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 
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n个数

+ a代表把a加入字典

- a代表把b移出字典

? a要求输出字典中与a异或最大的


思路:

套路神似,保留所有的二进制位数,从高位开始去寻找相反的,贪心找最高位,答案会最大

注意cnt的问题,因为我们要保证当前这个是字典中某个数的二进制数 而不是 二进制数的前缀. 所以,我们全部cnt++. 在query的时候,如果p->next==NULL,说明根本没出现过.如果p->cnt==0说明出现过,但是已经被删完了,下面全都是空的.

假如,我们的cnt++放在最后.而不是对于所有前缀进行cnt++.那么会发生什么问题呢?对于很多节点我们的cnt都是0,无法得知下面到底有没有.那么在p->[t]还是p->[1-t]就无法判断.出错. 对于每个节点cnt++,这样就完美地联系了前缀和完整二进制的关系.

#include <stdio.h>
#include <malloc.h>
#include <iostream>
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define bug1 cout << "bug1" << endl
#define bug2 cout << "bug2" << endl
#define bug3 cout << "bug3" << endl
#define bug4 cout << "bug4" << endl
#pragma comment(linker, "/STACK:102400000,102400000")


using namespace std;
typedef long long ll;

const int MAX_N=1e5+5;
struct node{
    node *next[2];
    int cnt;
    node(){
        cnt=0;
        next[0]=next[1]=NULL;
    }
};

node *root;
void buildtrie(ll x){
    node *p=root;
    for(int i=1;i<=32;i++){
        int t=(x>>(32-i))&1;
//        if(x==11)cout <<"t="<<t<<" i="<<i<<endl;
        if(!p->next[t]) p->next[t]=new node();
        p=p->next[t];
        p->cnt++;
    }
    return ;
}

void del(ll x){
    node *p=root;
    for(int i=1;i<=32;i++){
        int t=(x>>(32-i))&1;
        p=p->next[t];p->cnt--;
//        cout <<"bit="<<33-i<<" p->cnt="<< p->cnt << endl;
    }
    return ;
}

ll query(ll x){
    node *p=root;
    ll sum=0;
    for(int i=1;i<=32;i++){
        int t=(x>>(32-i))&1;
        if(!p->next[1-t] || !p->next[1-t]->cnt) p=p->next[t];
        else{
            sum+=1<<(32-i);
            p=p->next[1-t];
//            cout <<"need="<<1-t<<"bit="<<33-i<<" p->cnt="<< p->cnt << endl;
        }
    }
    return sum;
}

int main(void){
    int t;
    root=new node();
    buildtrie(0);
    cin >> t;
    for(int i=1;i<=t;i++){
        char op[2];
        ll num;
        scanf("%s",op+1);
        scanf("%I64d",&num);
        if(op[1]=='+')  buildtrie(num);
        else if(op[1]=='-') del(num);
        else if(op[1]=='?') printf("%I64d\n",query(num));
    }
    return 0;
}
/*
*/