We will try to figure out the values of $$$p_0, p_1, \ldots, p_{n - 1}$$$ while not overshooting the allowed limits on the number of times we are allowed to use each. We maintain a set of equations which represent the stuff we know. Initially, all we know is $$$p_0 = P0$$$.

The only information that we actually use from the queries is the sum of the elements that were chosen (which we can easily find). It is a bit disappointing (to me) that nothing smarter is needed.

Suppose we have an equation $$$p_{x_1} + p_{x_2} + \ldots + p_{x_k} = s$$$ where $$$x_i$$$ is increasing (assume $$$k \gt 1$$$). The important idea here is to query $$$\frac{s}{k}$$$ (rounded down), and then there are $$$2$$$ important properties.

1. $$$x_1$$$ or any number smaller than $$$x_1$$$ cannot appear in that equation because by PHP, $$$p_{x_1} \gt \frac{s}{k}$$$, and thus all $$$p_i$$$ with $$$i \lt x_1$$$ can also not appear as they will be larger.

2. It is still a valid query, because again by PHP, $$$p_{x_k} \le \frac{s}{k}$$$.

When $$$k = 1$$$, we can instead query $$$s - 1$$$ if needed to get info about the smaller values.

This is actually sufficient to solve the problem, because this gives us a descending set of equations which we can keep on using to find more and more equations till we eventually go into the base case of $$$k = 1$$$ where we figure out the value of $$$p_i$$$.

For the set of equations that you have (the ones where you have not yet figured out everything), order them by the minimum element, and then look at the last equation, and break it into subcase by doing the above.

Note that there is some algebra ahead, because that is how I think of things. It is useful to work out the algebra yourself, reading it doesn't have the same effect imo.

Let's casework on which of the values among $$$h_i, h_j, h_k$$$ is the largest value, i.e. $$$k - i$$$.

When $$$h_i = k - i$$$, let us fix $$$i$$$, and this fixes $$$k = i + h_i$$$. Further, either $$$h_k = j - i$$$ or $$$h_k = k - j$$$ and in both cases, $$$j$$$ also gets fixed. Thus, there are potentially $$$2 \cdot N$$$ triples, and we can verify all easily in $$$O(n)$$$.

Same analysis holds for $$$h_k = k - i$$$. Let's move on to the more challenging $$$h_j = k - i$$$ case.

We further subdivide into $$$2$$$ cases:

Case 1: $$$h_i = j - i, h_k = k - j$$$ : Again this case is pretty simple. Bruteforce on $$$i$$$, it fixes $$$j$$$. Fixing $$$j$$$ in turn fixes the value of $$$k$$$. At most $$$n$$$ triples, verify in $$$O(n)$$$ time.

Case 2 : $$$h_i = k - j, h_k = j - i$$$ : This is a hard case, and the only case where there are more than $$$O(n)$$$ triples. However, the idea is still the same, enumerate all triples and verify.

We have the equations $$$h_i = k - j, h_j = k - i$$$. Subtract these to get $$$h_i - h_j = i - j$$$ or $$$h_i - i = h_j - j$$$. This already imposes a constraint on the pair $$$(i, j)$$$ to possibly be extended to form a valid triplet. We can now solve this problem in $$$\sum f_x^2$$$ where $$$f_x$$$ is the number of occurences of $$$h_i - i = x$$$.

Let us group things by the same value of $$$x = h_i - i$$$. Now, if we fix $$$k$$$ instead of $$$(i, j)$$$, we get the following: $$$k = h_i + j$$$ and $$$h_k = j - i$$$. Subtract to get $$$k - h_k = h_i + i = x + i + i$$$. So, this fixes $$$i$$$ and hence, it fixes $$$j$$$.

In summary, every pair of $$$(k, x)$$$ where $$$x$$$ is the required difference of $$$h_i - i$$$ has exactly one suitable pair $$$(i, j)$$$ and we can enumerate all possible $$$k$$$ to solve each group in $$$O(n)$$$. However, this still has a worst case complexity of $$$O(n^2)$$$ when there are a lot of distinct values of $$$h_i - i$$$.

As a final trick, we use square root decomposition, breaking groups into $$$ \gt sqrt(n)$$$ size or $$$\le sqrt(n)$$$ size. For the smaller groups, we use the bruteforce $$$f_x^2$$$ method, for the larger groups, we use the $$$O(n)$$$ method. It is not hard to show that the total runtime is $$$O(n^{1.5})$$$

Obviously, if the graph is not connected, it is impossible. Assume it is connected. Let's go step by step. Actually, when you solve the first step, the rest falls into place.

Looking at the final construction visually is useful if my explanation sucks. You can run my code.

Let us go step by step. Once we fix a set of items to buy, what is the optimal order? We can easily find this with exchange argument, take $$$i$$$ before $$$j$$$, and take $$$j$$$ before $$$i$$$, and compare which leaves us with more money.

Now, let us assume items are sorted in this order. What do we do? Write DP ofcourse. $$$dp_{i, j} = $$$ maximum money after buying $$$j$$$ items from $$$[1, i]$$$. We get a working $$$n^2$$$ solution after preventing overflow.

Now, we move on to the subtask with $$$(A - p_i) \cdot t_i \lt A$$$. Ignore $$$t_i = 1$$$ for now. We can actually prove we will take only $$$log(A)$$$ many other items. This is because $$$(A - x - p_i) \cdot t_i \lt A - x \cdot t_i$$$, i.e. if the starting money was $$$A - x$$$, the current money will be less than $$$A - x \cdot t_i$$$, the difference from $$$A$$$ gets multiplied by at least $$$t_i$$$ each step.

So we can write the same dp, but limit $$$j$$$ to $$$log(A)$$$, and then bruteforce over the number of $$$T_i \gt 1$$$ and look at how many $$$T_i = 1$$$ you can take.

The idea for the full problem is the same as above, but some prefix will be non-negative value (i.e. increasing your coins), which you should process at the start, and then do the above solution, carefully handling overflow. The proof that it will only be a prefix which is non-negative value comes from the optimality of the exchange order.

We will implement a solution with $$$Z = 4$$$ and $$$21$$$ messages. Ignore the first $$$N - 1 - 21$$$ calls, only focus on the last $$$21$$$. Note that there are actually $$$5$$$ allowed states, not just $$$4$$$ since you can use $$$S_i = 0$$$ as an information source too.

The idea is to send the pair $$$(u, v)$$$ where $$$u$$$ and $$$v$$$ are diameter endpoints. We can encode them easily, but the problem is when a new node arrives and is a new diameter endpoint.

Let us first focus on transmitting $$$u$$$. We transmit it in base $$$4$$$ using $$$0 \le S_i \le 3$$$, but if the current node $$$i$$$ becomes an endpoint of a diameter at any time, we transmit $$$S_i = 4$$$ to indicate that. In this way, at the end of this process, we will know one endpoint of a diameter for sure. This takes at most $$$log_4(10^4) = 7$$$ steps.

Now, we want to transmit the other endpoint $$$v$$$. When a new node $$$i$$$ arrives, the new diameters can be $$$(u, v), (u, i)$$$ or $$$(v, i)$$$. We will use $$$0 \le S_i \le 1$$$ to indicate the first, $$$S_i = 2$$$ to indicate the second, and $$$3 \le S_i \le 4$$$ to indicate the third. In the case of first and third, we can transmit one bit of $$$v$$$ because we have $$$2$$$ options. In the case of second, we already know our current diameter because we knew $$$u$$$ before, and we don't need $$$v$$$ anymore. This takes at most $$$log_2(10^4) = 14$$$ steps.

Let us solve $$$L = 0, R = M - 1$$$ for now. The important property is that the allowed segments in a row are always subsets or supersets of each other (which is easy to show).

You can travel from column $$$S$$$ to column $$$D$$$ only in the row that has both of them in the same segment. It is not hard to show that infact, the first row satisfying this property is optimal.

Let us discuss a process by which we can decide how far you can travel from some column $$$S$$$. At row $$$0$$$, there is some segment of allowed columns $$$[L_1, R_1]$$$ with $$$L_1 \le S \le R_1$$$. Look at the first row where this segment is larger, say row $$$x$$$ has a segment $$$[L_2, R_2]$$$ ($$$L_2 \le L_1 \le R_1 \le R_2$$$). We need to check if our segment can reach this larger segment. How to do that?

We can reach the larger segment iff there is some column which is completely free for all rows $$$0, 1, \ldots x$$$. This is due to the subsets, supersets property. It is easy to check if such a column exists, because it is simply the column with the smallest value of $$$H_j$$$.

In general, the problem can be thought about as jumping to increasing segments (and checking if these jumps are valid). For a query $$$(S, D)$$$, we need to check if the largest segment that $$$S$$$ can reach is the same as the largest segment that $$$D$$$ can reach. Implementation details of finding the segments and validity of jumps are left to the reader.

Let us move on to general $$$L$$$ and $$$R$$$. The core idea remains the same, but now for each jump to a larger segment, we actually care about what column we use. Also, now we will travel from $$$S$$$ and $$$D$$$ to their LCA in this tree of segments. Suppose $$$S \lt D$$$. Then, the path from $$$S$$$ to the LCA will have to use columns $$$\ge L$$$ ($$$\le R$$$ is pointless, because we never cross $$$D$$$ anyways), and vice versa.

For each jump to an increasing segment, we will now have to find the "maximum index" where such a jump can be made. This can be done with a Segment tree walk. Then, we build a binary lifting structure to answer path max/min queries, and check if $$$minjump(S, lca) \ge L$$$ and $$$maxjump(D, lca) \le R$$$