Hello Codeforces!
CQXYM and I are glad to invite you to Codeforces Round 745 (Div. 1) and Codeforces Round 745 (Div. 2), which will be held on Sep/30/2021 13:15 (Moscow time). Note the unusual time of the round.
Each division will have 6 problems and 2 hours to solve them. All problems were written and prepared by CQXYM and me. The round will be rated for both divisions.
We would like to thank:
- Aleks5d for his excellent round coordination and many useful suggestions that greatly help with preparation.
- Monogon, voidmax, alexX512, ijxjdjd, wxhtzdy, nogokama, 16204, Hemose, Hinai_Paulette, H4ckOm, 4qqqq, FBKdaisuki, kassutta, Mo2men and pkpawan for testing the round and providing useful feedback.
- MikeMirzayanov for great platforms Codeforces and Polygon.
- And you, for participating!
This is our first round, and great efforts have been put into preparing this round. Were you to kindly participate in this round, we would be very grateful and hope you will enjoy it.
Good luck!
UPD: Here are the scoring distributions:
Div. 1: $$$500$$$ — $$$1000$$$ — $$$1750$$$ — $$$2000$$$ — $$$3500$$$ — $$$3500$$$
Div. 2: $$$500$$$ — $$$1000$$$ — $$$1500$$$ — $$$1750$$$ — $$$2500$$$ — $$$2750$$$
And we would also like to thank Um_nik for testing the round.
UPD2: Editorial is out!
UPD3:
Winners
Congratulations to the winners:
Div1:
Div2:
As a tester I would encourage you to participate in this round and assure you that the problems are interesting and good .
Update :- My review to testers
Thanks.
As a participant I will never trust your word again :D
As a Chinese participant I will never participate in Chinese rounds again :D
my face after reading problem A :
You should use a spoiler
it is surely the worst contest I have taken part in
You were wrong
Here is my review I gave to problem setters . Name is blurred due to security reasons .review
Then you shouldn't have said all all these things (your comment)in the first place !
your review is good.
problem B has too many situations and what makes it worse is K (Diameter) might be negative.
and presets weren't sufficient to show wrongs during contest time (intended).
wtf???forces
Here it is!
Mathforces?
Mathforces!
It was FSTforces!
Sorry but I don't understand.
Just asking
I think he might be implying that cf contests prepared by Chinese authors tend to have more math problems. :)
20k+ users registered and only around 3500 submissions, weird...
They didnt have noticed the unusual time
You were so ******* right my guy. Damn
Hardforces!
Another Chinese round and friendly time for Chinese! Looking forward to great problems and having fun!
Sad, why isn't Monogon a VIP tester :(
I'm only moderately important this time.
What? 1-gon's comment didn't hit 100 upvotes in a day?? Impossible!)
This time is good for me.
Orz CQXYM!
Orz,2000 scores for a round!
Orz purple Hearth
Orz zqs!
I hope.
I don't have anything against Chinese rounds, but... really, morning on a weekday?
Probably people in different time zones exist.
That's why Saturdays exist.
ICPC World Finals I guess.
In China it is 6:05p.m. on the day before the National Day holiday. You can imagine the experience of participating a CF round after school :)
But (maybe?) most middle school students have to stay at school and study at night.
It often happens on weekdays. But if tomorrow is holiday, students can go back home in the afternoon.
I think that it's because the National Day holiday in China is coming.
However authors don't know 6 p.m. is even worse than 10 p.m. for many Chinese because most of schools end at 9 :(
However we must take the fact that ICPC World Final will be held today into consideration. Thus we choose 6 p.m. over 9 p.m.
Rare time....but maybe just right for Chinese contestants like me. I'm a little EXCITED. Hope all of us can enjoy it!
I finnish school 5 minutes before the round, can't join, but I wish those who read the greatest of luck.
I managed to convince the teacher to let me compete at school.
Big thanks for the time of the round from Asia! :)
Great, 3:05 AM.
i guess that's an i for an i for us having to participate from 10:30pm to almost 1am before...
Phantasmagorias I think we need to update this blog to make this appear on top of the div.3 contest. Many people might not see it. Especially the unusual start time of the round. I didn't see this blog existed. Because the first post from the top was round 744. thanks for considering. :) good luck to the people who are participating.
.
Although no, it's not funny. You idiot.
You High on Weed . :) __ | | | | | | | | ( | | | | )
Hope you Understand my feeling.
That's not a good idea to play jokes like this on the Internet. We can sure that there was something wrong with the education you get from your parents.
Although no, it's not funny. You idiot.
Although no, it's not funny. You idiot.
Although no, it's not funny. You idiot.
let me stop you guys :)
Imagine having school online, and not offline
Now that the contest is over, are you happy with your decision?
i guess yes as he got a very good rank in this one
I have a test in my class today that completely overlaps with the contest timings. Cant skip them even if I want to
:sobb:
Another great round!! Hope I can do better than before in it.
Hi there. Could the contest be at 18.35?
The round authors are Chinese; I assume starting at 22:35 is not comfortable to them.
Oh, I am sorry I didn't think that. Thanks.
Hoping to become Pupil after this round...Best of Luck everyone!!
you did it! Congrats :D
Thanks a lot!!!!
As a Chinese,I usually leave school at 6:30 PM,so I cannot enjoy this round :(
Leave school, but don't leave a contest. (Don't take it seriously.)
And now I think I am lucky :D.
Can you somehow postpone the competition for 2 — 3 hours? I think it would be more convenient to write everything in the evening than in the afternoon. usually the competitions are held at 17:30 Moscow time and this is very convenient.
Is 1900-2100 no longer in Div 2?
People rated <2100 can participate in div2, when it is div2Only round(so no parallel div1), but when there is a parallel div1 round, 1900+ have to participate there. It has been like this for a long time
Oh! I see
A way to increase contribution from 0 to 100:
0-20: leave stupid comments
20-40: write a blog on how to increase contribution from 0 to 20
40-60: write a blog on how to increase contribution from 20 to 40
60-80: write a blog on how to increase contribution from 40 to 60
80-100: write a blog on how to increase contribution from 60 to 80
A way to increase contribution from 0 to 126:
0-126: leave stupid comments
that leads in either ways UP or DOWN (+126 || -126)
Big thanks for the friendly time of the round:)
Actually, 9 p.m. is better than 6 p.m. for Chinese students.
My first div2 round here. Hopefully it doesnt go downhill. fingers crossed. I like this time slot :)
This ages very well
From contest authors and testers, I'm waiting really good contest!
Idk what's up with everyone, so retracting it for the better!
I think the content you posted is not appropriate, because we all know that the Great Firewall in Mainland China blocks almost all "user-generated content" websites, and codeforces as a "user-generated content website" is not blocked by it because There is very little political content on codeforces. If you post more content in this area, the result is likely to be that codeforces is blocked by the Great Firewall of China. As a result, it will cause great inconvenience to users in mainland China. They need to cross the Great Firewall to access codeforces.
Good.
However, according to some netizens in the QQ group Universal OJ Users, almost everyone in China mainland who use Codeforces can across the Great Firewall
But anyway, I don't think any political content posted on codeforces is appropirate because codeforces is a website for people who love programming. I think whether it will make codeforces blocked or not it's not good to post content which is not about algorithms on codeforces
Although I heard some netizens say, and I also think whether Codeforces is blocked or not doesn't matter to many codeforces users, I think after all using websites not blocked is more convenient.
Lol, this is pretty good, don't know why it's being downvoted
can anyone tell me what is the process of penalty in DIV 2 round
Technically, there are no 'penalties' on time count. But yes, you get less points if you solve a problem after incorrect attempts.
Each problem is assigned points (e.g., Problem A in this round is 500), the faster you solve a problem, the more points you get. The amount of points you can achieve gradually decreases over time.
Problem A points deteriorates at a rate of 2 points per minute (you get 498 points if you solve after 1 minute, 496 points if you solve after 2 minutes, 494 points if you solve after 3 minutes and so on)
Plus, every incorrect attempt costs you 50 points.
Now let's see, if you solve that problem A after 10 minutes on your 2nd attempt, you get
(500) - (10*2) for 10 minutes to solve - (50) for one incorrect attempt = 430.If you solve that problem A after 15 minutes on your 3rd attempt, you get
(500) - (15*2) for 15 minutes to solve - (50*2) for two incorrect attempt = 370.Additionally, the minimum points you get on any problem is lower bounded by 30% of the max points on that particular problem. e.g., irrelevant of how many incorrect attempts or how much time, you will get a minimum of
500 * 30% = 150for that problem A.Round delayed by 10 minutes
deleted...
Time added 10 minutes!
contest extended by 10 min!
does it change anything?
Why this sudden addition of ten minutes in starting the round??
To reach 20K+ participations maybe.
Hopefully the delay doesn't signify an unrated round.
now what to do with 10 min of my life??
Think about what to do with 10 min of life
Listening to your words is just like listening to your words lol
Listen to your words, so I heard your words
how many point i need to join to div 1 ???
1900 and above
1900
I hope it'll be the only problem of the contest.
Wish all the best!!!
P.S. The "problem" here doesn't mean the problems that we're going to solve in the contest.
This doesn't really matter, but why would you use "obsidian" instead of "black"?
Minecraft
Because author wanted to complicate the problem statement without any reason.
If you can't write a nice problemset for cf, just move the problemset to other websites.
Terrible contest.
edit: Chinese slang
?______?
It means he hates the idea provider due to hard problems, but it may be Chinese slang that is very disruptive and not appropriate to show in public.
First time not being able to solve (or not wanting to implement such an ugly solution for) even the first problem. Might just be me, but I think this contest needed a lot more tuning.
Thought it was one of best contests I competed in, besides B and C out of order. Finally some actual implementation and more ideas than just greedy/math with one for loop required...
If you can't solve a 30 line solution due to implementation I'm not sure you should be in div1. I've long said there is severe lack of implementation skill for most low div1 users and below.
Hardforces (: (at least for me )
Coding is very hard
div.1 too hard for me :(
Why would you write statement like this:
"and the diameter of the graph must be less than $$$k-1$$$?"
Couldn't you write "diameter of the graph must be less than $$$k$$$"?
Or even better, "diameter of the graph must be less than or equal to $$$k$$$?
For real, my code was so messed up I had to rewrite because I got confused in the middle.
I'm sorry but i am happy because today it's not only me who can't solve a single problem.
How to unregister?
Not my cup of tea...
Damn bro you didn't have to kill us with those problem statements!
Man!! explain me the problems. Cant understand anything from the statements
its my 2nd chinese round and i find it harder the usual div2 coz i can solve avg 2 problem but in this round i can't do even A. Chinese loves maths too much ≧ ﹏ ≦
only 3.7k submissions to problem A, btw WAonpretest2forces :(
Getting demoralized when not able to solve any problem.
then coming to comments-->turns out to be a natural therapy
Not even able to solve one quest, very tough round.
but atleast doing no submission won't affect my rating ^-^
The wording of some of the problems are hard to understand :(
In the last 12 rounds held over the past 3-4 months, this is the only round where I had to spend ~10 minutes on both A and B just to be able to interpret properly what the question wants to say, never mind attempt to solve them. Statements were weirdly put, and maybe difficulty wasn't properly calibrated going by the meagre amount of submissions ~4k for div2 A even after an hour.
one silly mistake costed 4 penalties on problem B :(
Same :(
Thanks a lot for clarifying this, before the round I was worried that my solution might also be tested on a negative number of test cases.
I am even not able to solve B this time, my brain exploded thinking of it :(
I am not even able to solve A :/
A flaw in the rating system exposed brutally by this round: in div 1, those who do manage to solve A, but do so late and/or with penalty attempts, will see their rating fall significantly, while those who bail out completely on solving A will see no change to their rating.
that's why A should always be comparatively easy in both divisions, can you imagine less than 6k participation in div2 :(
100% agree
Click this
Damn, That's the hardest A and C I've ever seen
For me B Totaly was pain in ass!
HardForces :(
UPD: Sad, Got FST on Div.2 B :(
It`s really good, but is so hard:(
20000 registers 6000 contestants hmmm makes sense
Didn't participate in a Div. 0 before, but today i did.
Is O(N^3) the intended solution for Div1 A or are you expecting O(N^2)?
$$$O(N^3)$$$.
Amazing. I had an O(N^3) time with O(N^3) space initially which got MLE.
Optimized it to O(N^3) time with O(N^2) memory and still got TLE.
Tried optimizing it further and got WA.
What a waste of 2 hours for me.
I solved it O(N^3 * 16) it gave me TLE but i break when i find solution with value greater than 16 and passed the pretests
Atleast you were writing codes. I was just crying for 2 hrs T_T
Yes, my O(n^3) passed
I think O(n^3)
Are the CF div2 rounds are getting harder or me getting dumber?
You just happened to be in the wrong contest at the wrong time
A contest with really high difficulty...
So is there anyone tell me how to solve Div.1A/Div.2C. I got WA on pretest 18 :(
You can discover the property that the answer will not be more than 16. and just do it in brute force. it will be O(n * m * 16 * 16). Hint : try to use the number larger than 16, maybe 17.
I found that but failed to have further a further thought. Could you please tell me what the (16 * 16) is?
Any O(nlogn) solution for Div1C/Div2E? I only came up with sqrt decomposition XD
wasn't time limit a little too tight for sqrt decomp? :(
O(nsqrt(nlogn)) should work for n=100k.
Example: 1540D - Inverse Inversions
It has 5 sec limit. 1 sec limit kept me under the impression that nsqrt(n) is being punished here. anyways my fault :(
Not in this task, not with 1 second time limit, IMO. 1.5 seconds would probably be enough.
I passed system tests with $$$O(n\cdot \sqrt{n}\cdot \log{n})$$$.
I used a fenwick tree for big values of $$$x[i]+y[i]$$$ instead of just maintaining when to add or subtract something.
Weak systests :/ But I see that this solution was already hacked after contest. My solution with FT was a bit too slow too.
I can only come up with $$$O(n\sqrt{n})$$$ solution too.
The TL may be a bit tight for O(N sqrt(N))
In fact the constant of the solution is very small. I think even $$$O(n\sqrt{n}log_n)$$$ solution can also pass.
And O(n^5) can pass B with some optimization...
$$$O(n^5)$$$ is the expected complexity.
In fact you will find the constant is extreme small. A lot of state is useless.
That's really unexpected.. I thought author may have come up with something smarter to get $$$O(n^4)$$$ solution and set $$$n=100$$$ to cut $$$O(n^5)$$$ one..
In fact there do have a solution works in $$$O(n^3log^2n)$$$, but it's too difficult for D1B.
Well, that's surprising. Why didn't you set $$$n \leq 50$$$? It would be much easier to understand that the solution is fast enough and it's worth coding. Was it that important to fail $$$O(n^6)$$$ solution?
Um_nik played these games.
Sorry but I didn't come up with any $$$O(n^6)$$$ solution.
Then my question remains: why $$$n \leq 100$$$? Only to make some people not implement a correct solution?
"In fact the constant of the solution is very small." -Phantasmagorias
I understand that the solution is fast enough. It would still be with $$$n \leq 50$$$. And this fact would be clear before implementing.
Just because the $$$O(n^5)$$$ solution can pass the task and it runs very fast. In fact you will find that the answer will be 0 if $$$m,k$$$ is larger than $$$\frac{n}{2}$$$.
Not always, my $$$O(n^5)$$$ solution got TL.
Are u serious? Only because of this, I didn't code the solution and thought for an O(n^4) solution for almost 1.5 hrs. And the constant isn't that small, recursion won't pass easily without optimizations (returning wherever you can).
However my O(N sqrt(N)) solution didn't pass.
I failed on both BC for TL
I think your solution for C is slow because of "cin and cout" .
I'm sorry but as a matter of fact,coder's submission has got another TL after replace cin/cout with faster IO streams. Isn't it obvious that this failure it due to implementing or Time Limit but not IO ways?
You are right..my bad
O(nsqrt(n)), you can operate trains whose x+y>=sqrt(n) and whose x+y<sqrt(n) in different way.
Perfect! Problems are challengable and of benifits. especially C, it`s great on both Thinking and Realization. All in all, great round.
Ooh man! one of the toughest A in div 2!!!
maybe you can just look at sample test cases to find the pattern without thinking much or proving anything . At least that's what I did XD
I figured it out with PnC but still I would say it shouldn't have had been an A problem, maybe B or B1 if possible!
So difficult.
UPD : also fstforces.
Why is the diameter of the graph given as <=k-1, why not <=k ?
It's < k-1, My guess is just to add more confusion.
it's like let's make the contest very hard and confusing for the people who decided to participate at 3 AM as they already have a brain fog at this time :D
Am I the only one who has a bad feeling about system tests?
Well at least the judge queue will be short lol
Please, if you will write a round again in the future think about the quality of statements twice before. Some things were not very clear, like in div2 B where you should have specified strictly less, as you did while the round was going. Also, talking about B, it wouldn't be a bad idea to also write in bold important words, not just numbers and variables(here I'm talking about the word CONNECTED); you can see in other past problems that this is sometimes done, especially for important conditions like the connectivity one in B(for example for saying that all the numbers in an array need to be different). One last thing about this problem would be the limits for k, like were those special cases of k <= 1 really that important to be placed in the tests, ore are they even in the tests?
I am really confused why the diameter of the graph must be strictly less than k−1. other than k, it causes unnecessary confusion.
hardforce
I feel like today's contest is suitable for Div 1.5 and Div 0.5
The round had nice problems (I read only div1A-C tho), though ridiculously hard. Could've easily been split into two separate div1 rounds with some easier, filler problems. I want to express my deepest sympathy to all div2 participants, who had those super tough problems given as C-E. Div1A should've been div2E IMO. To the authors, you did very good job with inventing interesting problems. Only that you underestimated their hardness and the overall contest turned out too hard.
That is the job of coordinators and testers but I don't know how can they all agree to serve all these problems together.
Some solutions for Div2 B will fail for this test: n = 1, m = 0, k = 1
is it NO ?
Yeah, some solutions didn't consider that the given diameter could be negative as well.
In first sample he told u diameter of the graph is zero, its obv. not strictly less than 0
Although I honestly don't see why the authors would want to make k start from 0 and not 1.
I felt,The definition of diameter is valid only when there are atleast 2 vertices. Is this assumption wrong?
Honestly how to iterate on this map in c++ for Div1 C ? map[(xi + yi)][day % (xi + yi)][xi].push_back({st_j, end_j}) Plus is it really the solution ?
Just use map.insert.
The solution is about O(n sqrt m)
Maybe the pretest is too weak?
system testing is super fast today!
Despite so many testers spanning across different ratings, a pretty unbalanced round.
Saw a pattern in the ans of A for n = 2 , half of the total-permutations and it actually worked for larger n
You could bold "connected" in B :/
k = 0 on problem B. Damn you guys got me!!!
Awesome difficulty Forces+FST Forces = Codeforces Round #745
Did someone mistakenly swap B and C?
Before the contest I thought B is easier but it turns out that I was wrong.
Who actually proved their solutions for A lol
I thought of something like, if a permutation is valid, it's reverse is always invalid. So half would be valid and half invalid
Proof by symmetry is valid, and very simple.
I think you can make a one-to-one mapping from valid permutations to invalid ones; let's say you want to count permutations which have exactly $$$n$$$ indices such that $$$p[i] \lt p[i+1]$$$; it's not hard to show that by negating all numbers and then adding $$$2n$$$ to all of them, you are going to have a permutation with exactly $$$n$$$ indices such that $$$p[i] \gt p[i+1]$$$ (note that we don't have any repeated numbers, so this condition is the exact opposite of the one we consider valid.) Let's assume $$$\Pi(i)$$$ is the count of all permutations that have $$$i$$$ consecutive increasing pairs, and $$$\neg\Pi(i)$$$ the same for decreasing pairs; we know a pair is either increasing or decreasing; so we have: $$$\Pi(i)=\neg\Pi(2n-1-i)$$$; so we have: $$$\neg\Pi(i)=\Pi(2n-1-i)$$$; therefore, we have $$$\sum_{i=0}^{n-1}\Pi(i)=\sum_{i=n}^{2n-i}\Pi(i)$$$; I think by now you can see why $$$\frac{n!}{2}$$$ works in this problem.
What is test 4 for problem Div2B ?
I came up with a (n^2)m solution for Div1 A but it gave TLE. Is there a more efficient solution?
I also have this exact complexity and it passed. I used only vectors and pairs, maybe you used some more complex data structures?
Hmm. Seems like the constraints were tight. I see many solutions giving TLE in the system test.
Agreed, If the intended solution was O(n*n*m) it's just sad.
yep. FST for $$$O(n^2 m)$$$
130355128
update: constant optimization (using tuple instead of vector) passed with 139ms.
130386010
But it actually feels good to see a tough round now and then.
How to solve Problem C ?
hardforces + clumsyforces + (1.5 + 0.5 ) forces +system test fail forces + ....
so, bad round. lol
It is so reassuring to see that the problems were hard for everyone! Nevertheless, they were also really engaging.
important rule : don't join chinese contest LOL
FSTForces :(
shitty contest
I think the pretest in Div 2 B could have been a little better. Other than that, contest had really interesting problems, except C. I didn't like C, as the only solution that I thought of was implementation heavy, so I simply skipped it and moved ahead. E was good. Just missed the window for submission.
Annoucement-cum-Hardforces
Is it possible to solve Div1B faster than $$$O(n^5)$$$?
With 2D FFT we can solve it in $$$O(n^3log^2n)$$$.
I wondered if it's possible during the round. Nice :|
How? Don't you need division for that?
let dp[ k ][ i ][ j ] denote number of sequences of length i that have exactly j elements with k maximums (e.g. as in the number of maximums in the process defined in statement, e.g. the variable m). Then dp[ k ][ i ][ j ] is the sum over all i1 , i2, j1 , j2 , such that i1 + i2 == i-1 , j1 + j2 == j of ((i-1 choose i1) * dp[k-1][i1][j1] ) * dp[k-1][i2][j2]. This thing can be done with 2d FFT. E.g. we are fixing the position of the maximum element.
Edit: right, the modulo isnt guaranteed to be prime, not too sure.
No, its wrong.
1) it should be i1+i2=i-1, though that's pretty easy to handle
2) you forgot an (i-1 choose i1) factor, which requires dividing by i! before the convolution.
ohh right
I am not too sure i understand part 2). Do you mean calculation of (i-1 choose i1) mod p needs division ? If so, we can pre calculate that with pascal's triangle equality.
Edit : I see, makes sense.
No; you can divide by i! before convolution so you just do m convolutions of size nk; unless I'm missing something, directly calculating (i-1 choose i1) requires you to do mn^2 convolutions of size k, which is slower than the complexity interlude mentioned.
How to solve A ?? Intution ??
Most people just deduced it from the given sample cases
why ?? i need intution not just the plain answer :_:
You can find this rule by directly calculating the result when $$$n=3,4,5$$$
It's in fact $$$\frac{(2n)!}{2}$$$. For any permutation with $$$k(k \gt =n)$$$ such $$$i$$$, you can reverse the permutation and get another unique one with $$$2n-1-k(2n-1-k \lt n)$$$ such $$$i$$$. So answer should be half of number of permutations.
the answer is (2*n)!/2, because each permutation whose the number of pi < pi+1 is no less than n corresponds to another permutation whose the number of pi < pi+1 is less than n after you reverse this permutation
Tight ML in div2 C ;(
Sometimes I miss the old times when this assumption at least sometimes was incorrect.
The data of this task is actually difficult to make, and I can't come up with all the wrong solution too. Maybe we could communicate more about this task.
maroonrk what's incorrect in your code?
My solutions is like this:
The essential part of this problem is to make in-degrees of all vertices except the root no less than 2. For each vertex $$$i$$$, we know to which vertex $$$j$$$ we can span an edge. Valid $$$j$$$'s form a set of ranges if we order vertices by their distances from the root, and the number of ranges is at most the degree of the vertex $$$i$$$. My solution consumes O(number of ranges of the queried vertex) time, so it's a $$$O(NQ)$$$ time algorithm. However, it passed somehow.
Well, doesn't look like "we can't come up with all the wrong solution".
I didn't enjoy this round very much,not only because I got a negative delta.
(For div2) The problems are way harder than a typical Div2 round,Problem ABC are not very suitable for their place and Problem DEF are so hard that very few people can solve them during the 2h contest.
Also the statement is kinda hard to understand(maybe it's my fault for not reading the statement that carefully but).You have to read it mutiple times to find certain words like "connected" or figure out what the problem is asking.
Contests are used to seperate contestants into two parts:Those who are very good at Problem Solving and those who are not that good.They are NOT used to find out who are better at English reading and typing.
upd: obviously the pretests are weak too but I blame myself
one of the worst contests with worst pretests with hard problems
at all, you won the worst contest
true!
I agree with you.
Every Time I think I am now good enough to become CM , My Rating Starts Falling Until it reaches Specialist Range
Sincerely, this round is not a good round. I have taken nearly 40 contests, and this contest is the first contest that I downvote.
1, the problem are so unbalanced, for div1, so many people are not able to solve A, and for div2, difficulty gap between B and C are too large, so as C and D.
2, div2 B is the worst problem I have ever met. It is so tricky, and description are soooooooo uncomfortable. What is "less than k-1" mean? Could you just write less than k" or even better "no greater than k". "n=1" is meaningless because we cannot pick two points from n=1, and "k=0 or k=1" is even more meaningless because, who want to discuss a graph with negative k?
3, div2 A are so unfriendly to newbies.
4, weak pretests
5, Except for the balance and description, a good contest should be educated. Even I cannot solve the problem during the contest and get negative delta, it doesn't matter. We can learn something after the contest, and improve our ways of thinking or coding skills.
But in this contest, I just want to say, what the hell of this ?? just a bunch of mind tricks or brain teaser? Can I learn something in the contest?
In all, getting up at 5:30 AM and take part in such contest is the most wasteful thing I have ever done.
systest is pretty weak too? I uphacked myself with a case with obvious 32-bit integer overflow...
130380959
I'm wondering if the pretest is too weak. I see many people got FST on 1C
In Div2 B, what is the expected answer for case: n = 1, m = 0 and any k. I guess according to definition of diameter given in the problem statement, diameter is not defined for n = 1 (we need at least 2 nodes to define diameter). I think the answer for this case is ambiguous and the problem should accept both "Yes" and "No" as the correct answers for the above-mentioned case.
totally argee
I think they made it quite clear the if n=1 the diameter is 0 by putting an example with n=1 and a note that says in this case the diameter is 0.
I agree but this thing should be mentioned in the problem statement itself, most people don't see the explanation of the sample test case.
Why condition in div2 B use less than k-1 instead of less than just k? It just creates some confusions.
something like
cin>>k--;
what is test 61 in DIV2 C/ DIV1 A ?
This contest has :
2.Super weak pretests.
Thanks for the undeserving downfall.
Why do you set n=100 for n^5?
I cannot agree more.
This contest feels harder than other div 2. It was really hard to understand the problems for me.
Nothing hurts more than "FAILED SYSTEM TESTING" utmost useless pretests after utmost annoying explanation!
THCE (The hardest contest ever)
FSTForces helped increase my delta from -3 to +30.
And it helped decrease my delta from +1 to -108. :)
And it helped decrease my delta from +40 to -45 :)
And it helped increase my delta from -10 to +3 :)
very weak testcases on Div2C, if you implement just some search with prefix sums beginning from every index and trying just a = 5 -> 20 and b = 4 -> 15 a x b matrices plus considering only cases when first and last rows of given matrix are only 11111... and others 100........001, it will be accepted :(
Allowing $$$p=1$$$ in B is just too evil, while it seems like there is no case where this really matters in the testdata.
As i said before with my other alt account named Akbar Covid you always make trouble and you caused to CF ban my other account ! This contest was a shity hard contest in the other words hackforces round 745 ! ! ! Please don't make more contests thanks :) And also again i have to notice that this contest was the worst contest in the year :)
You deserve it.
Your comment exhibits an acute amount of immaturity. Competitive programming is hard. Don't inculpate your own incompetence on the problem setters. The round was studied and tested by multiple people. You make it seem like they purposely decided to set a high threshold for the contestants.
amine woke up and chose violence
I only read the problems in div2 and found out that it's too hard for me. :( It took me a long time to solve A in div2 so I gave up submitting the code.
that's better thank solving A quick :D
As a Chinese, I think I'd better stay up late for a 22:30 contest in order to avoid a 18:30 Chinese Round next time.
Pretests in Div2 B be like: This
What is the intended time complexity for D1B? I could only come up with a O(n^5) solution.
The constraint $$$n\le 100$$$ in d1b is really frustrating. I like B, but don't like its constraints. Also I have no idea why C's point value is so great, as I think it's much easier than B.
I need a help in Question B my submission 130361322 . I am unable to figure out where my code fail. Thanks for the help
You should do long long z = ((long long)n)*((long long)(n-1)))/2; problem with long long conversion. because n*(n-1) convert into int only before giving value to z. better way declare n as long long.
when you run: n*(n-1)/2, n is a int, so it overflows. Then you assigned an overflowed result to int64, it brings wrong result.
Wow!!! Scored a lot, happy!
Orz.
The contest problems were good and tricky. However, there was no relevant difficulty jump between problems B and C.
I am literally crying （>﹏<）
Problem A didn't even have a N = 400, M = 400 max pretest, and I ended up with a MLE in system tests. I don't know the exact line of thought of the setters, but I'd personally feel that a max-bound test is among the most important things to put in pretests.
Even polygon will notify writers if test cases don't contain maximum data ranges. I really wonder how they make pretests for this contest lol
I went through a painful time during the contest. Looking forward to the next div2.
Goodbye Grandmaster,too sad,hahahahahahahaha
Odd contest, A and B were harder than usual, while C and D after them were much easier. I think I've seen a problem with the same idea as C before (square root decomposition on length of cycle), and D is just the standard $$$\mathcal{O}(n^2)$$$ tree merge with the tree hidden and a bit unusual cost function. I wish I had time to think about E since it looks more interesting, but I was too slow this time :(
At least I got to LGM somehow? I feel pretty bad getting there by just solving the easy problems slightly faster than others for many rounds in a row :/
How do you solve D? What do you mean by "$$$O(N^2)$$$ tree merge"?
So say you have a tree, and some random condition like "What is the maximum sum you can get by taking at most k nodes in the tree, such that no two of them are adjacent" to make the problem not completely trivial. Let's handle a binary tree since we have one in this problem, but it doesn't really matter.
You can do this in $$$\mathcal{O}(N^2)$$$ by computing for every subtree, for every amount of selected nodes in the subtree, the maximum sum (for both if you take and don't take the root of the subtree). Then, you can merge children in $$$\mathcal{O}(L \cdot R)$$$ time where $$$L$$$ is the size of the left subtree, and $$$R$$$ the size of the right subtree by looping how many you take from the left and right subtrees and taking / not taking the root node itself.
This runs in $$$\mathcal{O}(N^2)$$$ time since every node multiples the sizes of the other subtrees on the path from it to the root, and the sum of those sizes is at most $$$\mathcal{O}(N)$$$. I think this is a pretty standard technique.
In this problem, we take the minimum as root, then build the trees for both the left and right sides and take their roots as the left and right children of the root. To merge, we loop over the amounts $$$x$$$ and $$$y$$$ of nodes we take from the left and right subtree, and do
as those are the updates when we do not take and take the root of the current subtree (which is the minimum of its interval) to the set.
Thanks very much. If someone explains such way for every question than we noobs(as I think I'am) may learn a lot faster ... respect :D
Thanks! 1499 F also uses that same complexity analysis you're talking about, and the editorial has a very nice explanation of it. Unfortunately I forgot about it.
let dp[i][x]=maximum sum of pairwise distances between x number of nodes in the subtree rooted at node i,
any idea on how to merge left and right subtree in this case?
shared the same feeling while became expert:\
I still don't know the logic behind Div.2A. I only know that the answer is (2N)! / 2 (figured this from the inputs), have no idea how...
imagine if the problem was a[i]>a[i+1] instead of a[i]<a[i+1], the answer is still the same.
Great logic. Thanks
Thanks, understand it now.
This is the diameter of a graph according to the question. And for the case n == 1, its not clear at all what the diameter should be. So, I think there should be no constraints on k(which is not defined at all) and hence every k should be valid
Can you please look into this Phantasmagorias CQXYM and see if tests could be ran once again on B. Surely, this is much ambigous
In the case of n == 1, the diameter of the graph is 0 so any K <= 1 isn't valid.
DarkTigerFAF Its good to read the comment well before replying. In short, I want to say that the diameter is not defined at all when n == 1 and hence, there should be no constraints on something which is not defined or exists.
Oh sorry about that, thought u were just asking why it's an invalid answer. But yeah I agree with you.
The sentence you've copied well defines diameter for n=1 as 0. I don't see anything wrong with it.
[Deleted]
Yes I meant picking two same nodes. If the statement didn't specify that two nodes should be distinct, you don't add such condition. I do not believe the author is at fault for participants' inability to understand mathematical statements.
Diameter is maximum distance between any pair of nodes. When N==1, we cant form any pair.
Sorry but this problem is prepared by CQXYM. Maybe you shold ask him.
Thanks! Edited my comment to tag him too!
All problems were written and prepared by CQXYM and me
No I don't mean that. I would like to bear all the criticism or something like that. But I really can't answer his question so I suggest him to ask CQXYM.
Only 6 persons passed 2D in div2. I passed 2D but didn't pass 2C, then I'm rk 185. Is this really resonable???
because you take too much time and also 2 wrong submissions and since 1500 and 1750 are near so 2 wa makes it equal to those who submit it in one go.
20,000 thousands registered for contest but only 5600 participants solved A because many of the registered participants couldn't solve A and many participant's rating will be badly effected for this but they don't deserve it. I think A should be more easy to make sure the actual number of participants because the rating change depends on number of participants I guess.
I don't think this A is difficulter than it should be. It's very easy to find that the answer is A(n,n)/2.
very poor round, div 2 c is much harder than normal, D is near impossible for div 2 writing problem B is very confusing, why make it <k-1 why not <=k and just use a proper k? because the author wanted to confuse people to make it seem like it's a harder problem
Why "less than k-1"?Why not just "less than k" or "no more than k" or anything readable?
XD I understand.
Problems themselves were not bad, rather good. But too unkind to participants
Meaningless 2E/1C with the time limit of 1 sec.
All in all, I think this round is not good.
The description of div.2 B is not clear. Why not use "not less than k" rather than "less than k-1"?
I think the case that k<=2 in this problem is meaningless, although I used this to hack someone.
For div.2 D I couldn't find any solution under $$$O(n^3)$$$ or $$$O(n^3 \log n)$$$. In fact, I felt confused when I know that $$$O(n^5)$$$ is acceptable. How the $$$O(n^5)$$$ solutions works in $$$n=100$$$?
For div.2 E, I was troubled with the data range, too. The solution works in $$$O(n \sqrt n)$$$, but I think it is too close to the time limit so I gived up to solve this problem :( . And what's more, will players who use python language got $$$TLE$$$ because of this? (python runs slowly than other languages)
All in all, I don't like this round very much although I got a nice ranking.
And hope the next round will be better.
Sorry for my terrible English.
For div.2 E, I can't agree more. LOL
Just two days ago, I got TLE in a contest, because I used an O(n sqrt n) complexity algorithm (the solution is O(n log n)).
In div.2 E, I got a O(n sqrt n) complexity alogirhtm and of course I gave up to use it... What a pity!
It's normal to have $$$O(n^4)$$$ Solution when $$$n=100$$$.$$$O(n^3)$$$ is for $$$n=500$$$.And this $$$O(n^5)$$$ Solution have constant factor no more than 1/16(maybe have lower bound).I think it's important to analyze the constant factor.
Why the solution of B is $$$O(n^5) (n \le 100)$$$ and the solution of C is $$$O(n \sqrt n) (n \le 2 \times 10^5)$$$? I think many contestants may be afraid to submit their solutions.
For C it's completely normal for $$$O(n \sqrt{n})$$$ to pass, I'd be afraid to submit $$$O(n \sqrt{n} \log)$$$. Sqrt solutions are common stuff. The constraints in B are a shitty jebait where I only decided to try and constant-optimise based on a lot of experience with constant optimisation.
.
Oh no, 10^8 simple operations. It might run in more than 0.1 seconds on a modern CPU, the tragedy.
Can anyone tell me whats wrong in my solution to Div1 A. The link is https://mirror.codeforces.com/contest/1580/submission/130356049.
I tried stress testing it with a correct solution but couldn't find any failing test of reasonable size.
CodeFSTes Round #745
In div2B
testcase : 5 4 4 i think ans should be NO but everyone code is giving yes. could someone please explain ?
1-2 1-3 1-4 1-5
4 edges and the diameter is 2
K=4 Which means you can have the diameter be 2 You can achieve that by making a star graph Connect all nodes to node number 1 You will use only 4 edges so answer is yes
It's a terrible round for me :(
I still have a question!!!!
For Div1 D, note on sample 1.
No matter how I calculate the answer, subsequence [15,2,18,12] has the sum of $$$97$$$, so it's not "also a possible answer".
Can anyone help me?
Sorry. I make a mistake in the note. Subsequence [15,2,8,12] does have the sum 97. I'm very sorry for that.
I spent nearly 30 minutes reading the problem over and over again to check if I read the problem wrongly:(
I can't understand why the author doesn't directly say "diameter is less than or equal to $$$k$$$" but "diameter is strictly less than $$$k-1$$$" in Div.2B. It can only be said that it is purely used to torture people.
Can a strong public opinion make this round unrated?
To not keep you waiting, the ratings are updated preliminarily. In a few hours/days, I will remove cheaters and update the ratings again!
I think this round should be unrated for having so many problems including vital ones (like the wrong notes during the whole contest). The vote and comments of this blog also tell how contestants judge this contest.
I also think this round should be unrated. Unlike Div2 contests, problems were harder and the statements of some problems were confusing. In Div1, the situation was even worse, there were wrong ordering and weak pretests. Also some users reported that Div1A or Div2C constraints were misleading.
Overall, this was a badly authored contest. (We can see the effect on the announcement votes)
I'm missunderstanding the problem div2C/div1A The portal could be in any place? or its fixed on the (1,1)?
If it's fixed, where it's the portal of the second testacase?
If not, why it's say M_(1,1) can be of any type?
I also had this misunderstanding, too
There is a room for missunderstanding, it can be read "sub rectangle start with(1,1)
conditions for the portal part made us to understand "subrectangle start with point (1,1). many people understood this problem like me. please consider this issue.
Never Have I ever Seen so many Downvotes on an Announcement of a Round
My screencast https://youtu.be/E6RXjEaf3W4 (solved A, C, then tried F and B).
I like B because for a long time I couldn't solve it in polynomial time. And I appreciate the non-trivial-to-understand time complexity in D. I didn't like constraints in C and A (would be easier to test with $$$a, b \leq 3$$$ instead of staying true to Minecraft).
F smelled like Berlekamp-Massey with FFT. The coefficients are indeed very easy to guess and compute (some binomial coefficients). Sadly, I couldn't solve the problem fast enough for $$$m \leq 50\,000$$$.
I managed to solve it without even knowing Berlekamp-Massey, with OEIS and matrix analysis.
I am so happy because today I became pupil. it really feels good 😭 .
Any way, good luck for the next contest for every one.
Div. 1 A told us how the remaining problems look like:
The Nether.
I have proof for div2-A. First let's introduce some definitions:
For a given length of permutations ($$$2n$$$ for the problem), Let $$$X[i]$$$ be the number of permutations having exactly $$$i$$$ indices $$$j$$$ such that $$$P_j \lt P_{j+1}$$$
And $$$Y[i]$$$ be the number of permutations having exactly $$$i$$$ indices $$$j$$$ such that $$$P_j \gt P_{j+1}$$$
First observation: for any $$$0 \leqslant i \leqslant 2n - 1$$$ we have $$$X[i] = Y[2n-i-1]$$$ (pretty obvious if you have $$$i$$$ indices lesser than the next element then the remaining indices will be greater than their next element)
Second observation: for any $$$0 \leqslant i \leqslant 2n - 1$$$ we have $$$X[i] = Y[i]$$$ (why ? if you reverse any permutation having $$$i$$$ indices lesser than their next element you will get a permutation having $$$i$$$ indices greater than their next element)
Therefore we have $$$X[i] = X[2n - i - 1]$$$ for all $$$i$$$. Hence $$$ \sum_{i=0}^{2n - 1} X[i] = 2 * \sum_{i=n}^{2n - 1} X[i] $$$
The left part is the total number of permutaions which is $$$(2n)!$$$ and our answer is $$$\sum_{i=n}^{2n - 1} X[i] $$$, so $$$ANS = (2n)! / 2$$$
i did the same thing and got tle in pretest 1.130344427
I'm not so sure why this happens but I think calculating the factorial without MOD is consuming a lot of time, since the number is very very large.
Based and relatively easy to understand.
Thanks
Can anyone hack this submission (or prove that it actually works)?
Submission: 130359855
Basically, the code above fixes two columns, and then greedily expands the rows one by one. It breaks whenever a worse answer is found. But even if the current answer is worse, maybe the future answer is actually better and hence this greedy approach fails? Please help.
In D1E, what is the intention of such a large constraint on $$$x_i$$$? It's really useless and forces us to use unsigned 64bit integers, which cost me 2 incorrect attempts.
I was wondering why CQXYM didn't post this blog. Now I know...
I was about to upvote this announcement earlier today because it looked short and clean, but luckily something happened with my brain and decided to wait and decide whether to upvote or downvote after the contest.
Ironically enough, I misclicked when trying to hit the red button, and upvoted it by mistake :(
Though I didn't participate in the contest, those downvotes show me the deep anger of the participants, considering the fact that competitive programmers usually upvote contest announcements because at least they know that it is very demanding to prepare a contest.
Sorry, not a fan of this contest tbh:
My totally legit solution to A in Java ($$$n^3$$$, low overhead) TLEs. I have no idea why there couldn't have just been a 3s timelimit on this problem, there's no way an $$$n^4$$$ will finish in 3s. It took me about 15% of the contest to do what should be unnecesssary speedups and convert this to C++ to get it to pass, when I'm basically 100% sure my solution is what authors intended. This is an indication of both bad time limit intuition and not having enough testers in other languages; having either of those things could have addressed this.
I was legit paranoid that C wouldn't be possible in java, since it had a sqrt factor on a 1s timelimit on $$$n = 2*10^5$$$. Again, I see no good reason why this time limit couldn't have been either 3s, or at the very least made $$$n = 10^5$$$ if you really need a 1s time limit.
For B, the author says: "$$$O(n^5)$$$ is the expected complexity." Seriously??? Why not just make bounds equal to at the most $$$40$$$ then? In contest I suspected I would have trouble with an $$$n^4$$$ even, because that's already $$$10^8$$$. $$$n^5$$$ with bounds of $$$100$$$ is way too big to be reasonable here imo.
I've heard people say that D was interesting, but I didn't even come close to getting a chance to work on it.
Finally, I had witnessed SecondThread solving a problem in C++ ( *_* ) !!!
Let me add some more points to this.
In problem A, I would emphasize in bold that the rules in this problem are not like they are in actual minecraft, as I suppose in minecraft only $$$5\times 4$$$ rectangles are portals. Usually something like this is written in problems based on some well-known game.
Pretests in A would be much worse without 17-th and 18-th, but it would be perfect if they were added before the contest.
As there is still no editorial, I wonder which cutoffs authors used in B. I guess I was quite optimal in terms of memory leaps, and I also used Barrett reduction, so it was enough for me to skip states with $$$n \lt k + m - 1$$$ (without it my solutions worked for 9s, with this it became 1.2s).
Actually I do not feel anything wrong with $$$n\sqrt{n}$$$ in C. I think it was quite obvious that this must pass. My solution worked about 300ms in custom invocation, and I didn't even adjust my block size. However, I don't know about other languages and $$$n\sqrt{n}\log{n}$$$. I still believe it should be fine if $$$\log{n}$$$ is from Fenwick.
C < B.
I have not read problem F till the end, but when I did, it reminded me some problem from project euler. I do not claim anything, but I wouldn't be surprised if it turned out to be a problem from pe.
FYI, in recent versions of minecraft, you can indeed make portals of larger sizes than 5x4 (you can go up to 23 by 23, but nobody really knows that rule. It was only in old versions that only exactly 5x4 worked.
Yes, problem F seems to be a harder version of this. Looking at the comments there may probably help (there is a cute characteristic polynomial in the second comment of the thread).
Golovanov399 Hey, how is n.root(n) without any pruning supposed to pass? Can you clarify a bit? I am a little bit confused, as I remember the problem (https://mirror.codeforces.com/contest/1561/problem/D1), in this n.root(n) solution, was desired, but many test cases are taking time > 1s, hence accordingly time limit of 6s was given.
Let $$$K$$$ be a fixed constant. Divide all trains into big (with $$$x + y \geq K$$$) and small (all others). We will maintain the following:
for all $$$s \lt K$$$ and $$$r \lt s$$$ we store the number of (small) trains with $$$x + y = s$$$ which are going to be in maintenance on any day which is $$$r$$$ modulo $$$s$$$;
an array $$$p$$$ of size $$$m$$$ so that for every day $$$t$$$ we have $$$p_1 + p_2 + \ldots + p_t$$$ is the number of big trains in maintenance on day $$$t$$$;
current prefix sum of $$$p$$$ till today.
When we add/remove a small train, we need to update at most $$$x + y \lt K$$$ values. When we add/remove a big train, we need to update at most $$$m/(x+y)\cdot 2\leq 2m/K$$$ values. To compute an answer, we need to add $$$K$$$ numbers for all small sums to the current prefix sum of $$$p$$$. Therefore if $$$K\approx\sqrt{2m}\approx 650$$$ then this should be good.
In my solution $$$K$$$ was $$$300$$$, and it still was fast enough.
Got it! Thank you for the nice explanation.
.
Downvoted because of "A few pretests were added, submissions were rejudged"
Why are people so afraid of hacks today and try to prevent them even by rejudge? Ridiculous.
I heard an opinion that the site will lose some traffic otherwise, but think who it will lose? I can't believe Mike can bend to satisfy this potentially lost traffic so he gives coordinators orders to prevent hacks.
Please, don't downvote me, but I think that hacks doesn't make a contest more interesting.
For example, I have participated in Technocup 2020. I have solved tasks A-E, and was about to solve G. Imagine my surprise when my solution failed system tests...
Well, you can say "lol, just wrong solution". I haven't used double hashes, so I thought at first that I had got WA just for this reason. But imagine my shock when I had found out, that I have written SQRT-decomposition wrong, and this solution worked in O(N^2)!!!
($$$N \lt = 2 * 10^5$$$)
Are you serious? One could pass pretests using just a bruteforce and one if statement to speed this up.
I think, this is not normal.
But, what about hacks now? Hacks are the same thing. I am here to solve tasks. I think that we need pretests as a 99.9% guarantee, that your solution is right. And running full system tests is just impossible during the contest because there are too much tests and there will be very big queue.
I think that wee need hacks only to make sure, that there aren't any wrong solutions that are accepted after system tests, because every successful hack is going straight to the system tests.
As a conclusion: yes, I am one of those persons who are afraid of hacks and weak pretests.
Agreed.
So many grammar mistakes:
Thanks for quick system testing and rating updates, even if I had a huge negative delta haha. I actually liked these tricky problems that are harder than usual (maybe unpopular opinion?). The only relevant problem with this round was a bad pretests set for some tasks/problems/questions/bugaboos.
Painforces
Div.2A is quite hard since you should use some mathematical knowledge to solve it.
How to solve it?
ans is ((2*n)!)/2 it want pi<pi+1 no less than n which is also no more than n pi>pi+1. intersection of both of these is null. total cases are (2*n)!. so and is ((2*n)!)/2
hey i tried this solution however, it gives wrong answers any help? https://ideone.com/z6ACC0
(2*n)! = (2*n)*((2*n-1)!)
(2*n)!/2 = n*((2*n-1)!)
try doing this way here is my solution 130384244 pls upvote if this helped
The answer is $$$\frac{(2n)!}{2}$$$.You can try to reverse $$$p$$$ and count the number of $$$i$$$ satisfying $$$p_i \lt p_{i+1}$$$.
It's so ridiculous that my $$$a=5,b=4$$$ solution to div.1 A passed the previous pretests. The pretests are likely to be generated by the problem setter's feet I guess. I will think twice before participating in a Chinese round again (last time when I participated in a Chinese round I said so. My rating went down in every Chinese round before lol).
i wish i knew some ordure in Chinese to comment here
I don't want to be rude or anything like that but I just wanted to know what is the job of a coordinator?.
All in all,it's interesting but really hard.
Congratulations mango_lassi for LGM.
$$$O(n^4)$$$ can pass div1 A and $$$O(n^5)$$$ can pass div1 B!
My O(n^3) solution runs in 998 ms and you are saying O(n^4) passes.
my pruned brute-force solution is probably still O(n^4), but runs in 44ms
can you share O(n^4) of div 1A
Here's my $$$\mathcal O(N^4)$$$ solution.
https://mirror.codeforces.com/contest/1580/submission/130354161
.
Why no stress tests in pretests on Div1C? I belive, that there should be some tests that at least will break some popular SQRTs, like 500.... In task which main solution I belive is SQRT decomposition. It's pretty disapointing, when you change root after contest, and it works.
To sum up my feelings about the contest in the nicest way I can, let's just say I didn't have fun today.
Make this contest unrated!
Why? Any reasons to do that?
Come on man, problem A is not of 800 Rating.
It was an interesting contest. It will be really fun to upsolve this one. :)
i understand and agree on the fact that contest was too tough than it should be,but still the kind of comments that we are posting is not at all healthy from a community point of view,so lets give feedback's but in a constructive way as everyone is a human here,so we can at least give one more chance to the author's.Hope u understand my point:)
Nice Contest!
It would be great if they can provide the editorial very soon.
I don't understand the point of having so many testers!
Did none of them mention issues like too hard Div2 A, unbalanced problemset, $$$T \geq 1$$$ constraint xD, unnecessarily strict TLs, negative diameter,
< K - 1condition, unintelligible grammar etc...I'm sure they do provide feedback, as I can see from pkpawan's comment at top. In that case why are their suggestions not heeded?
Unintelligent grammar for sure,
U forgot the fact that they are chineese
Yeah, but the testers are not
just curiously asking..
last education round 113, my rank was about 2303 and got +63 (1325-1388)
today in this contest my rank is around 2068 but got just +22 (1349-1371)
isn't it something weird....
i mean, i thought i will get at least 40-50,
No. It's not so weird to get less rating change while you have more rating on the almost same place
number and rank of ppl you passed in contest also matters
because educational round is rated for purle, but this div.2 round is not.(maybe)
Problem div1A has very weak tests.
The following solution gets AC.
And there's obviously a very simple counter-test: A very big portal that don't touch the borders of the matrix.
Can anyone explain how to solve problem Div1.B?
C is 1700 rating???
Див2 индусы сегодня не на шутку разозлились
.
Well, when will the editorial be available?
when will editorial out??? Some recent rounds were hard for me. I am so bored now and hope next round will be more interesting
I think downvotes for this contest are a little unfair. And downvotes are bridges for upvotes. I hope that interlude have great contests in future.
Phantasmagorias now has the 3rd least contribution, with only Good.Stone and Sparky_Master_WCH1226 beating him.
MikeMirzayanov do smth about this bulling, community is too cruel. Yeah, of course round was not very pleasant by opinion of many people, but cmon authors do not deserve such treatment. Feels bad for them
A really amazing set of problems to think on. It was a nice and different experience, to have a rather difficult round than usual. Some amazing work by authors to make a contest that no one could complete XD.
Thanks a lot to the authors for the round and some interesting problems.
No way A is 800. I only solved it bc I noticed a pattern of n!/2 in the examples. But proving the correctness is at least a 1000+ problem.
every div 1. participant: eh, 800, 1000, what's the difference
why does this post have so many dislikes? because of hard problems, that lot of participants didn't solve? is is the problem of author, or a problem of participants?
guys, do not dislike. better try to improve yourself
The problems were harder than they should be, but this is extreme overreaction IMO. The authors do not deserve this kind of treatment.
Yeah I agree, I think people got frustrated due to the statement "strictly less than k-1" which was kind of vague but I think authors gave that just to confuse(or help?) as people who would have basic ideas of trees and graphs would tend to pause and think in direction of n nodes, n-1 edges(what I feel).
All in all, I guess mostly people are frustrated due to FST, odd times, n^5 solution passing but seeing at the number of downvotes, it made me quite sad as preparing DIV 1 problems are not everyone's cup of coffee :(
I upvoted though :)
the problems are great (and I like it). But there are a lot of thing little things that can be done to make the contest much better - no reason to have n = 100 while the intended solution O(n^5) (some people said that they did not pass with O(N^5))
O(n sqrt n) solution for n = 200000 in 1s very tight. I don't understand why the time limit is not 3s or n = 100000.
wrong ordering for 1B and 1C.
super hard 2A (and quite hard 1A) A(s) problem should not be hard so people are encouraged to participate)
very high difficulty jump from 2B to 2C
rejudging of 1A during contest
not to mention some grammar and note error(s) made.
Creating too difficult/unbalanced problem is okay to be criticized or else someone can just make contest with 5 3000-rated problems and say "hahaha, get gud brrrrrrr" when everyone had a hard time solving it
I dont get why people are downvoting.
Can anyone help me find what's wrong with my solution for Div2B ? Here
T.C: 1 0 1 Ans should be NO, your's is YES.
thx
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Don't be toxic
Please stop this guys....Everyone is criticizing to0 much and if this goes on many authors will be afraid to make contests on codeforces....I know how it feels.I performed badly too but why criticize this much in first place?? is it because they wasted 2 hr of your time.You can manage 2 hr but dont bring his morale down... $$$No$$$ $$$one$$$ $$$deserves$$$ $$$negative$$$ $$$votes$$$ $$$for$$$ $$$making$$$ $$$a$$$ $$$contest!!!! $$$.Please show your dissatisfaction in comments but atleast upvote the announcement.Its request from one coder to another
Div2D's data range is too large for O(n^5). Finally I removed an extra %MOD and runtime reduced significantly. Does that impact so much?
I don't know why many people criticize 2A. It's quite straightforward to know that for any pair of permutation and its reversed version there must be one and only one in the final result. I agree that the description < k — 1 of 2B is too confusing. I read more than 10 times to confirm I have understood it. The data limitation of 2D and 2E is too large.
A is easy for expert but not for newbie
F**k, I submit a wrong code of A and get a TLE then I headed to an Expert.
Reminded me of div1A.
This round is much better than round #810.