modulo

问题 I need to perform a real mathematical modulo in C. It makes sense for me to allow negative numbers for the moduled argument, since my modular calculations can produce negative intermediate results, which must be put back into the least residue system. But it makes no sense to allow negative module, therefore i wrote unsigned int mod( int x, unsigned int m ) { int r = x % m; return r >= 0 ? r : r + m; } However calling such function with negative number and positive module printf("%u\n", mod(

问题 have been searching for a mod operator in ios, just like the % in c, but no luck in finding it. Tried the answer in this link but it gives the same error. I have a float variable 'rotationAngle' whose angle keeps incrementing or decrementing based on the users finger movement. Some thing like this: if (startPoint.x < pt.x) { if (pt.y<936/2) rotationAngle += pt.x - startPoint.x; else rotationAngle += startPoint.x - pt.x; } rotationAngle = (rotationAngle % 360); } I just need to make sure that

问题 I need to perform modular exponentiation on quite large numbers on both python3 and javascript. I have functions that do the task, but they give me different outputs. Python (all of the three work in the same way): pow(176672119508, 55, 200000023499) def expmod_iter(a,b,c): x = 1 while(b>0): if(b&1==1): x = (x*a)%c a=(a*a)%c b >>= 1 return x%c def pow_mod(x, y, z): number = 1 while y: if y & 1: number = number * x % z y >>= 1 x = x * x % z return number # The result is always 124912252967 and

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 I was wondering how JavaScript handles modulo. For example, what would JavaScript evaluate 47 % 8 as? I can’t seem to find any documentation on it, and my skills on modulo aren’t the best. 回答1: Exactly as every language handles modulo: The remainder of X / Y . 47 % 8 == 7 Also if you use a browser like Firefox + Firebug, Safari, Chrome, or even IE8+ you could test such an operation as quickly as hitting F12. 回答2: Modulo should behave like you expect. I expect. 47 % 8 == 7 Fiddle Link 回答3:

问题 I am trying to compute a^b^c mod p for some positive integers a,b,c,p. One possible (and obvious) way is to use fast modular exponentiation which will run in O(log(b^c))=clog(b) . While I don't mind the efficiency here, the obvious downfall of this method is that you need an explicit binary representation of b^c which in itself is already exponential. So the question for me is, if I can not represent b^c as a binary representation, is there a way I can compute a^b^c mod p from the binary

问题 Hashing using division method means h(k) = k mod m . I read that m should not be power of 2. This is because if m = 2^p, h becomes just the p lowest-order bits of k. Usually we choose m to be a prime number not too close to a power of 2. Could someone explain with a small example the lowest order bits part? I thought all (mod m) does is that it wraps the result around a range m. Somehow cant see the issue if m was power of 2. 回答1: All data in the computer is stored as binary data. A binary

问题 I want to run a function every 5 minutes and have it synced with the clock. If I use time.sleep(60*5), the time starts to drift because my function adds a tiny bit of processing time. Is this a good way of running my function synced with the clock or is there a better way in python? def run(condition): def task(): #run data here pass runOnce = True while condition: if dt.datetime.now().minute % 5 == 0 and dt.datetime.now().second == 0 and runOnce: runOnce = False task() elif dt.datetime.now()

问题 In MIPS, I am confused on how to get the mod to work. Below is the code I have come up with thus far. I may have more errors besides the mod, but I feel those errors are a result of the mod misunderstanding. All I'm trying to do is to get the working code (python) here: i = 1 k = 0 while i < 9: if i % 2 != 0: k = k + i i += 1 print(k) to be correctly translated into MIPS. This is my first shot at assembly, so there may be more than mod errors that are tripping me up in the code below: # Takes

问题 Until recently (i.e. C99), the behavior of the modulo operator was implementation defined in C. Since Perl 5 is written in C, is it reliant on the behavior of the C compiler used to build it? 回答1: No, Perl 5 defines the modulo operator in perlop and even has tests to ensure it works as documented. from perl/t/op/arith.t tryeq $T++, 13 % 4, 1; tryeq $T++, -13 % 4, 3; tryeq $T++, 13 % -4, -3; tryeq $T++, -13 % -4, -1; However, if you use the integer pragma, you are at the mercies of the C