We say that vaults with a common x-coordinate form a column. There are ~A+B+1~ columns and the limits on ~A~ and ~B~ are relatively small, so we will solve the problem by counting vaults each of the guards sees in a particular column.

Let:

• ~a(x)~ be the total number of vaults in column ~x~ seen by the guard at ~(-A, 0)~;
• ~b(x)~ be the total number of vaults in column ~x~ seen by the guard at ~(B, 0)~;
• ~ab(x)~ be the total number of vaults in column ~x~ seen by both guards.

Knowing these numbers, we can calculate:

• The number of super-secure vaults in column ~x~ as ~ab(x)~;
• The number of secure vaults in column ~x~ as ~a(x) + b(x) - 2 \cdot ab(x)~;
• The number of insecure vaults as ~L - a(x) - b(x) + ab(x)~.

How do we calculate the number of vaults in a column? Let ~f(dx)~ be the number of vaults the guard can see in the column at distance ~dx~ from his position. For some ~y~, the guard can see the vault at ~(x+dx, y)~ only if ~y~ and ~dx~ are relatively prime. If the greatest common divisor between ~y~ and ~dx~ is greater than ~1~, then ~y~ and ~dx~ can be divided by the divisor to get the coordinates of another point blocking the view.

The number of integers relatively prime with ~dx~ that are at most ~L~ can be calculated by finding the prime factors of ~dx~ and using the inclusion-exclusion principle. Take for example ~L = 15~ and ~dx = 6~. Then the number of vaults the guard can see in a column at distance ~6~ from his position is:

$$\displaystyle f(6) = \left\lfloor \frac{15}{1} \right\rfloor - \left\lfloor \frac{15}{2} \right\rfloor - \left\lfloor \frac{15}{3} \right\rfloor + \left\lfloor \frac{15}{6} \right\rfloor = 15-7-5+2 = 5$$

A special case is ~f(0) = 1~, because the guard can only see the vault directly ahead of him in his column.

Now it is obvious that ~a(x) = f(x-A)~ and ~b(x) = f(B-x)~. It is less obvious that we can calculate ~ab(x)~ by applying the inclusion-exclusion principle to the union of prime factors of ~x-A~ and ~B-x~.