modulo

I currently have code in objective C that can pull out an integer's most significant digit value. My only question is if there is a better way to do it than with how I have provided below. It gets the job done, but it just feels like a cheap hack. What the code does is that it takes a number passed in and loops through until that number has been successfully divided to a certain value. The reason I am doing this is for an educational app that splits a number up by it's value and shows the values added all together to produce the final output (1234 = 1000 + 200 + 30 + 4) . int test = 1; int

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center . Closed 6 years ago . Ruby modulo rules with negative numbers are unclear. In IRB: -7 % 3 == 2 Should be 1 ! Why? Because -7/3 is -3 under Ruby's integer division semantics. 3*-3 is -9, so that would leave a remainder of 2. According to the docs , x modulo y is defined as: x-y*(x/y).floor andrewdotn When one of the operands to % is

Recently I discovered that C#'s operator % is applicable to double. Tried some things out, and after all came up with this test: class Program { static void test(double a, double b) { if (a % b != a - b * Math.Truncate(a / b)) { Console.WriteLine(a + ", " + b); } } static void Main(string[] args) { test(2.5, 7); test(-6.7, -3); test(8.7, 4); //... } } Everything in this test works. Is a % b always equivalent to a - b*Math.Round(a/b) ? If not, please explain to me how this operator really works. EDIT: Answering to James L, I understand that this is a modulo operator and everything. I'm curious

I try to calculate with JS' modulo function, but don't get the right result (which should be 1). Here is a hardcoded piece of code. var checkSum = 210501700012345678131468; alert(checkSum % 97); Result: 66 Whats the problem here? Regards, Benedikt A bunch of improvements to Benedikt's version: "cRest += '' + cDivident;" is a bugfix; parseInt(divisor) makes it possible to pass both arguments as strings; check for empty string at the end makes it always return numerical values; added var statements so it's not using global variables; converted foreach to old-style for so it works in browsers

Someone does 20 Hours 42 Minutes & 16 Seconds in one shift totaling 74536 seconds. How do I get the hours from number of seconds the person has done for that shift? 20 * 60 * 60 = 72000 42 * 60 = 2520 16 = 16 + ----- Total = 74536 ____________________________ Total % 60 = Seconds (16) Total % ? = Minutes (42) Total % ? = Hours (20) Tried 84600 already; turns out when a number is lower the modulus, it really is not very helpful, and something I am going to have to catch should someone only sign in for a few seconds ... You need to use both modulus and division: t = seconds_in_shift; secs = t %

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. 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**32 yield 2863311531 (0xAAAAAAAB) and 3067833783 (0xB6DB6DB7) respectively. There's an online calculator here

How do I perform a mod operation between two integers in C++? Like this: x=y%z In c++, use % operator More Help As the other answers have stated, you can use the C++ % operator. But be aware that there's a wrinkle no one has mentioned yet: in the expression a % b , what if a is negative? Should the result of this operation be positive or negative? The C++ standard leaves this up to the implementation. So if you want to handle negative inputs portably, you should probably do something like r = abs(a) % b , then fix up the sign of r to match your requirements. C++ has the % operator,

I'm trying to compute the following number in a C program : result = (3 * pow(2,500000000) - 2 ) % 1000000000 The power of 2 is way to large to be handled properly => I'm under the impression I could split the calculation in many steps using the modulo to reduce the result size. Does someone has a strategy for doing so ? Any other idea ? Thanx in advance Manu The simplest method is exponentiation by repeated squaring reducing by the modulus in each step. unsigned long long mod_pow(unsigned long long base, unsigned long long exponent, unsigned long long modulus) { if (exponent == 0) return 1;