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.

土曜日に開催される「第五回日本最強プログラム学生セレクション-プレ選考-」がABCの時間を割いてくれた。

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

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.

??? Just brute force!

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!

We have noticed that some contestants' submissions for Problem B were judged as WA instead of AC due to violations of the input constraints we prepared. We are currently discussing internally how to address this issue, so please wait a few days for the rating update.

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

We have noticed that some contestants' submissions for Problem B were judged as WA instead of AC due to violations of the input constraints we prepared. We are currently discussing internally how to address this issue, so please wait a few days for the rating update.