There's ARC on Saturday.

Don't know their reason but for me makes sense having the ARC on Saturday instead of doing it right before div1+div2 round on Sunday.

ABC is on Saturday in China......

grass（

They are updated late, probably after 1 month ig after the contest.

ABC rounds are usually held on Saturdays but this time ARC round was scheduled on Saturday hence this was scheduled on Sunday.

Assuming your username in AtCoder is SkyWave2022, you didn't even submit anything for Problem FG. What "solution of F" are you referring to?

Try this case. I found my wa*1 solution output 0 0 during the contest.

input:
1
999999999 1000000000
output:
0 1

Same here

I have $$$O(log(k))$$$ (which come from power).

The formula is $$$\frac{\frac{n(n + 1)}{2}a + (n^2 - 2n)^{k}}{n^{2k}}$$$ where $$$a=\frac{n^{2k} - (n^2 -2n)^{k}}{n}$$$. Actually number of cases, that the black ball end up at position $$$i$$$ is $$$a$$$ if $$$i$$$ is not $$$1$$$. For $$$1$$$, it is $$$a + (n^2 - 2n)^{k}$$$. Since the number of possible cases for doing the operation $$$k$$$ times is $$$n^{2k}$$$, we can get the mentioned value of $$$a$$$.

I got $$$\mathbb{E}[\text{X}] = \left(1 - \frac{2}{N}\right)^k + \left(1-\left(1 - \frac{2}{N}\right)^k \right)\left(\frac{N+1}{2}\right)$$$

I first solved for $$$k=1$$$, then for the general case, assume $$$a_{t,k}$$$ be the probability of reaching $$$t$$$ in $$$k$$$ operations, then got the recursion (inspired from calculations of $$$k=1$$$ case) : $$$a_{t,k} = \frac{2}{N^2} + \left(1-\frac{2}{N}\right)(a_{t,k-1})$$$ with initial conditions as $$$a_{t,0} = 0$$$, if $$$t \neq 1$$$; $$$a_{t,0} = 1$$$ else.

Now, writing the expression of the expectation and the probability derived from above, remains long amounts of simplifications T_T.

please see my comment

I was about to submit, luckily went back to confirm this from the constraints!

Also wondering. Seems kind of slow this time.

Not sure if the editorial of ABC 355D can answer your question.

Intervals have same lenth, so if you maintain # of intervals of one direction and count pairs when you meet one with another direction, can prove that you will not count the same pair at different end points.Here is my Code

If you go on clarification tab they have said it will take few days for rating to be updated as they have provided wrong constraints on problem B due to which many participants were getting wrong answer.so they are looking into it

Let $$$A=[1,2,3,1,5]$$$. If you set $$$A[4] := 4$$$ you get an answer of $$$5$$$.

But the LIS of $$$A=[0,2,3,1,5]$$$ gives you an answer of $$$4$$$.

(1) 3, 1 3 3 and (2) 5, 1 2 3 5 4 Expected : (1) --> 3, (2) --> 5 Got : (1) --> 2, (2) --> 4 :)

If you go on clarification tab they have said it will take few days for rating to be updated as they have provided wrong constraints on problem B due to which many participants were getting wrong answer

According to the clarification, rating updation will be delayed due to B's incorrect problem statement.

If your rating color is $$$C$$$, let $$$l=$$$ the lowest rating of $$$C$$$. Then

• If your rating is in $$$[l+0,l+99]$$$, you have $$$1$$$ Λ.
• If your rating is in $$$[l+100,l+199]$$$, you have $$$2$$$ Λs.
• If your rating is in $$$[l+200,l+299]$$$, you have $$$3$$$ Λs.
• If your rating is in $$$[l+300,l+399]$$$, you have $$$4$$$ Λs.

off-topic 1: I want a similar feature on CF too. Especially red, $$$[2400,2999]$$$ have the same visual and I want to distinguish them in one look on the standings.
off-topic 2: How to represent the mark? ^(XOR), ∧(logical and), or Λ(Large lambda)?

(1)4, 1 3 2 5 Expected : 4, Got : 3 :)

Yes. Per https://atcoder.jp/contests/abc360/clarifications

There was a mistake in the constraints. We have already fixed the statement. It was written as |T| < |S| but the correct constraints is |T| <= |S|.

Also, a detailed explanation of solution of G would be much appreciated. There is a Japanese language editorial with two different solutions, but the machine translation of it does not satisfy me.

(1) 5 $$$\newline$$$ 1 4 3 2 6 $$$\newline$$$ Expected : 4 $$$\newline$$$ Got : 3 $$$\newline$$$ :)

I found the bug.

the main problem is that when starting with dp0[0] dp1[0], the corner case was not handled correctly, so I directly calculate dp value when n = 1, then start loop with i = 2.

here's my ac sumbission. link

How high the user's rating is in their name colors, I guess. Though some tampermonkey scripts had already implemented them before.

How do you know the increment you'll be getting?

No.