We will hold ユニークビジョンプログラミングコンテスト2023 春 (AtCoder Beginner Contest 300).

- Contest URL: https://atcoder.jp/contests/abc300
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20230429T2100&p1=248
- Duration: 100 minutes
- Number of Tasks: 8
- Writer: Nyaan, physics0523, evima
- Tester: nok0, m_99
- Rated range: ~ 1999

The point values will be 100-200-300-400-500-500-600-600.

For more information on ABC300 support, please click here. https://atcoder.jp/posts/1028

A pdf of the problem statement will be distributed here. https://img.atcoder.jp/abc300/tasks.pdf

We are looking forward to your participation!

I am looking forward to AtCoder server stability.

UPD: yeah finally an ABC without site-down. Sincerely thanks to the AtCoder team!

Please no more ddos!

it's a milestone , 300th contest , keep going my favourite OJ

Hope there will be no more DDoS on Atcoder, Codeforces and all the websites.

Finally a rated round without DDoS came!

Problem statement of C gave me a headache.

Why so many IELTS reading tests on competitive programming platforms. I think we should keep them simple.

Hooray! The round ends in successful! Happy centennial!

In Problem E,

1 and 6 (**inclusive**)made my contest bad :(Never noticed about it before :(

I assumed the second one and could not solve E and moved to F :(

PS: Solved A-D and F, feels good after solving F just before contest ended.

But here's the thing. It really

isthe second one. You just need to solve a dumb equation instead.can you elaborate?

I Aced with the first one.

Consider that $$$f(n)$$$ equals to the prob of $$$n$$$.

then $$$f(n)=\dfrac{f(n)+f(2\times n)+\ldots+f(6*n)}{6}$$$. find the rest first, then multiply by $$$\dfrac{5}{6}$$$. We r done:)

AC submission

ohh got it. Thanks you.

Hey can u please explain your solution from scratch please

Problem G is the same as subgroup number 6 of problem I of the long tour of the Moscow Open Olympiad. Constraints there are $$$n \leq 10^{18}$$$ and $$$p \leq 500$$$, which probably requires a lot more local optimizations.

It's also similar to this problem on Codeforces

Thank the contest!

I reached 4kyu in this contest!!!

Hello, Atcoder Grand Contest!

Congratulations for the 300th ABC! And without loser ddosers around!

In Problem C, is the 3rd condition

`At least one of C[a+n+1][b+n+1],C[a+n+1][b−n−1],C[a−n−1][b+n+1], and C[a−n−1][b−n−1] is ..`

redundant? I can't make sense of it.For instance in sample 1, for the cross of size 2 centered at cell (3,7), n is 2, the 4 cells, at least one of which should be '.' are (in 1 based indexing) : (6,10), (6,4), (0,10) and (0,4) all of which lie outside the grid. How is this condition satisfied then?

The 3rd condition is there to avoid overcounting. Imagine you have a cross of dimension 2. Without the 3rd condition, one would count there are 2 crosses. One with dimension 1. The other with dimension 2. After adding this 3rd condition, we know only the cross with dimension 2 should be counted.

But how is it doing that? Besides, the cells mentioned in 3rd condition can lie outside the grid like mentioned in later part of my previous comment.