UVa 369 - Combinations

Problem Number - Problem Name
[ 369 - Combinations]

Summary
You are required to evaluate $$ (^nC_m) $$ which is equal to $$ \frac{n!}{m! * (n-m)!}$$

Explanation
Consider the following example: $$ (^{100}C_6) $$

$$ \frac{100!}{(100-6)! * 6!}=$$ $$ \frac{100!}{94! *6!}=$$ $$ \frac{94! * 95 * 96 * 97 * 98 * 100}{94! * 6!}$$

Cancel 94!

$$ \frac{95 * 96 * 97 * 98 * 99 * 100}{2 * 3 * 4 * 5 * 6}$$

Cancel 96 with 2*6 = 8 and 99 with 3 = 33 and 100 with 4*5 = 5

$$ 95 * 8 * 97 * 98 * 33 * 5 = 1192052400 $$

Gotchas
You should cancel and simplify the formula as much as possible, to avoid overflow.

Implementations
Using long int in storing the result should be enough.

Optimizations
Just Simplify the formula to fit in a 32-bit int.

Alternative solution
Generate Pascal's triangle by using addition and then just do a simple lookup for each of the queries. It's a simpler method which is also much faster.

Input
20 5 18 6 15 7 10 5 100 100 40 5 0 0

Output
20 things taken 5 at a time is 15504 exactly. 18 things taken 6 at a time is 18564 exactly. 15 things taken 7 at a time is 6435 exactly. 10 things taken 5 at a time is 252 exactly. 100 things taken 100 at a time is 1 exactly. 40 things taken 5 at a time is 658008 exactly.