division

问题 Consider the following code snippet in C++ :(visual studio 2015) First Block const int size = 500000000; int sum =0; int *num1 = new int[size];//initialized between 1-250 int *num2 = new int[size];//initialized between 1-250 for (int i = 0; i < size; i++) { sum +=(num1[i] / num2[i]); } Second Block const int size = 500000000; int sum =0; float *num1 = new float [size]; //initialized between 1-250 float *num2 = new float [size]; //initialized between 1-250 for (int i = 0; i < size; i++) { sum

问题 If you have a list of integers in python, say L = [4,8,12,24] , how can you compute their greatest common denominator/divisor (4 in this case)? 回答1: One way to do it is: import fractions def gcd(L): return reduce(fractions.gcd, L) print gcd([4,8,12,24]) 来源： https://stackoverflow.com/questions/3640955/computing-greatest-common-denominator-in-python

问题 What's the difference between / and \ for division in VB.NET? My code gives very different answers depending on which I use. I've seen both before, but I never knew the difference. 回答1: There are two ways to divide numbers. The fast way and the slow way. A lot of compilers try to trick you into doing it the fast way. C# is one of them, try this: using System; class Program { static void Main(string[] args) { Console.WriteLine(1 / 2); Console.ReadLine(); } } Output: 0 Are you happy with that

问题 What's the difference between / and \ for division in VB.NET? My code gives very different answers depending on which I use. I've seen both before, but I never knew the difference. 回答1: There are two ways to divide numbers. The fast way and the slow way. A lot of compilers try to trick you into doing it the fast way. C# is one of them, try this: using System; class Program { static void Main(string[] args) { Console.WriteLine(1 / 2); Console.ReadLine(); } } Output: 0 Are you happy with that

问题 What's the difference between / and \ for division in VB.NET? My code gives very different answers depending on which I use. I've seen both before, but I never knew the difference. 回答1: There are two ways to divide numbers. The fast way and the slow way. A lot of compilers try to trick you into doing it the fast way. C# is one of them, try this: using System; class Program { static void Main(string[] args) { Console.WriteLine(1 / 2); Console.ReadLine(); } } Output: 0 Are you happy with that

问题 How does one divide numbers but exclude the remainder in PHP? 回答1: Just cast the resulting value to an int. $n = (int) ($i / $m); Interesting functions (depending on what you want to achieve and if you expect negative integers to get devided) are floor(), ceil() and round(). 回答2: PHP 7 has a new built-in function for this named intdiv . Example: $result = intdiv(13, 2); The value of $result in this example will be 6 . You can find the full documentation for this function in the PHP

问题 This question already has answers here : What is the behavior of integer division? (5 answers) Closed 3 years ago . I entered the following code (and had no compiling problems or anything): float y = 5/2; printf("%f\n", y); The output was simply: 2.00000 My math isn't wrong is it? Or am I wrong on the / operator? It means divide doesn't it? And 5/2 should equal 2.5? Any help is greatly appreciated! 回答1: 5 is an int and 2 is an int . Therefore, 5/2 will use integer division. If you replace 5

问题 My question is very simple, suppose that I have an array like array = np.array([1, 2, 3, 4]) and I'd like to get an array like [1, 0.5, 0.3333333, 0.25] However, if you write something like 1/array or np.divide(1.0, array) it won't work. The only way I've found so far is to write something like: print np.divide(np.ones_like(array)*1.0, array) But I'm absolutely certains that there is a better way to do that. Does anyone have any idea? 回答1: 1 / array makes an integer division and returns array

问题 Is there any faster method for division of large integers(having 1000 digits or more) other than the school method? 回答1: Wikipedia lists multiple division algorithms. See Computational complexity of mathematical operations which lists Schoolbook long division as O(n^2) and Newton's method as M(n) where M is the complexity of the multiplication algorithm used, which could be as good as O(n log n 2^( log* n)) asymptotically. Note from the discussion of one of the multiplication algorithms that

问题 ( note : not the same as this other question since the OP never explicitly specified rounding towards 0 or -Infinity) JLS 15.17.2 says that integer division rounds towards zero. If I want floor() -like behavior for positive divisors (I don't care about the behavior for negative divisors), what's the simplest way to achieve this that is numerically correct for all inputs? int ifloor(int n, int d) { /* returns q such that n = d*q + r where 0 <= r < d * for all integer n, d where d > 0 * * d = 0