After each submission you know how much points you received ( full testing is during the contest). For each correct testcase you receive some points. You can submit solutions as many times as you want. For each task time is equal with time when you submited the first solution for the maximum points which you had received. Total penalty is equal with maximum of those 4 petalties ( because we have 4 tasks ).

We have several subtasks ( the main subtasks are mentioned in statements), so if you don't have solution for whole score, you can write some brute force and receive positive score :)

Also I want to mention one more thing. The tasks will be great challenge for everybody. The first one is very easy, for the last I hope it will be great challenge even for Legendary Grandmasters in one hour :)

First, that problem is HackerEarth problem, and this is contest on HackerRank :) I think each problem on some bigger contest has editorial, I really don't have a lot of experience about HackerEarth contests ( I think for each problem they have setter, tester and editoralist).

For the HourRank contest you have editorial for each problem, and codes from setter and tester. I am not sure when you can see the codes of other contestants, somethime that option is allowed sometime it isn't :)

The site is bit slow. I have the same problem with Chrome, but please try Mozilia. Also you can reload everything one-two times :)

UPD: Now is 100% everything fine :)

Maybe this: 14 6 6 3 3 3 3 3 3 1 5 1 5 1 5. The answer is 197.

6
1 2 3 6 6 6
?

I agree that the last problem is too difficult for that place, particularly it has a couple of corner cases which are not caught in samples. The other problems were good, as they can be good for an 1-hour contest.

When you click on the some task, you have at the top several topics. The last one is editorial (it is well written).

BIT is binary indexed tree. The python code is just simulation, uncomment the #print org line, and run the code with sample cases.

If someone has been bribed more than 2 times, terminate the algorithm.
Why terminate the algorithm? The counter example to that statement:
23451
Clearly, 1 has been bribed by four different people and that is valid.

Inversion of array A is such pair of indexes (i,j) where A_i>A_j. You have different approaches for all the task, so you don't need to think about BIT ;)

Because adding or subtracting a number which is divisible by 3 will not impact |Sb — Sk| mod 3. So, selecting such a number can be seen as like changing turns.

I was looking a lot of codes for this task after contest. There are some other dp approaches for this task, very interesting. For my opinion, this task is was solved by good amount of coders. I didn't expect so big number of correct dp approaches. At least I hope you find something interesting in this task.

Also I was analysing a lot of things on the contest ( that is my favourite part, when I am setter :D ). Kostroma solved the last task for 45 minutes, so I think that winner of contest and LayCurse had enough time to solve the task :) Yes, the task has several cases which aren't wrriten in samples and it isn't so easy for codinig, but Nikoloz solution had 40 lines of code, so I beleive with great thinking and fast realizing of some things you can solve it for one hour. Personally I expeced 2-3 good solutions on the contest. On the other hand, this is the last task and I think that everybody should prefer one extra hard for contest ( this is extra hard, because contest lasts 1 hour ). This contest, should be graat challenge for everyone ( read this sentences I didn't want to allow Tanya's boyfriend to solve everything in 30 minutes :P ). Also I didn't see easier O(n^3) solutions for 50% of socre. Such solutions could be written in smaller amount of time and yesterday it would be win for that coder ;)

I hope you had great fun, and you were happy after contest. I know name of next setter, I am sure he will prepare very good and intertesting tasks :)

No. You have at the begginng sorted array ( 1,2...,n ), after it you are interested how many at least bribes you perform to gen this array.

1-step 1 2 3 4 5 6 7 8

2-step 1 2 3 5 4 6 7 8 (5 bribed 4)

3-step 1 2 5 3 4 6 7 8 (5 bribed 3)

4-step 1 2 5 3 6 4 7 8 (6 bribed 4)

5-step 1 2 5 3 6 7 4 8 (7 bribed 4)

6-step 1 2 5 3 6 7 8 4 (8 bribed 4)

7-step 1 2 5 3 7 6 8 4 (7 bribed 6)

8-step 1 2 5 3 7 8 6 4 (8 bribed 6)

Answer 7

4 has been bribed 4 times, but 4 didn't bribe anyone else. Condition of task — some person can not bribe more that two other persons.