题干：
DZY has a sequence a[1..n]a[1..n]. It is a permutation of integers 1∼n1∼n.
Now he wants to perform two types of operations:
0lr0lr: Sort a[l..r]a[l..r] in increasing order.
1lr1lr: Sort a[l..r]a[l..r] in decreasing order.
After doing all the operations, he will tell you a position kk, and ask you the value of a[k]a[k].
Input
First line contains tt, denoting the number of testcases.
tt testcases follow. For each testcase:
First line contains n,mn,m. mm is the number of operations.
Second line contains nn space-separated integers a[1],a[2],⋯,a[n]a[1],a[2],⋯,a[n], the initial sequence. We ensure that it is a permutation of 1∼n1∼n.
Then mm lines follow. In each line there are three integers opt,l,ropt,l,r to indicate an operation.
Last line contains kk.
(1≤t≤50,1≤n,m≤100000,1≤k≤n,1≤l≤r≤n,opt∈{0,1}1≤t≤50,1≤n,m≤100000,1≤k≤n,1≤l≤r≤n,opt∈{0,1}. Sum of nn in all testcases does not exceed 150000150000. Sum of mm in all testcases does not exceed 150000150000)
Output
For each testcase, output one line - the value of a[k]a[k] after performing all mmoperations.
Sample Input
1
6 3
1 6 2 5 3 4
0 1 4
1 3 6
0 2 4
3
Sample Output
5
Hint
1 6 2 5 3 4 -> [1 2 5 6] 3 4 -> 1 2 [6 5 4 3] -> 1 [2 5 6] 4 3. At last $a[3]=5$.
题目大意：
解题报告：
两个log的做法展现了二分答案的强大功能。首先二分枚举第 k 位的值x，然后将大于等于x的数都变为 1 ，小于x的数变为 0 ，这样这数字序列就变成了01序列，只有这两种性质。我们用线段树不难实现对 01 序列按要求进行排序，然后如果第 k 位为 1 说明x可以是ans但是太小了，要调整下界。就这样不断二分下来，得到的边界值就是第 k 位真实的值。这个做法是离线的，有两个log ，但代码好实现。
这题巧妙之处：第一在于只有一次查询，第二在于是n的全排列。（但是貌似不是全排列也可用类似的方法做。）
nlogn的神仙方法：https://www.cnblogs.com/Paulliant/p/10185235.html（线段树分割）
AC代码：
#include
#include
#include
#include
#include
