Если запросы оффлайн, можно сделать сканлайн с фенвиком.

Если онлайн, то MergeSortTree.

I think you can create for example fenwick tree or segment tree of order statistic trees and then sum up orders of x in all order statistic trees in [l, r].

EDIT: Or simply a merge sort tree.

Можно решать через ДО. В вершинах будем хранить отсортированные отрезки массива, которые отвечают за соответствующие отрезки в ДО. Для запроса будем обращаться к соответствующим вершинам для нужных отрезков и будем делать бинпоиск в них.

You can solve offline the queries. 1) sort the queries in nondecreasing order regarding x; 2) build an array V with 2 parameters: the value of A[i] and i, the position, then sort V regarding first value; 3) after these operations, keep 2 pointers: you will start i from the first value from V and start with j from the first element in Q (sorted); update the values in the binary indexed tree unless V[i].first_parameter will be >= the x from actual j query, then calculate with BIT the answer like this: sum of first r values in BIT — sum of first l-1 values in BIT, then print this value.

For offline:Sort the queries by x,then for each query,add the elements(weren't added before)less than x into the i-th position of a BIT.Then check the sum in interval l and r.Complexity is .

For online:Discrete the array first.Then build a persistent segment tree.The i-th version contains the first i elements in the array.The difference between i and i-1 is the a[i]-th position is increased by 1.For a query,the answer is:the sum in [l,r] in the r-th tree minus the sum in [l,r] in the l-1-th tree.The Complexity is also .