可持久化数据结构题目泛做。

 1 #include <cstdlib>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <cstdio>
 5 #include <cstring>
 6 
 7 using namespace std;
 8 const int N = 100000 + 5;
 9 
10 struct SegTree {
11   int l, r, size;
12 }Node[N * 30];
13 
14 struct data {
15   int v, pos;
16   bool operator < (const data &k) const {
17     return v < k.v;
18   }
19 }a[N];
20 
21 int n, m, rank[N], l, r, k, root[N], tot;
22 
23 void build(int &o, int l, int r) {
24   o = ++ tot;
25   Node[o].l = Node[o].r = Node[o].size = 0;
26   if(l >= r) return;
27   int mid = (l + r) >> 1;
28   build(Node[o].l, l, mid);
29   build(Node[o].r, mid + 1, r);
30 }
31 
32 void update(int pre, int &o, int l, int r, int kth) {
33   o = ++ tot;
34   Node[o] = Node[pre];
35   Node[o].size ++;
36   if(l >= r) return;
37   int mid = (l + r) >> 1;
38   if(kth <= mid)
39     update(Node[pre].l, Node[o].l, l, mid, kth);
40   else
41     update(Node[pre].r, Node[o].r, mid + 1, r, kth);
42 }
43 
44 int query(int r1, int r2, int l, int r, int kth) {
45   if(l >= r) return l;
46   int lc = Node[Node[r2].l].size - Node[Node[r1].l].size;
47   int mid = (l + r) >> 1;
48   if(lc >= kth)
49     return query(Node[r1].l, Node[r2].l, l, mid, kth);
50   else
51     return query(Node[r1].r, Node[r2].r, mid + 1, r, kth - lc);
52 }
53 
54 int main() {
55   scanf("%d%d", &n, &m);
56   for(int i = 1; i <= n; ++ i) {
57     scanf("%d", &a[i].v);
58     a[i].pos = i;
59   }
60   sort(a + 1, a + n + 1);
61   for(int i = 1; i <= n; ++ i)
62     rank[a[i].pos] = i;
63   build(root[0], 1, n);
64   for(int i = 1; i <= n; ++ i)
65     update(root[i - 1], root[i], 1, n, rank[i]);
66   for(int i = 1; i <= m; ++ i) {
67     scanf("%d%d%d", &l, &r, &k);
68     printf("%d\n", a[query(root[l - 1], root[r], 1, n, k)].v);
69   }
70   return 0;
71 }