Ok, So the editorial uses some DP, let me mention my combinatorical solution here. The main observation is that the xor of an array is 0, if all bits are set even number of times. So, the task reduces to finding the number of arrays for each length from 1 to N, such that each bit is set even number of times through the array.

For each i from 1 to N, perform the following procedure , and let's call the number obtained as res[i] :

For each bit from 0 to 8 ,create a polynomial of size N, where the co-efficient of x^j, is if 2·l·2^currbit divides j. This indicates that we set the bit in 2·l items in current series, that contributes 2·l·2^currbit to the total sum.

Now, multiply all these polynomials. Let res[i] = coefficient of xⁿ after this. Now, only problem is that we could still have empty elements in arrays. However, notice that if array has k empty elements out of i, then it should be counted in res[i - k]. Then we can build new array ans[], where

. Build this in left to right fashion. The meaning of this is that each way we can have i - j zeros in the array of length i, we create array of length j, deleting them

Note that this is  > O(N³) overall, we can optimize, by noticing that for an i, for any bit j, only N / 2·2^j entries are not 0, iterating only over them during polynomial multiplication.

My Code