用二叉搜索树写的，想试试指针。样例能过没问题，自己做数据测了测也没毛病，交上去全部RE，不知道为什么。

#include <bits/stdc++.h>
using namespace std;

struct node {
	int val, rec, siz;
	node *fa, *son[2];
};

class Tree {
	public:
		node *rt = NULL;
		void insert(int);
		int find(int);
		int kth(int);
		int pre(int);
		int nxt(int);
} tr;

void Tree::insert(int x) {
	if (rt == NULL) {
		rt = new node;
		rt->val = x;
		rt->siz = rt->rec = 1;
		rt->fa = rt->son[0] = rt->son[1] = NULL;
		//cout << "new root" << endl;

		return;
	}

	node *now = rt, *f = NULL;

	while (1) {
		if (now->val == x) {
			now->rec++;
			now->siz++;
			return;
		}

		f = now;
		now = f->son[f->val < x];
		f->siz++;

		if (now == NULL) {
			now = new node;
			now->fa = f;
			now->val = x;
			now->son[0] = now->son[1] = NULL;
			f->son[f->val < x] = now;
			now->siz = now->rec = 1;
			return;
		}
	}
}

int Tree::find(int x) {
	node *now = rt;
	int ans = 0;

	while (1) {
		if (now->son[0] != NULL && now->val > x) {
			now = now->son[0];
			continue;
		}

		if (now->son[0] != NULL) {
			ans += now->son[0]->siz;
		}

		if (now->val == x) {
			return ans + 1;
		}

		ans += now->rec;
		now = now->son[1];
	}
}

int Tree::kth(int x) {
	node *now = rt;

	while (1) {
		if (now->son[0] != NULL && now->val > x) {
			now = now->son[0];
			continue;
		}

		if (now->son[0] != NULL) {
			x -= now->son[0]->siz;
		}

		if (x <= now->rec) {
			return now->val;
		}

		x -= now->rec;
		now = now->son[1];
	}
}

int Tree::pre(int x) {
	node *now = rt;
	int ans = -2147483647;

	while (1) {
		if (now->val >= x) {
			if (now->son[0] != NULL) {
				now = now->son[0];
				continue;
			} else {
				return ans;
			}
		} else {
			ans = max(ans, now->val);

			if (now->son[1] == NULL) {
				return ans;
			} else {
				now = now->son[1];
				continue;
			}
		}
	}
}

int Tree::nxt(int x) {
	node *now = rt;
	int ans = 2147483647;

	while (1) {
		if (now->val <= x) {
			if (now->son[1] != NULL) {
				now = now->son[1];
				continue;
			} else {
				return ans;
			}
		} else {
			ans = min(ans, now->val);

			if (now->son[0] == NULL) {
				return ans;
			} else {
				now = now->son[0];
				continue;
			}
		}
	}
}
int n, x, y;

int main() {
	scanf("%d", &n);

	while (n--) {
		scanf("%d%d", &x, &y);

		switch (x) {
			case 1: {
				printf("%d\n", tr.find(y));
				break;
			}

			case 2: {
				printf("%d\n", tr.kth(y));
				break;
			}

			case 3: {
				printf("%d\n", tr.pre(y));
				break;
			}

			case 4: {
				printf("%d\n", tr.nxt(y));
				break;
			}

			case 5: {
				tr.insert(y);
				break;
			}

			default: {
				break;
			}
		}

	}

	return 0;
}