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Recall that for M = [[1, 1], [1, 0]], M^N is [[F_N + 1, F_N], [F_N, F_N - 1]].

Then we make an array of A[i] * Mⁱ and build a segment tree on top of that. Also we precompute powers of the inverse of M (the inverse is M^- 1 = [[0, 1], [1,  - 1]]).

In the segment tree for combining two intervals we just sum the respective matrices.

Modifying a single element is straightforward.

If we are queried fibsum(l, r), then we query the tree on this interval and get res = A[l] * M^l + A[l + 1] * M^l + 1 + .... We then multiply by M^{- l + 1}. Because of linearity, we will get res * M^{- l + 1} = A[l] * M + A[l + 1] * M² + .... The answer is in res[0][1].

In short, instead of working with Fibonacci numbers, we work with the matrices. This allows us to easily shift the Fibonacci sequences in any direction.

PS: Fenwick Tree would also work.

Let S1, S2, ..., Sn be the set S. Define P = S1#S2#S3#...#Sn. Now we want to check if string T is a substring of P. We can build suffix automaton for string P and then proccess query in O(|T|) time.

If you choose row i and col j and flip a "cross of tiles" with a center (x,y) for every pair (x,y) such that x=i or y=j, every tile except (i,j) will be flipped an even amount of times, while the tile (i,j) will be flipped n+m-1 times which is odd. So doing this will flip only one tile (i,j). This means that any state of tiles could be reached from any other.

Now let's point out the order of flips doesn't matter and there is no point of flipping the same cross more than once, so there are at most 2^mn different positions (mn is the number of different crosses) which can be obtained from the starting one. Also there are exactly 2^mn positions total. That means that for every position there is exactly one way of getting to it (not counting the order) without flipping the same cross twice.

Now we only have to find it. So we will get to all white tiles and then remove all pairs of repeating flips of the same crosses. This can easily be done in O(mn) doing the combo-flip from the first paragraph and storing for each row and col the number of times each cross from it have to be flipped.

Final algorithm:

for each black tile (i,j)
    rowFlipCount[i]++
    colFlipCount[j]++
    tileFlipCount[i][j]++

ans = 0
for each tile (x,y)
    if (rowFlipCount[x] + colFlipCount[y] + tileFlipCount[x][y]) is odd
        ans++