online distinct elements queries need persistent segtree i think

1. Your code is invisible for us. We can't see your submission on spoj.

2. 1st and 2nd problems where discussed on codeforces — search it (1st using bit or segtree here).

Hello, I once created tutorial of sweep line or mainly sweep concept in general, in that I have explained dquery, kquery spoj, hope you will like it.
I have used Fenwick tree for range addition point update, but you can also use segment tree for the same.
https://youtu.be/_McXnKpGLhg If you like the video share it with your friends.

Solution for the first problem

Let's consider the array is $$$a_1, a_2, \dots, a_n$$$.

For each position $$$i$$$, consider $$$prev[i]$$$ to be the last $$$j < i$$$ such that $$$a_j = a_i$$$, (or $$$0$$$ if such $$$j$$$ does not exist). Then, for a range, you can count each distinct element the first time it appears on that range; therefore, the number of distinct elements in a range $$$[l, r]$$$ is equal to the number of $$$i\in[l, r]$$$ such that $$$prev[i] < l$$$.

Now, the problem becomes, given an array (in this cases is $$$prev[\ ]$$$), and given some queries consisting of a range $$$[l, r]$$$ and a number $$$x$$$, count the number of elements in the array that are less than $$$x$$$. For sure this problem is easier. You can have a segment tree, and keep on each node of the segment tree a sorted vector with all the elements belonging to the range that node represents. This Segment Tree can be constructed easily in $$$\mathcal{O}(n\cdot \log n)$$$ time, and it consumes $$$\mathcal{O}(n\cdot \log n)$$$ space, because each element of the array has a copy in its $$$\mathcal{O}(\log n)$$$ parents. Then, for answering a query, just go through the segment tree, and for each of the important nodes you can do a binary search over its sorted vector for finding how many elements $$$< x$$$ it contains. Each query's answer is computed in $$$\mathcal{O}(\log^2 n)$$$, although you can do it in $$$\mathcal{O}(\log n)$$$ with a technique called Fractional Cascading. You could have also used a Persistent Segment Tree. Be aware that this is an online solution, you could have easier solutions if you preprocess all the queries; indeed, the Persistent Segment Tree approach is the same as the offline approach but memorizing the state of each position to make it online.

I suggest this article.

Why are you solving problems involving segment trees? Go solve some constructive/math/implementation 1200, not this garbage.