sinus_070's blog

By sinus_070, history, 10 months ago, In English,
inline int inv(int a, int m) {
  a %= m;
  if (a < 0) a += m;
  int b = m, u = 0, v = 1;
  while (a) {
    int t = b / a;
    ux++;
    b -= t * a; swap(a, b);
    u -= t * v; swap(u, v);
  }
  assert(b == 1); 
  if (u < 0) u += m;
  return u;
}

I bumped into this while reading tourist's code for today's Atcoder's F. It finds the inverse of a modulo m if it exists. All I could understand is that, b has to be 1 in the end for the inverse to exist. Could anyone help me understand this code?

TIA.

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