Consider the solution L ≤ D x mod M ≤ RL\le Dx \bmod M \le RL≤DxmodM≤R，其中 0 ≤ L ≤ R < M , D < M 0 \le L \le R < M, D < M 0≤L≤R<M,D<M。
Suppose there is a solution, then there is yyy， 使得L ≤ D x - M y ≤ ML \ le Dx -My \ le ML≤Dx−My≤M . Deformed to get
D x − R ≤ M y ≤ D x − L Dx-R\le My \le Dx-LDx−R≤My≤Dx−L
To DDD 取模，得到
− R m o d D ≤ ( M m o d D ) y m o d D ≤ − L m o d D -R \bmod D \le (M \bmod D) y \bmod D\le -L \bmod D −RmodD≤(MmodD)ymodD≤−LmodD
So it becomes a smaller form, which can be recursed according to this form. The complexity is the same as the division method, which is O (log n) O(\log n)O(logn ) . The solution obtained in this way is the smallest non-negative integer solution.
As for why the solution after such a transformation must be the same as the original question, and why the solution must be the smallest, I have not found a convincing answer.
Take POJ 3530 as an example, give the realization:
ll modinq(ll m, ll d, ll l, ll r){
// 0 <= l <= r < m, d < m, minimal non-negative solution
if (r < l) return -1;
if (l == 0) return 0;
if (d == 0) return -1;
if ((r / d) * d >= l) return (l - 1) / d + 1;
ll res = modinq(d, m % d, (d - r % d) % d, (d - l % d) % d);
return res == -1 ? -1: (m * res + l - 1) / d + 1;
}