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:

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

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 $B-x$.