There is no documented way to do this. But set is a RB-tree, so you probably can get access to tree's structure and get i-th element in by yourself. I don't think it's a good way

The trouble is that none of the tree-like data structures available in the standard libraries support a fast k-th element query ("given k, find the k-th smallest of the stored numbers").

The thing you have to add to a canonical balanced tree structure is information about subtree sizes: for each node in the tree, keep the count of elements in the subtree rooted at this node. For a particularly easy-to-implement balanced tree structures, check out splay trees and treaps.

(c) http://community.topcoder.com/tc?module=Static&d1=match_editorials&d2=srm310

There is a cheat to access Parent, Left and Right nodes. But i have no idea how to use this information "online". There is problems because of tree modifications — rotations and other balancing stuff while inserting or erasing nodes.

Hey! Looks like there's a way to do it after all. Please see http://codeforces.com/blog/entry/11080

You can use C++ SGI STL tree.

For example:

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> //required
#include <ext/pb_ds/tree_policy.hpp> //required
using namespace __gnu_pbds; //required 
using namespace std;
template <typename T> using ordered_set =  tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; 

ordered_set <int> s;
/* or:
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
ordered_set s;
This works in C++98 but the above version only works in C++11
*/
int main(){
	// It has C++ `set` functions:
	for(auto i: {1,3,5,8})
		s.insert(i);
	s.erase(8);
	s.erase(s.find(8));
	cout << * s.begin() << endl;
	if(s.begin() == s.end() or s.empty())
		cout << "empty" << endl;
	s.lower_bound(3);
	s.upper_bound(1);
	cout << s.size() << endl;
	// special `tree` functions:
	cout << s.order_of_key(3) << endl;	// the number of elements in s less than 3 (in this case, 0-based index of element 3)
	cout << *s.find_by_order(1) << endl; // 1-st elemt in s (in sorted order, 0-based)
}

How about binary search? You have lower_bound() for set. Please correct me if I am wrong. The complexity is log(n)*log(L) though. L= range of values in set,n=no. of values in set.

int idx=(int)(S.lower_bound(start,end,value)-S.begin()) if idx>k , search again.

Does anyone know of something similar in Java?