1. Thanks for pointing it out. I have added the schedule of the contest.
2. Yes, they can participate in the contest.

1. This is a Practice contest for ICPC so ,except for the onsite round, contestants are allowed to use 3 computers.
2. No, onsite and mirror finalist will compete in different contests and the prizes for both the contests will be different. Onsite Round will be Offline and contestants will not be allowed to access Internet. As mirror round contestants have access to Internet and 3 computers, the prizes for the mirror round are less than that of Onsite Round.
3. Yes. First round is very easy to pass but teams have to solve at-least one problem in the period of 2 hours.

As you know ICPC is only for college teams, so, as per the ICPC rules, only college teams are allowed to participate and we would stick to their rules as its a practice contest for the same

Yes. Please forward me the team name you want to delete and also the team name you want to connect to.

I used Meet in the middle + Trie.

since n <= 40, Split the problem into two (each n <= 20 ), Solve them individually and then merge them.

You can apply brute force( all the subsets of size <= k ) for n <= 20 and Merging can be done using TRIE.

Trie part is similar problem to this "Given an array of integers, we have to find two elements whose XOR is maximum".

Note: To get AC , you will need few optimisations.

My AC Code.

We will announce the results tomorrow and send invitation to all the teams who are qualified.

We have sent the mails to the teams who are qualified.

I created the same team few hours back and currently the website is just not loading....

I did the same and got ac. Maybe you are forgetting the case when the same element is swapped two times

I solved it using "Suffix Arrays".

Let the two strings be S and T. For each query, let the two substrings be S1 of S and T1 of T.

We need to calculate quickly for each query the "Longest common Prefix(LCP) of S1 and T1" as well as the "Longest common Suffix(LCS) of S1 and T1".

We can do this using Suffix arrays and LCP array. We can find LCP of two Suffixes in O(1) constructing RMQ sparse table on LCP array.

For LCP , make a string = concatenate S and T using any special character in between and just locate where is S1 and T1 in the lcp array and find LCP of those two suffixes in O(1).

For LCS , just make another string = "reverse(S) + "$" + reverse(T)", and do the same as LCP.

Now why do we need this?

There are 3 cases to consider.

1. K = 0 | In this case the LCP(S1,T1) = length(S1) since both must be equal , if they are not then we can't make S1 and T1 same.
2. K = 1 | In this case we will calculate LCP(S1,T1) and LCS(S1,T1) using our Suffix array Pre-Computation. Now Think like this, traverse LCP(S1,T1) from left and traverse LCS(S1,T1) from right. We can only make both S1 , T1 equal using only 1 adjacent swaps , if after swapping 2 characters( which are left after traversing both ways) we get equal strings.

3. K = 2 | This case is a little bit tricky. We have two sub-cases. Think like this, Traverse LCP(S1,T1) from left and traverse LCS(S1,T1) from right.

• now we are left with only 3 characters in between. ( 2 character case is done when K was 1). So just check if you can make S1 = T1 by doing 2 swaps in this 3 length string.
• If there are more than 3 characters in between then how to do?

let the position where first character is different while traversing from left side be id_l.

let the position where first character is different while traversing from right side be id_r.

Now we can do at most 2 swaps. 1st swap can be done on (id_l,id_l+1) and the second swap can be done on (id_r-1,id_r). Also we have to check that the substring from [id_l+2,id_r-2] are equal in both S1 and T1.We can check this using same concept of LCP array discussed earlier( when k was 0 ). If either of this fails then ans = "NO".

If in a query | k = 1, check from k = 0 -> 1. If k = 2, then start checking from k = 0 — > 1 — > 2.

My AC Code using same Algorithm.

If anything is unclear, feel free to ask!

Hi, I was the setter of problem C. Intended solution was the idea you tried, so perhaps you missed something (see Morphy comment). The easiest way was to use hashing + binary search to find LCP. Suffix Array was also possible of course, but seems harder to implement, at least from scratch in my opinion. My solution is uploaded in Author solution section on HackerEarth. I hope you enjoyed the problem :)

I can't get the formula of f(wi,y)? Where can I find the proof?

Did you had any submission from Team Leader account? I also experienced this same thing in the qualifying round and today too. Both times, i was able to see the team on leader-board only after there was a submission from Team Leader account.

Same thing happened to my team too. Team members have got autogenerated,single person teams assigned.

Anyone can participate except for those who are qualified for the second elimination.

http://mirror.codeforces.com/blog/entry/47449 Here is a post regarding the Wild card which has all the rules of the round.

I will update it today.

Here are the results of the first elimination round: http://mirror.codeforces.com/blog/entry/47455