OJ Board

I'm hoping someone can suggest an algorithm for creating soccer teams based on some criteria..

1) Players suggest 2 or 3 'friends' whom they would like to be on a team with.
2) Teams would be evened out by player weight.
3) Players suggest their favorite position, and would be assigned to teams ...

Hi guys.. I'm having trouble trying to find the intersection point of two line segments. I found a webpage:

http://mathforum.org/library/drmath/view/62814.html

that seems helpful, but the last bit is hard to understand. Particularly, the part that says "Once we have checked this, we can solve for ...

This was actually a problem with 132 for me.

What I was trying to do is draw a line from the centre of mass to each end, then check that the angles of these lines are <= 90 degrees. That way I can check if the point lies between the endpoints.

Anyway, it worked and I got accepted. Does anyone know ...

I fixed it with
if(Math.abs(end1Angle) > 180) end1Angle += (end1Angle < 0) ? 360 : -360;
if(Math.abs(end2Angle) > 180) end2Angle += (end2Angle < 0) ? 360 : -360;

It took me a while to figure it out, but it works fine now.

As for the reference to P109... I solved that one no problem with the same ...

Joseph, thanks for trying to help.

Unfortunately, my problem is not just with negative angles... It is more with consistency.

If I get the angle (2,2)(1,1)(3,1) it should be 45.0
If I get the angle (2,2)(3,1)(1,1) it should be -45.0
(one turns right, the other left)

The 4 angles I gave as ...

Sorry about the last post, I was rushing so I didn't get to give a good reply.

Basically, I want a function that can take three points, A,B,C and return the angle formed by ABC. So I am looking for angle B.

My current way is this:
[java]
public static double getAngle(Point p1, Point p2) {
return ...

Farid, I found your post a little hard to understand. What is tg() ??

Joseph, the angle I want to find... It may change. Can you tell me a way to find an angle and I'll adapt my program to it?

I want to find the angle (in degrees) of three points.

I am trying to use the java Math.atan2(p1, p2) method, but am not getting what I want.

Take for example the points

(2, 2)(1, 1)(3, 1) the angle can be found by doing this:

angle1 = atan2( (1-1), (1-3) )
angle2 = atan2( (2-1), (2-1) )

angle ...

I converted your funky diagram into an input file:
Which gives the following output:
You did not say what the cost of each bumper was, so I assumed 0.

stcheung:

Again sorry for the size of the message. Your problem starts on line #160 (PRIDE and prejudice)


the world ACCORDING to garp
the ADVENTURES of huckleberry finn
out of AFRICA
the day AFTER tomorrow
AIRPORT
the sum of ALL fears
the sun ALSO rises
AMERIKA
farewell ANATOLIA
ANIMAL dreams ...

boatfish,

your problem exists for this input:


the
and
as
for
this
that
with
of
is
a
i
an
at
at
an
it
on
of
in
to
::
Fire in the Streets
Chaos
A troubled Feast
Amerika
A World of Ideas
Out of Africa
Death of a Salesman
Farewell Anatolia
Paradise Lost
Paradise Regained
The Descent of Man
The ...

Why not implement your own stack? I did, and it only took around 10 lines. That way you are sure that you are doing only what you need.

Really, all you need is a push, pop, and top method. Plus, you can statically allocate the memory needed for the stack of cards, whereas if you use the STL stack ...

I don't know exactly what classes the judge supports, but you can get the same compiler.

If you are using windows, go download CYGWIN and install it. Then, just download the GCC / GCJ compiler to use with it.

I believe you can download older versions of GCC, so if you want your compiler to match ...

Sometimes the trivial problems are the hardest. Especially when they word them in such a way as to make you think you have to do something spectacular.

This problem has a special judge, so it doesn't really matter HOW you sort them, as long as you do.

The strategy that is popular here is find the biggest, flip it to the top, then flip it to its final position.

So, find the largest, flip it to the top, then flip the entire stack over so the top is ...