1 // Persistant segment tree
2 struct Node {
3 int val;
4 Node *ch[2];
5 };
6
7 int a[maxn], n;
8 Node *root[maxn];
9
10 Node* sgtIns(Node* q, int pos) { // 在q的基础上将pos这个位置的数+1, 并返回一个新的线段树的根
11 int l = 0, r = n;
12 Node *s = new Node, *p = s;
13 while (l < r) {
14 p->val = q ? q->val + 1 : 1;
15 int mid = (l + r)>> 1;
16 if (pos <= mid) {
17 p->ch[1] = q ? q->ch[1] : 0;
18 p = p->ch[0] = new Node;
19 q = q ? q->ch[0] : 0;
20 r = mid;
21 } else {
22 p->ch[0] = q ? q->ch[0] : 0;
23 p = p->ch[1] = new Node;
24 q = q ? q->ch[1] : 0;
25 l = mid + 1;
26 }
27 }
28 p->val = q ? q->val + 1 : 1;
29 return s;
30 }
31
32 int sgtKth(Node* p, Node* q, int k) {
33 int l = 0, r = n;
34 while (l < r) {
35 int cntleft = (q && q->ch[0] ? q->ch[0]->val : 0) - (p && p->ch[0] ? p->ch[0]->val : 0);
36 int mid = (l + r)>> 1;
37 if (cntleft >= k) {
38 r = mid;
39 p = p ? p->ch[0] : 0;
40 q = q ? q->ch[0] : 0;
41 } else {
42 k -= cntleft;
43 l = mid + 1;
44 p = p ? p->ch[1] : 0;
45 q = q ? q->ch[1] : 0;
46 }
47 }
48 return l;
49 }
50
51 int main() {
52 cin >> n;
53 for (int i = 1; i <= n; ++ i) {
54 cin >> a[i];
55 }
56 root[0] = 0;
57 for (int i = 1; i <= n; ++ i) {
58 root[i] = sgtIns(root[i - 1], a[i]);
59 }
60 int q;
61 cin >> q;
62 while (q --) {
63 int l, r, k;
64 cin >> l >> r >> k;
65 cout << sgtKth(root[l - 1], root[r], k) << endl;
66 }
67 }
