division

Is divmod() faster than using the % and // operators?

狂风中的少年 提交于 2019-12-17 23:09:58
问题 I remember from assembly that integer division instructions yield both the quotient and remainder. So, in python will the built-in divmod() function be better performance-wise than using the % and // operators (suppose of course one needs both the quotient and the remainder)? q, r = divmod(n, d) q, r = (n // d, n % d) 回答1: To measure is to know (all timings on a Macbook Pro 2.8Ghz i7): >>> import sys, timeit >>> sys.version_info sys.version_info(major=2, minor=7, micro=12, releaselevel='final

Divison and remainder in Prolog

人走茶凉 提交于 2019-12-17 21:08:20
问题 Trying to figure out how to write a recursive predicate divide_by(X, D, I, R) that takes as input a positive integer X and a divisor D, and returns the answer as the whole number part I and the remainder part R, however, I can't seem to get my head around Prolog. How would I go about doing this? 回答1: There are predefined evaluable functors for this. (div)/2 and (mod)/2 always rounding down. Recommended by LIA-1, Knuth etc. (//)/2 and (rem)/2 rounding toward zero (actually, it's implementation

Assembly mod algorithm on processor with no division operator

一曲冷凌霜 提交于 2019-12-17 18:39:16
问题 I need to implement a simple macro that finds the modulo of two numbers on a processor that doesn't have a division operator (think ARM). I could use division by repeated subtraction, but I don't know if this was the most efficient or easiest to work with. Any suggestions? Code would be even more helpful. This particular class has us using a subset of SPARC, so most operations look like this: add r1, r2, rdest . This particular assignment calls for checking that a mod b == 0 or that the

What's wrong with this division? [closed]

拥有回忆 提交于 2019-12-17 14:55:49
问题 This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. Closed 7 years ago . I think there's a lot for me to learn about data types. Why this happens double result = ((3/8)*100).ToString(); it gives zero .. should be 37,5 ... :(

Fast multiplication/division by 2 for floats and doubles (C/C++)

落爺英雄遲暮 提交于 2019-12-17 10:58:35
问题 In the software I'm writing, I'm doing millions of multiplication or division by 2 (or powers of 2) of my values. I would really like these values to be int so that I could access the bitshift operators int a = 1; int b = a<<24 However, I cannot, and I have to stick with doubles. My question is : as there is a standard representation of doubles (sign, exponent, mantissa), is there a way to play with the exponent to get fast multiplications/divisions by a power of 2 ? I can even assume that

Division without using '/'

可紊 提交于 2019-12-17 02:35:09
问题 Can anyone tell me an efficient approach to perform the division operation without using '/'. I can calculate the integer value in log(n) steps using a method similar to binary search. 115/3 57 * 3 > 115 28 * 3 < 115 47 * 3 > 115 . . . 38 * 3 is quotient value ..... But is there any other more efficient method? 回答1: The typical way is to shift and subtract. This is basically pretty similar to long division as we learned it in school. The big difference is that in decimal division you need to

Divide a number by 3 without using *, /, +, -, % operators

橙三吉。 提交于 2019-12-17 02:16:11
问题 How would you divide a number by 3 without using * , / , + , - , % , operators? The number may be signed or unsigned. 回答1: 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; x = t; } return y; } int divideby3(int num) { int sum = 0; while (num > 3) { sum = add(num >> 2, sum); num = add(num

Why is division in Ruby returning an integer instead of decimal value?

时间秒杀一切 提交于 2019-12-16 20:04:12
问题 For example: 9 / 5 #=> 1 but I expected 1.8 . How can I get the correct decimal (non-integer) result? Why is it returning 1 at all? 回答1: It’s doing integer division. You can make one of the numbers a Float by adding .0 : 9.0 / 5 #=> 1.8 9 / 5.0 #=> 1.8 回答2: It’s doing integer division. You can use to_f to force things into floating-point mode: 9.to_f / 5 #=> 1.8 9 / 5.to_f #=> 1.8 This also works if your values are variables instead of literals. Converting one value to a float is sufficient

Division in Variable Template Returns Zero in Visual Studio 2017

断了今生、忘了曾经 提交于 2019-12-14 01:29:01
问题 This is probably a visual-studio-2017 bug related to this question: Templated Variables Bug With Lambdas in Visual Studio? And as mentioned in the comments there seems to be optimizer related. Division in the definition of a variable template seems to have a bug in Visual Studio 2017. So this code for example: template <typename T> const T PI = std::acos(static_cast<T>(-1)); template <typename T> const T ONE_EIGHTY = 180; template <typename T> const T DEG_TO_RAD = PI<T> / ONE_EIGHTY<T>; int

Python Division Of Complex Numbers Without Using Built In Types and Operators

前提是你 提交于 2019-12-14 01:17:53
问题 I have to implement a class called ComplexNumbers which is representing a complex number and I'm not allowed to use the built in types for that. I already have overwritten the operators ( __add__ , __sub__ , __mul__ , __abs__ , __str_ which allows to perform basic operations. But now I'm stuck with overwriting the __div__ operator. Allowed to use: I'm using float to represent the imaginary part of the number and float to represent the rel part. What I have already tried: I looked up how to