Only the following pairs of numbers are possible: $$$(x, x)$$$, $$$(x, 2 \cdot x)$$$, and $$$(x, 3 \cdot x)$$$

Notice that only the following pairs of numbers are possible: $$$(x, x)$$$, $$$(x, 2 \cdot x)$$$, and $$$(x, 3 \cdot x)$$$.

The number of the pairs of the first kind is $$$n$$$, of the second kind is $$$\lfloor \frac{n}{2} \rfloor$$$, and of the third kind is $$$\lfloor \frac{n}{3} \rfloor$$$. Therefore, the answer to the problem is $$$n + \lfloor \frac{n}{2} \rfloor + \lfloor \frac{n}{3} \rfloor$$$.

Solve the problem for $$$k = 2$$$.

Solve the problem without the "$$$(r, c)$$$ must be colored" limitation.

Try to paint diagonals

No, you don't need casework

Notice that the answer to the problem is at least $$$\frac{n^2}{k}$$$, because you can split the square into so many non-intersecting rectangles of dimensions $$$1 \times k$$$. So let's try to paint exactly so many cells and see if maybe it's always possible.

For simplicity, let's first solve the problem without necessarily painting $$$(r, c)$$$. In this case, we're looking for something like a chess coloring, which is a diagonal coloring.

Let's number the diagonals from the "lowest" to the "highest". Notice that every $$$1 \times k$$$ and $$$k \times 1$$$ subrectangle intersects exactly $$$k$$$ consecutive diagonals, so we can paint every $$$k$$$-th diagonal to obtain the required answer: every such subrectangle will contain exactly one painted cell.

To add the $$$(r, c)$$$ requirement back, notice that $$$(r, c)$$$ lies on the diagonal number $$$r + c$$$. (Because if you trace any path from $$$(0, 0)$$$ to $$$(r, c)$$$ with non-decreasing coordinates, going one cell upwards or rightwards increases exactly one of the coordinates by one, and also increases the number of the diagonal by one). Therefore, all we need to do is paint the cells whose coordinates satisfy $$$(x + y) \% k = (r + c) \% k$$$

We've got a problem if $$$b_{i} \ge b_{i + 1} + 1$$$

It's alright if $$$a_i = b_i$$$

Programmer aren't mathematicians, you don't need to prove the solution

Firstly, we obviously require $$$a_i \le b_i$$$ to hold for all $$$i$$$. With that out of our way, let's consider non-trivial cases. Also let $$$a_{n+1} = a_1, b_{n+1} = b_1$$$ cyclically.

For each $$$i$$$, we require that either $$$a_i = b_i$$$ or $$$b_i \leq b_{i + 1} + 1$$$ holds. That's because if we increment $$$a_i$$$ at least once, we had $$$a_i = b_i - 1$$$ and $$$a_{i + 1} \le b_{i + 1}$$$ before the last increment of $$$a_i$$$, and from here it's just a matter of simple algebraic transformations.

Now let's prove these two conditions are enough. Let $$$i$$$ be the index of the minimal element of $$$a$$$ such that $$$a_i < b_i$$$ (i.e. the smallest element that's not ready yet). Notice that in this case we can, in fact, assign $$$a_i := a_i+1$$$, because $$$a_i \leq b_i \leq b_{i + 1} + 1$$$ holds, and now we're one step closer to the required array. It's easy to continue this proof by induction.

Reformulate this problem in terms of a complete binary tree.

It would be strange for the sponsors to changes two nodes of the same depth.

The problem can be reformulated as follows. We've got a complete binary tree with $$$2^n$$$ leaves. There's a marked edge from each intermediate node to one of its children. The winner is the leaf reachable from the root via marked edges. Changes modify the outgoing marked edge of a node.

Now it should be fairly obvious that there's no reason to change more than one node per level, because only one node matters per level--the one on the path from the root to the answer node. So, the winner only depends on the subset of levels we perform changes on, and vice versa: different subsets always yield different winners.

Sponsors can change exactly $$$i$$$ nodes in $$$\binom{n}{i}$$$ ways. Summing this over $$$i$$$, we get $$$\sum_{i=0}^{\min(n, k)} \binom{n}{i}$$$. Call this number $$$m$$$. $$$m$$$ is the number of winners the sponsors choose between--let's call them candidates for brevity. It's easy to see that $$$m$$$ is the answer to the problem, because a) sponsors can guarantee the winner is at least $$$m$$$, as, independent of the list of candidate winners "provided" by Madoka, at least one of them must be at least $$$m$$$, and b) Madoka can guarantee the winner is at most $$$m$$$ by firstly marking edges arbitrarily, then computing the list of candidate nodes, and only then fill them with numbers from $$$1$$$ to $$$m$$$ (and the other nodes arbitrarily).

Bruteforce $$$c$$$

$$$gcd(a, b)$$$ divides $$$a + b$$$.

Recall the existence of the Euler's totient function

Let's bruteforce $$$c$$$, then we have $$$gcd(a, b) = gcd(a, a + b) = gcd(a, n - c)$$$. This means that $$$gcd(a, b)$$$ divides $$$n - c$$$, so let's just go through all divisors of $$$n - c$$$. For every factor $$$d$$$, the count of pairs $$$(a, b)$$$ satisfying $$$a + b = n - c$$$ and $$$gcd(a, b) = d$$$ is $$$\phi (\frac{n - c}{d})$$$, because we need $$$d$$$ to divide $$$a$$$ and be coprime with $$$\frac{n - c}{d}$$$, so that the $$$gcd$$$ is equal to $$$d$$$.

Therefore, the answer to the problem is $$$\sum{lcm(c, d) * \phi{\frac{n - c}{d}}}$$$, where $$$1 \leq c \leq n - 2$$$ and $$$d$$$ is a factor of $$$n - c$$$.