# Codeforces 842 D. Vitya and Strange Lesson （trie）

## Description

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 ≤ 3·10^5) — number of elements in array and number of queries.

Next line contains n integer numbers ai (0 ≤ ai ≤ 3·10^5) — elements of then array.

Each of next m lines contains query — one integer number x (0 ≤ x ≤ 3·10^5).

## Output

For each query print the answer on a separate line.

## Sample Input

5 4
0 1 5 6 7
1
1
4
5


## Sample Output

2
2
0
2


## AC 代码

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e7+10;
const int maxm = 20;
int n,m,tot;
int a[maxn];

struct node
{
int cnt,res;
node *next[2];
} pn[maxn],*root;

void insert(int x)      //trie树构建
{
node *p=root;
p->cnt++;
for(int i=maxm; ~i; i--)
{
int k=(x>>i)&1;
if(!p->next[k])p->next[k]=&pn[tot++];
p=p->next[k];
p->cnt++;
}
}

void push_down(node *p,int d)       //异或时反转左右孩子
{
for(int i=0; i<2; i++)
if(p->next[i])
p->next[i]->res^=p->res;
if((p->res>>d)&1)
swap(p->next[0],p->next[1]);
p->res=0;
}

void solve()
{
int ans=0;
node *p=root;
for(int i=maxm; ~i; i--)
{
push_down(p,i);
if((p->next[0]?p->next[0]->cnt:0)!=(1<<i))  //寻找mex
p=p->next[0];
else
p=p->next[1],ans|=1<<i;
if(!p)break;
}
cout<<ans<<endl;
}

int main()
{
cin>>n>>m;
for(int i=0; i<n; i++)
cin>>a[i];
sort(a,a+n);
n = unique(a,a+n)-a;
root = &pn[tot++];
for(int i=0; i<n; i++)
insert(a[i]);
for(int i=0; i<m; i++)
{
int x;
cin>>x;
root->res^=x;
solve();
}
return 0;
}