如何计算反向模量

Question

现在我有一个公式:

int a = 53, x = 53, length = 62, result;
result = (a + x) % length;

但如果我已知结果,如何计算反向模量以获得最小的"x"

(53 + x) % 62 = 44
//how to get x

我的意思是获得x的公式或逻辑是什么

Answer 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;
}

例如:

// (53 + x) % 62 = 44
var res = ReverseModulus(62,53,44); // res = 53

// (2 + x) % 8 = 3
var res = ReverseModulus(8,2,3); // res = 1

Answer 2

它可能不是最初在模数中使用的X,但如果你有

(A + x) % B = C

你可以做

(B + C - A) % B = x