codeforces 367 D. Vasiliy's Multiset

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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

Copy

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

output

Copy

11
10
14
13

题意：初始有个一个空集，n此操作，操作分三种

1 + x，将一个x加入集合

2 - x，删除集合内的一个x

3 ？x，询问集合中与x异或的最大值

求集合内异或最大值一般用字典树来求，所以建立一个字典树维护一下。

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
struct node
{
	int cnt;
	node *next[2];
	node()
	{
		cnt = 0;
		memset(next, 0, sizeof(next));
	}
};
node *root = new node();
void insert(int x)
{
	node *p = root;
	for (int i = 30;i >= 0;i--)
	{
		int num = x&(1 << i) ? 1 : 0;
		if (p->next[num] == NULL)
			p->next[num] = new node();
		p = p->next[num];
		p->cnt++;
	}
}
void del(int x)
{
	node *p = root;
	for (int i = 30;i >= 0;i--)
	{
		int num = x&(1 << i) ? 1 : 0;
		p = p->next[num];
		p->cnt--;
	}
}
int query(int x)
{
	node *p = root;
	int ans = 0;
	for (int i = 30;i >= 0;i--)
	{
		int num = x&(1 << i) ? 0 : 1;
		if (p->next[num] != NULL&&p->next[num]->cnt > 0)
			ans += 1 << i, p = p->next[num];
		else p = p->next[num ^ 1];
	}
	return ans;
}
int main()
{
	int n, i, x;
	char str[2]; 
	insert(0);
	scanf("%d", &n);
	for (i = 1;i <= n;i++)
	{
		scanf("%s%d", str, &x);
		if (str[0] == '+')
			insert(x);
		else if (str[0] == '-')
			del(x);
		else
			printf("%d\n", query(x));
	}
	return 0;
}