Here is my solution to the problem. The key idea is as follows: if we can calculate the beauty of each number of array $$$A$$$ fast enough, then we can just simply calculate the beauty of each element of array $$$A$$$ and check if it's greater than $$$C$$$. So, the question is how to find the number of subarrays of $$$U$$$ with sum equal $$$X$$$?

Imagine that we have subarray $$$[a, a + 1, a + 2, \dots , b - 1, b]$$$ of universal array. One can notice that this sequence of numbers is an arithmetic progression where the first element is equal to $$$a$$$, common difference is equal to $$$1$$$ and the length is equal to $$$b - a + 1$$$. Because of that, we can use a specialized formula to calculate the sum of the subarray: $$$S=\frac{(b - a + 1)(a + b)}{2}$$$. Now it's obvious that to calculate the beauty of the number $$$X$$$ we need to find the number of pairs $$$(a, b)$$$ where $$$-10^{18} \le a \le b \le 10^{18}$$$ such that $$$(b - a + 1)(a + b)=2X$$$.

Since $$$a, b$$$ were integers, values of $$$b - a + 1$$$ and $$$a + b$$$ are also integers, so $$$a + b$$$ is one of the divisors of $$$2X$$$. Imagine that we have a list of all divisors of $$$2X$$$. We can use this list to brute force the value $$$a + b$$$. Then we can use it to calculate $$$a-b=1-\frac{2X}{a+b}$$$. The only things remains is to check that $$$a, b$$$ satisfy the following requiremenets: $$$-10^{18} \le a \le b \le 10^{18}$$$. And now the last question: how to find all divisors of $$$X$$$?

The naive approach with the complexity of $$$O(\sqrt{X})$$$ should work just great for your constraints.

Hope it helps.