how to calculate reverse modulus
11,513
Solution 1
private int ReverseModulus(int div, int a, int remainder)
{
if(remainder >= div)
throw new ArgumentException("Remainder cannot be greater than or equal to divisor");
if(a < remainder)
return remainder - a;
return div + remainder - a;
}
e.g. :
// (53 + x) % 62 = 44
var res = ReverseModulus(62,53,44); // res = 53
// (2 + x) % 8 = 3
var res = ReverseModulus(8,2,3); // res = 1
Solution 2
It may not be the X that was originally used in the modulus, but if you have
(A + x) % B = C
You can do
(B + C - A) % B = x
Solution 3
x = (44 - 53) % 62
should work?
x = (44 - a) % length;
Solution 4
how about
IEnumerable<int> ReverseModulo(
int numeratorPart, int divisor, int modulus)
{
for(int i = (divisor + modulus) - numeratorPart;
i += divisor;
i <= int.MaxValue)
{
yield return i;
}
}
I'm now aware this answer is flawed because it does not gice the smallest but a .First()
would fix that.
Author by
Ivan Li
Updated on June 04, 2022Comments
-
Ivan Li almost 2 years
now I have one formula:
int a = 53, x = 53, length = 62, result; result = (a + x) % length;
but how to calculate reverse modulus to get the smallest "x" if I known result already
(53 + x) % 62 = 44 //how to get x
i mean what's the formula or logic to get x