Let us take a look only at the cuts ~x = c~ (solution for cuts ~y = c~ is analogous).

Notice the pastries that are cut by a line ~x = c~ are not completely left nor completely right to the line. The solution for this line is therefore:

$$\displaystyle N - \operatorname{number\_of\_pastries\_left}(c) - \operatorname{number\_of\_pastries\_right}(c)$$

We will calculate the function values of ~\operatorname{number\_of\_pastries\_left}(x)~ and ~\operatorname{number\_of\_pastries\_right}(x)~ before reading the cuts (therefore answering to each cut in a constant time complexity). We calculate the values using the following relation:

$$\displaystyle \operatorname{number\_of\_pastries\_left}(x) = \operatorname{number\_of\_pastries\_left}(x-1) + \operatorname{number\_of\_pastries\_with\_a\_rightmost\_point\_equal\_to}(x)$$

and an analogous relation for the second function. We read the values of the auxiliary function ~\operatorname{number\_of\_pastries\_with\_a\_rightmost\_point\_equal\_to}(x)~ from an array whose elements we increase during the input of the pastries.

An alternative solution uses a sweep-line algorithm and is left as an exercise to the reader.