Such kind of ratio problems can be solved by using a niche trick. I think it's hard to come up with this line of thinking unless you have encountered it before.

Let's say the optimal ratio is $$$Q$$$. Thus, $$$\frac{\sum R_i}{\sum T_i} = Q$$$. What we can do is binary search on $$$Q$$$, and check if a ratio is feasible or not. So, for a given $$$Q$$$, we have to check whether $$$\frac{\sum R_i}{\sum T_i} \ge Q$$$ is possible, which is the same thing as $$$\sum R_i - Q*\sum T_i \ge 0$$$ (let's call the L.H.S. $$$P$$$). The constraints on $$$R_i, T_i$$$ seem to be quite high to do any form of DP while constraints on $$$R, C$$$ are quite low although obviously not permitting brute force solutions. However, meet-in-the-middle looks promising since we can store $$$P$$$ while doing brute force from one corner of the grid and checking from another (meeting in cells lying on some sort of diagonal of the grid).

It's in fact the solution and we also have to take care of the costs along with $$$P$$$ to actually know if any of the options is actually feasible. So, when doing brute from one side, we store $$$(P_i, Cost_i)$$$ pairs on those diagonal cells and later sort them according to $$$Cost_i$$$ to ease our process in the next brute. While doing brute from the other end, we have $$$(P_j, Cost_j)$$$ and we want the highest $$$P_i$$$ whose cost satisfies $$$Cost_i \le K - Cost_j$$$. By storing $$$P_i$$$ in prefix maximum manner, we can binary search on the highest index option with cost $$$\le K - Cost_j$$$ and take its $$$P_i$$$ (since we already sorted) to get our answer.

A number can be expressed as a sum of two sqaures only if all its prime factors of the form $$$4k+3$$$ occur even times. So while taking the product of few numbers, we are only interested in its prime decomposition hence, we can store a bitvector for each $$$A_i$$$ denoting odd/even occurrence of such prime factors.

Now, $$$A_i \le 10^7$$$ so #primes is around $$$3* 10^5$$$. Normal gaussian elimination would easily TLE, even with bitsets. Still, let's imagine doing gaussian elimination traversing from the highest bit to the lowest and xoring when necessary. Notice that, when you are traversing, if any prime factor > $$$\sqrt A_i$$$ exists (*at max one can exist*), the moment you xor it, no more bits > $$$\sqrt A_i$$$ will be on anymore (since only one such was on earlier and now you have xored it). Now, you can simply simulate gaussian elimination on lower order primes (ie. $$$primes \le \sqrt A_i$$$) and handle that one higest bit (if it exists) separately. This way, combined with bitset for lower order primes solves the problem without queries.

For queries, let's process them offline. Let $$$queries[l]$$$ store all the queries which have their left end point as $$$l$$$. We will insert $$$A_i$$$ into our basis from the right side, $$$i$$$ going from $$$N-1$$$ to $$$0$$$. After we have inserted the element at index $$$i$$$, we will answer all the queries which begin their ie. the ones in $$$queries[i]$$$.

We will use a simple fact about linear combination of vectors but in a clever manner here to answer the queries. If $$$v = b_1 + b_2 + \cdots b_k$$$ (where $$$b_i \in Basis$$$), then we can remove one of the $$$b_i$$$ and instead insert $$$v$$$ into its place, keeping the basis intact. So, while inserting what we will try is to keep such elements in the basis which have leftmost index so as to answer as many queries as possible. We can do this greedily: while inserting $$$v$$$, whenever we are about to xor it with some $$$b_i$$$, we will first check which has the least index and $$$swap(b_i, v)$$$ if $$$v$$$'s index is smaller. And from then onwards, $$$v$$$ is a part of the basis (in place of $$$b_i$$$) and we progress doing Gaussian Elimination with $$$b_i$$$ instead.

Now, when we answer queries which have left index $$$i$$$, we will have the basis vectors from $$$A[i:N]$$$ having least possible index. So for a query $$$i\text{ }R$$$, we simply need to count how many vectors in our basis have index $$$\le R$$$. This can be easily done with a BIT or set.