Subtask 1

The condition ~A \subseteq X \subseteq B~ is equivalent to ~(a \mathbin{\&} x) = a~ and ~(b \mathbin{\&} x) = x~, which can be checked on ~\mathcal{O}(1)~ time (here, ~\&~ denotes the bitwise AND operation). For each type ~2~ query, we can loop through all ~2^N~ subsets and add its value to the total if it satisfies this condition.

Time Complexity: ~\mathcal{O}(Q\cdot 2^N)~

Subtask 2

If we treat the ~2^N~ identifiers as vertices of an ~N~-dimensional hypercube, we can create an ~N~-dimensional prefix sum array in ~\mathcal{O}(N\cdot 2^N)~ by calculating a prefix sum in each dimension. More specifically, for each ~i~ ~(0 \le i < N)~, for all integers ~x~ with ~i^{th}~ bit ~0~, add arr~[x]~ to arr~[x + 2^i]~. Now arr~[a]~ will be the sum of the values of all ~X~ such that ~X \subseteq A~.

Split the queries into ~\sqrt{Q}~ blocks of size ~\sqrt{Q}~. At the start of each block, initialize the prefix sum array as above. Keep a list of the updates that have occurred, and when querying a set ~A~, return arr~[a]~ with adjustments for any set ~X \subseteq A~ whose value was updated (there are at most ~\sqrt{Q}~ of them if you reset the list of updates each block).

Time Complexity: ~\mathcal{O}((N\cdot 2^N+Q)\sqrt{Q})~

Subtask 3

Let ~x[i]~ denote the ~i^{th}~ bit of ~x~. For each query ~(A, B)~, we need ~(a \mathbin{\&} b) = a~, otherwise the answer to the query would be ~0~. This means there are ~3~ possibilities for ~(a[i], b[i])~: either ~(0,0)~, ~(0,1)~, or ~(1,1)~. Let ~x_1~ and ~x_2~ be the first and second halves of the binary representation of an integer ~x~. We can 'join' the ~a_1~ and ~b_1~ to get a ternary number ~\text{join}(a_1, b_1)~, where the ~i^{th}~ digit of ~\text{join}(a_1, b_1)~ is ~0~ if ~(a_1[i], b_1[i]) = (0, 0)~, ~1~ if ~(a_1[i], b_1[i]) = (0, 1)~, or ~2~ if ~(a_1[i], b_1[i]) = (1, 1)~. Similarly, we can convert back, going from any ~N/2~-digit ternary integer ~c~ to a corresponding pair of ~N/2~-bit integers ~a~ and ~b~, with ~(a \mathbin{\&} b) = a~ and ~\text{join}(a, b) = c~.

Now, define the array arr~[c][d_2]~ as the sum of the values of all ~N~-bit integers ~d~, whose second half (in binary) is ~d_2~, and whose first half ~d_1~ satisfies ~(a \mathbin{\&} d_1) = a~ and ~(b \mathbin{\&} d_1) = d_1~, where ~a~ and ~b~ are the pair of integers such that ~(a \mathbin{\&} b) = a~ and ~\text{join}(a, b) = c~. When the value of ~s~ changes by ~v~, we can update this array in ~\mathcal{O}(2^{N/2})~ by adding ~v~ to arr~[\text{join}(a, b)][s_2]~, for all ~a, b~ such that ~(a \mathbin{\&} s_1) = a~ and ~(b \mathbin{\&} s_1) = s_1~ (there are exactly ~2^{N/2}~ such pairs ~a, b~). To answer the query ~(a, b)~, we add up arr~[\text{join}(a_1, b_1)][x_2]~ for all ~x_2~ such that ~(a_2 \mathbin{\&} x_2) = a_2~ and ~(b_2 \mathbin{\&} x_2) = x_2~, which is also ~\mathcal{O}(2^{N/2})~. Note that we can precalculate ~\text{join}(a, b)~ in ~\mathcal{O}(N\cdot 3^{N/2})~, and array arr can be efficiently initialized in ~\mathcal{O}(6^{N/2})~ (alternatively, you can initialize with all ~0~'s and update ~2^N~ times).

Time Complexity: ~\mathcal{O}(Q\cdot 2^{N/2} + 6^{N/2})~