Calculating the snail's position day by day would be too slow, since the expected result can be very large.

During one day and one night, the snail climbs exactly ~A-B~ meters. His entire trip consists of ~X~ days and ~X~ nights, plus the last day in which he reaches the top of the pole. Therefore, our solution is the smallest integer ~X~ for which ~X \times (A-B) + A \ge V~ holds.

Solution

It's now easy to find a direct expression for ~X~:

$$\displaystyle \begin{align*} X \times (A-B) + A &\ge V \\ X \times (A-B) &\ge V-A \\ X &= \left\lceil \frac{V-A}{A-B} \right\rceil \\ X &= \left\lfloor \frac{V-A+A-B-1}{A-B} \right\rfloor \\ X &= \left\lfloor \frac{V-B-1}{A-B} \right\rfloor \end{align*}$$

We use integer division, and print ~X+1~ as a result.

Alternative solution

We can use binary search to solve the problem. We start with some potential solution ~X'~, and check if the inequality holds. If the inequality doesn't hold, we must use a smaller ~X'~. If it holds, we can increase ~X'~ and try again. This gives us an ~\mathcal O(\log V)~ solution.