Vasiliy's Multiset CodeForces - 706D（异或&01字典树）

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本文介绍了一种使用01字典树解决带有增删操作的最大异或值查询问题的方法，通过维护一个能记录元素存在状态的字典树来高效处理各种查询。

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:

"+ x" — add integer x to multiset A.
"- 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.
"? x" — you are given integer x and need to compute the value $\text{[math]}$ , 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 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 set A.

Output

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

Example

Input

10
+ 8
+ 9
+ 11
+ 6
+ 1
? 3
- 8
? 3
? 8
? 11

Output

Note

After first five operations multiset A contains integers 0, 8, 9, 11, 6 and 1.

The answer for the sixth query is integer $\text{[math]}$ — maximum among integers $\text{[math]}$ , $\text{[math]}$ , $\text{[math]}$ , $\text{[math]}$ and $\text{[math]}$ .

题意：有一个集合，初始时只有0，三种操作+ x:意思是往集合中加入x，- x：从集合中删除x，？ x：询问x与集合中的所有数字的最大异或值。需要明确的一点是，若向集合中添加了两个数字x，然后删除了一个x，那么此时集合中还是存在一个x的。

思路：同样的是一道01字典树，只是多了一个删除操作，我们只需再开一个二维数组将每个数字是否存在统计一下即可。详细请看代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<stack>
using namespace std;
#define inf 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long LL;
const int mod=1e9+7;
const double pi=acos(-1);
const double eps=1e-8;
const int N=2e5+10;
int trie[N*32][2],tot;
int sum[N*32][2];//统计二进制位是否存在，为0是表示不存在，>0存在
void sert(int x,int y)
{
    int root=0;
    for(int i=31;i>=0;i--)
    {
        int id=(x>>i)&1;
        sum[root][id]+=y;
        if(!trie[root][id]) trie[root][id]=++tot;
        root=trie[root][id];
    }
}
void finf(int x)
{
    int root=0,ans=0;
    for(int i=31;i>=0;i--)
    {
        int id=(x>>i)&1^1;
        if(trie[root][id]&&sum[root][id]) ans+=1<<i;
        else id^=1;
        root=trie[root][id];
    }
    printf("%d\n",ans);
}
char s[2];
int main()
{
    int t,n;
    sert(0,1);
    scanf("%d",&t);
    while(t--)
    {
        mem(s,0);
        scanf("%s%d",s,&n);
        if(s[0]=='+') sert(n,1);
        else if(s[0]=='-') sert(n,-1);
        else finf(n);
    }
}

没用位运算的代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<stack>
using namespace std;
#define inf 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long LL;
const int mod=1e9+7;
const double pi=acos(-1);
const double eps=1e-8;
const int N=2e5+10;
int trie[N*32][2],tot,s[33],ans[33];
int sum[N*32][2];
void sert(int x)
{
    int root=0;
    for(int i=31;i>=0;i--)
    {
        int id=s[i];
        sum[root][id]+=x;
        if(!trie[root][id])
            trie[root][id]=++tot;
        root=trie[root][id];
    }
}
void finf(int x)
{
    int root=0,k=0;
    for(int i=31;i>=0;i--)
    {
        int id=s[i];
        if(trie[root][id^1]&&sum[root][id^1])
        {
            ans[k++]=id^1;
            root=trie[root][id^1];
        }
        else
        {
            ans[k++]=id;
            root=trie[root][id];
        }
    }
    int tmp=0,t=1;
    for(int i=31;i>=0;i--)
    {
        tmp+=ans[i]*t;
        t*=2;
    }
    printf("%d\n",tmp^x);
}
char ss[2];
int main()
{
    int t,n;
    mem(s,0);
    sert(1);
    scanf("%d",&t);
    while(t--)
    {
        mem(s,0);
        scanf("%s%d",ss,&n);
        int k=0,tt=n;
        while(n)
        {
            s[k++]=n%2;
            n>>=1;
        }
        if(ss[0]=='+') sert(1);
        else if(ss[0]=='-') sert(-1);
        else finf(tt);
    }
}