OJ Board

It seems my algorithm can't be fixed.

I think there is an O(n^3) algorithm. (mine) But at the moment it has some bugs. And now it's TLE. (but it's definilty O(n^3). To be more precise O(m^3*n^3) where m is the number of different letters in the alphabet that is == 3)

Try this input:
Your prgram make a step that is wrong: I hope it helps!

Helal Md. Morshed Alam wrote:ab 1
abd 3
abcd 1

What is the *correct* output for abc ab abcd abe abd #?
I know that this *may* be the correct one...
(so simple task, and 7WA!)

uva could change the data, but that wouldn't change the fact that the solution is available at the same place as the data.
Uva don't mind if one knows the judge data, but change it if somebody boosting with precalculated data for all answers.
(The limits was changed in 10303)
So the boosting is ...

Well, but littledump asked an other thing.
So i took my output from my brute forcing output & determined a relationship & now I can make the program run in less time.
He asked if it is honourable.
I think it is. Especially in a contest. Sometimes it's very difficult to prove something, but you can ...

Hi! My previous post is bad, you are right. So, the correct one:
You shuold calculate two bisector's intersection.
Every bisector passing through the corresponding midpoint,
and every bisector's direction vector is the normal vector
of the triangle's corresponding side. So you can easily
write the ...

You should change the judge data a bit. The time 0.00 is very annoying at the top of the ranklist.

first case:
Center's x-coordiante: (7.0 + -1.0 + 0.0) / 3 = 2.00
Center's y-coordiante: (-5.0 + 1.0 + -6.0) / 3 = -3.33

second case:
Center's x-coordiante: (1.0 + 8.0 + 7.0) / 3 = 5.33
Center's y-coordiante: (7.0 + 6.0 + -2.0) / 3 = 3.66

Thank you! (AC again)

The two chiefs have decided to divide the village into two parts by digging a straight ditch through the middle of the village so that the houses of the supporters of Majestix lie on one part and those of the followers of Cleverdix lie on the other.
If there is no supporters for Majestix, then ...

ntrhieu wrote:(for example, he said, if the size of either sets is <= 1, then the solution is Yes, which is obviously wrong

why? I think it's right.
Now my soulution rejudged to wrong. Is this the only modification?

The yellow colour means a special correction program for that problem. It's because there are more solution.

X(O) = (X(A)+X(B)+X(C))/3
Y(O) = (Y(A)+Y(B)+Y(C))/3

It's funny! Try this: 3, 3, 5, 4, 4, 3, 5, 5!