Observe that the even-indexed character of the string can be transformed from A-Z to Z-A. E.g. for the first example:

• AA → AZ
• AB → AY
• BB → BY
• BA → BZ
• AZ → AA

Now, you can use any known algorithms to sort the string as usual. You can sort it in linear time with trie, or std::sort in $$$O(nm \log n)$$$ time.

Time Complexity : $$$O(nm)$$$ or $$$O(nm \log n)$$$

We can binary search the answer $$$r$$$ in this case. Here, bird's habitats are referred as points. First of all, define a function $$$c(x)$$$ as the maximum number of points that can be covered with a circle of radius $$$x$$$ through the origin.

Define the park as a circle with radius $$$x$$$ and $$$\theta$$$, in a polar coordinate representation. Observe that each points have a radial/angle segment of which the point $$$p_i$$$ will be inside the circle if and only if $$$\theta$$$ belongs to the radial segment of $$$[L_{p_{i}}, R_{p_{i}}]$$$, where $$$-\pi < L_{p_{i}}, R_{p_{i}} \leq \pi$$$.

E.g for $$$x = 4$$$, Observe the $$$L_{p_{3}}$$$ for the $$$p_3 = (1, 5)$$$.

The green radial segment $$$e$$$ represents the $$$[L_{p_{3}}, R_{p_{3}}]$$$. Now, to find the two end points $$$B_i$$$ and $$$C_i$$$ of the arc for each point $$$p_i$$$. Because the triangle that is made by those 3 points are an isosceles triangle, simply find the angle where the distance of $$$p_i$$$ and $$$B_i$$$ equals to $$$x$$$, that is $$$\Delta = \cos^{-1}\dfrac{||p_i||}{2r}$$$. Now the segment can be found by calculating the angle of $$$\tan^{-1}p_i \pm \Delta$$$. Do a radial sweep to find the maximum number of points.

Time complexity is $$$O(n \log n \cdot \log(\text{MAX_R} \cdot \epsilon^{-1}))$$$.

We can optimize the binary search part further, since we only need $$$\log(\epsilon^{-1})$$$ most significant digits. We can binary search the position of the first non-zero digit in $$$O(\log\log(\text{MAX_R}))$$$, then use a normal binary search with $$$O(\log(\epsilon^{-1}))$$$ steps. In practice, this improves the time by around a factor of 2.

Time complexity: $$$O(n \log n \cdot \log(\text{MAX_R} \cdot \epsilon^{-1}))$$$ or $$$O(n \log n \cdot \log(\log(\text{MAX_R}) \cdot \epsilon^{-1}))$$$.

Let a valid segment $$$[l, r]$$$ be a segment in $$$b$$$ where the sum of elements in the segment is divisible by $$$k$$$.

We can try to solve a simpler problem: find the number of valid segments such that the right endpoint ends at $$$1$$$. That is, the valid segments $$$[l, 1]$$$ ($$$1 \leq l \leq n \cdot m$$$).

Let $$$prefix(p) = \sum_{i=1}^{p} {b[i]}$$$ and $$$cnt$$$ be an array where the $$$cnt[i]$$$ denote the number of $$$p$$$ ($$$1 \leq p \leq n \cdot m$$$) such that $$$i \equiv prefix(p) \mod k$$$.

Notice that the number of valid segment $$$[l, 1]$$$ is $$$cnt[prefix(n \cdot m) + b[1]]$$$. Furthermore, the number of valid segments $$$[l, 1 + x \cdot n]$$$ ($$$0 \leq x \leq m-1$$$) is the same as the number of valid segment $$$[l, 1]$$$.

Thus, we only need to calculate the number of valid segments for $$$[l, r]$$$ with $$$1 \leq l \leq n \cdot m$$$ and $$$1 \leq r \leq n$$$, then multiply the final result by $$$m$$$.

First we need to find the array $$$cnt$$$. Let $$$sum = prefix(n)$$$.

When $$$sum \equiv 0 \mod k$$$, we can find $$$cnt$$$ in a straightforward manner.

Now assume $$$sum \not\equiv 0 \mod k$$$. For a fixed $$$i$$$, let's try to find the contribution of $$$prefix(i + x \cdot n)$$$ for all $$$0 \leq x \leq m-1$$$ to $$$cnt$$$ at once. Observe that if one make a directed graph with $$$(i, \ (i + sum) \bmod k)$$$ for $$$0 \leq i < k$$$ as the edges, one will get a cycle of length $$$k$$$ (since $$$k$$$ is prime) as the result. To find the contribution of $$$prefix(i + x \cdot n)$$$, we can do a range add operation on this cycle. This can be done with offline prefix sums (prefix difference) in $$$O(k)$$$ total.

Now that we have the array $$$cnt$$$, we can find the number of valid segments that ends at $$$1$$$ easily. To find valid segment that ends at index $$$2$$$, we can modify $$$cnt$$$ by adding $$$prefix(n \cdot m) + b[1]$$$ to the counter and removing $$$b[1]$$$. We do this for all $$$1 \leq r \leq n$$$.

This solution is also applicable for arbitary $$$k$$$, albeit multiple cycles will be generated and must be handled separatedly.

Time complexity: $$$O(n + k)$$$

There are no dirty tricks to solve this problem. Brute force all possible number between $$$i \in [10^{|s| - 1}, 10^{|s|} - 1]$$$, with step $$$i := i + 25$$$. You might want to handle when $$$|s| = 1$$$, because $$$0$$$ is a valid $$$s$$$, if possible. For easier implementation, you can use the std::to_string(s) in C++.

It is also possible to solve it in $$$O(|s|)$$$ by case analysis.

Time complexity: $$$O(\frac{1}{25} \cdot |s| \cdot 10^{|s|})$$$ or $$$O(|s|)$$$.

We can use centroid decomposition to solve this problem.

Suppose we find the centroid $$$cen$$$ of the tree, and root the tree at $$$cen$$$. We consider each subtree of the children of $$$cen$$$ as different groups of vertices. We want to find the sum of $$$f(u,v)$$$ for all valid tours, such that $$$u$$$ and $$$v$$$ are from different groups.

We can solve this with basic inclusion-exclusion. We count the sum of $$$f(u,v)$$$ where the path $$$u \to cen \to v$$$ uses less than $$$k$$$ tickets, without caring which group $$$u,v$$$ belongs to. Then, we can subtract it by only considering $$$u \to cen \to v$$$ where $$$u,v$$$ belongs from the same group.

Define $$$cost(u)$$$ as the number of tickets you need to go from $$$u$$$ to $$$cen$$$. For a fixed set of vertices $$$S$$$, you can count $$$f(u,v)$$$ where $$$cost(u) + cost(v) + z \leq k$$$ with prefix sums. Note that $$$z$$$ depends on whether the last edge of the path from $$$u \to cen$$$ and $$$v \to cen$$$ has different colors. We can do all of these in $$$O(|S|)$$$.

We use the solution above while setting $$$S$$$ as the set of all vertices in $$$cen$$$'s subtree, or the set of vertices with the same group.

Because the depth of a centroid tree is $$$O(\log n)$$$, the overall complexity of the solution is $$$O(n \log n)$$$.

We can use this trick, which is also explained below.

Suppose $$$a_i \neq -1$$$ for now. We want to find a function $$$F(a)$$$ such that $$$\mathbb{E}(F_{t + 1} - F_t | F_t) = -1$$$, where $$$F_t$$$ is the value of $$$F(a)$$$ at time $$$t$$$. If we can find such a function, then the expected stopping time is equal to $$$F(a_0) - F(a_T)$$$, where $$$a_0$$$ is the initial array before doing any operation, and $$$a_T$$$ is the final array where we don't do any more operation (that is, all elements of $$$a_T$$$ are equal).

Suppose $$$occ(x)$$$ is the number of occurrences of $$$x$$$ in the current array, for some $$$0 \leq x < k$$$. It turns out we can find such $$$F$$$ satisfying $$$F = \sum_{x = 0}^{k - 1} f(occ(x))$$$ for some function $$$f$$$. We now try to find $$$f$$$.

Suppose we currently have $$$a_t$$$, and we want to find the expected value of $$$F(a_{t + 1})$$$. There are two cases to consider:

• $$$\forall x, occ_{t + 1}(x) = occ_t(x)$$$ if $$$a_i$$$ doesn't change when doing the operation. This happens with probability $$$\frac{1}{k} \cdot \frac{occ_t(x)}{n}$$$ for each $$$x$$$.
• Otherwise, there exist some $$$x, y$$$ ($$$x \neq y$$$) such that $$$occ_{t + 1}(x) = occ_t(x) - 1$$$ and $$$occ_{t + 1}(y) = occ_t(y) + 1$$$. This happens if initially $$$a_i = x$$$, then by doing the operation we change it to $$$y$$$. This happens with probability $$$\frac{1}{k} \cdot \frac{occ_t(x)}{n}$$$ for each $$$x,y$$$.

Thus,

Suppose $$$a = occ_t(x)$$$. If we can find $$$f$$$ such that

then $$$f$$$ satisfies $$$F$$$.

So we can set $$$f$$$ to any function that satisfies the recursive formula above, and then derive $$$F$$$.

To handle $$$a_i = -1$$$, note that $$$F$$$ depends only on the occurrence of each value $$$x$$$ ($$$0 \leq x < k$$$), and each of them is independent. Therefore, we can count the contribution for each $$$x$$$ towards all possible final arrays separately. This is easy to do in $$$O(n)$$$.

Moreoever, there is only $$$O(\sqrt{n})$$$ values of $$$occ(x)$$$ in the initial array (before changing $$$a_i = -1$$$), and each $$$x$$$ with the same occurrences contribute the same amount. Therefore, we can solve the problem in $$$O(n \sqrt{n})$$$.

Define: - $$$d(n)$$$ as the set of all divisors of $$$n$$$; - $$$\phi(x)$$$ as the euler totient function of $$$x$$$; - $$$d(a, b)$$$ as the set of all divisors of both $$$a$$$ and $$$b$$$; or equivalently, $$$d(\gcd(a, b))$$$; - $$$a \in d(b)$$$ means $$$a$$$ divides $$$b$$$ or $$$b$$$ is divisible by $$$a$$$.

Observe that $$$\sum_{x \in d(n)}\phi(x) = n$$$. This implies $$$\sum_{x \in d(a, b)}\phi(x) = \gcd(a,b)$$$

If we only iterate $$$y$$$ where $$$y$$$ is a divisor of one of $$$a_{ix}$$$, we can compute the above summation in $$$O(n \log n \max_{i=1}^n(|d(a_i)|))$$$.

Do a dynamic programming with three states: - Position in $$$s$$$ - Position in $$$t$$$ - How many matches left.

define the dynamic programming of $$$dp[a][b][rem]$$$ as the minimum cost of having the string $$$p = s[1..a]$$$, $$$rem$$$ matches left, and the longest prefix match between $$$s$$$ and $$$t$$$ is at $$$b$$$. The answer will be at $$$dp[n][c][0]$$$ for any arbitrary $$$c$$$.

The transition can first be precomputed with brute force in $$$O(n^3)$$$ or with Aho-Corasick.

Time complexity: $$$O(n^3)$$$ Space complexity: $$$O(n^2)$$$

Note that $$$\max(|a_x + a_y|, |a_x - a_y|) = |a_x| + |a_y|$$$.

Now the problem can be reduced to updating a vertex's value and querying the sum of values of vertices in a path.

This can be done in several ways. One can use euler tour tree flattening method, as described in Euler Tour Magic by brdy blog, or use heavy-light decomposition.

Time complexity : $$$O((q + n) \log^2 n)$$$ or $$$O((q + n) \log n)$$$

Naively simulating the ball's path is enough, and runs in $$$O(nm + nk)$$$. Note that if we visit a non-$$$2$$$ cell, then the path length of the current ball is increased by $$$1$$$, and then the cell turns into $$$2$$$. So the total length of all paths can be increased by at most $$$O(nm)$$$ times. In addition, each ball needs at least $$$O(n)$$$ moves to travel, so we get $$$O(nm + nk)$$$.

We can improve this further. You can speed up each drops by storing consecutive $$$2$$$-cell segments in the downwards direction for each column. Using a Disjoint-Set Union data structure, for each cell $$$a_{x,y} = 2$$$, join it with its bottom cell if $$$a_{x + 1, y} = 2$$$.

Time complexity: $$$O(k + rc\cdot\alpha(rc))$$$

Observe that only some several non-intersecting part of $$$nm - rc$$$ that is independent in the grid. Simple casework shows that the answer is $$$k^{nm}$$$ if $$$a = b$$$, and $$$k^{nm - rc}$$$ otherwise.

Time complexity: $$$O(\log nm)$$$