\(\text{Solution}\)

非常巧妙的二分答案
因為只有一個詢問，且答案滿足單調性，所以二分答案 \(mid\)
原序列中大於等於 \(mid\) 的數的位置賦為 \(1\)，否則賦為 \(0\)
然後走一遍排序操作，用線段樹賦值完成排序
最後檢查 \(q\) 位置上的情況，\(0\) 就不行，\(1\) 就可以
\(mid\) 越小越容易，越大越困難，臨界點即為答案

\(\text{Code}\)

#include <cstdio>
#include <cstring>
#define ls (p << 1)
#define rs (ls | 1)
#define re register
using namespace std;

const int N = 1e5 + 5;
int n, m, q, a[N];
struct node{int op, l, r;}b[N];

inline void read(int &x)
{
	x = 0; char ch = getchar();
	while (ch < '0' || ch > '9') ch = getchar();
	while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
}

int tag[N << 2], sum[N << 2];

inline void pushup(int p){sum[p] = sum[ls] + sum[rs];}
inline void pushdown(int p, int l, int r)
{
	if (tag[p] == -1) return;
	int mid = (l + r) >> 1;
	if (!tag[p]) sum[ls] = mid - l + 1, sum[rs] = r - mid;
	else sum[ls] = sum[rs] = 0;
	tag[ls] = tag[rs] = tag[p], tag[p] = -1;
}

void update(int p, int l, int r, int x, int y, int v)
{
	if (x > r || y < l) return;
	if (x <= l && r <= y) 
	{
		tag[p] = v, sum[p] = (v ? 0 : r - l + 1);
		return;
	}
	pushdown(p, l, r);
	int mid = (l + r) >> 1;
	if (x <= mid) update(ls, l, mid, x, y, v);
	if (y > mid) update(rs, mid + 1, r, x, y, v);
	pushup(p);
}

int query_0(int p, int l, int r, int x, int y)
{
	if (x > r || y < l) return 0;
	if (x <= l && r <= y) return sum[p];
	pushdown(p, l, r);
	int mid = (l + r) >> 1, res = 0;
	if (x <= mid) res = query_0(ls, l, mid, x, y);
	if (y > mid) res += query_0(rs, mid + 1, r, x, y);
	return res;
}

inline int check(int mid)
{
	memset(tag, 255, sizeof tag), memset(sum, 0, sizeof sum);
	for(re int i = 1; i <= n; i++) update(1, 1, n, i, i, a[i] >= mid);
	for(re int i = 1, l, r, num; i <= m; i++)
	{
		num = query_0(1, 1, n, b[i].l, b[i].r);
		if (!b[i].op) l = b[i].l, r = b[i].l + num - 1,
				update(1, 1, n, l, r, 0), update(1, 1, n, r + 1, b[i].r, 1);
		else l = b[i].l, r = b[i].l + (b[i].r - b[i].l + 1 - num) - 1,
			update(1, 1, n, l, r, 1), update(1, 1, n, r + 1, b[i].r, 0);
	}
	return !query_0(1, 1, n, q, q);
}

int main()
{
	read(n), read(m);
	for(re int i = 1; i <= n; i++) read(a[i]);
	int l = 1, r = n, mid, ans = 0;
	for(re int i = 1; i <= m; i++) read(b[i].op), read(b[i].l), read(b[i].r);
	read(q);
	while (l <= r)
	{
		mid = (l + r) >> 1;
		if (check(mid)) ans = mid, l = mid + 1;
		else r = mid - 1;
	}
	printf("%d\n", ans);
}