There are some stations on the X-axis. The distance between the i-th and i+1-th station is i+1. It takes one second to move one unit of distance, also it takes one second to move from one station to the next. We want to go from point 0 to point X in the minimum time.
Let di denote the position of the i-th station. The distance between station i and i-1 is equal to i. Therefore di = di-1 + i.
If we apply the recurrence recursively we get that di is equal to the sum of integers from 1 to i. In closed form: di = (i*(i+1))/2.
Note that triangular numbers increases quadratically. There are not many of them below 109, actually there are only 44720.
Chef wants to go to position X. So he can walk to station 1, then go by trains to the nearest station to X, and finally go walking the remaining distance.
Now the problem is to find the leftmost and rightmost station to the cinema, and check which one is more convenient. Is possible to binary search the answer, but since constraints are small, we can just iterate over all possible triangular numbers.