division

问题 How can I multiply and divide using only bit shifting and adding? 回答1: To multiply in terms of adding and shifting you want to decompose one of the numbers by powers of two, like so: 21 * 5 = 10101_2 * 101_2 (Initial step) = 10101_2 * (1 * 2^2 + 0 * 2^1 + 1 * 2^0) = 10101_2 * 2^2 + 10101_2 * 2^0 = 10101_2 << 2 + 10101_2 << 0 (Decomposed) = 10101_2 * 4 + 10101_2 * 1 = 10101_2 * 5 = 21 * 5 (Same as initial expression) ( _2 means base 2) As you can see, multiplication can be decomposed into

问题 This question already has answers here : Python division (12 answers) Closed 5 years ago . How can I divide two numbers in Python 2.7 and get the result with decimals? I don\'t get it why there is difference: in Python 3: >>> 20/15 1.3333333333333333 in Python 2: >>> 20/15 1 Isn\'t this a modulo actually? 回答1: In python 2.7, the / operator is integer division if inputs are integers. If you want float division (which is something I always prefer), just use this special import: from __future__

问题 I don\'t really understand how modulus division works. I was calculating 27 % 16 and wound up with 11 and I don\'t understand why. I can\'t seem to find an explanation in layman\'s terms online. Can someone elaborate on a very high level as to what\'s going on here? 回答1: The result of a modulo division is the remainder of an integer division of the given numbers. That means: 27 / 16 = 1, remainder 11 => 27 mod 16 = 11 Other examples: 30 / 3 = 10, remainder 0 => 30 mod 3 = 0 35 / 3 = 11,

问题 This question already has an answer here: Dividing 1/n always returns 0.0 [duplicate] 3 answers I got this C code. #include <stdio.h> int main(void) { int n, d, i; double t=0, k; scanf(\"%d %d\", &n, &d); t = (1/100) * d; k = n / 3; printf(\"%.2lf\\t%.2lf\\n\", t, k); return 0; } I want to know why my variable \'t\' is always zero (in the printf function) ? 回答1: because in this expression t = (1/100) * d; 1 and 100 are integer values, integer division truncates, so this It's the same as this

问题 Python 3.1 (r31:73574, Jun 26 2009, 20:21:35) [MSC v.1500 32 bit (Intel)] on win32 Type \"help\", \"copyright\", \"credits\" or \"license\" for more information. >>> 2/2 1.0 Is this intended? I strongly remember earlier versions returning int/int=int ? What should I do, is there a new division operator or must I always cast? 回答1: Take a look at PEP-238: Changing the Division Operator The // operator will be available to request floor division unambiguously. 回答2: Oops, immediately found 2//2 .

问题 Does anyone know why integer division in C# returns an integer and not a float? What is the idea behind it? (Is it only a legacy of C/C++?) In C#: float x = 13 / 4; //imagine I used have an overridden == operator here to use epsilon compare if (x == 3.0) print \'Hello world\'; Result of this code would be: \'Hello world\' Strictly speaking, there is no such thing as integer division (division by definition is an operation which produces a rational number, integers are a very small subset of

问题 I have two integer values a and b , but I need their ratio in floating point. I know that a < b and I want to calculate a / b , so if I use integer division I\'ll always get 0 with a remainder of a . How can I force c to be a floating point number in Python in the following? c = a / b 回答1: In Python 2, division of two ints produces an int. In Python 3, it produces a float. We can get the new behaviour by importing from __future__ . >>> from __future__ import division >>> a = 4 >>> b = 6 >>> c