Math.Floor rounds down, Math.Ceiling rounds up, and Math.Truncate rounds towards zero. Thus, Math.Truncate is like Math.Floor for positive numbers, and like Math.Ceiling for negative numbers. Here’s the reference. For completeness, Math.Round rounds to the nearest integer. If the number is exactly midway between two integers, then it rounds towards the even one. Reference. See also: … Read more
You can use full datetime variables with timedelta, and by providing a dummy date then using time to just get the time value. For example: import datetime a = datetime.datetime(100,1,1,11,34,59) b = a + datetime.timedelta(0,3) # days, seconds, then other fields. print(a.time()) print(b.time()) results in the two values, three seconds apart: 11:34:59 11:35:02 You could … Read more
NOTE – See Edit 4 if 3 digit precision is important var discount = (price / listprice).toFixed(2); toFixed will round up or down for you depending on the values beyond 2 decimals. Example: http://jsfiddle.net/calder12/tv9HY/ Documentation: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Number/toFixed Edit – As mentioned by others this converts the result to a string. To avoid this: var discount = … Read more
In Python 3.x, 5 / 2 will return 2.5 and 5 // 2 will return 2. The former is floating point division, and the latter is floor division, sometimes also called integer division. In Python 2.2 or later in the 2.x line, there is no difference for integers unless you perform a from __future__ import … Read more
For ruby >= 2.4 you can use sum: array.sum For ruby < 2.4 you can use inject: array.inject(0, :+) Note: the 0 base case is needed otherwise nil will be returned on empty arrays: > [].inject(:+) nil > [].inject(0, :+) 0
There’s a simple trick for this problem: bool IsPowerOfTwo(ulong x) { return (x & (x – 1)) == 0; } Note, this function will report true for 0, which is not a power of 2. If you want to exclude that, here’s how: bool IsPowerOfTwo(ulong x) { return (x != 0) && ((x & (x … Read more
This is a simple function which performs the desired operation. But it requires the + operator, so all you have left to do is to add the values with bit-operators: // replaces the + operator int add(int x, int y) { while (x) { int t = (x & y) << 1; y ^= x; … Read more
Since I’m the current world record holder for the most digits of pi, I’ll add my two cents: Unless you’re actually setting a new world record, the common practice is just to verify the computed digits against the known values. So that’s simple enough. In fact, I have a webpage that lists snippets of digits … Read more
Just a clarification Although the previous answers are right whenever you try to spot the randomness of a pseudo-random variable or its multiplication, you should be aware that while Random() is usually uniformly distributed, Random() * Random() is not. Example This is a uniform random distribution sample simulated through a pseudo-random variable: BarChart[BinCounts[RandomReal[{0, 1}, 50000], … Read more
You didn’t say what kind of language they expected… Here’s a static solution (Haskell). It’s basically messing with the 2 most significant bits: f :: Int -> Int f x | (testBit x 30 /= testBit x 31) = negate $ complementBit x 30 | otherwise = complementBit x 30 It’s much easier in a … Read more