UVa 10006 - Carmichael Numbers

Summary
Follow the instructions, and realize that you can compute $$a^n \bmod n$$ in $$O(\log n)$$ time, and without needing BigNum. The rest is simple.

Explanation
The key is realizing that you can compute $$a^n \bmod n$$ in $$O(\log n)$$ time, and without needing BigNum. This means for a given $$n$$, the total complexity is $$O(n \log n)$$. (Note that the actual complexity relative to the input size is $$O(n2^n)$$) For small $$n$$'s, this is good enough.

Optimizations
Note that there are only 15 Carmichael numbers within the range given, so you can actually precalculate them and check them with a simple statement.

Input
1729 17 561 1109 431 0

Output
The number 1729 is a Carmichael number. 17 is normal. The number 561 is a Carmichael number. 1109 is normal. 431 is normal.

Solutions
C++: http://codealltheproblems.blogspot.com/2015/06/uva-10006-carmichael-numbers.html