link

思路：

求异或的最大值容易想到01字典树，只需要再维护删除操作就好。
加一个\(a\)数组记录一下有多少个节点经过了这个点，删除的时候如果这是最后一个经过该点的节点，就删除他的父节点向下指的指针。用map维护一下每个数出现的次数，字典树只维护种类。
注意要先将0插入字典树，集合里没有数的时候，x的异或最大值依旧是x。

代码：

// Problem: CF706D Vasiliy's Multiset
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/CF706D
// Memory Limit: 250 MB
// Time Limit: 4000 ms
// Author:Cutele0626
//
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
typedef pair<double, double>PDD;
#define I_int ll
inline ll read()
{
    ll x = 0, f = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-')f = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}

inline void out(ll x){
	if (x < 0) x = ~x + 1, putchar('-');
	if (x > 9) out(x / 10);
	putchar(x % 10 + '0');
}

inline void write(ll x){
	if (x < 0) x = ~x + 1, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
	puts("");
}

#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define perr(i,a,b) for(int i=(a);i>(b);i--)
ll ksm(ll a, ll b, ll p)
{
    ll res = 1;
    while(b)
    {
        if(b & 1)res = res * a % p;
        a = a * a % p;
        b >>= 1;
    }
    return res;
}
const ll inf =1e15;
#define PI acos(-1)
const int maxn=100000+10;

const int N = 200000+10, M = 3000000;
int n;
int son[N*32][2], idx,a[N*32];
void insert(int x)
{
    int p = 0;
    for (int i = 31; i >= 0; i -- )
    {
        int &s = son[p][x >> i & 1];
        if (!s) s = ++ idx;
        p = s;
        a[p]++;
    }
}
int search(int x)
{
    int p = 0, res = 0;
    for (int i = 31; i >= 0; i -- )
    {
        int s = (x >> i) & 1;
        if (son[p][!s])
        {
            res |= (1 << i);
            p = son[p][!s];
        }
        else p = son[p][s];
    }
    return res;
}

void del(int x){
	int p = 0;
    for (int i = 31; i >= 0; i -- )
    {
        int &s = son[p][x >> i & 1];
        int tt=p;
        p = s;
        a[p]--;
        if(a[p]==0) son[tt][x>>i&1]=0;
    }
}

int main(){
	n=read;
	map<int,int>mp;
	insert(0);
	while(n--){
		char op[2];cin>>op;
		int x=read;
		if(op[0]=='+'){
			mp[x]++;
			if(mp[x]==1) insert(x);
		}
		else if(op[0]=='-'){
			mp[x]--;
			if(mp[x]==0) del(x);
		}
		else printf("%d\n",search(x));
	}
	return 0;
}