Easy programming-language-knowledge check typical of the first task of AtCoder beginner contests. The following is a passing Kotlin submission:

fun main() {
    val k = readLine()!!.toInt()
    val ans = "ACL".repeat(k)
    println(ans)
}

Assume there is an integer in both ranges and call it $$$x$$$. Note that both $$$x \ge \max(A, C)$$$ and $$$x \le \min(B, D)$$$ must therefore be true.

Thus $$$x$$$ exists if and only if $$$\max(A, C) \le \min(B, D)$$$ is true. This is the standard formula for finding the overlap of two ranges.

Note that the initial network of cities can be divided into one or more components, where any city of a component can reach any other city in that component following roads, but there are no roads connecting components together.

If there are $$$c$$$ components, Snuke can connect two of them in one operation, which combines them to a single component, so the answer would be $$$c - 1$$$.

Components can be found using breadth-first/depth-first searches, however there's an easier way if you already have a special data structure known as DSU (disjoint-set union). And there is an implementation available right in the AtCoder library. So start with $$$c := N$$$, for each road given in the input, you can use same to check if the vertices are already connected; if there isn't, merge them and decrement $$$c$$$.

Note that though the input is $$$1$$$-indexed, the AtCoder DSU implementation is $$$0$$$-indexed.

Can be done with basic dynamic programming. Let $$$dp_i$$$ represent the number of sequences whose sum is $$$i$$$. $$$dp_0 = 1$$$, the empty sequence. Then $$$\displaystyle dp_i = \sum_{j=0}^{i-3} dp_j$$$, i.e. the sum of $$$dp_j$$$ for all integers $$$j$$$ between $$$0$$$ and $$$i-3$$$ inclusive, representing the addition of a number $$$\ge 3$$$ to each of the previous sequences. The final answer, naturally, is $$$dp_s$$$.

This can be easily calculated in $$$O(s^2)$$$, which is good enough to pass, however it can be improved to $$$O(s)$$$ by accumulating the prefix sum.

An $$$O(\log s)$$$ solution is possible via matrix exponentiation.

We can take the following approach — for each $$$i$$$ from $$$2$$$ to $$$n$$$, try to match $$$(x_i, y_i)$$$ with the "best" previous point $$$(x_j, y_j), 1 \le j < i$$$, then update the answer.

Testing all previous points is too slow ($$$O(n^2)$$$). However, it is sufficient only to keep up to $$$4$$$ previous points to test against. $$$2$$$ of those points are the maximal and minimal of the "score" $$$x_i + y_i$$$, while the other $$$2$$$ maximize and minimize $$$x_i - y_i$$$

Why? Note that taking any point as the center, the set of points with Manhattan distance some arbitrary constant $$$d$$$, forms a "diamond" shape, i.e. a square rotated 45 degrees. Thus the furthest point must lie on at least one of the four sides of this square when $$$d$$$ is maximized. As the sides are on lines with equations of $$$x+y = C$$$ or $$$x-y = C$$$, the maximums and minimums of these values are sufficient to find the furthest point in each of the 4 possible directions.

Thus we have a solution in $$$O(n)$$$.

Let $$$C$$$ be the array that holds the answer sequence. At first, we can greedily put any value of $$$B$$$ that doesn't match $$$A_i$$$ for each position from left to right.

If this strategy fails, it means that we are in a position where $$$A_i = z$$$ and all remaining members of $$$B$$$ are $$$z$$$. If we put all the remaining $$$z$$$ values into $$$C$$$, the situation looks like this:

A = [   < z   ][           z           ][    > z    ]
C = [ maybe z ][  not z  ][            z            ]
                          {TROUBLE ZONE}

So to fix the "trouble zone", we need to swap the $$$z$$$ values in it with some other values. It can be seen that the only values we can swap with lie in the "maybe $$$z$$$" zone; we can't swap with "not $$$z$$$" as that will cause another collision.

Thus we can iterate through the "maybe $$$z$$$" zone to try to find enough non-$$$z$$$ values to swap with. If we run out of values to swap with without eliminating all collisions, the answer is No, as there are no more values to swap with.