Codeforces Round #430 (Div. 2) D.Vitya and Strange Lesson 异或 01字典树补集最小

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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

题意：求区间没有的最小的那个数，可以异或x。

思路：首先明确对[1,2^n]区间内的所有数亦或x，仍然是[1,2^n]的所有数。

设给定集合为A，全集C，A的补集为B。
易知每次亦或A无需真的改动A数组，A^x1^x2 = A^(x1^x2)。
每次query对A中每个数亦或x, 然后查询min(C-A^x)。
又A^x+B^x = A + B = C，得所求即为min(B^x)。用字典树来求解亦或最值问题。

//china no.1
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
typedef long long LL;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=3e5+10;
const int maxx=1e6+100;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

int next[maxn*20][2];
int val[maxn*20];
int st;
void init()
{
me(next[0],0);
me(val,0);
st=1;
}
void insert(LL x)
{
int u=0;
for(int i=20;i>=0;i--)
{
int c=((x>>i)&1);
if(!next[u][c])
{
me(next[st],0);
next[u][c]=st++;
}
u=next[u][c];
++val[u];
}
}
LL query(LL x)
{
int t=0;
LL ans=0;
for(int i=20;i>=0;i--)
{
int k=((x>>i)&1);
if(!next[t][k])
return ans;
if(val[next[t][k]]==(1<<i))
{
ans|=(1<<i);
t=next[t][1-k];
}
else
t=next[t][k];
}
return ans;
}
int n,m,vis[maxn];
int main()
{
scan_d(n);
scan_d(m);
init();
int x;
FOR(1,n,i)
{
scan_d(x);
if(vis[x])
continue;
else
{
vis[x]=1;
insert(x);
}
}
int y=0;
FOR(1,m,i)
{
scan_d(x);
y^=x;
print(query(y));
}
}

#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <algorithm>
#include <functional>
#include <utility>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <cmath>
#include <cctype>
#define CLEAR(a, b) memset(a, b, sizeof(a))
#define IN() freopen("in.txt", "r", stdin)
#define OUT() freopen("out.txt", "w", stdout)
#define LL long long
#define maxn 524288
#define maxm 6000005
#define mod  10007
#define INF 1000000007
#define EPS 1e-7
#define PI 3.1415926535898
#define N 4294967296
using namespace std;
//-------------------------CHC------------------------------//
struct Node {
Node *next[2];
Node() { next[0] = next[1] = NULL; }
}node[maxm];
int total;

void insert(int num, Node *root) {
Node *cur = root;
for (int i = 19; i >= 0; --i) {
int id = num >> i & 1;
if (!cur->next[id]) cur->next[id] = &node[total++];
cur = cur->next[id];
}
}

int query(int num, Node *root) {
int ret = 0;
Node *cur = root;
for (int i = 19; i >= 0; --i) {
int id = num >> i & 1;
if (!cur->next[id]) id = !id, ret |= (1 << i);
cur = cur->next[id];
}
return ret;
}

bool vis[maxn+10];

int main() {
int n, q;
scanf("%d%d", &n, &q);
int x;
for (int i = 0; i < n; ++i) scanf("%d", &x), vis[x] = 1;
Node *root = &node[total++];
for (int i = 0; i <= maxn; ++i) if (!vis[i]) insert(i, root);
int y = 0;
while (q--) {
scanf("%d", &x);
y ^= x;
printf("%d\n", query(y, root));
}
return 0;
}