[−][src]Function rdcl_aoc_helpers::math::solve_crt
pub fn solve_crt((n_1, a_1): (u64, u64), (n_2, a_2): (u64, u64)) -> u64
Solve the chinese remainder theorem for (n1, a1) and (n2, a2). We assume that:
- n1 and n2 are coprime
- n1 and n2 are no more than 63 bits (as they are converted to i64)