题目链接

  因为这道题没有删除修改之类的,所以很多人会用离散化之后的线段树来做,但是实际上(可能是我懒得去做离散化这个操作了),然后就是直接写可持久化线段树,区间的长度就是int的从最小到最大的长度,然后记录的是size,我们最后直接返回的是对应的位置即可。


 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <string>
 5 #include <cstring>
 6 #include <algorithm>
 7 #include <limits>
 8 #include <vector>
 9 #include <stack>
10 #include <queue>
11 #include <set>
12 #include <map>
13 #define lowbit(x) ( x&(-x) )
14 #define pi 3.141592653589793
15 #define e 2.718281828459045
16 #define INF 0x3f3f3f3f
17 #define HalF (l + r)>>1
18 #define lsn rt<<1
19 #define rsn rt<<1|1
20 #define Lson lsn, l, mid
21 #define Rson rsn, mid+1, r
22 #define QL Lson, ql, qr
23 #define QR Rson, ql, qr
24 #define myself rt, l, r
25 using namespace std;
26 typedef unsigned long long ull;
27 typedef long long ll;
28 const int maxN = 1e5 + 7, _UP = 2147483647, _DOWN = -2147483648;
29 int N, M, root[maxN], tot, siz[40 * maxN], lc[40 * maxN], rc[40 * maxN];
30 inline void insert(int &rt, int old, ll l, ll r, int qx)
31 {
32     rt = ++tot;
33     siz[rt] = siz[old] + 1; lc[rt] = lc[old];   rc[rt] = rc[old];
34     if(l == r) return;
35     ll mid = HalF;
36     if(qx <= mid) insert(lc[rt], lc[old], l, mid, qx);
37     else insert(rc[rt], rc[old], mid + 1, r, qx);
38 }
39 inline ll query(int now, int old, ll l, ll r, int k)
40 {
41     if(l == r) return l;
42     ll mid = HalF;
43     if(k <= siz[lc[now]] - siz[lc[old]]) return query(lc[now], lc[old], l, mid, k);
44     else return query(rc[now], rc[old], mid + 1, r, k - siz[lc[now]] + siz[lc[old]]);
45 }
46 inline void init()
47 {
48     for(int i=0; i<=tot; i++) root[i] = 0;
49     tot = 0;
50 }
51 int main()
52 {
53     while(scanf("%d%d", &N, &M) != EOF)
54     {
55         init();
56         for(int i=1, val; i<=N; i++)
57         {
58             scanf("%d", &val);
59             insert(root[i], root[i - 1], _DOWN, _UP, val);
60         }
61         int l, r, k;
62         while(M--)
63         {
64             scanf("%d%d%d", &l, &r, &k);
65             printf("%lldn", query(root[r], root[l-1], _DOWN, _UP, k));
66         }
67     }
68     return 0;
69 }
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