On x86 (and other processor architectures), integer division and modulo are carried out by a single operation, idiv (div for unsigned values), which produces both quotient and remainder (for word-sized arguments, in AX and DX respectively). This is used in the C library function divmod, which can be optimised by the compiler to a single … Read more
Does this help 22 % 5 = 2 17 % 5 = 2 12 % 5 = 2 7 % 5 = 2 2 % 5 = 2 Maybe this 22 / 5 = 4 + 2/5 17 / 5 = 3 + 2/5 12 / 5 = 2 + 2/5 7 / 5 = … Read more
Both variants are correct, however in mathematics (number theory in particular), Python’s modulo is most commonly used. In C, you do ((n % M) + M) % M to get the same result as in Python. E. g. ((-1 % 10) + 10) % 10. Note, how it still works for positive integers: ((17 % … Read more
The % does two things, depending on its arguments. In this case, it acts as the modulo operator, meaning when its arguments are numbers, it divides the first by the second and returns the remainder. 34 % 10 == 4 since 34 divided by 10 is three, with a remainder of four. If the first … Read more
Okay, so you want to calculate a^b mod m. First we’ll take a naive approach and then see how we can refine it. First, reduce a mod m. That means, find a number a1 so that 0 <= a1 < m and a = a1 mod m. Then repeatedly in a loop multiply by a1 … Read more
% is for int or long, not float or double. You can use fmod() or fmodf() from <math.h> instead. Better is <tgmath.h> as suggested by the inventor of CGFloat.
Math.Pow works on double-precision floating-point numbers; thus, you shouldn’t expect more than the first 15–17 digits of the result to be accurate: All floating-point numbers also have a limited number of significant digits, which also determines how accurately a floating-point value approximates a real number. A Double value has up to 15 decimal digits of … Read more
First of all I’d like to note that you cannot even rely on the fact that (-1) % 8 == -1. the only thing you can rely on is that (x / y) * y + ( x % y) == x. However whether or not the remainder is negative is implementation-defined. Reference: C++03 paragraph … Read more
Here’s an answer from the MSDN documentation. When you divide two integers, the result is always an integer. For example, the result of 7 / 3 is 2. To determine the remainder of 7 / 3, use the remainder operator (%). int a = 5; int b = 3; int div = a / b; … Read more
Both definitions of modulus of negative numbers are in use – some languages use one definition and some the other. If you want to get a negative number for negative inputs then you can use this: int r = x % n; if (r > 0 && x < 0) { r -= n; } … Read more