Codeforces 706D Vasiliy's Multiset（异或字典树）

转自：传送门
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:
"+ x" — add integer x to multiset A.
"- 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.
"? 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 integerxi 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

Note

After first five operations multiset A contains integers 0, 8, 9, 11, 6 and 1.

The answer for the sixth query is integer 

 — maximum among integers 

, 

, 

, 

 and 

.

题目大意：
    有一个开始只有0的集合，有3种操作，向集合中添加一个数字，删去集合中的一个数字，寻找输入数字与集合中的所有数字的最大异或值。

解题思路：
    新姿势get√
    据说这是一个经典的异或字典树，我们把每个数字的每一个二进制位拆开处理。最高位作根节点，插入和删除和普通字典树相同，对于查询，我们可以贪心的从高位往下找，如果这一位存在可以异或得到1的儿子，则沿这个儿子的方向向下走，否则沿另一个儿子的方向。
    O(Q*log len)。

#include<stdio.h>
#include<iostream>
#include<string.h>

#define me(x) memset(x,0,sizeof(x))
#define LL long long
#define close() ios::sync_with_stdio(0); cin.tie(0);
using namespace std;

const int maxn=2e5+10;
int next[maxn*32][2];
LL val[maxn*32];
int st;
void init()
{
me(next[0]);
me(val);
st=1;
}
void insert(LL x)
{
int u=0;
for(int i=32;i>=0;i--)
{
int c=((x>>i)&1);
if(!next[u][c])
{
me(next[st]);
next[u][c]=st++;
}
u=next[u][c];
++val[u];
}
}
void _delete(LL x)
{
int u=0;
for (int i=32; i>=0; --i)
{
int c=((x>>i)&1);
u=next[u][c];
--val[u];
}
}
LL query(LL x)
{
int t=0;
LL ans=0;
for(int i=32;i>=0;i--)
{
int k=((x>>i)&1);
if(k==1)
{
if(next[t][0]&&val[next[t][0]])
{
ans+=1<<i;
t=next[t][0];
}
else t=next[t][1];
}
else
{
if(next[t][1]&&val[next[t][1]])
{
ans+=1<<i;
t=next[t][1];
}
else t=next[t][0];
}
}
return ans;
}
int main()
{
int n;
close();
cin>>n;
init();
insert(0);
char op[5];int x;
while(n--)
{
cin>>op>>x;
if(op[0]=='+')
insert(x);
if(op[0]=='-')
_delete(x);
if(op[0]=='?')
cout<<query(x)<<endl;
}
}