线性基求交（2019牛客国庆集训派对day4）

题意：https://ac.nowcoder.com/acm/contest/1109/C

问你有几个x满足A，B集合都能XOR出x。

思路：

就是线性基求交后，有几个基就是2^几次方。

#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
#include <cstdio>//sprintf islower isupper
#include <cstdlib>//malloc  exit strcat itoa system("cls")
#include <iostream>//pair
#include <fstream>//freopen("C:\Users\13606\Desktop\草稿.txt","r",stdin);
#include <bitset>
//#include <map>
//#include<unordered_map>
#include <vector>
#include <stack>
#include <set>
#include <string.h>//strstr substr
#include <string>
#include <time.h>//srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
#include <cmath>
#include <deque>
#include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
#include <vector>//emplace_back
//#include <math.h>
//#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
#include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
//******************
int abss(int a);
int lowbit(int n);
int Del_bit_1(int n);
int maxx(int a,int b);
int minn(int a,int b);
double fabss(double a);
void swapp(int &a,int &b);
clock_t __STRAT,__END;
double __TOTALTIME;
void _MS(){__STRAT=clock();}
void _ME(){__END=clock();__TOTALTIME=(double)(__END-__STRAT)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;}
//***********************
#define rint register int
#define fo(a,b,c) for(rint a=b;a<=c;++a)
#define fr(a,b,c) for(rint a=b;a>=c;--a)
#define mem(a,b) memset(a,b,sizeof(a))
#define pr printf
#define sc scanf
#define ls rt<<1
#define rs rt<<1|1
typedef long long ll;
const double E=2.718281828;
const double PI=acos(-1.0);
//const ll INF=(1LL<<60);
const int inf=(1<<30);
const double ESP=1e-9;
const int mod=(int)1e9+7;
const int N=(int)1e6+10;

struct node
{
    long long base[62];

    void Init()
    {
        for(int i=0;i<61;++i)
            base[i]=0;
    }

    bool check(long long x)
    {
        for(int i=61;i>=0;--i)
        {
            if(x&(1LL<<i))
            {
                if(!base[i])return 0;
                else x^=base[i];
            }
        }
        return 1;
    }

    void Insert(long long x)
    {
        for(int i=61;i>=0;--i)
        {
            if(x>>i&1)
            {
                if(base[i])
                    x^=base[i];
                else
                {
                    base[i]=x;
                    break;
                }
            }
        }
    }
}BASE[5],temp,v,ans;

ll a[N],b[N];
void merge(const node &a,node &b,node &ans)
{//  https://blog.csdn.net/qcwlmqy/article/details/97584411
    temp=v=a;
    fo(i,0,60)
    {
        if(b.base[i])
        {
            long long x=b.base[i],now=0;
            int g=0;
            for(int j=60;j>=0;j--)
            {
                if(x>>j&1)
                {
                    if(!temp.base[j])
                    {
                        g=1;
                        temp.base[j]=x;
                        v.base[j]=now;
                        break;
                    }
                    x^=temp.base[j];now^=v.base[j];
                }
            }
            if(!g)
                ans.Insert(now);
        }
    }
}
int main()
{
    int n;
    while(~sc("%d",&n))
    {
        BASE[1].Init();
        BASE[2].Init();
        ans.Init();
        for(int i=1;i<=n;++i)
            sc("%lld",&a[i]),BASE[1].Insert(a[i]);
        for(int i=1;i<=n;++i)
            sc("%lld",&b[i]),BASE[2].Insert(b[i]);
        merge(BASE[1],BASE[2],ans);
        int cnt=0;
        for(int i=0;i<61;++i)
            if(ans.base[i])
                cnt++;
        ll ANS=1;
        for(int i=1;i<=cnt;++i)
            ANS*=2;
        pr("%lld
",ANS);
    }
    return 0;
}

/**************************************************************************************/

int maxx(int a,int b)
{
    return a>b?a:b;
}

void swapp(int &a,int &b)
{
    a^=b^=a^=b;
}

int lowbit(int n)
{
    return n&(-n);
}

int Del_bit_1(int n)
{
    return n&(n-1);
}

int abss(int a)
{
    return a>0?a:-a;
}

double fabss(double a)
{
    return a>0?a:-a;
}

int minn(int a,int b)
{
    return a<b?a:b;
}