Hello Codechef Community, Warm greetings from Team Technocracy, National Institute Of Technology, Raipur. We are glad to invite you all to participate in ALQUORA'18. It is our annual programming competition, conducted on the eve of AAVARTAN, our annual Tech Fest. It will be a single round IOI style Individual contest, hosted on Codechef. It will be consisting of 6 to 7 challenges, to be solved in 3 hours duration. The problems are expected to be engaging for everyone participating. Ranking is based on IOI style, with penalty of 10 minutes for every rejected solution (except Compile Errors). Contest Details: Contest link: Contest Link Date and Time: 4th October'18 22:00 hrs to 5th October'18 01:00 hrs (IST) Languages allowed: C, C++, Java and Python. Problem Setting Panel: Amarnath Sharma , Nagendra Jha Prizes: Registration is Required for receiving below mentioned prizes. Top 3 Indian performers will get Codechef Laddus, Exciting Goodies and Certificates from NIT Raipur. Top performer from NITRR will also get Exciting Goodies and Certificate. Programming Partner: Codechef Registration Link for Prizes: Register Here. For more details visit Aavartan Events or follow Alquora. Good Luck to every participant. Hope you all enjoy the problemset!! See you guys on the leader board. For any queries: njhabot@gmail.com asked 04 Oct '18, 03:33

contest timer is on but the problems are not visible. Please look into it! answered 04 Oct '18, 22:03
+1. Facing same problem
(04 Oct '18, 22:05)
2
the contest is postponed to 10:15 PM as per the announcement
(04 Oct '18, 22:06)
The same. what a pity.
(04 Oct '18, 22:06)
@jjtomar +1
(04 Oct '18, 22:07)

not at all organized one! The solutions of the other contestants are visible clearly! :( answered 05 Oct '18, 01:11
Yes, it was visible after 1:00 am for last 15 minutes :(
(05 Oct '18, 01:21)
Wtf!!,honest people :)
(05 Oct '18, 01:22)
@vivek_1998299 Congrats! Btw can you please explain your solution to the last problm?
(05 Oct '18, 08:20)
1
Thanx For last sum ,we consider each node as B type node,obviously now if we wanna find C node for that node v ,we should just take maximum in subtree of v. Now v is B node,max is C node So only thing left is A node,if we have a map containing elements of path root to v,then we can find easily the largest value less than v among the parents of v,which is A. These all can be done in a single dfd(i dont know y i used 2,should've done with 1). Solution:
(05 Oct '18, 10:22)

The contest is postponed to 10:15 PM guys as per the announcement. answered 04 Oct '18, 22:10

My submission is running for a several minutes https://www.codechef.com/submit/complete/20463819 answered 04 Oct '18, 23:06

@manaranjanfav you have to just partition the entire array into 3 partition and then for that specific partition you'll have the answer (1^x * 2^y * 3^z) where(x+y+z = n) and now the problem boils down to find a way to partition at and then add the corresponding number of ways to the answer. So, for now suppose we have only 2 values i.e. we need to partition the array into two half (suppose they are of the kind 2nd and 3rd). So, for a particular length n we have total of n+1 partitions possible (1st > include 0 2nd type and n 3rd type, 2nd > include 1 2nd type and n1 3rd type and so on...). So, total number of ways for a particual length (such that we are allowed to use only 2nd and 3rd type) $t$ is sum over all the above mentioned cases $(2^0 * 3^t + 2^1 * 3^(t1) + .....)$ and we know it's a GP with sum $(3^(t^+1^)  2^(t^+1^))$. Now, we know the number of ways we can make ornaments using 2nd and 3rd type and only case to include to get the answer is to include 1st type ornament and for that also we can do in same way. Suppose we have a total length of '$n$' and so the total number of ways = sum of((0 length of 1st type and n length from remaining two) + (1 length of 1st type and (n1) length of remaining two) + ........ ) = now as we know that (number of ways to select 1st type ornament of length x is 1) and so, we can reduce the above equation to the sum of expression $(3^(t^+1^)  2^(t^+1^))$ from $0$ to $N$. answered 05 Oct '18, 20:20
