【 Codeforces Round #367 (Div. 2) D】 Vasiliy's Multiset （Trie 按数位建字典树）

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

11101413

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 .

#include <bits/stdc++.h>using namespace std;#define maxn 200005int ch[maxn * 32 * 2][2], sz[maxn * 32 * 2], tot;void insert(int x, int v){    int k = 0, y;    for(int i = 30; i >= 0; --i){        y = (x >> i) & 1;        if(ch[k][y] == 0){            ch[k][y] = ++tot;            ch[tot][0] = ch[tot][1] = 0;        }        k = ch[k][y];        sz[k] += v;    }}int query(int x){    int k = 0, y, ans = 0;    for(int i = 30; i >= 0; --i){        y = (x >> i) & 1;        if(sz[ch[k][y ^ 1]] > 0){            y ^= 1;            ans |= 1 << i;           }        k = ch[k][y];    }    return ans;}int main(){    int n, x;    cin >> n;    char s[2];    tot = ch[0][0] = ch[0][1] = 0;    memset(sz, 0, sizeof(sz));    insert(0, 1);    for(int i = 1; i <= n; ++i){        scanf("%s %d", s, &x);        if(s[0] == '+'){            insert(x, 1);        }        else if(s[0] == '-'){            insert(x, -1);        }        else{            printf("%d\n", query(x));        }    }} /*题意：给2e5次操作，每次询问给一个整数x，要么将其加入集合（multiset），要么从集合删除，要么找出集合中和它异或最大的答案，集合中保证一定有0。思路：建Trie，插入和删除操作用sz[i]来维护，大于0说明这个结点没被删掉。在Trie上找与x异或最大的值，实际上就是对于x，假设某一位上值为y，我们优先访问y^1即可。因为最高位能异或得到1一定让它优先为1。遍历时维护一下答案即可。*/