Codeforces Round #367 (Div. 2)：Vasiliy's Multiset（01字典树）

D. Vasiliy's Multiset

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Author has gone out of the stories about Vasiliy, so here is just a formal task description.

You are given q queries and a multiset A, initially containing only integer 0. There are three types of queries:

1. "+ x" — add integer x to multiset A.
2. "- x" — erase one occurrence of integer x from multiset A. It's guaranteed that at least one x is present in the multiset A before this query.
3. "? x" — you are given integer x and need to compute the value , i.e. the maximum value of bitwise exclusive OR (also know as XOR) of integer x and some integer y from the multiset A.

Multiset is a set, where equal elements are allowed.

Input

The first line of the input contains a single integer q (1 ≤ q ≤ 200 000) — the number of queries Vasiliy has to perform.

Each of the following q lines of the input contains one of three characters '+', '-' or '?' and an integer xi (1 ≤ xi ≤ 109). It's guaranteed that there is at least one query of the third type.

Note, that the integer 0 will always be present in the set A.

Output

For each query of the type '?' print one integer — the maximum value of bitwise exclusive OR (XOR) of integer xi and some integer from the multiset A.

Example

input

10
+ 8
+ 9
+ 11
+ 6
+ 1
? 3
- 8
? 3
? 8
? 11

output

11
10
14
13

问题概述：你有一个的集合，一开始里面只有1个0，现有3种操作：

①向集合中添加一个数字；②删去集合中的一个数字（必须存在）；

③给一个k，并在集合中找到一个数字使得它异或k最大，求这个最大异或值

http://codeforces.com/contest/706/problem/D

和 http://blog.youkuaiyun.com/jaihk662/article/details/53914302 大同小异，只不过多了添加和删除操作

#include<stdio.h>
typedef struct Trie
{
	int sum;
	Trie *next[2];
}Trie;
void Delete(Trie *root)
{
	if(root->next[0])
		Delete(root->next[0]);
	if(root->next[1])
		Delete(root->next[1]);
	delete root;
}
void Create(Trie *root, int a, int c)
{
	int now, i;
	Trie *p = root;
	for(i=30;i>=0;i--)
	{
		now = 0;
		if(a&(1<<i))
			now = 1;
		if(p->next[now]==NULL)
		{
			Trie *temp = new Trie;
			temp->next[0] = temp->next[1] = NULL;
			temp->sum = 0;
			p->next[now] = temp;
		}
		p = p->next[now];
		p->sum += c;
	}
}
int Query(Trie *root, int k)
{
	int now, i, ans = 0;
	Trie *p = root;
	for(i=30;i>=0;i--)
	{
		now = 0;
		if(k&(1<<i))
			now = 1;
		if(p->next[now^1] && p->next[now^1]->sum>0)
		{
			ans += (1<<i);
			p = p->next[now^1];
		}
		else
			p = p->next[now];
	}
	return ans;
}
int main(void)
{
	int q, a;
	char ch;
	scanf("%d", &q);
	Trie *root = new Trie;
	root->next[0] = root->next[1] = NULL;
	root->sum = 0;
	Create(root, 0, 1);
	while(q--)
	{
		scanf(" %c%d", &ch, &a);
		if(ch=='+')
			Create(root, a, 1);
		if(ch=='-')
			Create(root, a, -1);
		if(ch=='?')
			printf("%d\n", Query(root, a));
	}
	return 0;
}