动态字典树

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);}
struct node
{
    node * child[2];
    LL idx;
    int cnt;
    bool flag;
};
LL ans,query;
int N,M,cas;
bool found;
node * root;
node* addnew()
{
    node * u = (node*)malloc(sizeof(node));
    for (int i = 0 ; i < 2; i++)
        u -> child[i] = NULL;
    u -> flag = false;
    u -> cnt = 0;
    return u;
}
void Insert(LL val,LL id)
{
    node *u = root;
    for (LL i = 0 ; i <= 32 ; i++)
    {
        int tmp;
        if ((val & (1LL << (32 - i))) == 0)   tmp = 0;
        else tmp = 1;
        if (u -> child[tmp] == NULL) u -> child[tmp] = addnew();
        u = u -> child[tmp];
        u -> cnt++;
    }
    u -> cnt++;
}

void del(LL val) {
    node * u = root;
    for (LL i = 0 ; i <= 32 ; i++) {
        int tmp;
        if ((val & (1LL << (32 - i))) == 0)   tmp = 0;
        else tmp = 1;
        if (u -> child[tmp] == NULL) continue;
        u = u -> child[tmp];
        u -> cnt--;
    }
    u -> cnt--;
}

void calcu(LL val,node * cur,int depth,LL sum)
{
    if (depth >= 33)
    {
        ans = max(ans,sum);
        return;
    }
    int tmp;
    if (val & (1LL << (32 - depth))) tmp = 1;
    else tmp = 0;
    if (tmp == 1)
    {
        if (cur -> child[0] != NULL && cur -> child[0] -> cnt > 0)
            calcu(val,cur -> child[0],depth + 1,sum + (1LL << (32 - depth)));
        else if (cur -> child[1] != NULL &&  cur -> child[1] -> cnt > 0)
            calcu(val,cur -> child[1],depth + 1,sum);
    }
    else
    {
        if (cur -> child[1] != NULL && cur -> child[1] -> cnt > 0)
            calcu(val,cur -> child[1],depth + 1,sum + (1LL << (32 - depth)));
        else if (cur -> child[0] != NULL && cur -> child[0] -> cnt > 0 )
            calcu(val,cur -> child[0],depth + 1,sum);
    }
}

int main()
{
    cin >> N;
    root = addnew();
    found = false;
    cas = N + 1;
    Insert(0,0);
    for (int i = 0 ; i < N ; i++) {
        char op[5];
        cin >> op;
        if (op[0] == '+') {
            LL x;
            cin >> x;
            Insert(x,x);
        }
        else if (op[0] == '-') {
            LL x;
            cin >> x;
            del(x);
        }
        else {
            LL x;
            cin >> x;
            ans = 0;
            found = false;
            calcu(x,root,0,0);
            cout << ans << endl;
        }
    }
    return 0;
}

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 .