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=(<<);
 const double ESP=1e-;
 const int mod=(int)1e9+;
 const int N=(int)1e6+;

 struct node
 {
     long long base[];

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

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

     void Insert(long long x)
     {
         for(int i=;i>=;--i)
         {
             if(x>>i&)
             {
                 if(base[i])
                     x^=base[i];
                 else
                 {
                     base[i]=x;
                     break;
                 }
             }
         }
     }
 }BASE[],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,,)
     {
         if(b.base[i])
         {
             long long x=b.base[i],now=;
             int g=;
             for(int j=;j>=;j--)
             {
                 if(x>>j&)
                 {
                     if(!temp.base[j])
                     {
                         g=;
                         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[].Init();
         BASE[].Init();
         ans.Init();
         for(int i=;i<=n;++i)
             sc("%lld",&a[i]),BASE[].Insert(a[i]);
         for(int i=;i<=n;++i)
             sc("%lld",&b[i]),BASE[].Insert(b[i]);
         merge(BASE[],BASE[],ans);
         int cnt=;
         for(int i=;i<;++i)
             if(ans.base[i])
                 cnt++;
         ll ANS=;
         for(int i=;i<=cnt;++i)
             ANS*=;
         pr("%lld\n",ANS);
     }
     return ;
 }

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

 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-);
 }

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

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

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