Hello Codeforces,
Competitive Programming Club, AIT Pune is excited to invite you to CoDeft 2.0 which will take place on Monday, April 18, 2022 at 21:00 IST. This contest will be an ICPC styled contest to provide an exciting learning platform to showcase your CP skills.
Many thanks to all the people who made this contest possible:
- Problem Setters: RohitRKS, marksman_, _AbRn_ and Free_Electron_.
- unstoppable_N, codeict, fake_name and _hecxoc for testing and providing detailed feedback that improved the quality and balance of the round significantly.
- CodeSlayer007 for coordinating the event!
This event will be held in 2 stages:
Qualifiers
- Date & Time: Monday, April 18, 2022 at 21:00 IST
- Participation: Individual
- Contest Link: CoDeft 2.0
- Registration Link: Click
Finals
- Date: Saturday, April 30, 2022
- Participation: Individual
- Mode: Onsite, Top 100 participants from online qualifiers will be called for onsite finals at AIT, Pune.
- Participants reporting for onsite finals will have three events CodeRed, Bug-Off and Retracer to take part.
Prizes and Goodies
Qualifiers
- Top 25 Indian participants will receive goodies, delivered to their addresses.
- 5 Random Indian participants from the top 300 will receive goodies, delivered to their addresses.
- 5 Random 1st and 2nd-year students from AIT will receive goodies.
Finals
All the onsite finalists will receive a CoDeft T-shirt + Goodies.
-
Event Winner Runner Up CodeRed 10,000/- INR 6,000/- INR Retracer 5,000/- INR 3,000/- INR Bug-Off 5,000/- INR 3,000/- INR Goodies for the top 7 participants on the leaderboard in all the 3 events separately.
Don't forget to register to be eligible for goodies and prizes — Link
Join Discord Server for regular updates — Link
Good luck, see you on the leaderboard!
UPD : Contest starts in 15 mins. All the best!!
Auto comment: topic has been updated by unstoppable_N (previous revision, new revision, compare).
Where will I get the link for the contest?
You can find the contest link in the blog. Also, don't forget to register before taking part in the contest.
Contest Link: Link
Registration Link: Click
Auto comment: topic has been updated by unstoppable_N (previous revision, new revision, compare).
Reminder : Online Qualifier starts in 15 mins. Hoping to see you all in the leader-board.
tfw when u are doing multiset.erase(variable) for 45 minutes
Was F related to matrix exponentiation?
Yes, just you have to take care that if you have no moves, the number of ways will be 0, as stated in the announcement....
Hey, why is my solution for F giving WA? I have handled the case as told in the announcements.
I have gone through your solution, it was stated in the announcement that when the moves will be 0, the answer will be 0. So when the home node is not connected to any other node , you can not perform any move so the answer will be 0, but your code is giving 1.
Can you provide the testcase? I think my code handles that scenario
.
This testcase has Home node (60) connected to node (21) (line 4)
when was this announced? am I the only one who missed it? very unusual corner case.. (´。＿。｀)
Infact, initially in constraints, K was > 0. They changed it later.
Why is the form not accepting responses?
Was D related to some kind of dp?
Yes....there are five states x, y, dx, dy and turn(alice or bob)
In country planning, we just had to find the diameter of each component, sort them in reverse order and then had to make 4 cases right? cnt(no of components)= n, 1,2 and >2 for cnt == n ans=1, for cnt == 1 ans = (d1+1)/2, for cnt == 2 ans = (d1+1)/2 + (d2+1)/2+1, for cnt > 2 ans = max((d1+1)/2 + (d2+1)/2+1,(d2+1)/2 + (d3+1)/2+2). where d1, d2 and d3 are the largest diameters(d1>d2>d3)
When the count of the components is n then
ans=0 if n==1
ans=1 if n==2
ans=2 if n>=3 (because the graph can be made like a star graph) What is Star Graph
Here n is no. of nodes or no. of components?
If number of components is 1 and we have more than 2 nodes isn't the component just a tree which already has defined edges? how can we convert it to star graph.
And if you meant n is count of centers of the components, I did the same case work as you but we need to take account of the diameters of largest components as well right for end points of each component?
When the number of the components is n, it means there is no edge (m=0), then the answer can be greater than 1 , but in your previous comment you mentioned that is the number of components is n , the answer would be 1.
Sorry I misread your previous comment. Got my mistake. Thanks!
Looks like E was this problem verbatim and F was this but with implementation issues in the intended solution. I wonder how the sole person who got AC on F solved it, since most people were getting multiple WAs (it was somewhat suspicious too).
I studied a few codes and found that everyone missed this corner case, please have a look at this comment Comment Link
Still ambiguous. I had a couple of submissions where I ignored self-loops, and it really doesn't make sense to ignore self-loops for the home vertex but not for the remaining vertex.