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Starts in 4.5 hours!
No, is starts in ~70 minutes!
Nop, 30 minutes!
People at codeforces are amazing, aren't they?
Since the link to your pic starts with
googleusercontent.com
, it looks like you're linking to Google user content (thanks, Captain Obvious). Trying to view it directly tells me I don't have permissions to view it.My bad, tried to upload to Google Drive for reusability lol
Can't see the picture.
I really don't understand the reason for so many down votes. I only commented to bring this post closer to the top so more could notice it and suddenly -18, WTF?
Because CF voting system.
It needs to be done in the opposite way to Disqus: people who downvoted need to be public.
dont need to get angry, go eat a banana and calm down
Racist.
Fuck. I am late again T.T
Editorial of Div1-Easy and Div1-Hard: http://codeforces.net/blog/entry/17341
250 was a really funny problem for me... At first I thought that the intended solution is to check whether there is a monochromatic k*k square and whether there is a monochromatic segment of length k in each row and in each column. But I couldn't prove it's correct and after some time I found a counterexample. So I left the 250 be and moved to the other two problems.
In the challenge phase I used my counterexample to challenge two solutions and I think that I could challenge ~4 more, if I was fast enough (it's quite hard to challenge when in same room with tourist). So, at the end, I would gain more than 250 points for 250, without even solving it :)
What was the counterexample for your initial idea?
k = 2
A more symmetrical example:
My room had only 5 wrong 250s, 3 of which used a greedy "paint the square with whatever color it is" algorithm and 2 used this row checking idea, so there were only 100 points for the picking with that case :(
How to solve div2 250 pt?
My approach —
There are only 2 possible patterns for the board, either it starts with 'W' or 'B'. After that we can fill the board according to the question.
So there we obtain 2 possible patterns, say C and D.
Now loop through the given board and wherever the character is not '?', compare it with C(pattern).
If the condition is true, means board is not equal to pattern C.
SImilarly do for pattern D.
If both C and D does not match then answer is "Impossible" otherwise "Possible".
I ran into a bit of trouble too while trying to figure out the logic, so I ran a bfs starting from the first square on the board that isn't a '?' (that is, starting from the first coloured square you come across on the board).
Suppose your current starting square is white, check the 4 adjacent squares. If any of them is the same colour, return -1. If it is a question mark, colour it the opposite colour(black). Then push all four into the queue, keeping track of the visited squares.