Hello Codeforces!
Computer Club, MNNIT Allahabad, India is glad to invite you to the annual programming competition of MNNIT, ɪɴᴤᴏᴍɴɪᴀ, which is an ACM-ICPC style team programming contest of 3 hours duration held on CodeChef during its annual technical fest Avishkar. The team can consist up to 3 members.
Contest Details:
- Start Time: Friday, September 6, 2019, at 21:30 IST
- Duration: 3 hours
- Contest Link: https://www.codechef.com/INQU2019
- Languages allowed: C, C++, Java and Python.
For the online round, top few global and top few MNNIT teams will get Codechef goodies. For the onsite round, prizes worth 31000 INR to be won. Top Indian teams from the online round will be called for the onsite round to be held at MNNIT Allahabad.
Problem setting panel includes chinmay0906, smtcoder, GreedyMind (myself), ram_24, xian02, gamer_coder13, acIN1go, ayushghd and mayankrana00.
Past few year's problemset: 2016, 2017, 2018
Join our Facebook event for further information.
We have an exciting problemset awaiting for you. Good luck and have fun!
UPD1: The contest was extended by 30 mins.
UPD2: Congratulations to the global winners:
Previous year problem statements are really good.
I m scared that servers may go down :(. I wish it was on Codeforces.
Bump! The contest begins in a day.
Are there any travel reimbursements for teams going to Onsite?
Currently, we don't have any plans to provide the travel reimbursements.
So how many top global teams will get Codechef goodies exactly?
That depends on the number of teams registering for the contest. Last year top 2 Global teams and top 2 MNNIT teams got the CodeChef laddus.
10 minutes to go, 1000 teams already registered.
How to solve "zero the path"?
Ok so firstly we observe that that the question is equvalent to finding the maximal path but you can throw away the values in the middle. We can do a DP on tree but we need to memo a few things.
Firstly, lets consider 3 types of path. A full path is a path where you traverse down and take eveerything until a certain point. A bottom path is when you have a full path buy your zeroed path must end at the node you are computing for. A split path is when you have a path where you have an arbituary segment that is zeroed. It turns out you can compute these three things in O(1) based on the values of your children.
Now to compute the maximal path that involves you and your subtree you must consider the 2 best full path of your children and your value, 1 full path and 1 split path, and 2 bottom paths.
If you compute everything for each node the maximum of those is your answer.
I cant explain this very well so ill link my code.
Problem Bonds is exact copy of problem B from Grand Prix of Japan XVII OpenCup (for those who have opentrains login: http://opentrains.snarknews.info/~ejudge/team.cgi?contest_id=001489 )
It was mere coincidence..sorry for the inconvenience caused.
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How to solve Triangles problem:)
Answer is
min(8, n+2).Given
Ntriangles, there will beN + 2vertices and let there beSedges, we need to travel atleastS - 1edges. If you observe, one can only travel all theSedges without repeating any edge if one starts from the top left vertex or the bottom right vertex of the structure formed. So the answer is always greater>= 2.Now, it is easy to understand that if we start from the vertices directly connected to any of these 2 vertices through an edge, we can always travel atleast
S - 1edges since we could travelSedges starting from those 2 vertices. Hence, the required answer is2+ the number of such vertices. ForN <= 6, the number of such vertices isN. Hence, the answer isN + 2.For
N > 6, there are only 6 such vertices (3 each connected to top left and bottom right vertex). So, the answer is6 + 2 = 8.How to solved Bonds?
I did the following approach cannot find what is wrong in that.
The total number of components with odd number of vertices should be exactly 1.
For that odd component if a vertex is cut vertex then now all new forming components should be even size.
My Code: Link
Accepted code of someone else with a similar type of logic:
I looked at this code but was not able to understand that for a Cut vertex, why they considered the size of last new-formed component only instead of all new-formed components.
That's our code. We considered only the last one because for a cut vertex, there are exactly two newly formed components.
Yeah so why you did not check both of them and checked only 1 instead?
The size of the component is odd, so if we remove one vertex and know that one part is even, then the other one is necessarily even because subtracting an even number from an even number produces an even number
Thanks :)
How many teams would be selected for the onsite round?
Top 15 teams will be selected for the onsite round.
Problems were really good, you could have organised a Div2 on CF.
Thanks :)
Can someone share promblems like https://www.codechef.com/INQU2019/problems/INQU1904 Thank you
The same problem on arrays .
The contest problem was actually an extension of the problem on arrays itself :)
will editorials be posted?
We'll try to do that soon.
yeah plz, it will help
the problems where really good btw
Nice problem set! :)
Thank you :)
Great contest!! What was the approach to the last problem?
Main catch was to sort the toys according to their velocity, then to find the range in which each toy will immune others on getting immuned,and finally find the best intersection of ranges that cover all and demand least energy. Problem setter will upload editorial shortly in more detail.
Any info about date and time of onsite round?
It will be held on 20th of September in the afternoon.
can u explain your logic for que. Zero the Path !! i read your solution but didn't understand your approach
The first problem was very similar to: https://mirror.codeforces.com/problemset/problem/579/A
I commented the test case part and fix X=2 and my same code accepted on CF.
Codeforces Submission
Codechef Submission
where are the editorials?
Where are the editorials of the contest ??