Suppose that you can use $$$x$$$ operations of type $$$1$$$ and $$$y$$$ operations of type $$$2$$$. Try to reorder the operations in such a way that $$$a$$$ becomes the minimum possible.

You should use operations of type $$$2$$$ first, then moves of type $$$1$$$. How many operations do you need in the worst case? ($$$a = 10^9$$$, $$$b = 1$$$)

You need at most $$$30$$$ operations. Iterate over the number of operations of type $$$2$$$.

Notice how it is never better to increase $$$b$$$ after dividing ($$$\lfloor \frac{a}{b+1} \rfloor \le \lfloor \frac{a}{b} \rfloor$$$).

So we can try to increase $$$b$$$ to a certain value and then divide $$$a$$$ by $$$b$$$ until it is $$$0$$$. Being careful as not to do this with $$$b<2$$$, the number of times we divide is going to be $$$O(\log a)$$$. In particular, if you reach $$$b \geq 2$$$ (this requires at most $$$1$$$ move), you need at most $$$\lfloor \log_2(10^9) \rfloor = 29$$$ moves to finish.

Let $$$y$$$ be the number of moves of type $$$2$$$; we can try all values of $$$y$$$ ($$$0 \leq y \leq 30$$$) and, for each $$$y$$$, check how many moves of type $$$1$$$ are necessary.

Complexity: $$$O(\log^2 a)$$$.

If we notice that it is never convenient to increase $$$b$$$ over $$$6$$$, we can also achieve a solution with better complexity.

You can make a $$$k$$$-similar array by assigning $$$a_i = x$$$ for some $$$l \leq i \leq r$$$ and $$$1 \leq x \leq k$$$.

How many $$$k$$$-similar arrays can you make if $$$x$$$ is already equal to some $$$a_i$$$ ($$$l \leq i \leq r$$$)?

How many $$$k$$$-similar arrays can you make if either $$$x < a_l$$$ or $$$x > a_r$$$?

How many $$$k$$$-similar arrays can you make if none of the previous conditions holds?

Let's consider each value $$$x$$$ from $$$1$$$ to $$$k$$$.

• If $$$x < a_l$$$, you can replace $$$a_l$$$ with $$$x$$$ (and you get $$$1$$$ $$$k$$$-similar array). There are $$$a_l-1$$$ such values of $$$x$$$.
• If $$$x > a_r$$$, you can replace $$$a_r$$$ with $$$x$$$ (and you get $$$1$$$ $$$k$$$-similar array). There are $$$k-a_r$$$ such values of $$$x$$$.
• If $$$a_l < x < a_r$$$, and $$$x \neq a_i$$$ for all $$$i$$$ in $$$[l, r]$$$, you can either replace the rightmost $$$b_i$$$ which is less than $$$x$$$, or the leftmost $$$b_i$$$ which is greater than $$$x$$$ (and you get $$$2$$$ $$$k$$$-similar arrays). There are $$$(a_r - a_l + 1) - (r - l + 1)$$$ such values of $$$x$$$.
• If $$$x = a_i$$$ for some $$$i$$$ in $$$[l, r]$$$, no $$$k$$$-similar arrays can be made.

The total count is $$$(a_l-1)+(k-a_r)+2((a_r - a_l + 1) - (r - l + 1))$$$, which simplifies to $$$k + (a_r - a_l + 1) - 2(r - l + 1)$$$.

Complexity: $$$O(n + q)$$$.

Let $$$\lfloor \frac{a}{b} \rfloor = a~\mathrm{mod}~b = k$$$. Is there an upper bound for $$$k$$$?

$$$k \leq \sqrt x$$$. For a fixed $$$k$$$, can you count the number of special pairs such that $$$a \leq x$$$ and $$$b \leq y$$$ in $$$O(1)$$$?

We can notice that, if $$$\lfloor \frac{a}{b} \rfloor = a~\mathrm{mod}~b = k$$$, then $$$a$$$ can be written as $$$kb+k$$$ ($$$b > k$$$). Since $$$b > k$$$, we have that $$$k^2 < kb+k = a \leq x$$$. Hence $$$k \leq \sqrt x$$$.

Now let's count special pairs for any fixed $$$k$$$ ($$$1 \leq k \leq \sqrt x$$$). For each $$$k$$$, you have to count the number of $$$b$$$ such that $$$b > k$$$, $$$1 \leq b \leq y$$$, $$$1 \leq kb+k \leq x$$$. The second condition is equivalent to $$$1 \leq b \leq x/k-1$$$.

Therefore, for any fixed $$$k > 0$$$, the number of special pairs ($$$a\leq x$$$; $$$b \leq y$$$) is $$$max(0, min(y,x/k-1) - k)$$$. The result is the sum of the number of special pairs for each $$$k$$$.

Complexity: $$$O(\sqrt x)$$$.

Brute force doesn't work (even if you optimize it): there are relatively few solutions.

There may be very few possible values of $$$b_{i,j}$$$, if $$$b_{i-1,j}$$$ is fixed. The problem arises when you have to find a value for a cell with, e.g., $$$4$$$ fixed neighbors. Try to find a possible property of the neighbors of $$$(i, j)$$$, such that at least a solution for $$$b_{i,j}$$$ exists.

The least common multiple of all integers from $$$1$$$ to $$$16$$$ is less than $$$10^6$$$.

Build a matrix with a checkerboard pattern: let $$$b_{i, j} = 720720$$$ if $$$i + j$$$ is even, and $$$720720+a_{i, j}^4$$$ otherwise. The difference between two adjacent cells is obviously a fourth power of an integer. We choose $$$720720$$$ because it is $$$\operatorname{lcm}(1, 2, \dots, 16)$$$. This ensures that $$$b_{i, j}$$$ is a multiple of $$$a_{i, j}$$$, because it is either $$$720720$$$ itself or the sum of two multiples of $$$a_{i, j}$$$.

Complexity: $$$O(nm)$$$.

Solve the problem with $$$d=2$$$, it can be generalized easily.

What happens if you can't swap coins?

Let $$$dp_i$$$ be the maximum score that you can reach after $$$dist(1, i)$$$ moves if you put a coin on node $$$i$$$. If, after step 2, the coin on node $$$i$$$ is red, the transitions are easy.

Considering only that case solves the problem if there are no swaps. Instead, if the coin on node $$$i$$$ is blue after step 2, you have to calculate $$$\max(dp_{parent_j} + |a_j - a_i|)$$$ for each $$$i$$$ with a fixed $$$dist(1, i)$$$. How?

Divide the nodes in groups based on the distance from the root and sort them in increasing order. Then, calculate $$$dp_i$$$ — the maximum score that you can reach after $$$dist(1, i)$$$ moves if you put a coin on node $$$i$$$. You can calculate $$$dp_i$$$ if you know $$$dp_j$$$ for each $$$j$$$ that belongs to the previous group. There are two cases:

• if before the eventual swap the coin on node $$$i$$$ is red, the previous position of the red coin is fixed, and the blue coin should reach either the minimum or the maximum $$$a_j$$$ among the $$$j$$$ that belong to the same group of $$$i$$$;

• if before the eventual swap the coin on node $$$i$$$ is blue, you should maximize the score $$$dp_{parent_j} + |a_j - a_i|$$$ ($$$j$$$ and $$$i$$$ in the same group). This can be done efficiently by sorting the $$$a_i$$$ and calculating the answer separately for $$$a_j \leq a_i$$$ and $$$a_j > a_i$$$; for each $$$i$$$ in the group, the optimal $$$j$$$ is either the same of last processed node or the last processed node.

The answer is $$$max(dp_i)$$$.

Complexity: $$$O(n\log n)$$$.

Why isn't the answer $$$2^{n-1}$$$? What would you overcount?

Find a dp with time complexity $$$O(n^2\log n)$$$.

Is it really necessary to copy all the pairs $$$(i, j)$$$ to $$$(i+1, j+b_i)$$$?

For each $$$i$$$, you can choose either $$$a_i = b_i$$$ or $$$a_i = b_i - \sum_{k=1}^{i-1} a_k$$$. If $$$\sum_{k=1}^{i-1} a_k = 0$$$, the two options coincide and you have to avoid overcounting them.

This leads to an $$$O(n^2\log n)$$$ solution: let $$$dp_i$$$ be a map such that $$$dp_{i, j}$$$ corresponds to the number of ways to create a hybrid prefix $$$[1, i]$$$ with sum $$$j$$$. The transitions are $$$dp_{i, j} \rightarrow dp_{i+1, j+b_i}$$$ (if you choose $$$b_i = a_i$$$, and $$$j \neq 0$$$), $$$dp_{i,j} \rightarrow dp_{i+1,b_i}$$$ (if you choose $$$b_i = \sum_{k=1}^{i} a_k$$$).

Let's try to get rid of the first layer of the dp. It turns out that the operations required are

• move all $$$dp_j$$$ to $$$dp_{j+b_i}$$$
• calculate the sum of all $$$dp_j$$$ in some moment
• change the value of a $$$dp_j$$$

and they can be handled in $$$O(n\log n)$$$ with "Venice technique". It's not necessary to use a multiset (a map is enough) because you don't have to handle min / max queries.

$$$dp$$$ is now a map such that $$$dp_j$$$ corresponds to the number of ways to create a hybrid prefix $$$[1, i]$$$ such that $$$\sum_{k=1}^{i} a_k - b_k = j$$$. Calculate the dp for all prefixes from left to right. if $$$b_i = a_i$$$, you don't need to change any value of the dp; if $$$b_i = \sum_{k=1}^{i} a_k$$$, you have to set $$$dp_{\sum_{k=1}^{i} -b_k}$$$ to the total number of hybrid arrays of length $$$i-1$$$.

Complexity: $$$O(n\log n)$$$.