modulus

问题 I'm looking for a better algorithm than one I found on stackoverflow to handle 4096 byte numbers, i'm hitting a maximum recursion depth. Code from stackoverlow post, i copy/pasted it but lost the original link: def linear_congruence(a, b, m): if b == 0: return 0 if a < 0: a = -a b = -b b %= m while a > m: a -= m return (m * linear_congruence(m, -b, a) + b) // a This works fine for smaller numbers, for example: In [167]: pow_mod(8261, 63, 4033) 63 1 8261 4033 31 195 1728 4033 15 2221 1564 4033

问题 This question already has an answer here : C++ size_t modulus operation with negative operand (1 answer) Closed 16 days ago . #include <iostream> #include <string> #include <vector> using namespace std; int main() { vector<int> v = {1, 2, 3, 4, 5, 6, 7}; int i = -4; cout << i << endl; cout << v.size() << endl; cout << i % v.size() << endl; cout << -4 % 7 << endl; } The above code prints: -4 7 5 -4 Can someone please explain why i % v.size() prints 5 instead of -4 ? I'm guessing it has

问题 If I have the following code var num = 15 % 2 + 6 * 4; for example... I'd like to know what the output will be, specifically I would like to know the order of precedence for modulus (the operation executed by the % symbol). Will the modulus be performed before or after the addition and multiplication operations? Edit : I have already looked at the article people are linking me to MDN Operator Precedence And had done so before asking the question, but unfortunately it didn't contain enough

问题 How do I compute C's % using Python's % ? The difference between the two is in the way they handle the case of negative arguments. In both languages, the % is defined in such a way that this relationship ( // being integer division) holds: a // b * b + a % b == a but the rounding of a // b is different in C and in Python, leading to a different definition of a % b . For example, in C (where integer division is just / with int operands) we have: int a = 31; int b = -3; a / b; // -10 a % b; //

问题 I'm trying to test if a float i.e( 54.10 ) is divisible by 0.10 . 54.10 % .10 returns .10 and not 0 , why is that and how can I get it to do what I want it to do? 回答1: You can use the decimal module to avoid the floating point precision problem: >>> import decimal >>> decimal.Decimal('54.10') % decimal.Decimal('0.10') Decimal('0.00') 回答2: The tried and true method here is to multiply your divisor and dividend by a power of 10. Effectively, 54.10 becomes 541 and 0.10 becomes 1. Then you can

问题 I am about to create a function in matlab which will accept multiple modulo and their corresponding remainders then it will determine the least possible value that will fit the given modulo conditions. Major trouble is that I am not allowed to use mod() and rem() built-in function in matlab. Can you help me with this? 回答1: You can easily create custom my_mod and my_rem functions without using mod and rem , and you can use these as you would use mod and rem . function modulus = my_mod(X, Y) if

问题 How can I write a modulus that will select the following iterations in sequence? 1, 4, 5, 8, 9, 12, 13 etc (+3+1r) I'm working within a loop and counting posts (iterations). So for example, I could catch every third post (1, 4, 7, 10) by:- if ($i % 3 == 1) { echo 'something here'; } But how can I write one that will catch 1, 4, 5, 8, 9, 12, 13? 回答1: I'm not quite sure about your algorithm, but it seems like you try to get every 3rd and 4th post not (starting at 0). The fitting code would be:

问题 How do I find the modulo (%) of two long values in Java? My code says 'Integer number too large' followed by the number I'm trying to mod. I tried casting it to a long but it didn't work. Do I have to convert it to a BigInteger and use the remainder method? Thanks. 回答1: The % operator does work for longs. It sounds like you may have forgotten to stick L at the end of a numeric literal, as in 123456789L . Can we see your code? 回答2: You can only have an integer up to 2 147 483 647. If you want

问题 I originally tried this, however the % operator isn't defined for float64. func main(){ var a float64 a = 1.23 if a%1 == 0{ fmt.Println("yay") }else{ fmt.Println("you fail") } } 回答1: Assuming your numbers will fit into an int64 , you can just compare the float value with a converted integer value to see if they're the same: if a == float64(int64(a)) { fmt.Println("yay") } else { fmt.Println("you fail") } Otherwise you can use the math.Trunc function detailed here, with something like: if a ==

问题 pow(a,b,c) operator in python returns (a**b)%c . If I have values of b , c , and the result of this operation (res=pow(a,b,c)) , how can I find the value of a ? 回答1: Despite the statements in the comments this is not the discrete logarithm problem. This more closely resembles the RSA problem in which c is the product of two large primes, b is the encrypt exponent, and a is the unknown plaintext. I always like to make x the unknown variable you want to solve for, so you have y = x b mod c