codeforces 706D D. Vasiliy's Multiset(trie树)

题目链接：

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

题意：

 

+表示吧这个数加到集合中,-表示把这个数从集合中减去一次,？表示集合里面的一个y使的x^y最大;

 

思路：

 

trie树,把这些数都转化成01串插入到trie树种,？的时候从高位到低位,是0的时候就尽量取1，是1的时候就尽量取0这样贪心就好了;比赛的时候数组开小了;

 

AC代码;

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e18;
const int N=2e5+10;
const int maxn=5e3+4;
const double eps=1e-12;

char s[100];
int x;
int ch[32*N][2],sz=1,val[32*N][2],temp[50];
void in(int num)
{
    int u=0;
    for(int i=31;i>=0;i--)
    {
        int c=temp[i];val[u][c]+=num;
        if(!ch[u][c])
        {
            ch[sz][0]=ch[sz][1]=0;
            ch[u][c]=sz++;
        }
        u=ch[u][c];
    }
}
int query()
{
    int u=0,ans=0;
    for(int i=31;i>=0;i--)
    {
        int c=temp[i];
        if(ch[u][c^1]&&val[u][c^1])
        {
            ans|=(1<<i);
            u=ch[u][c^1];
        }
        else u=ch[u][c];
    }
    return ans;
}

int main()
{
        mst(temp,0);
        in(1);
        int q;
        read(q);
        while(q--)
        {
            scanf("%s%d",s,&x);
            mst(temp,0);
            int cnt=0;
            while(x)
            {
                temp[cnt++]=x%2;
                x>>=1;
            }
            if(s[0]=='+')in(1);
            else if(s[0]=='-')in(-1);
            else  printf("%d\n",query());
        }
        return 0;
}