Codeforces Round #367 (Div. 2)D. Vasiliy's Multiset (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

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 .

 得学姐指导----涉及到XOR的建树都建字典树。

哇哈哈~

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn = 5e6+5;
struct node
{
    int next[5];
    int v;
};
node tree[maxn];
int sz = 1;
void build(int x,int v)
{
    int root = 0;
    for(int i=31;i>=0;i--)
    {
        int id = (x>>i)&1;
        if(tree[root].next[id]==0)
        {
            memset(tree[sz].next,0,sizeof(tree[sz].next));
            tree[sz].v = 0;
            tree[root].next[id] = sz++;
        }
        root = tree[root].next[id];
        tree[root].v+=v;
    }
}
void match(int x)
{
    int root = 0;
    x = ~x;
    int ans = 0;
    for(int i=31;i>=0;i--)
    {
        ans *= 2;
        int id = (x>>i)&1;
        if(tree[root].next[id]&&tree[tree[root].next[id]].v)
        {
            ans++;
            root = tree[root].next[id];
        }
        else
        {
            root = tree[root].next[1-id];
        }
    }
    printf("%d
",ans);
}
int main()
{
    int n,x;
    char s[3];
    cin>>n;

  /*  for(int i=0;i<=maxn-1;i++)
    {
         tree[i].v = 0;
         memset(tree[i].next,0,sizeof(tree[i].next));
    }*/
    build(0,1);
    for(int i=1;i<=n;i++)
    {
        scanf("%s %d",s,&x);
        if(s[0]=='+')
        {
            build(x,1);
        }
        else if(s[0]=='-')
        {
            build(x,-1);
        }
        else
        {
            match(x);
        }
    }
    return 0;
}