1. if $$$B > A$$$, then $$$A \% B = A$$$
2. if $$$B \leq A$$$, then $$$A \% B \leq A/2$$$

For each trip, then we'll do $$$O(log(10^9))$$$ operations of type $$$2$$$, and all the rest will be of type $$$1$$$. But operations of type $$$1$$$ don't do nothing, the number of alfajores left doesn't change when applying them, so we should try to come up with a way of completely avoiding them.

A way of avoiding the useless operations would be by being able to solve the following sub-problem: "find the next office where some alfajores will be given away", i.e. assuming we are at office $$$i$$$ and we have $$$A$$$ alfajores left, find the left-most index $$$j$$$ such that $$$E_j \leq A$$$ and $$$j > i$$$.

This sub-problem can be solved in different ways. One example is using segment tree or similar data structure. Another way is to modify the array $$$E$$$ into a decreasing array at the beginning, and then do binary search over it every time we need to solve the sub-problem.

The improved solution is then $$$O(N \cdot log(M) \cdot log(10^9))$$$

$$$S = x_1 \cdot n + D \cdot \frac{n \cdot (n-1)}{2}$$$

From the hint above, we know the value of $$$S$$$ and we have $$$3$$$ integer variables $$$(x_1$$$, $$$n$$$ and $$$D)$$$, each of which could take $$$O(10^5)$$$ different values.

If we iterate $$$2$$$ of those variables, then we can solve the equation and find the value of the third one, if exists. At first this is looks like a quadratic time solution, which is too slow, but we can speed this up if we carefully use the restriction of the statement saying that $$$x_i = A$$$ for some $$$i$$$.

Let's define $$$K = R-L$$$

We can iterate $$$D$$$ and $$$x_1$$$ in the following way

for(D = 1; D <= K; D++)
  for(x1 = A; x1 >= L; x1 -= D)
    check if there is a valid value for n

The check will then be done approximately the following number of times: $$$\frac{K}{1} + \frac{K}{2} + \frac{K}{3} + ... + \frac{K}{K}$$$

And this number is $$$O(K \cdot log(K))$$$ (why?)

The check could be done using the Bhaskara's formula or binary search (be careful with the overflow). And the final time complexity of the solution would then be $$$O(K \cdot log(K))$$$ or $$$O(K \cdot log(K) \cdot log(S))$$$

One more detail, be careful with the case $$$A=S$$$