We will hold AtCoder Regular Contest 222.
- Contest URL: https://atcoder.jp/contests/arc222
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20260614T2100&p1=248
- Duration: 120 minutes
- Writer: maspy
- Tester: Nyaan
- Rated range: 1200 ~ 2799
- Point Values: 300-500-700-700-700-700
We are looking forward to your participation!
Is this not ARC--? I always assumed ARC++ is explicitly mentioned, so this would be the ARC-- edition.
Please read Changes to ARC in 2026 and in case you are considering this as ARC-- because of 300 pointer A then, please note that we have four problems of 700 points; thus increasing overall difficulty.
The Problem A of this ARC contest is only worth 300 points and Problem F is just 700 points. Would it be more appropriate to classify this contest as an ARC--?
Dear Atcoder Official, can you show the code the previous winner atgc(banned) submitted for each problem? 'atgc' is a strong and famous Chinese Informatics competitor. I am wondering if he was banned by mistake.
To be honest, not just me—I don't think any Chinese competitor believes he would cheat.
This is not the first time when a legit player is banned. User MCYYDS (AtCoder) competed in arc218, solved F2, then banned, without any reason. The user himself asked AtCoder via zendesk but the response is "violated the rules" WITHOUT ANY clear/detailed information.
I believe AtCoder is having cheater-PTSD so every strong participant, once "suspected" as AI, will be banned.
On the issue of AI cheating, Codeforces is too lenient, while AtCoder is too strict.
In fact, AtCoder only randomly bans low-rated users on the first two pages of the scoreboard.
And, in fact, Atcoder never checks comments under its posts.
UPD: False.
You mean Codeforces is too lenient? I don't think so.
A similar issue has occurred on Codeforces and has yet to be resolved.
can someone share the code for the problem B please with explanation if possible ?
https://atcoder.jp/contests/arc222/submissions/76699276
Is this simple solution of E intended?
We add values to a set $$$S$$$ in descending order of occurrences on the input. For each new value $$$a$$$, we multiply its number of occurrences by $$$\sum_{i \in S} 10^{a \oplus i}$$$ and add it to the answer.
Brute force would be $$$O(4^N)$$$, so we use a meet-in-the-middle approach. Let's denote the lower and upper $$$M/2$$$ bits by subscripts $$$\cdot_{lo}$$$ and $$$\cdot_{hi}$$$ respectively. We store the sums $$$\sum_{i \in S} 10^{a_{lo} \oplus i_{lo}}$$$ for each possible combination of $$$a_{lo}$$$ and $$$i_{hi}$$$. Inserting $$$i$$$ into $$$S$$$ requires only updating those values for fixed $$$i_{hi}$$$. A query requires looking up only values for fixed $$$a_{hi}$$$; we multiply them by $$$10^{a_{hi} \oplus i_{hi}}$$$ since $$$a \oplus i = a_{hi} \oplus i_{hi} + a_{lo} \oplus i_{lo}$$$. Complexity: $$$O(2^{3M/2})$$$ time, $$$O(2^M)$$$ memory.
We can also use the fact that $$$N \ll 2^M$$$ and stop when the number of occurrences of the currently processed value reaches $$$0$$$, bringing the complexity down to $$$O(N 2^{M/2})$$$.
In what world is that simple?
There's no xorvolution involved. There's no threshold involved. It's a single linear thoughtstream, not a combination of two.