​As a first-time tester, I hope you enjoy the round!​

As a tester, I lose a chance to get negative rating cuz I solved 5 problems with totally +8 :(

Hope everyone be careful and good luck and have fun (:

First time to be a tester, wish you good luck & huve fun! :)

As a tester,I like this round which is very interesting! Wish you can enjoy it :)

As a tester, I test.

As a tester, I won't say that the problems are perfect, but I still enjoyed the round. Hope you guys do too :)

Btw, rui_er is kawaii >w<

Upd. my testing experince and comment on each problem goes below.

During testing, I solved all six problems, but just barely. (I made my last submission at 01:50:15) Out of 7 total submissions, I got a runtime error on problem D but only due to insufficient array size. All my submissions run in one-third of the time limit.

• A: Nice easy problem, seems a bit hard for its place. A little similiar to CNOI2016 Grid, but I think it is OK since Grid is much much much harder.
• B: Nice easy problem, fits its place.
• C: I personally do not approve of placing an interactive problem at C — for the unexperinced participants, the main difficulty will be to understand what is interaction, and a lot will likely spend the entire round wondering why they got "Idleness limit exceeded on test 1". The problem itself is OK though.
• D: I don't quite get the point. It seems like it is just implementing the process with a balanced BST (something like std::set in C++), but somehow it also looks enough difficult for D? I don't know why. Still I think it is ok because at least it is not annoying to solve.
• E: Quite standard, I got the solution 5 minutes after reading the problem, while Div.2 E usually cost me 15 minutes or more (not including implementation). Perhaps not that standard for non-Chinese participants? I don't know.
• F: My favourite problem in this round. I was getting prepared to implement 300 lines of code at the first sight of the problem. And after getting nowhere with centroid decomposition, I came up with the intended solution which is just so elegant. I felt brilliant after solving it.

As a tester, I think this contest is very Chinese characteristics and worth attending.

Good luck to all of you.

As a tester, The problemsetter rui_er is very cute! The problems are interesting and challenging. Wish you can enjoy her round!

As a tester, I enjoyed the Chinese problems, and I wish you can enjoy them too! :)

Problems are great. Good luck!

(When will Li Hua stop his foolish behavior of writing letters? lol)

As a tester, I recommend everyone to participate in this round! The problems are interesting. Good luck and positive delta!

As a first-time tester, I hope you can enjoy the fun problems!

As a tester, the problems are interesting but some are trivial.

[EDIT] Sorry. I didn't want to tell you not to participate in this round. I think one should participate by himself and has his evaluation.

Love rounds with Chinese authors. Let's do more in the coming days :)

OMG CN ROUND

Why did all these testers get downvote? bcz meaningless?

hope you enjoy this round.

the contest starts only 5 minutes after the end of ARC :(

Li Hua is the main character in English writing exams for Chinese students. He always asks you to write letters for him.

Expecting string problems O_o

I can't wait to enjoy this round cause rui_er is so cute!

Those who think MagentaCobra is newbie tester :o

My first unrated Div2 :)

No offense, but I'm scared of problems from Chinese high school...

Please do not downvote .Apologize for my illogical comment.

Please test more about problem D.

This contest is a bit earlier than usual (For Chinese people)

Usually I perform badly in CNR. But still willing to give it a try(bcz rui_er and DitaMirika are cute qwq).

Hope this round won't betray my trust.

I bet something is gonna be wrong with problemset (problems are too mathy, unbalanced, annoying). Let's see after the round.

we will solve Li Hua's problems.

Hope to become an expert...

The nice contest

Chinese writers. Looking forward to Mathforces

Hope to become an Pupil ...

I have solved a good problem raised by rui_er in Luogu, and I believe these six questions will be very interesting.

good luck

Goo luck everyone!

Yet another stupid mathforces round

its normal time!Good luck everybody,i wish to all get positive delta!Thanks for doing contest!

I think it will be stringForces

There is a high chance that there will be some math problem on crt

stringforces?

Writing letters and not sending them?

There will be a finite number of $$$graph$$$ problems.

this will be my first round,Hopefully to get positive delta, Good luck everybody!i wish to all get positive delta.Thanks for doing contest!!!Sorry for my poor english.

Maybe the author's writing task for the next English exam will be like this.

Did any one notice Last line of I would like to thank ...

I guess this will be a mathforces round. ಠ_ಠ

Mazeforces :) || Patternforces :)

I think I made a mistake by giving priority to this contest over my tomorrow's exam ;(

WAforces

stupid contest. most of these problems are well known.

C was nice for its position! glad that they atleast made strong pretests, WAforces > FSTforces

Problem D is very interesting — literally do what is written in the statement

iq test round

Is...Isn't E just segment tree?

Cool round! I enjoyed the problems!

The problems are very good.I am glad I participated in this round.

Div √(2*√2)

D???

Thank god, my Last minute submission of C saved me :>

Problem D is the worst possible div2D. Thank you for this problem.

Why is this giving wrong answer on pretest 1 for C? I clearly get (5,1) as the answer for the second test of pretest 1.

link to code: https://codeforces.net/contest/1797/submission/201318204

Hmm why does'nt this code work for B? Did i misunderstand the question?

https://codeforces.net/contest/1797/submission/201297818

I think the example of D is too weak.

Question about Problem C: I ask (1,1), (1,2) first. If the answer to the two times is the same, I can get x. If it is not the same, I can get y. And then I ask (1,y) or (x,1) to get the other coordinate. Is it wrong?

Original statement of B was very very bad, nothing said that you are not allowed to recolor a red cell to a red cell.

It sometimes makes sense in real world. Like in my room all of the walls are painted green. And if I want to freshen up the look and I like the same color, I will recolor green walls to green.

Idk maybe I am insane

Good problems. liked the idea of C and D. Good contest overall.

ABC were very decent, but problem D is very surprising to see on Codeforces in 2023, this is just implementation exercise, isn't it? Could've say the same about E if it was Div1 round, but in Div2 it's probably alright

Denote the heavy son of a non-leaf vertex as the son with the largest subtree size

RIP, I've been doing heavy son of non-leaf vertex as the son with the largest subtree importance

B and C are too simple, but annoying to debug.

Я снял видео-разбор задач A, B, C, D. Кому интересно, разбор можно посмотреть по ссылке: https://youtu.be/DjynOzLa818

How to solve Problem C

who the hell starts numbering from top left T_T

But the problems were good ^_^

Solutions looking for problems. Great examples.

The problems are good. But for me, problem D is a bit annoying, because I can't write the code correctly. Hope to enhance my coding skills.

Gridforces!

Update got the mistake:)

Debugforces

Can we solve $$$E$$$ using segment tree?

I firstly build a tree with $$$C$$$ vertices and prepare lca there (though is consumes much TL and ML).

I store in node the number of numbers, the lca of numbers, the minimum depth of number and the sum of distances from all numbers to that lca. On merge I do $$$c.lca = lca(a.lca, b.lca)$$$ and $$$c.res = a.res + b.res + a.cnt \cdot (depth[a.lca] - depth[c.lca]) + b.cnt \cdot (depth[b.lca] - depth[c.lca])$$$. On move up I do $$$mindepth--$$$ and if $$$depth[lca] > mindepth$$$ then $$$lca = euler[lca]$$$ and add to result $$$\pm 1 \cdot cnt$$$. But there can be case, when there are numbers $$$1$$$, and seems I have to count their number. This killed me.

Great contest, I had a lot of fun solving the problems (until i got a bug in D that i couldn't figure out)

Can anyone please help me find the flaw in this code for C? Thanks!

201335214

My solution process is to first look for the top left and bottom right corners, and then all the possible answers left will either be a horizontal line, vertical line, or two points. I check this solution for half an hour but honestly had no clue of why its wrong...

Nice contest, I like the problems a lot

I made 2 wa 2 re 1 iq
looks like I need work on it.

Bad problemset — at least for 2A ~ 2D. In my opinion, 1797D - Li Hua and Tree is the worst div2D I have ever seen. I can't imagine why the authors put such a trivial problem for this position. And Celtic you are right. :)

Why am i getting WA on pretest 3 in this submission for problem D : 201343094

how to solve problem b?? i was checking v[i][j]==v[n-1-i][n-1-j] for every i, j it should satisfy but getting wrong answer on pretest 3 please help

To those who feel confused about 'Li Hua'（李华）:

Li Hua frequented in many important English exams in China, including NEMT. So all students in China who take the exams are very familiar with him(or her?).

Here is the writing section in NEMT 2019:

translation:

Suppose you are Li Hua, your friend Terry in New Zealand asked you about the local customs, please reply the letter....

I think problem C is very similar to a previous problem from an ICPC mirror.

good contest.>.<

Could someone please tell me why WA1 happens? I can get the correct answer when I run it on my computer. I really don't understand

#include <bits/stdc++.h>
using namespace std;

int main()
{
	std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int t;
	cin >> t;
	while (t--)
		{
			int n,m;
			cin>>n>>m;
			cout<<'?'<<' '<<1<<' '<<1<<endl;
			int d1,d2,d3;
			cin>>d1;
			if(d1<=n-1)
				{
					cout<<'?'<<' '<<1+d1<<' '<<1<<endl;
					cin>>d2;
					int y=(d1+d2+3-n)/2;
					cout<<'?'<<' '<<1<<' '<<y<<endl;
					cin>>d3;
					int x=1+d3;
					cout<<'!'<<' '<<x<<' '<<y<<endl;

				}
			else
				{
					int tmp=d1-(n-1);
					cout<<'?'<<' ' <<n<<' '<<1+tmp<<endl;
					cin>>d2;
					int y=(d1+d2+3-n+tmp)/2;

					cout<<'?'<<' '<<1<<' '<<y<<endl;
					cin>>d3;
					int 	x=1+d3;
					cout<<'!'<<' '<<x<<' '<<y<<endl;

				}

		}

	return 0;
}

I actually like problem D lol

bruh nvm im stupid

If only I had used "set" instead of "priority_queue" on D...

apparently problem C appeared here:

https://codingcompetitions.withgoogle.com/codejam/round/0000000000051708/000000000016c77c (subtask 1)

probably just a coincidence though

A: We need to block all adjacent cells of start or end cell. If one of them is on the corner, answer is 2; if one of them is on the edge, answer is 3; otherwise answer is 4.

B: We check each pair of symmetric cells ((x, y) and (n+1-x, n+1-y)), each pair of different color needs 1 operation. So we first adjust these pairs, and if there are extra operations, we do all of them on any single cell. If the amount of extra operations is even, we are done, otherwise we need to do the final operation on the center cell (if n is odd). If there's no center cell (when n is even) there's no solution.

C: For a single query (r, c) the set of valid answers is the boundary of a square center at (r, c) with length of side 2*query(r, c). We can first query on 2 corners, and the intersection of their set of valid answers will be a segment or 2 different points, then we can query for the final answer.

D: Implementation. We need to maintain the set of children of each node and sort them by (size[u], -u), and maintain the parent of each node. For rotating x, we first find it's parent p and it's heavy son h (which is the maximum in the child set of x), then we do (size[h], size[x]) := (size[x], size[x]-size[h]), similar for their importance. Finally we set parent of h to p, parent of x to h, and modify the child set of p, x, h.

E: Let dp[i]=the number of operation we need to change i into 1. By some pre-calculation we can see dp[i]<=23. So we can store all elements of a in 23 different sets, where set[i]= (set of indexs j where dp[a[j]]=i), and we can do range updates in 23*n set operations. Also we can use segment tree to maintain the sum of dp[i] and LCA of a[i] in range [l, r] (we assume all numbers are on a tree rooted at 1 and has an edge between each pair of (i, phi[i])).

Why the color of "You" (in the last entry of thanks list) same as the color of a LGM?

Thanks for this great round, with +109 I'm now a cyan with 1435 rating.

GridForces

.

Ratings updated preliminary, it will be recalculated after removing the cheaters.

i really liked the problems A to D, but i am not able to figure out what was that edge case in pretest 3 of D

Can we determine the answer by asking for three vertices For problem C

Not solving D with over an hour left is so disappointing. I'm too slow at implementation

E is nice, but constraints on values up to 10^6 adds too much difficulty

It was the best Chinese codeforces round (in my opinion), thanks to rui_er and AC-Automation!

Hello Codeforces, In todays contest,in Ques B,my code got TLE on test case 10,but when I used ios_base::sync_with_stdio(false); line it got accepted after the contest. It is my request to plz accept this code as this code deserve to be right. Submission Id : https://codeforces.net/contest/1797/submission/201286585