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:

Multiset is a set, where equal elements are allowed.

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 x_i (1?≤?x_i?≤?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 setA.

For each query of the type '?' print one integer — the maximum value of bitwise exclusive OR (XOR) of integerx_i and some integer from the multisetA.

10+ 8+ 9+ 11+ 6+ 1? 3- 8? 3? 8? 11Output
11101413Note
After first five Operations multiset A contains integers0, 8, 9, 11, 6 and 1.
The answer for the sixth query is integer  — maximum among integers,,, and.
題目大意：
一共有三種操作：
+ x 表示我們新添加一個元素x（可以重復添加）.
- x 表示我們刪除一個元素x.
？ x 詢問此時有的元素可以和x亦或的最大值，（最終值可能是本身）；
思路：
按位壓縮為01串，然后將其按照三種操作處理：
+x -x分別就是增添和刪除操作。
？x我們就貪心去查詢即可。
1、我們按照長度為35位的二進制01字符串建樹。將所有數據都建到樹中。
因為最終解可能是本身，那么最初的時候添加到樹中一個全是0的字符串。
2、接下來對于每個查詢，同樣處理成35位二進制01字符串，對應進行查詢，如果當前位子是0，那么盡量往1那邊走，同理，如果當前位子是1，那么盡量往0那邊走即可。
Ac代碼：
#include<stdio.h>#include<string.h>#include<stdlib.h>using namespace std;#define maxn 2typedef struct tree{    tree *nex[maxn];    int v;    int val;}tree;tree root;void init(){    for(int i=0;i<maxn;i++)    {        root.nex[i]=NULL;    }}void creat(char *str,int va){    int len=strlen(str);    tree *p=&root,*q;    for(int i=0;i<len;i++)    {        int id=str[i]-'0';        if(p->nex[id]==NULL)        {            q=(tree *)malloc(sizeof(root));            q->v=1;            for(int j=0;j<2;j++)            {                q->nex[j]=NULL;            }            p->nex[id]=q;        }        else        {            p->nex[id]->v++;        }        p=p->nex[id];        if(i==len-1)        {            p->val=va;        }    }}void del(char *str){    int len=strlen(str);    tree *p=&root;    for(int i=0;i<len;i++)    {        int id=str[i]-'0';        p->nex[id]->v--;        tree *tmp=p->nex[id];        if(p->nex[id]->v==0)        {            p->nex[id]=NULL;        }        p=tmp;    }    return ;}void find(char *str,int query){    int len=strlen(str);    tree *p=&root;    for(int i=0;i<len;i++)    {        int id=str[i]-'0';        if(p->nex[1-id]!=0)        {            p=p->nex[1-id];        }        else        p=p->nex[id];        if(p==NULL)        return ;        if(i==len-1)printf("%d/n",p->val^query);    }}int main(){    int n;    while(~scanf("%d",&n))    {        init();        char s[50];        s[36]='/0';        for(int i=0;i<=35;i++)s[i]='0';        creat(s,0);        while(n--)        {            char tmp[5];            char s[50];            scanf("%s",tmp);            int val;            scanf("%d",&val);            int a=val;            s[36]='/0';            for(int j=35;j>=0;j--)            {                if(a)                {                    s[j]=a%2+'0';                    a/=2;                }                else                {                    s[j]='0';                }            }            if(tmp[0]=='+')creat(s,val);            if(tmp[0]=='?')find(s,val);            if(tmp[0]=='-')del(s);        }    }}