UVa 107 - The Cat in the Hat

Summary
We need to find two integer numbers N and M, where N is the number of cats inside each (non-smallest) cat's hat and M is the number of different heights in all cats.

Explanation

 * In order to solve this problem we have to find the two integers N and M. We consider the height of the first cat = A and the number of working cats = B. Using some simple math logics we can write this equation: $$\left( \frac{N}{N + 1} \right)^{M - 1} = \frac{B}{A}$$


 * We search M in the interval (2, HighValue) and binary search for R such that $$R^{M - 1} = B$$. We do the same for T such that $$T^{M - 1} = A$$. If T = R + 1 then we find N = R. With the curent values of N and M we compute the output numbers:


 * $$1 + N + N^2 + \ldots  + N ^{M - 2}$$ AND $$A\left(1 + \frac{N}{N + 1} + \frac{N}{N + 1}^2 + \ldots   \frac{N}{N + 1}^{M - 1}\right)$$

Gotchas

 * Take care of "1 1" input case. Solving this case doesn't work with the upmentioned algorithm.
 * I used HighValue = 200 in AC code.
 * Take care of precision issues in the use of log.
 * Consider modulus operator (%) to test A/T is an integer or not in integer-only implementation.

Input
1 1 216 125 5764801 1679616 1024 243 2 1 4 1 1024 1 371293 248832 11 10 1048576 59049 483736625 481890304 125 64 64 1 81 64 0 0

Output
0 1 31 671 335923 30275911 121 3367 1 3 2 7 10 2047 22621 1840825 1 21 29524 4017157 615441 1931252289 21 369 6 127 9 217

Solution

 * C++: http://acm-solution.blogspot.com/2010/11/acm-uva-107-cat-in-hat.html
 * Code All the Problems