CodeForces 706D 字典树+贪心

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 Abefore 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

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 .

题意

给一个集合，三种操作：1.添加一个数2.删除一个数3.求集合内与给定的数异或最大值

题解

维护一个字典树就行，每次贪心，这东西貌似在noi之前写过，放了一年什么都忘了

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int ch[3200000][2];
int n,x,tot=1,ans,sum[3200000];
char s;
void add(int x)
{
    int now=1;
    for(int i=30;i>=0;i--)
    {
        if(x&(1<<i))
        {
            if(!ch[now][1])ch[now][1]=++tot;
            now=ch[now][1];
            //cout<<x<<" "<<1<<" "<<now<<" "<<sum[now]<<endl;
            sum[now]++;
        }
        else
        {
            if(!ch[now][0])ch[now][0]=++tot;
            now=ch[now][0];
            //cout<<x<<" "<<0<<" "<<now<<" "<<sum[now]<<endl;
            sum[now]++;
        }
    }
}
void del(int x)
{
    int now=1;
    for(int i=30;i>=0;i--)
    {
        if(x&(1<<i))
        {
            if(!ch[now][1])ch[now][1]=++tot;
            now=ch[now][1];sum[now]--;
        }
        else
        {
            if(!ch[now][0])ch[now][0]=++tot;
            now=ch[now][0];sum[now]--;
        }
    }
}
void query(int x)
{
    int now=1;
    for(int i=30;i>=0;i--)
    {
        if(x&(1<<i))
        {
            if(sum[ch[now][0]])
            {
                ans+=(1<<i);
                now=ch[now][0];
            }
            else now=ch[now][1];
        }
        else
        {
            if(sum[ch[now][1]])
            {
                ans+=(1<<i);
                now=ch[now][1];
            }
            else now=ch[now][0];
        }
    }
}
int main()
{
    cin>>n;
    add(0);
    while(n--)
    {
        cin>>s>>x;
        if(s=='+')
        {
            add(x);
        }
        if(s=='-')
        {
            del(x);
        }
        if(s=='?')
        {
            ans=0;
            query(x);
            printf("%d\n",ans);
        }
    }
    return 0;
}