| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | tourist | 4009 |
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| 3 | Benq | 3738 |
| 4 | Radewoosh | 3633 |
| 5 | jqdai0815 | 3620 |
| 6 | orzdevinwang | 3529 |
| 7 | ecnerwala | 3446 |
| 8 | Um_nik | 3396 |
| 9 | ksun48 | 3390 |
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| Страны | Города | Организации | Всё → |
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|---|---|---|
| 1 | cry | 167 |
| 2 | Um_nik | 163 |
| 3 | maomao90 | 162 |
| 3 | atcoder_official | 162 |
| 5 | adamant | 159 |
| 6 | -is-this-fft- | 158 |
| 7 | awoo | 157 |
| 8 | TheScrasse | 154 |
| 9 | Dominater069 | 153 |
| 9 | nor | 153 |
CodeAgon is on 27th September 2020 at 20:00IST.
And I'm sure almost all Indian coders will participate in CodeAgon rather than CF Div1,Div2.
Please change the date for round #673. Otherwise many people will have to skip this round.
Please look into this, MikeMirzayanov .
antontrygubO_o , if you are the coordinator for the round. Please look into this matter.
for beginners it is better to participate in DIV2 rather than CodeAgon
CodeAgon me kuch khas ukhad nahi pauge meri baad suno sirf baap coders ke liye hai vo contest chup chap apna DIV 2 do aur codeagon ka traffic kam karo, pata chalo starting me hi site down ho gayi xD
Most beginners don't even know about codeagon.
Right ! So better they give DIV2 and enhance their skills
Well, I can't deny this. But I feel like all expert+ coders should definitely try it out. ¯_(ツ)_/¯
I have never given CodeAgon, should cyans even try it?
Sure try.
Let me rephrase. For a cyan, Div2 or CodeAgon?
i know a coder with rating 1440 (almost) got selected in codeagon and got the job offer
You are wrong ,instead beginners are the one ,who are sharing codeagon links throughout the WhatsApp, social platforms.
Just to get T-shirt :)
I agree upto some fact, but atleast you will get experience for next time, as there are yearly two contest,
also this time sadly platform has changed from hackerank to interviewbit, where we cannot use our own editor, so at least participation will get you some experience.
I think only 2021/2022 graduates will participate in codeAgon, others are ineligible for prizes and job offers
2020/2021/2022 and that is current passouts, 3rd years and 4th years. Majority of good competitive programmers(Div1 and experts ) are in these batches only. So they will prefer attending codeAgon
I dont think 2020 grads are eligible https://www.interviewbit.com/contest/codeagon-2020/?rce=a7f2e8898e23&rcy=1
Read it again.
Its clearly written,its only for 2021/2022 grads. https://ibb.co/qpJ0Djd
Am I missing something ?
Where did you get it from ? I have attached the image from official announce where it doesn't mention 2020
https://ibb.co/r2jKYcD This is from the link you shared.
Oh,thanks it seems they have made 2 condratictory statements in the announcement.
Eventhough its unlikely, I do hope it gets postponed by a day. I hate missing cf rounds lol
I don't think timing will be changed anymore. Guess we'll have to miss it.
yea, please do something i have not missed a round in more than two months and I also have to give codeagon. There may be many like me
For people wanting link codenation
Nice way to spread your referrals lol.
i think you should personally message those officials of codeforces.
Mike don't see my messages. :(
And I don't know who are the coordinators of the round.
Please Don't postpone it. just prepone it to today or tomorrow. MikeMirzayanov
When CF clashes with a long contest and you really want to do both, do both.
It's not a long contest, it's a short contest
It gives opportunity for job/intern at great company Codenation .
It is super important exam. I know there is going to be a huge cheating scandal in this contest .Hope officials will do whatever possible to stop it . Hope ,Only deserving will get a chance.
Someone recently asked to vary Codeforces round times but they were told that Codeforces focuses on the European and Asian users, so I don't think the time should be changed. You only have to skip one round, at least you live in a good time zone. Many people in Japan/North America have to skip rounds all the time because it is so early, and on the weekend at least there won't be school or work for them. Remember Codeforces is international not just India.
Like its a loss from both side, isn't it. Lots of people won't be able to participate and its' a loss for both codeforces and the people. Don't deny the fact that most of the users are from russia and india, so they have customized the time in convinience of both of these countries. Therefore, japan/NA people asking for time change cannot be heard but indian people demanding may be heard. Think practically
+1
Why not have contests at different times rather than a standard time, this probably won't affect the quality participants(Not me) cause they love cp so much to give contest at any hour, competition is maintained and every nation might get a equal chance.
Think positive: there will be much less clarification requests. Help CF coordinators and round authors and participate in CodeAgon!
Why not ask to postpone codeagon.
[Deleted]
Doesn't mean it has more registers, so it is more important. For example, in IOI, only like top 350-400 qualify to the finals from the whole world. So that is only 350-400 registered users. But in fact, it is a very important contest.
you are comparing apples with bananas.
CF round is for personal improvement and satisfaction mostly, and one can get most of that experience by doing a virtual contest, while this test carries a lot more weigh for a lot of people since it can potentially get them internships and full time roles.
Just skip the round and do it as a virtual contest later? Less load on servers, smaller queues and less clarification requests
Me who rage quit the contest within 10 mins - 
) Translation — This is a high-quality insult.
Nice Div 1 round .
How to solve problems 3, 5, and 6?
Same doubt XD
5 was dp
dp[i][1024]
what is dp[i][j]?, One observation that I made was that for all k consecutive subsegments to have xor sum of 0, the k subsegment should repeat i.e. 1,2,6,1,2,6,.., where k = 3, somehow we have to choose the first k values such that total cost is minimum.
I= guess dp[i][j] is minimum changes required to make the sequence k xor till i if I set value j at index i. Nice thought by someone who wrote this solution.
How will you compute the transitions?
That is the thing Bajrang_Pandey can answer I am thinking over it!
Yes... actually iterate i till B, then if you set j at position i, then calculate the change by subtracting no of numbers equal to j at index i
here i is effectively i%B
The table is too large. 10^4*10^3 , 10^7 ordered table??!!! And also computing the transitions will timeout I guess?
No it worked ! did you try it?
No, i thought it may time out or show MLE.
I also had the same doubt I don't know were tests weak or memory limits were not more also there were some details hidden in any problem b was becoming y sometimes d was becoming x and also they even can't clear that your solution run on multiple tests . I got confused for hours that Why I was getting WAs on 2nd direct and easy problem. As I was not clearing adjacentcy list thinking of only one Test Case or may be some other error.
Not your fault brother. These problem statements clearly show how they didn't even bother to verify the problems.
Was Q.1 Rerooting on dfs tree or something tricky I wasn't able to approach 1st
It was greedy.
Thanks I will try approaching it greedily
Nice to see that everybody is on the same boat
How to solve 1 and 2 ? Please 1 first.
2nd was easy, count nodes with depth < B, say
cnt... answer is $$$2^{cnt}$$$ % modFor 2nd all the nodes which are at max distance B from the root can chose their own colour because they have no Bth parent above them. Remaining nodes will just chose the colours based on the rule of alternate colours thus ans wer will be 2^(total number of nodes which are at most B distance from the root )%P;
For problem 3:
Let F(n,k) denote the number of distinct necklaces of n beads that can be made with at most k colours.
Let $$$X_i$$$ be an Indicator Random variable which is equal to 1 if the $$$i^{th}$$$ colour is included in a necklace. So for a given necklace number of distinct colours = $$$\sum_{i=1}^{i=k}{X_i}$$$.
We need to calculate $$$E[\sum_{i=1}^{i=k}{X_i}] = \sum_{i=1}^{i=k}{E(X_i)} = \sum_{i=1}^{i=k}{P(X_i=1)} = k*P(X_1 = 1)$$$.
Now $$$P(X_1 = 0) = \frac{F(n,k-1)}{F(n,k)}$$$
$$$ Ans = k * (1 - \frac{F(n,k-1)}{F(n,k)})$$$
The value of F(n,k) can be be found here: https://en.wikipedia.org/wiki/Necklace_(combinatorics)