is even. If $$$x+y=3\bmod 4$$$ the first wins too. Otherwise the second wins.
I got 1588 rating. When will I get blue name?
I got 1192 rating. When will I get cyan name?
I got 1660 rating in contest ABC274!
I opened my computer at 20:13(utc+8) and want to take part in this contest, because my rating fell in the last contest. But when I registered, I discovered that I was considered unrated immediately. I was very sad, so I didn't do even one problem.
I think it is too early to set unrated automatically (only 5 minutes after contest started). If this can be 15 minutes, I could do the problem happily.
I know that I will receive many downvotes in Codeforces, but I still want to say it.
Why don't you just resister earlier?
Can anyone share how they solved F? The editorial looks ridiculous.
Can anyone share the proof of Grundy number for different segments in problem C?
Can anyone explain me about grundy number or attach some link to follow ? Its a new term for me , thanks in advance
https://cp-algorithms.com/game_theory/sprague-grundy-nim.html
For problem C, the editorial say "If the component touches just one digit, the Grundy number is n."
I can't really understand how is this true. Can anyone please help me?
Wish you have understood it. If not, you can consider the definition of Grundy number. Let $$$f_x$$$ be the Grundy number when the segment has limit on only one side. Then $$$f_x = \operatorname{mex} ( f_y \operatorname{xor} 1 (y < x), f_y (y \leq x)) $$$
$$$f_y \operatorname{xor} 1$$$ is for when we put a number same from the limit on the side, it would split into two games where one has length $$$y$$$ and limit on one side, one has the same limits on both sides whose Grundy number is just $$$1$$$. $$$f_y (y \leq x)$$$ is for when we put a number different from the limit, and we can solve it like the situation upside.
Hello, in the editorial of B (link), it says
the number of A’s such that A′<A lexicographically and the number of those such that A′>A are equal from the symmetrywhat is the symmetry here? how are they equal from symmetry? Can someone provide proof that they are equal using symmetry?