how to calculate reverse modulus

问题:

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

回答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

回答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

回答3:

x = (44 - 53) % 62 should work?

x = (44 - a) % length;

回答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.

回答5:

Who needs a computer? If 53 + x is congruent to 44, modulo 62, then we know that for integer k,

53 + x + 62*k = 44

Solving for x, we see that

x = 44 - 53 - 62*k = -9 - 62*k

Clearly the smallest solutions are -9 (when k=0) and 53 (when k=1).