Hello, Codeforces!

Daniel_yuan, waaitg, smg23333 and I are glad to invite you to Codeforces Round #706 (Div. 1) and Codeforces Round #706 (Div. 2), which will take place on 10.03.2021 15:05 (Московское время). **Note the unusual time of the round.** In both divisions, you will be given **6 problems** and **2 hours** to solve them all.

We would like to thank:

- isaf27 for the great coordination and discussion.
- KAN for the great help with round preparation.
- Wolam for
~~all his rejected problems and~~discussing the problems with us. - tiger0132, EmptySoulist, CYJian, parallelist, Sunward_z and ODT for testing the problems in advance and sharing their ideas.
- dXqwq, LTb, rui_er, CSP_Sept, Qiuly.qwq, mac, dapingguo8, mazihang2022, Ashishgup, aajisaka, i_wyxkk_ak, marX, Mitsukasa_Ayase, Oak_limy, sidiangongyuan, zscoder, tomato_sans, lgswdn, tianbu, orzdevinwang, happydef, Kewth, Widowmaker, star_xingchen_c, 3.141592653, Aaeria and 4c5948 for testing the round and their invaluable feedbacks.
- MikeMirzayanov for amazing platform Codeforces and Polygon.

Score distribution will be announced before the round.

Hope you all gain positive ratings $$$\Delta$$$ in this round!

**UPD1**: Score distribution is

**Div. 2**: $$$500-1000-1500-2000-2500-3000$$$

**Div. 1**: $$$500-1000-1500-2000-2500-3250$$$

**UPD2**: Editorial

**UPD3**: Congratulations to the winners:

**Div 1:**

**Div 2:**

As a problem setter, I set some problems.

As a problem setter, I set some problems.

Tao God, eternal God! fkami, my superhero!

As a tester, the statements are short and pretests are strong and give me contribution.

As a contestant, I will take back my up-vote, if the tests turn out to be weak.

You can't do that.

I can,by down-voting from an alt account.

how does one become tester?

As a contestant, this claim was incorrect.

pretests are strong!codeforces × FSTForces √ :(

I love this round and wish I could participate in this nice round (but sadly as a tester I can't, but can I get some contribution?) XD

Good Luck!

It was good

As a problem setter, I want some contribution, too. XD

"In the upcoming round, you can use only Kotlin."is this notification for this round?Apparently no. That's the notification for

Kotlin Heroes: Episode 6. This round is a normal Codeforces round in which C++, Java, Python, etc. can be used.As a contestant, I will solve some problems. :-)

As a contestant, I will try to solve more problems than create?..

Sure.. you can but for this, you have to solve problem sets for past contests. :)

I just meant I don't want to be additional problem for the contest creators :D

As a contestant, i will try to solve all problems:)

BEST OF LUCK BRO. ;)

Okay, I am a bit serious now, how to be a tester and what are the perks of being a tester ?

It is NOT funny anymore, stop repeating this stupid flat joke 100th time in a row after every announcement.

As a participant I solved some problems.

Why still no rating changes for div2? Over 16 hours from the end of the contest passed and still nothing

As a tester,this round is really nice,statements are short and clear and problems are interesting.

Wish you can enjoy the problems and get high rating!

Don't forget to upvote me XD

upd:Imakf why you steal the first comment >_<

as a participant i hope for +ve change in rating.

As a tester, the statements are short and pretests are strong and give me contribution XD

Not so strong pretests apparently :/

Why doesn't B have multiple testcases (like A and C have) to check for edge cases in the pretests?

As a tester, the statements are short and pretests are strong and give me contribution XD

soulist /se

why downvote me? i do nothing wrong!

Another Chinese round! It's a pity that I won't be able to participate in it because of my classes :(

Anyway, I wish good luck to all participants!

Me too :(

If a $$$\text{Div-1 Round}$$$ and $$$\text{The Whole Education System}$$$ were dangling from a cliff, I would sacrifice $$$\text{The Whole Education System}$$$ to save the $$$\text{Div-1 Round}$$$

Yeah, as the saying of one of the problem setter, Chinese schools should be destroyed(Just for joking)

As a tester, I want contribution!

The problems are fun and interesting, so don't forget to participate!

`Hope you all gain positive ratings Δ in this round!`

It is theoretically impossible sadly :/

It's not sad . If everyone gets positive rating then it won't remain fun and competitive.

You could have it both ways.

Create rolling sum that increases after each round, add the max rank to it each time. Each rank would be than sum + actual rank. Increases every time!. And yet distances between users stay the same! :D

Yeah I know its silly and impractical. But it is possible.

What if there was only 1 participant (initially rated 0) who solved all questions during the round? What about this "theoretical possibility"?

If it's his "2nd " account on CF

As a tester, the statements are short and pretests are strong and give me contribution XD

please don't keep downvoting me!

As a tester, I really recommend you to check your code carefully before submitting.

(btw : it's the first time for me to test in a CodeForces round)

rui_er! /se

upd: Why I got so many downvotes? I'm just a classmate of rui_er :(

Well I really wonder why I got so many downvotes. Maybe you disagree with me, but this is just a kind suggestion, so please be more friendly to a person who is still a new hand but always tries his best to do every job. :(

You are right!

"

I really recommend you to check your code carefully before submitting."Well, you are right. But why you guys didn't make the pretests of B strong? I think it's easy to be done. :)

I'm really sorry if this makes you confused, but I'm only a tester (not problem setter), so I don't know whether the pretests are strong enough. I started my virtual participation and tested some solutions and the constraints. This is the first time for me to do this job, so maybe I'm lack of experience. I'm sorry again and I'll try to work better next time. :)

Thanks for your reply. It's always hard to make something perfect in the first time. And I think you will do better in next contest.

Are contest timings decided by problem setters?

Yes. This unusual timing is because the problem setters are Chinese and they have to go to school the next day (maybe.

Chinese round!!! Chinese time!!! I can take part in the contest at school!!!

Chinese Round is coming!

Gaining positive rating $$$\Delta$$$ is not the most important thing that you got from a contest. Passing interesting problems and benefiting from them are the core purpose of participating in the contest. Otherwise, if everyone only want to increase rating, the utilitarians may fill the community.

Of course, getting high rating can be your motivation for progress. But I also wish you not go against your original intention of studying Computer Programing.

And, good luck! XD

Now found someone like me who also dislikes people yelling for rating here.Followed you buddy.

Why would you

dislikesomeone for having a different motive than you?I also don't have answer to this as somethings which are felt can't be described. It is a feeeeeling. I can't describe it.

Remember in my comment yelling is not meant as motive. motive is something different. some people just freak out on social media if they gain something. some people just cry here in the discussion when they suffer huge delta this deviates the purpose of this blog. 100 / 400 comments in every round discussion says about. "Hope I become expert/ specialist/master today!.

"Oh shit damn due to a silly mistake I lost 150 delta".

"OMG what a tough round".

"WTF the pretest 2".

This all things are fairly common in present comments in the announcement pages . And ya most popular trend.

"As a Tester ....shit"

Everything aside, you are a legend man. You are from Ukraine, study in China and have an Indian meme as your dp. XD

lmfao that said from your mouth lol

You are quite right

China is an actually a good country :D

I agree with you

Wow a newbie testers

Wow a pupil comment

Wow a Specialist reply

Wow a pupil comment

Attention, time is unusual !!!lollll :DD, I try to remind people but people don't like it =))

It is really unusual,I don't have any time to pretend to be weak to my friends!:)

what do you mean ?

Only a joke.Take it easy.

Maymay...[]()

Codeforces Supremacy

As a tester, give me upvote!

Hope you can enjoy it. When you get stuck by one, you can just read problems after. :)

I saw lots of brilliant coder testing the round, maybe this means the contest won't be easy.

I want to become purple after finishing the contest(Daydreaming, of course.), so I'll try my best.

Hope everyone enjoys the fun of coding, like I do. Stay Cheeki Breeki :-)

The contest being hard or easy doesn't makes much difference in ratings outcome maybe because The contest is same for everyone....

Yeah, you're right.

What does it means- " In the upcoming round, you can use only Kotlin. You can read the code examples here."

That is for kotlin heros contest you can use only kotlin

Why downvotes, it wasn't needed. I just asked.

Because your question was already answered. Plus, it's quite obvious that the announcement is for the Kotlin contest.

I think the notification was for Kotlin Heroes: Episode 6 not for Codeforces Round #706 .

As a tester, the statements are short and pretests are strong and give me contribution XD

This is my second time to be a problem tester.Thanks to Imakf for giving this valuable opportunity.

The problems are well-prepared with interesting stories.Hope you enjoy them.

At last,good luck and have fun and get high rating

As a tester,please give me some contribution XD

(σﾟ∀ﾟ)σ⁶⁶⁶⁶⁶⁶

Aaeria

ScarletS

Binod

This is my first time to be a codeforces tester, I feel glad. Thanks to all of the writers!

This round will be very interesting.

Wish everyone good luck!

(The statements are short and pretests are strong and give me contribution XD)

Colorful testers :D

Imagine getting + delta in a div1 as a low purple

you know you could at least read the comments before reposting a meme

Spoiler...

Contribution > problem solving

An

Orange-blueRound XD.why my contribution is in negative?

Feels great to see newbie in testers :)

As a grey colour, I'll solve A and attempt to solve B.

no, A will be too hard for you, sorry

Kuso!

just small predict

hahaha

You were right, A was too hard. But B wasn't. So I solved B

Good job!

Thanks :)

Believe me! Just try to solve A and B quickly and you'll become specialist like me not long after that :)

That's the plan :)

As a problem solver, I want some contribution, too. XD

I hope the difficulty level is according to the score

The start time is too early. Sad, Chinese Round should be destroyed.

Good luck to everyone!!!

As a problem solver, I want some contribution, too. XD

whenever you comment to get upvotes, every time get downvotes XD so please Stop now XD

May we know number of common problems between divisions?

As a resident of UTC+7 please give me more chinese rounds

More Chinese rounds — less rating, I warned you

Another 17:35 msk lover

Sorry guysAs a contestant didn't do anything I want some contribution

At long there is a game whose time is suitable for me.

vote for the Dogecoin

It will be evening in India and will be missing playing footbal.. Hope I give it my all and make it worthwhile

As a participant, the statements are short and pretests are strong and I want +ve delta XD

What mean rejected problems? see about it in almost all contest announcements. Why everybody write about it?

coordinator can reject problem, and it is often case if you are new in problemsetting

When you know you are going to lose rating, but still want to participate (Chinese round fact)

For me, it's just like when you know there is a round today, but you have classes :(

At least the timing is reasonable. Usually its late and uncomfortable. I like it when I can actually make it a round without loosing sleep or having to turn up to work a dead beat.

After reading prob D(lets go hiking)- others — Try to solve the problem. me — lets leave this and starts searching for hiking destinations to visit post Covid.

You must be kidding.

Joined about an hour late because of classes and then not able to solve even a single problem. :( feeling bad

I fucking hate double in problem C I have right algorithm and because of double it doesn't accept on first pretest

I also had the same problem.

mine failed at pretest 2

No one tester noticed that statement of div1B is confusing? Really? I lost 0.5 hour and made 2 excess submissions because of that, and I am angry now

I am in similar pain because of other problems.

Come on setters and tester could have done a better job.

How to solve B ?

And what's wrong with this solution ?? WA

if mex is equal to n then ans is n+k otherwise answer can atmost be increased by 1.

Your method for calculating mex only works for sorted array.

unique_numbers is a value of unique numbers in array on the start.

if mex < max then we have dont have a result of (Math.Ceiling((mex + max) / 2)) in our array then result would be (unique_numbers + 1).

if mex < max and we already have a result of Math.Ceiling((mex + max) / 2) in array or k == 0, then result would be (unique_numbers).

if mex > max then result would be (unique_numbers + k)

Your logic for finding the

mexis not correct because the input is not necessarily in sorted order. For example, if the input was`1 2 0`

, then themexyou find is`1`

. But the correctmexfor this input is`3`

. Try applying the same logic while iterating over the sorted set of elements after you have inserted every element into the set.Thanks mate. Now, I am Feeling like shit :(

I think Problem C should not be in the position of C...

Very easy or very hard?

I actually write a math proof to verify my algorithm, but I believe there are people who had their fingers crossed and passed XD.

Can you give proof?

In which part?

First of all you can move all the mines and miners at negative positions to positive positions, since this doesn't affect the distance formula.

Then I simply try to proof that when using a line connecting miners and mines, there should not have any intersection.

nooo... i mean the logic part of it...like what should we take first the miner with the highest distance and coal with shortest distance(antisorted) or both in sorted orders? I took minimum of both cases lol by maintaining 4 priority queues xD.

simply sort the miners and mines by their distance to the origin, and pair them one by one.

My submission https://codeforces.com/contest/1496/submission/109585265

give proof

This is not meant to be proofed. Just guess and submit works fine.

Asking for the proofe here is asking for a formular. Google it.

Editorial gives nice proof.

This is no proof. They just draw a picture and say "It is that we must pair them in order."

noooooooooo..

It takes some time to write down my proof at the below link

i did the same thing, but why did it fail on pretest 2?? https://codeforces.com/contest/1496/submission/109623427

I think it is probably because the square part gives you overflow

https://ibb.co/mHNCJQL

That's nice!

But you did not write that while contest, did you?

The last line should be $$$(c^2 - a^2)(b^2 - d^2) < 0$$$ , Hence (a) is better than (b).

actually I did xd, but with a lot of note and calculation.

My contest proof version https://ibb.co/bLDDx0C LOL

just use triangle inequality

https://ibb.co/Xp8yVnS

What a nice proof!!

f

Very dumb

Thanks authors for good problem setting. This time i haven't feeled myself as cryptographer who deciphers the cipher.

As a 中国人, I'm sorry I can't solve these problems. (Just started competitive coding with python)

What is in the 3rd pretest in div1C?

btw, div1B is elegant imo :)

Round A Video Tut:https://www.youtube.com/watch?v=ug-g4z1giEk

I want to cry, I want to rant, I want to do some things to some people.

I am sure I am not the only one with these feelings.

I think 1B should have highlighted the fact the two inequality signs are different for the two people's possible moves. Also I should learn to read better :(.

Turns out I wrote the wrong thing then ):

Same to me. It seems that having wrong interpretation can pass all the samples and the first few pretests, especially when there is no explanation of the samples.

Wtf. I actually realized this after reading your comment.

Same T_T

Same :(

Can Any one tell why i am getting TLE? I am just using sort and sqrt functions This is my submission:https://codeforces.com/contest/1496/submission/109622532

instead of double use long long for x, y because the comparison between doubles take more time than long long

multiset are very slow (something like 20 time slower than a sort), try with arrays of size n and counters to know where to add, or with vectors it should pass

Whats the approach for Div2 D. I am getting WA on pretest 9.

author should provide explanation for first test case of Div2D, I even asked them during the contest and they have responded as "no comments", this is not good.

Div 1 D has such a scary time limit. I spent a bunch of time doubting if $$$\mathcal{O}(n^2 m)$$$ would pass, and it turns out my $$$\mathcal{O}(n^2 m)$$$ passed pretests in 2.13s/2.5s. It doesn't have any optimisations, so the runtime should be fairly independent of the input, but still, I'm scared of the systests.

anybody who can explain what's the deal with using Fast io and setprecision together resulting in TLE for C ??

Because you read the input in doubles when the input is integers. Doubles read for a long time. It will be faster if you read ints and convert into doubles.

Also, I read at someplaces that sorting doubles is costly. Is that true?

I don't know

I think the differentiation of problem C is TOO low.

[Div1 B] Am I the only one to not notice that the conditions for x and y are opposite and made 3 WA submissions?

I also didn't notice that at first. Fortunately I spotted that before submitting anything.

I noticed it now :(

what's with the 4th pretest of div2 c?

Just don't take the container in which you are storing miners and mines of long double type, take it of int type. And rather take miner[i]*miner[i] + mine[i]*mine[i] as a long double variable to pass to sqrt function. And it passed for me after a number of Time Limit errors. Edit : also sqrtl() in place of sqrt()

and why is that?

Can anyone explain their approach for Problem C ? Thanx in advance .

separate the x axis and y axis values and take the absolute values only . After that sort both x-axis and y-axis arrays . Now we can pair same index values in both arrays .

Implemented the same ! but wrong answer :(

first square the coordinates then apply.

Int is not enough to store square of coordinates

just sort x axis values and y axis values choose max from both

How to D?

I am not sure if this correct approach , I think answer will be always $$$1$$$ or $$$0$$$ .Answer will be $$$1$$$ only when there exist some index $$$i$$$ such that array is decreasing in both right and left direction of index $$$i$$$ and size of the maximum decreasing sub array in both directions are equal,even and maximum in the whole array.

Update this approach is correct : 109629126

ohk cool, never thought like that..never thought of anything..the problem just seemed impenetrable..was sitting blank for an hour xD.

Rewrite input as mountain profile. Then first player only go down, and second player only go up. The players can't collide. First player should always start on peak. Otherwise, if he is in hole, he can't do any move, if he is on slope, second player can start under first player. Find for every peak its length to the left and to the right. If there are two peaks with same maximum length to the left/right, then first player always lose. Assume he starts on one peak, then second player will start on bottom of another peak, and since he is second, when they walk same distance, the first player will must do move. Now, we have the only peak with its length to left and right. This is the only position, where first player has chance to win. So the answer is either 0 or 1. First player wins if and only if the lengths of peak are equal in both sides and even.

Very elegant thanks..will try to implement!

First player wins if and only if the lengths of peak are equal in both sides and even.Shouldn't it be odd ??

Oh, yes, depends on your definition of length. I defined length as number of edges: here $$$[1, 2, 6, 5, 4, 3]$$$ length to the left is $$$2$$$ and to the right is $$$3$$$.

Hm, right

In, div2 C, using a vector of

doublecaused TLE, while a vector oflong longpassed the pretests :(same. wasted 1 hr :)

why is it that the vector of double givees tle?

I misread statement from problem B and I thought that both players could make the same movements (I didn't see that $$$p_{y'} > p_{y}$$$). Goodbye div1 :(

I'm really surprised that all my submissions reached pretest 7.

lol, it happened to me too :(

so unfortunate

When you read Div 2. D as py′<py and spend 1 hour solving the wrong problem ://

That is bad problem statement. I did read it like 25 times before noticing.

Providing the explanation in problem D before hand would have saved a lot of time and maybe that could have been utilized in writing better solution.

I was still solving it for the same until I saw your comment. Still getting a positive delta but could have done better.

What was Div1B Pretest 7?

Nice problem A.

Div.2

E/ Div.1Chas elegant solution, just think about a pattern.SpoilerWhat if you clear every $$$k$$$-th row? Think about suitable $$$k$$$.

Upd.Managed to construct a solution which uses regexes (Perl): 109637022Can someone explain why I got TLE in D2C? :((((

https://codeforces.com/contest/1496/submission/109611221

https://codeforces.com/contest/1496/submission/109611077

use long long int instead of long double, the sort will be faster (I know it's weird)

Why my code is wrong in Div2C?

Codehttps://pastebin.com/zwS43AAZ

firstly the value you set ans to is too low, it can be up to 2 10^13, define it by the first value and it should become correct (to my mind). However, according to others bad experience, you should use vector of long long int and not long double because elsewise sort will be too slow on test 4

In recent contests I spent most of the time in asking myself, "How could I missunderstood the problem statement so that my solution does not work?"

Of course it is not allways the reason for WA, but if something does not work as expected, than usually such missunderstanding is the reason.

Would've really appreciated if the sample explanation for B were added beforehand. Thought the p’ > p signs were same for both players and solved for that. I take full blame for my mistake, but would've been nicer if at least one simple simulation were shown in notes, just sucks to lose so much time on a misread. Gotta be more careful I guess

in problem C, why is (https://codeforces.com/contest/1496/submission/109623109) giving WA? answers almost same but can't understand what is wrong

you must sort it in ascending order of their absolute values

same with me too. https://codeforces.com/contest/1496/submission/109620775 here is my submission. can anyone help plz?

I dont know, can you use other functions for quicksort in c++, but you can put the absolute values to array. And your solve will become correct.

why will it make a major differnce? can someone tell a test case where this part code would be wrong?

The FST rate of div1B looks so scary...I am so curious about the pitfalls...

Why is this solution giving TLE for 1496C?

Question Link: 1496C Submission Link: 109603686

The solution is under the constraints and the approach and method is same as most of the accepted solutions that I have seen. Why is my particular solution giving TLE?

use vector type long long

Okay, using long long in vector worked, but why? Both long double and double give TLE, why? I have seen accepted answers using double in vector?

dont use long double in vector

Okay, using long long in vector worked, but why? Both long double and double give TLE, why? I have seen accepted answers using double in vector?

I really don't know much about others.. But I've experienced this problem several times

Please tell whats wrong with these solutions ?? (Today's div2-c)

Link1

Link2

I was at ~200 position after solving 2nd problem then there comes this problem :(.

you must sort it in ascending order of their absolute values

Use vector long long vect1,vect2 rather than long double vect1,vect2. I had the same problem made 8 submissions before making the right change. Sorting is O(nlogn) but there is considerable overhead sorting long doubles than integers. Stupid C++ :)

I just want to know, why i have got WA in Problem C :( 109590727

Problem statement for this problem "Let's Go Hiking" was very fuzzy. It could have been more clear with test case explanation.

I think the number of pretests of Div2D/1B are too few...

Damn D's are failing fast.

I can't understand why I have got TLE for problem C in Div.2 for this solution: https://codeforces.com/problemset/submission/1496/109589906

but OK for this solution: https://codeforces.com/problemset/submission/1496/109596686

Downvote me to hell but no more participation in Chinese rounds and Russian middle school rounds.

Indians tend to prepare much much better rounds. Racist? Yes. Factually correct? Also yes.

how to solve D div1 ?

A solution should be available on the editorial, but here's an outline of my thought process.

Precompute the pairwise distances between all nodes. Then, we'll compute the answer separately for each pair of nodes.

The key observation is that a pair of nodes has answer zero if there are multiple shortest paths between them. We can prove that any edge on a shortest path from i to j must be in a BFS tree that can be rooted at both i and j, and if multiple shortest paths exist, this will necessarily create a cycle.

The shortest path from i to j is unique if, for every d from 0 to dist[i][j], there exists a single node k such that dist[i][k] = d and dist[k][j] = dist[i][j] — d. We can check this criterion in O(n).

Then, the BFS tree becomes a forest containing a tree rooted at each node on the shortest path from i to j. We can then show that vertex k not on the shortest path such that dist[i][k] = a and dist[j][k] = b must be connected to a parent vertex p satisfying dist[i][p] = a-1 and dist[j][p] = b-1. Therefore, we can compute the answer by multiplying together the number of possible parents for each vertex.

We can thus deal with each pair of vertices in $$$O(n+m)$$$ time, so the total complexity is $$$O(n^2(n+m)).$$$

It is more clear than the tutorial, thanks.

Thanks for the explanation.

I am your fan xd

Thanks for this round!!! :>

There is a huge difference of who solved A,B,C and who solved A,B,C. :(

How ?

There are only 544 people Passed the D.So who passed A,B,C quickly will make great grade.

Hey could anyone explain when using long double rather than long long, its giving TLE on div2C?(https://codeforces.com/contest/1496/submission/109596148) (https://codeforces.com/contest/1496/submission/109613452) I know sorting has considerable overloads for long double but still wouldnt the complexity still be O(N*64*LOG(N*64))?

Got TLE in div2-C but don't know why . can anyone help?? 109609708

I did pretty much the same thing and got AC, not sure why your TLEs, let me know if you figure it out so I can avoid it in the future :D, anyway here's what I did:

SpoilerJust changed double x, y to long long x, y and it passed :)

Solution for Div.1 C : Consider partitioning about columns, that is mark complete columns as empty for columns 1, 4, 7 .... in 1 based indexing. Now between each 2 consecutive columns, if there is at least 1 cell that was originally empty, eg. in column 2 and 3 while considering columns 1 and 4, join both the columns using that row's elements in between them. Otherwise join them by the first row if no such empty cells exist. Finally if the number of columns are divisible by 3, we have to join the last column elements to the rest of the elements joined till now. To do that, for each such original empty cell in the last column, mark it's previous column (same row) element as empty.

Again, not good presets XD

Why is my code incorrect: https://codeforces.com/contest/1496/submission/109627982

At least, we can show that the solution must be 0 or 1, but your output for test 5 is 57.

Why is Problem E easier than D —— even Problem B?

Oh,I'd made a mistake.Now fixed.

You can just edit your comment...

:D finally fst on 1B, because of some small bugs.

But it's still a nice trip, isn't it?

In the end, my rank reduced by 60. Luckily I still become an IGM, with various motions in my heart.

It's an exciting and unforgettable day for me, although this round maybe my last time to participate in a codeforces round.

It's time for me to stop studying Computer Programing, because of too many hard competitions and the stress of studying subjects. Otherwise I must try my best to become a top OIer in China, it's too hard for me ...

Wish you all good luck! And, thanks. :D

Have a good trip and hope you'll achieve your target! :)

My code is printing 1 on my pc and ideone but on codeforces its printing 0.

Whats wrong?

mentioning: MikeMirzayanov

For div.2 C problem. Can someone tell me why is 109605593 getting TLE and 109609603 getting accepted when both are having the same logic?

Just printf statement is different.

%.15fin one while%.15lfin second."strong" pretests for B. I think that there aren't too many situations but the pretests didn't include those. Anyway, you should consider all the possible situations. :(

Pretests for problem D were very weak!! I mean look at the 30th case and 20th test case, how can one not include it in the pretests? Anyway it didn't affect my ratings but still those got WA on those test cases are unlucky :(

With many participants happened this: one of the solutions got TLE, the other one got OK, but they have the same logic (problem Div.2-C).

Why??

Thanks for the round! I thought the problems were very high-quality, and I enjoyed thinking through them for a few hours. (This was probably the most intense CF round I've ever taken--finishing C and solving D in the last ten minutes saved me on the order of 150 rating.)

One point of criticism: I'm not entirely sure why B did not use multitesting. There's no particular reason it would have been more annoying to write up if multiple test cases were included in each input, and this sort of problem (i.e., problems with only a handful of possible answers where determining which one is correct relies on checking some fixed criteria) always seem to lead some competitors to janky solutions that break down on relatively rare edge cases. The resulting FSTs would have been avoidable if multitesting was employed: for example, many solutions FSTed on a case with $$$n = 5$$$, and this could easily be prevented by a single batch test containing all inputs from $$$n = 2$$$ to $$$n = 7.$$$

Because of the pretest of div2D/div1B , lots of people are disappointed.

But you should consider all the details.

If you can't consider all the details , please improve yourself , do not blame the problem.

but a wrong solution which passed pretest will make people relax vigilance

how weak 2D's pretests are!

sorry I accidently downvoted you .... consider you upvotes to be (current_votes+2)

sorry I accidently downvoted you .... consider you upvotes to be (current_votes+4)

Proof of Problem C that we should not have intersecting lines in the midway somewhere.

EDIT : D

_{2}^{2}— D_{1}^{2}should be in the next equation. I hope you all can understand it. Please tell me in the comments if you have any issue in this. Also i missed to write +k when simplifying D_{2}^{2}.Well, i think it can be proved by

https://imgtu.com/i/6Y36II

Sorry i don't know how to insert a picture

Nice and short proof!!

$$$OA+AB \geq AB$$$

I think you want to say $$$OA+OB \geq AB$$$

Yes, you're right, it was a mistake

sto IGM MonkeyKing!!

Connect possible lines between two points ，It can also be proved by Geometric intuitionistic proof（The sum of the two sides of a triangle is greater than the third side）

It can also be proved by looking at the scoreboard.

Can't agree more :)

Nice contest, but author of problem F should learn the difference between a circle and a square.

Problem C

Can anyone proof it to me.

https://codeforces.com/blog/entry/88497?#comment-769345

It's clear now

Thank you.

Less participation = -vve rating

As a tester of some past contests, did the testers who say "pretests are strong" really check the pretests? I sometimes checked the (sys.) test during testing, but I never see the tasks on Polygon(because almost always testers don't access or don't have READ rights on there) during testing and then I couldn't check which tests are included in the PT.

This is a problem.. It seems like testers nowadays are just spamming "Statements are short and pretests are strong" to get contribution. Logically not every round they test can be that perfect, yet I see these comments on every round.

waiting for rating change:)

messed up C due to overflow issues

.

Can anyone tell me why this doesn't work for problem B? I think my idea is correct.

There is something called "Time Complexity". Learn it first...

Ok, I don't know much about time complexity, but my code gives me wrong output, not TLE.

There are cases where the max of array is not constant such as case#4 (0 1 2) so you need to update your max_ hence the ans every round of k loop. In your code you keep it constant.

I was thinking about this, but can you help me how to do it?

this is the "fixed" version of your code. it runs but this will TLE anyway because you actually don't need to go through this k loop. you can check my contest submission or the editorial.

Thank you very much!

i think change

to

Thank you, but it gives me exactly same output as before. On first test case: 4 4 3 3 3

I checked your code and found two problems. First mex_ can be from 0 to n but your code check 0 to n — 1 change

to