Hello, users.
I'm glad to invite you to take part in a Open Contest of the TAP 2012 (Torneo Argentino de Programación). The Real Contest was held on September 29, 2012. The official results are here. This Open Contest will be hosted on the Caribben Online Judge (COJ). You don't need registering for the contest, the participation is completely free, but you need to have an account in the COJ.
It starts on Tuesday, October 9, 2012 12:15:00 (CUBAN LOCAL TIME). The contest duration is 5 hours. The tasks are in Spanish. The allowed programming languages are C, C++ and Java. The contest will be held on the rules of ACM-ICPC competitions. If you have any questions, please feel free to ask on the FORUM.
Wish you good results.
ymondelo20 and COJ Team.
The Open Contest is about to start. But don't worry if you miss it, because you can use the Virtual Contest Module for reviving the contest again.
The links to the problems in English are here:
http://www.spoj.pl/problems/TAP2012A/ http://www.spoj.pl/problems/TAP2012B/ http://www.spoj.pl/problems/TAP2012C/ http://www.spoj.pl/problems/TAP2012D/ http://www.spoj.pl/problems/TAP2012E/ http://www.spoj.pl/problems/TAP2012F/ http://www.spoj.pl/problems/TAP2012G/ http://www.spoj.pl/problems/TAP2012H/ http://www.spoj.pl/problems/TAP2012I/
A little hint for every problem (as I don't have enough time to write a full editorial):
A: It's a simulation, you always have only one choice. B: Just three for cycles. C: Typical TopCoder-like DP. Be careful with digit 0. D: Greedy on a Trie. (this problem was mine). E: Shortest/Longest Path on a DAG. (also mine). F: Simulation. G: It's not a strings problem, but a graphs problem. H: This was the hardest problem, try to prove that there are lineal intersections that matter. I: The easy problem. (also mine).
I hope you all liked this contest. This was my first experience as problem setter and it was really funny and I learned a lot from it.
can i have a little more hint for TAP2012B
Just 3 fors will do, you have to ve very carefull with the limit of the iteration, I mean, for i: 0 -> n, for j: 0 -> n, for k: 0 -> n won't work, but if you are carefull with how much can the smallest number be and this kind of things it will work with three fors.