Good day!
We have got over two weeks of our new term. And we are glad to see you on our regular rating Codeforces round for Div.2 participants. And all who wants can take part in this competition.
This round has been prepared by a team of three people: Gerald, NALP and Polichka. We are grateful for their help in preparation and translation to Artem Rakhov(RAD), Mike Mirzayanov (MikeMirzayanov) and Maria Belova(Delinur).
Today Peter entangled in the tables:-( And you can help him! It's rather easy!
Score distribution: 500-1000-1500-2500-2500
Easy solutions and high rating to all!
UPD:
Thanks all for participation!
You can read the tutorial on this link: Tutorial.
Good luck to everybody :)
if it is Easy for all , it is not easy for anybody ! :P
good luck
I hope C won't be tricky!
It was my first contest , I really enjoyed it :)
To the team of three writers: you guys are really productive. Thank you all.
I think that C was easier than B :-/
I realized too late, that result for B is bigger than long (I'm dumb, message "Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator." in statement didn't help me too) :-/
I think they both are easy.
I'm not telling that they are not easy, but I think that it is more difficult to implement B than C — I used one loop and 7 conditions in B and just 4 loops without any condition in C (I do not like line metric).
"easy-ness" is a relative term btw.
That's why I wrote "I think...", it's just my opinion. Opinion of problem writers is clearly different. I just wanted to let them and others to know, nothing more, nothing less...
Cool.:D
but still u need some combinatorics to solve C
I don't think so, or maybe it's so trivial that I do not consider it as combinatorics, because it is such straight...
Yeah, of course C was easier than B, that's why some people submited it before B.
I failed here, because when I saw C first time, I didn't see it's ease.
So I submit it after debugging B and lost loads of points :(
Same here. I saw that coders are submiting C, but when I read the statement, I didn't see the solution in the moment.
500-1000-1000-2500-2500 would be better.
I made an unsuccessful hack to this code thanks to lack of syntax highlighter in the hacking window. :(
Yeah, it really sucks that there is no syntax highlighting while hacking...
On the other hand in last TC SRM my unseccessful challenge was because I missed that small L is same as 1 (and there is highlighting, but not for this case) :-/
The problem D only 5 AC during this 2 hours! :O
I think, maybe it's not difficult to solve the problem D , but it's hard to understand it's meaning. ( Maybe I have a big problem with English — -.)
Problem D: I thought it to solve like this, is it correct:
Find x number of potential vertical lines, and y number of horizontal lines. if x > 4 or y > 4 , return "NO"; else proceed as: take C(x,2) * C(y,2) possible combinations of those lines and test whether these lines can be part of frames in O(n^2) .
Overall complexity : C(4,2) * C(4,2) * O(n^2) .
The number of the potential vertical (or horizontal) lines can be more than 4, even if the answer is "YES".
For example: ('x' instead of '#' because of Markdown)
5 5
xxxxx
xxxxx
xx.xx
xxxxx
xxxxx
Here only 4 potentials are there. (Those having atleast 3 consecutive X)
May be my algo. is wrong, can there be more than 4 potential lines?
what about that:
****
****
****
****
****
****
answer is "yes"
Oh, yeah. My algo is wrong.
In this case we can check both rows and columns, if answer is yes then I think either rows or columns will have at most 4 lines containing potential sides, so that it can be rotated 90 degrees and solved in the same way.
Test for two rectangles could be done in O(n+m) with preprocessing O(n*m), since two following conditions seems to be enough:
1. all of points in rectangles are in our table [O(n+m)];
2. perimeter of rectangles' union equals to amount of #s in table [O(1)].
I guess because of my English, I got extremely confused with this line "You wonder how many different names Vasya can write instead of name number 1" (Problem C). What the problem writer meant with "instead of name number 1"?
On the place of the first name.
He meant: After Vasya performs all his operations, how many possible strings can be at the place where the first name was initially.
When will the safe mode end?!
Btw, it's going continiously since R107...
And safe mode still did not ended completely: for example, country rating and contests participated by particular coder still show "Do not panic" page...
This contest have 3 easy problem and 2 difficult problem. This contest doesn't have any medium problem. I think it's very bad.