modulus

问题 Why is this even possible? console.log('13' % 2); 1 I assume JavaScript just converts the string on its own. I would appreciate any info on this behaviour. 回答1: As a starting point: http://ecma-international.org/ecma-262/5.1/#sec-11.5 The ToNumber operation is performed on the left argument (which results in the left argument being treated as the number, as explained in http://ecma-international.org/ecma-262/5.1/#sec-9.3.1) 来源： https://stackoverflow.com/questions/17140073/remainder-modulus-of

问题 i encountered a bug in some c code i wrote, and while it was relatively easy to fix, i want to be able to understand the issue underlying it better. essentially what happened is i had two unsigned integers (uint32_t, in fact) that, when the modulus operation was applied, yielded the unsigned equivalent of a negative number, a number that had been wrapped and was thus "big". here is an example program to demonstrate: #include <stdio.h> #include <stdint.h> int main(int argc, char* argv[]) {

问题 I am learning to work with some math like PHP query and just got to the modulo, I am not quite sure in what situations to use this because of something i stumbled on and yes I did already read one of the posts here about the modulo : Understanding The Modulus Operator % (This explanation is only for positive numbers since it depends on the language otherwise) The quote above is in the top answer there. But if I focus on PHP only and i use the modulo like this: $x = 8; $y = 10; $z = $x % $y;

问题 I would like to use modulus to apply a CSS class to a grid of images in two different situations: Situation 1 I would like to apply a CSS class to all middle images indicated by the *. I cannot simply use nth-child as I do not know if there will always be 9 images in the grid. Sometimes there could be 5, sometimes 7, sometimes 8, etc...it's variable. [Img 1] [Img 2*] [Img 3] [Img 4] [Img 5*] [Img 6] [Img 7] [Img 8*] [Img 9] Situation 2 I would like to apply a CSS class again to the images in

问题 So, Im using java to find the public key of a RSA password. Right now Im unsure what I'm doing and also if it's correct. I have this information for the public key. C = 5449089907 n = p*q = 8271344041 q = 181123 p = n/q = 45667 d = 53 phi(n) = (p-1)(q-1) = 8271117252 What complicates things are the BigIntegers, the numbers are way to huge for int and longs, so I have to use the clumsy BigIntegers. As far as I understand I have the following equation to solve. e*5198987987 - x*8271117252 = 1 I

问题 I want to create a list of string that resemble the column letters in Microsoft Excel. For example, after 26 columns, the next columns become AA , AB , AC , etc. I have tried using the modulus operator, but I just end up with AA , BB , CC , etc... import string passes_through_alphabet = 0 for num, col in enumerate([_ for _ in range(40)]): if num % 26 == 0: passes_through_alphabet += 1 excel_col = string.ascii_uppercase[num%26] * passes_through_alphabet print(num, excel_col) 0 A 1 B 2 C 3 D ..

问题 I want to find a way to know if an integer is divided by 3 or 7 without using division, because it is very slow in MIPS assembly. I have done a lot of research but found nothing. 回答1: There's a method described by Granlund & Montgomery that requires the modular / multiplicative inverse of the (odd) divisor modulo 2**b . (Some parts of this paper have been improved recently) The divisors: (d) = 3, 7 (odd numbers) are an easy case. Assuming 32 bit (unsigned) arithmetic, the inverses modulo 2*

问题 I want to compute nCk mod m with following constraints: n<=10^18 k<=10^5 m=10^9+7 I have read this article: Calculating Binomial Coefficient (nCk) for large n & k But here value of m is 1009. Hence using Lucas theorem, we need only to calculate 1009*1009 different values of aCb where a,b<=1009 How to do it with above constraints. I cannot make a array of O(m*k) space complexity with given constraints. Help! 回答1: Just use the fact that (n, k) = n! / k! / (n - k)! = n*(n-1)*...*(n-k+1)/[k*(k-1)

问题 In C/C++ how can I calculate (a^b)%m where b does not fit into 64 bits? In other words, is there a way of calculating the above value using b%m instead of b ? And is there any algorithm that can compute the above result in O(log(b)) time or O(log(b%m)) time? 回答1: According to Euler's theorem, if a and m are coprime: a b mod m = a b mod phi(m) mod m so if b is large, you can use the value b % phi(m) instead of b . phi(m) is Euler's totient function, which can be easily calculated if you know

问题 I am trying to implement the SAFER+ algorithm. The algorithm requires finding the modulus of a power function as follows: pow(45, x) mod 257 The variable x is a byte, and thus can range from 0 to 255. Accordingly, the result of the power function can be VERY big resulting in incorrect values if implemented using 32- or 64-bit integers. How can I perform this calculation? 回答1: some pseudo code function powermod(base, exponent, modulus) { if (base < 1 || exponent < 0 || modulus < 1) return -1