Key observation is to notice that gcd is monotonic. (meaning if the gcd of elements between i to j is y, then the gcd of elements i to j+1 will be <= y.)

Then for each i from 1 to n, u can binary search for the rightmost index more than i (let's say j, such that the gcd of elements between i to j is <= x), after that, binary search for the leftmost index more than i (let's say k, such that the gcd of elements between i to k is >= x).

Then the number of subarrays with gcd of x starting from i = j-k+1.

And gcd from elements between i to j can be calculated in O(1) with sparse table.

Thus resulting complexity will be n * log2(n)

A similar question can be found here: https://dunjudge.me/analysis/problems/1121/