Codeforces Round #430 D. Vitya and Strange Lesson

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.

题目大意:
定义mex数为数组中第一个没有出现的非负整数.有m个操作,每个操作有一个x，将数组中所有的元素都异或x，然后询问当前的mex
解题报告:
考场上搞了一个小时，原来看错题了，其实只是简单的Trie树基本操作，
mex数:如果左子树没满直接走左子树，不然就走右子树.
异或操作:如果x的该位为1,交换该节点的左右子树,打上标记即可

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#define RG register
#define il inline
#define iter iterator
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
const int N=6e6+10,maxdep=21;
int gi(){
	int str=0;char ch=getchar();
	while(ch>'9' || ch<'0')ch=getchar();
	while(ch>='0' && ch<='9')str=(str<<1)+(str<<3)+ch-48,ch=getchar();
	return str;
}
struct node{
	int l,r,s,rev;
}t[N];
int n,root=0,tot=0,w[30],m;
void insert(int &rt,int x,int d){
	if(!rt)rt=++tot;
	if(d==-1){
		t[rt].s=1;return ;
	}
	if(x&w[d])insert(t[rt].r,x,d-1);
	else insert(t[rt].l,x,d-1);
	t[rt].s=t[t[rt].l].s&t[t[rt].r].s;
}
void pushdown(int rt,int d){
	if(!t[rt].rev)return ;
	int k=t[rt].rev;
	t[t[rt].l].rev^=k;t[t[rt].r].rev^=k;
	if(d>=1 && (k&w[d-1])){
		swap(t[t[rt].l].l,t[t[rt].l].r);swap(t[t[rt].r].l,t[t[rt].r].r);
	}
	t[rt].rev=0;
}
int query(int rt,int d){
	if(d==-1)return 0;
	pushdown(rt,d);
	if(!t[t[rt].l].s)return query(t[rt].l,d-1);
	return query(t[rt].r,d-1)+w[d];
}
void work()
{
	int x;
	n=gi();m=gi();
	w[0]=1;for(int i=1;i<=maxdep;i++)w[i]=w[i-1]<<1;
	for(int i=1;i<=n;i++){
		x=gi();insert(root,x,maxdep);
	}
	while(m--){
		scanf("%d",&x);
		t[root].rev^=x;
		printf("%d\n",query(root,maxdep));
	}
}

int main()
{
	work();
	return 0;
}