I gather you’re looking for an usable aspect ratio integer:integer solution like 16:9 rather than a float:1 solution like 1.77778:1. If so, what you need to do is find the greatest common divisor (GCD) and divide both values by that. The GCD is the highest number that evenly divides both numbers. So the GCD for … Read more
Converting the output is too late; the calculation has already taken place in integer arithmetic. You need to convert the inputs to double: System.out.println((double)completed/(double)total); Note that you don’t actually need to convert both of the inputs. So long as one of them is double, the other will be implicitly converted. But I prefer to do … Read more
It’s all a matter of adequate storage and algorithms to treat numbers as smaller parts. Let’s assume you have a compiler in which an int can only be 0 through 99 and you want to handle numbers up to 999999 (we’ll only worry about positive numbers here to keep it simple). You do that by … Read more
That’s the “forall” (for all) symbol, as seen in Wikipedia’s table of mathematical symbols or the Unicode forall character (\u2200, ∀).
It just so happens that I wrote a BigFraction class not too long ago, for Project Euler problems. It keeps a BigInteger numerator and denominator, so it’ll never overflow. But it’ll be a tad slow for a lot of operations that you know will never overflow.. anyway, use it if you want it. I’ve been … Read more
I remember reading many years ago why 1.5 is preferred over two, at least as applied to C++ (this probably doesn’t apply to managed languages, where the runtime system can relocate objects at will). The reasoning is this: Say you start with a 16-byte allocation. When you need more, you allocate 32 bytes, then free … Read more
Rounding is typically done on floating point numbers, and here there are three basic functions you should know: round (rounds to the nearest integer), math.floor (always rounds down), and math.ceil (always rounds up). You ask about integers and rounding up to hundreds, but we can still use math.ceil as long as your numbers smaller than … Read more
Just do something like value = value < 0 ? 0 : value; or if (value < 0) value = 0; or value = Math.max(0, value);
To get the logarithm with a custom base using math.log: import math number = 74088 # = 42^3 base = 42 exponent = math.log(number, base) # = 3 To get the logarithm with a custom base using numpy.log: import numpy as np array = np.array([74088, 3111696]) # = [42^3, 42^4] base = 42 exponent = … Read more
Modern Intel/AMD processors have instruction FSINCOS for calculating sine and cosine functions simultaneously. If you need strong optimization, perhaps you should use it. Here is a small example: http://home.broadpark.no/~alein/fsincos.html Here is another example (for MSVC): http://www.codeguru.com/forum/showthread.php?t=328669 Here is yet another example (with gcc): http://www.allegro.cc/forums/thread/588470 Hope one of them helps. (I didn’t use this instruction myself, … Read more