Codeforces Round #430 (Div. 2) D. Vitya and Strange Lesson

地址：http://codeforces.com/contest/842/problem/D

题目：

D. Vitya and Strange Lesson

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

 思路：

　　很明显的trie树题，本来以为写完就ac了的，没想到一直wa6.。。。

　　今天起来后发现是把重复数字插入了，应该hash去重的。

#include <bits/stdc++.h>

using namespace std;
int hs[1000000];
struct Trie
{
    int root, tot, next[1000005][2], cnt[1000005], end[1000005];

    inline int Newnode()
    {
        memset(next[tot], -1, sizeof(next[tot]));
        cnt[tot] = 0;
        end[tot] = 0;
        return tot ++;
    }

    inline void Init()
    {
        tot = 0;
        root = Newnode();
    }

    inline int Insert(int x)
    {
        int p = root;
        cnt[p] ++;
        for(int i = 31; i >= 0; i --)
        {
            int idx = ((1 << i) & x) ? 1 : 0;
            if(next[p][idx] == -1)
                next[p][idx] = Newnode();
            p = next[p][idx];
            cnt[p] ++;
        }
        end[p] = x;
        return 1;
    }
    inline int Search(int x)
    {
        int p = root;
        int sum[50],ret=0;
        sum[0]=1;
        for(int i=1;i<31;i++)
            sum[i]=sum[i-1]*2;
        for(int i = 31; i >= 0; i --)
        {
            int idx = ((1 << i) & x) ? 1 : 0;
            if(idx == 0)
            {
                if(next[p][0]==-1) return ret;
                else if(cnt[next[p][0]]==sum[i])
                    p = next[p][1],ret+=sum[i];
                else
                    p = next[p][0];
            }
            else
            {
                if(next[p][1]==-1) return ret;
                else if(cnt[next[p][1]]==sum[i])
                    p = next[p][0],ret+=sum[i];
                else
                    p = next[p][1];
            }
            if(p==-1) return ret;
        }
        return ret;
    }
}tr;

int main(void)
{
    int n,m;
    cin>>n>>m;
    tr.Init();
    for(int i=1,x;i<=n;i++)
        scanf("%d",&x),hs[x]?0:hs[x]=tr.Insert(x);
    for(int i=1,x,ls=0;i<=m;i++)
        scanf("%d",&x),printf("%d
",tr.Search(ls=(x^ls)));
    return 0;
}