题目来源：POJ 1442 Black Box

题意：输入xi 输出前xi个数的第i大的数

思路：试了下自己的treap模版

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <ctime>
using namespace std;

struct Node
{
	Node *ch[2];
	int r;
	int v;
	int s;
	Node(){}
	Node(int v): v(v) {
		ch[0] = ch[1] = NULL; r = rand(); s = 1;
	}
	bool operator < (const Node& rhs) const{
		return r < rhs.r;
	}
	int cmp(int x) const{
		if(x == v) return -1;
		return x < v ? 0 : 1;
	}
	void maintain(){
		s = 1;
		if(ch[0] != NULL) s += ch[0]->s;
		if(ch[1] != NULL) s += ch[1]->s;
	}
};

void rotate(Node* &o, int d){
	Node* k = o->ch[d^1]; o->ch[d^1] = k->ch[d]; k->ch[d] = o;
	o->maintain(); k->maintain(); o = k;
}

void insert(Node* &o, int x){
	if(o == NULL){
		o = new Node(x);
	}
	else{
		int d = (x < o->v ? 0 : 1);
		insert(o->ch[d], x);
		if((o->ch[d]->r) > (o->r)) rotate(o, d^1);
	}
	o->maintain();
}

void remove(Node* &o, int x){
	int d= o->cmp(x);
	if(d == -1){
		Node* u = o;	
		if(o->ch[0] != NULL && o->ch[1] != NULL){			
			int d2 = (o->ch[0] > o->ch[1] ? 1 : 0);
			rotate(o, d2); remove(o->ch[d2], x);
		}
		else{
			if(o->ch[0] == NULL) o = o->ch[1];
			else o = o->ch[0];
			delete u;
		}
	}
	else
		remove(o->ch[d], x);
	if(o != NULL) o->maintain();
}

int kth(Node* o, int k){
	if(o == NULL || k <= 0 || k > o->s)
		return 0;
	int s = (o->ch[0] == NULL ? 0 : o->ch[0]->s);
	if(k == s+1) return o->v;
	else if(k <= s) return kth(o->ch[0], k);
	else return kth(o->ch[1], k-s-1);
}

void removetree(Node* &x){
	if(x->ch[0] != NULL) removetree(x->ch[0]);
	if(x->ch[1] != NULL) removetree(x->ch[1]);
	delete x;
	x = NULL;
}
int n, m, a[30010];
Node *rt = NULL;
int main()
{
	while(scanf("%d %d", &n, &m) != EOF)
	{
		srand(time(0));
		for(int i = 1; i <= n; i++)
			scanf("%d", &a[i]);
		int l = 1;
		for(int i = 1; i <= m; i++)
		{
			int x;
			scanf("%d", &x);
			while(l <= x)
			{
				insert(rt, a[l]);
				l++;
			}
			printf("%d\n", kth(rt, i));
		}
		//removetree(rt);
	}
	return 0;
}

POJ 1442 Black Box treap求区间第k大