4 ms·
> Are you claiming this doesn't work? No, I am not. I misinterpreted the "larger than all coefficients" because I was sure that the condition that at least one
by nanis 8y ago
> Are you claiming this doesn't work?
No, I am not. I misinterpreted the "larger than all coefficients" because I was sure that the condition that at least one of the coefficients other than the constant term had to be positive. You pointed out my mistake, you are correct, thank you.
> Are you claiming it's trivial to accomplish?
It is easy to state what you need to do. Find largest a such that k^a ≤ p(k). Find p(k) / k^a using integer division. Because division is a function, there is unique quotient and unique remainder. The quotient is your first coefficient. Take the remainder, repeat.
- ColinWright 8y agoRight - that's one way of doing it. It's not the way I first thought of, and the other way I have has a connection to something else that I think makes it all kinda nice. That solution is elsewhere in the thread, but here it is in rot13: jevgrcxvaonfrxnaqgurqvtvgfnergurpbrssvpvragf I'm pleased we've got it sorted, though. Thanks for your responses, and I'd be happy to hear more thoughts.