题干：

Give you a sequence and ask you the kth big number of a inteval.

Input

The first line is the number of the test cases. 
For each test case, the first line contain two integer n and m (n, m <= 100000), indicates the number of integers in the sequence and the number of the quaere. 
The second line contains n integers, describe the sequence. 
Each of following m lines contains three integers s, t, k. 
[s, t] indicates the interval and k indicates the kth big number in interval [s, t]

Output

For each test case, output m lines. Each line contains the kth big number.

Sample Input

1 
10 1 
1 4 2 3 5 6 7 8 9 0 
1 3 2 

Sample Output

2

解题报告：

求区间第K小值。

第一行两个整数n和m (n, m <= 100000)
第二行n个整数
接下来m行，每行3个整数s,t,k.表示要查询区间[s,t]内第k小的。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 1e5 + 5;
struct TREE {int l,r;int val;
} tr[MAX*35];int tot;
int a[MAX],b[MAX];
int root[MAX];
int build(int l,int r) {int cur = ++tot;tr[cur].val = 0;if(l == r) {tr[cur].l = tr[cur].r = 0;return cur;}int m = (l+r)>>1;tr[cur].l = build(l,m);tr[cur].r = build(m+1,r);return cur;
}
int update(int pre,int tar,int l,int r) {int cur = ++tot;tr[cur] = tr[pre];tr[cur].val++;if(l == r) return cur;int m = (l+r)>>1;if(tar <= m) tr[cur].l = update(tr[pre].l,tar,l,m);else tr[cur].r = update(tr[pre].r,tar,m+1,r);return cur;
}
int query(int pl,int pr,int l,int r,int k) {if(l == r) return l;int m = (l+r)>>1;if(tr[tr[pr].l].val - tr[tr[pl].l].val >= k) return query(tr[pl].l,tr[pr].l,l,m,k);else return query(tr[pl].r,tr[pr].r,m+1,r,k - (tr[tr[pr].l].val - tr[tr[pl].l].val));
}
int main()
{int n,m;int t;cin>>t;while(t--) {cin>>n>>m;tot=0;for(int i = 1; i<=n; i++) scanf("%d",a+i),b[i] = a[i];root[0] = build(1,n);sort(b+1,b+n+1);int LEN = unique(b+1,b+n+1) - b - 1;for(int i = 1; i<=n; i++) {int pos = lower_bound(b+1,b+LEN+1,a[i]) - b;root[i] = update(root[i-1],pos,1,n);}while(m--) {int l,r,k;scanf("%d%d%d",&l,&r,&k);printf("%d\n",b[query(root[l-1],root[r],1,n,k)]);}}return 0 ;
}