I believe this to be an ACM question from either 96' or 97'
bool nonzerodigit(constunsigned int& n, unsigned int& result)
if (n > 100000) returnfalse;
unsignedint result = 1;
for (unsignedint i = 1; i <= n; ++i)
unsignedint f = i;
while (0 == (f % 5))
f /= 5;
result /= 2;
result = (result % 100000) * f;
result %= 10;
100000 is an upper sanity bound and also a quantizer for the minimum required 6 "least significant" digits.
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.
I'm putting together a Silverlight 2 demo application. It is going to be a simple 2-D tank game where you can use the arrow keys to rotate and move the tank. The tank can point in any direction (360 possible directions, right?). I've written the code to rotate the tank, but I'm not sure where to begin regarding moving the tank. How can I take the angle of the tank (say, 156 degrees for example) and use that to determine the next x,y position of the tank when it moves forward?
I'm sure there must be a simple equation for this, but I don't know what it would be called or even how to google this problem.
The only things you need to remember are:
> Math.Sin and Math.Cos take the angle argument in radians not degrees.
> Sin and Cos return double precision numbers: which may round to 0 or 1 if you are using integer positions and therefore the tank will not move as you want.
i have set of 2D points and i want to offset with a given distance (like offset command in AutoCAD)
i do not know how to deal with corners. i have searched on Internet, there are advanced methods like straight skeletons etc. but my polyline is not self crossing and no holes in it.
At the end, it is a translation however it must be implemented with respect to the normals, as far as i read through if it is a parametric curve then it is rather complicated. If you like to see how much it can further get complicated, you can have a look at the link below.