G - Vitya and Strange Lesson（数组-字典树）

G - Vitya and Strange Lesson

 CodeForces - 842D 

Today at the lesson Vitya learned a very interesting function — mex. Mex of a sequence of numbers is the minimum non-negative number that is not present in the sequence as element. For example, mex([4, 33, 0, 1, 1, 5]) = 2 and mex([1, 2, 3]) = 0.

Vitya quickly understood all tasks of the teacher, but can you do the same?

You are given an array consisting of n non-negative integers, and m queries. Each query is characterized by one number x and consists of the following consecutive steps:

• Perform the bitwise addition operation modulo 2 (xor) of each array element with the number x.
• Find mex of the resulting array.

Note that after each query the array changes.

Input

First line contains two integer numbers n and m (1 ≤ n, m ≤ 3·105) — number of elements in array and number of queries.

Next line contains n integer numbers ai (0 ≤ ai ≤ 3·105) — elements of then array.

Each of next m lines contains query — one integer number x (0 ≤ x ≤ 3·105).

Output

For each query print the answer on a separate line.

Examples

Input

2 2
1 3
1
3

Output

1
0

Input

4 3
0 1 5 6
1
2
4

Output

2
0
0

Input

5 4
0 1 5 6 7
1
1
4
5

Output

2
2
0
2

 

题意 + 思路：

这两篇博客解释的非常好，我就不多说了

https://blog.youkuaiyun.com/brazy/article/details/77841433

 

https://blog.youkuaiyun.com/xzzf1024/article/details/79181290

 解释：

  主函数里  再次求MEX的时候 和上次的MEX值异或

            num^=x;           //与上次得到的值异或
            printf("%d\n",num^Find_Trie(num));

这相当于再次输入的X与存在字典树的序列异或。（设存在字典树的序列a，则相当于a^x1^x2）;

个人理解+代码：

//MEX求数列中未出现过的最小自然数
//X异或任何一位结果不同
//与存在的数异或后 会再次成为这个N数组的数
//求MEX的值 会出现在 不在这N数组的数。
//再次求MEX的时候 和上次的MEX值异或 就可以了
//
#include <map>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <cstring>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int maxn=1e6;
int trie[5*maxn][2];
int vis[5*maxn];
int rt,tot;
void init()
{
    mem(vis,0);
    mem(trie,0);
    tot=0;
}
void Build_Trie(int y)
{
    rt=0;
    for(int i=31; i>=0; i--)
    {
        int x=(y>>i)&1;
        if(!trie[rt][x])
            trie[rt][x]=++tot;
        rt=trie[rt][x];
    }
    vis[rt]=y;
}
int Find_Trie(int y)
{
    rt=0;
    int ans=0;
    for(int i=31; i>=0; i--)
    {
        int x=(y>>i)&1;    //找到最小的异或值
        if(!trie[rt][x])
            rt=trie[rt][x^1];
        else
            rt=trie[rt][x];
    }
    return vis[rt];
}
map<int,int>mp;
int main()
{
    int n,m,y;
    while(~scanf("%d%d",&n,&m))
    {
        init();
        mp.clear();
        int x;
        for(int i=0; i<n; i++)
        {
            scanf("%d",&x);
            mp[x]=1;
        }
        for(int i=0; i<=600005; i++)
        {
            if(!mp[i])
                Build_Trie(i);
        }
        int num=0;
        for(int i=0; i<m; i++)
        {
            scanf("%d",&x);
            num^=x;           //与上次得到的值异或
            printf("%d\n",num^Find_Trie(num));
        }
    }
    return 0;
}