Welcome and good luck on the round!
I'd like to remind that if you have any questions on the problems, the best way to ask them is to use the web interface on the problems page.
Later in the same post we will discuss the round.
Wish you high rating,
MikeMirzayanov.
UPD2. And better late than never: the presence of English statements we are obliged only to Julia. Many thanks to her for 8 wonderful translations of 8 rounds.
It is very tough for me think of all the test cases....
It seems, that you can't scroll through all submissions, but you will find some "Accepted" for sure.
http://codeforces.me/contest/8/status/B?order=BY_ARRIVED_ASC
You can view any problem, just append order=BY_ARRIVED_ASC manually to the URL.
I think that open test would be useful.
Especially when an error is in technical details e.g. array sizes etc.
It is not differ much from WA.
Solution returned some time T, and some order P.
PE returned when total time, needed to collect objects in order P differ from time T.
UURRDL
Here the answer should be "BUG", but if you don't check the adjacent squares, you'd get "OK".
I am a new for Codeforces.
Non-rated users (newbies) or those having less than 1500 rating points belongs to second division. Others ( >= 1500 rating ) belongs to first division.
By the way at first contest you have rating 1500 and have changes with it participating in second division.
http://codeforces.me/blog/entry/259#comment-2894
galymzhan:
Problem B doesn't need BFS. I solved it in linear time using followin approach:
set all f[201][201] = 0;
set f[100][100] = 1; // start location
simulate moving and for each new location f[x][y], check the sum of four adjacent cells. If the sum > 1 then answer is BUG.
me too.I solve it without bfs.
UPD: And many thanks to Julia for her devoted help in translating the 8th round in a row.
Not modest, but true...
Can anyone tell me what can be bug with WA 6 in problem C?
Seriously, can anyone tell me what is the test case #6? I really want to know.
Still remember this problem. I only glanced at your code, so I can be wrong, but here is something that I really don't think should be in there:
The oddity of the number of the objects does not matter, also you I don't think you want to jump 2 objects at a time. It looks like you are trying to make as many groups as possible, when in fact some objects are better left alone. (Precisely, pairing up any 2 objects which make an obtuse angle with the purse will only increase the distance, given how it's calculated in this task).
Consider a bitmask dp where you either pick the leftmost zero alone or in a group with any other zero.
Oh, now I see. Thanks a lot. I understood my mistake.
(C.Looking for Order) My solution gets TL 18. It is working O((2^n) * n), I think it's OK for 4 sec. Can somebody help me?