CodeFest'11, the annual online international coding festival of Computer Engineering Society IT-BHU, presents Manthan - the algorithm intensive programming contest. The name Manthan or मंथन has been taken from the Hindi language and means Brainstorming. |
The contest aims to provide a platform for programmers round the globe to showcase their problem-solving and programming skills. Most problems of the event will be real life computing problems where coders will be required to prune existing standard algorithms to find the solution for the domain specific application of the algorithm.
- The main event is scheduled on March 13 at 2200 Hrs IST (UTC+5:30)
- Rules of contest similar to Codeforces Beta Rounds.
- The Contest will be of 3 hours consisting of 5-6 problems.
- Programing languages allowed: C, C++ , Pascal, Java, C#, Python, Ruby, PHP, F# and Haskell.
- Open to all, students as well as professionals.
- Declaration of Results on March 16 at 1800 Hrs IST.
Visit the Manthan event page for further details of the contest.
Registration for the competition will occur automatically on the fact of your registration on the official CodeFest site. If you have been properly registered there, then your Codeforces handle should appear in the list on http://codefest.org.in/codeforces-handles.php, and registration for the contest will take place
automatically.
1)For the registration form that pops up on clicking 'Register for Manthan', provide a new codeforces handle and new password.
2) Re-register on codefest website using the same email id as in codeforces. Then when you login and click on 'Register for Manthan' no extra form pops up.
What are the differences between the rules?
- There will be 40 participants in each room.
- Distribution in rooms will be done randomly
- Difference in number of problems
More detailed rules are here: http://codefest.org.in/event.php?name=manthan#rulesHope sooner or later problem writers will realize that this is not cool.
Some people earned 3000 points by hacking problem A. It is more than average person could achieve by solving C and D together. I don't know what it proves.
This is one on which I failed, and then I saw several people doing the same mistake.
My bad. I meant RRRRRL
You can see on which test your solution failed. Click on the submission ID on the status page.
Please do provide us your feedback!
Problems themselves are very good overall. I pretty much enjoyed solving all of them.
http://mirror.codeforces.com/blog/entry/1473#comments
but i'm not sure when the ratings will be updated.
2
1 2
1 2
Output
0
Answer
1
Checker Log
wrong answer 1st numbers differ - expected: '1', found: '0'
I'm not agree with this.
A ray doesn't intersect itself.
And from the sentence:
"Sir, there are some groups of rays such that all rays in that group intersect every other ray in that group.
We can conclude that there is at least one group with at least two intersecting rays,since "every other" is at least one ray.
I want to be rejudged:)
It seems nobody understands me:)
We have:
Statement says:
there are some groups (at least one SUCH group exists,do you agree?some>0) of rays such that all rays in that group intersect every other (THEY DO INTERSECT).
If we have only groups with 1 ray there are no intersections.
For every pair of rays from that group, thay are intersecting.
If there are no pairs, this statement is correct in any case.
If I have no friends than statement "all my friends are girls" is correct:)
This is different from the ordinary way we use words. In the ordinary sense, if we say “all the children playing in the park are boys,” it usually implicitly means that there is at least one child playing in the park (and he is a boy). But in mathematics, we use the terms in a slightly different way.
One may argue that it is unrealistic to assume that the student used the terms in the mathematical sense. But then what would be the correct answer? If you do not think that a set of one ray satisfies the condition, I cannot see why you think that the set of zero rays satisfies the condition, either.
Thanks.Your answer is the best for me.Vacous truth is not a new thing for me.And the problem here for me was really that the statement was written in ordinary language and i didn't think that the student could have used some math approaches to express his thoughts.
Really,the set of 0 rays also makes us believe that the student told a lie in the case of ordinary reasoning.There are no intersections in this case too.You 've almost convinced me that it was my fault too that i made this bug:)
Anyway, great contest with great problems. Thanks!
A = {3,1,5,2,4) and k = 1, this is how to calculate B:
B1 = 1 means there's 1 element >=1+k (3) standing before 1 in A
B2 = 2 means there're 2 elements >=2+k (3,5) standing before 2 in A
so on...
I understand now.
You(or others who knows it) know how to lock solution code?
Problems - Submit - ... - Room -....
Choose Problems, you will see locks on problems you submit and pass pretests. Click on the locks. Choose Room and click on other participants' submission to hack.
"Also, it is given that 2· te ≥ ti + td."
If this condition is not true, does anyone have an efficient way of solving this problem?
Many thanks in advance!
"References" section here http://en.wikipedia.org/wiki/Sequence_alignment
In this almost everyone got good ratings, mainly because there were lots of "Unrated" participants.
Is this OK?
Because I saw many people not even solving a single problem got increase in rating.
(I also think that my rating would not have gone that high by getting just getting 1200 pts if there would have been less "unrated" people.)
I consider this is mainly because there were lots of "Unrated" contestants . That was only my concern that whether the rating actually reflect the performance , because u see "not even solving a single question and getting rating increase" is in my opinion weird.
As far as new contestants are concerned , they are always Welcome . :)
In theory, if someone were desperate to keep the rating without actually improving his/her skills, he/she could use a silly “strategy”: participate only in the events where many first-timers are expected. But it is unlikely that many people would try to use this “strategy,” and I do not think that we have to worry about such a possibility because it will not harm the rating of other participants anyway.
I believe that screw_it is pt1989.
You can compare the ranks from codefest results to that of codeforces standings.