UVa 10079 - Pizza Cutting

10079 - Pizza Cutting

 * http://acm.uva.es/p/v100/10079.html

Summary
Given the number of straight cuts you can make on a pizza, find the maximum number of pieces in which the pizza can be divided.

Explanation
If we make no cuts at all, we still have the whole pizza to eat. Hence $$A(0)=1$$. Each new cut can possibly intersect all the previous cuts. After $$n-1$$ lines have been drawn, the nth line can intersect each of these, so it has $$n-1$$ intersections, and hence $$n$$ new regions, so the recurrence we get is $$A(n)=A(n-1)+n;\ A(0)=1$$. Solving this recurrence gives the formula $$A(n)=n+(n-1)+(n-2)+...2+1+A(0)$$ and hence $$A(n)=n(n+1)/2 + 1$$.

Implementation

 * Use 64-bit integers (long long in C++) to avoid overflow.

Input
0 5 10 210000000 -100

Output
1 16 56 22050000105000001