Hello Codeforces community!
I bring to you your Hogwarts letter.
I am glad to announce that Codecracker 2k17 will be held on 20th Jan 2017 at 16:00 UTC.
The problem statements will be available in English. The cash prize of the contest is Rs.20,000($300). Though what's priceless is solving interesting problems and the thrill of competition, prizes make the competition fierce. The contest will consist of 5 problems. We have tried our level best to set a balanced problem set. Hopefully, everyone will enjoy.
The theme of the contest is Harry Potter. So join us and feel the magic.
The contest is hosted on our own judge
Contest Link: http://codecracker.in/
Contest Details: http://mkti.in/events/codecracker/
Good luck and have fun!
I am unable to login on the site. Using Google+ login gives me an error message saying "invalid credentials"
Too bad descriptions, simple and non-original questions. Done 4 — leaving 3rd question since test data seems to be wrong.
True. Also, first question time limit is so strict! vector/map lovers are at a loss :(
unengineering or chocolate frog cards? — i had both AC in < 0.6 secs. You are kahakaharegisterkaru ? :P
Unengineering. And yes, thats my handle.
Are you talking about "Potions"? I think you forgot the point that "Single member is not considered as a team".
Yes. I took care of that. Can you share the judge solution?
Now, that the contest is over, can we see author's solution of p3?
PS: Don't make a Harry Potter fan write problems. The descriptions become too messy and hard to understand.
true that!
O(n) = O(n-1) + n * 7^n
How to compute O(10^9) ?
http://ideone.com/OrBTQ3
See the f(n) in my code. [WolframAlpha] https://www.wolframalpha.com/input/?i=sum(i%3D1..n,i*7%5Ei) can lead to this in 1 minute or you can derive on paper in 5 minutes.
Thanks. but how to prove this
Prove? — Induction. It holds for n=1, n=k implies it holds for n=k+1 for any k>1. Hence it holds for all n :)
That sequence is an AGP. The method of finding the sum of an AGP is given here.
You may also solve in this way :
You can create a 6 * 6 matrix, and solve normally. This got AC :)
how was omen done, i am getting an extra log factor for exponentiation?
please make all submissions visible
can someone show me the code for the first question i think i was getting tle for no apparent reason.thanks
http://ideone.com/EQkjti
thanks!!