Intersection（线性基）

题目描述：
Bobo has two sets of integers A = {a1,a2,…,an}and B = {b1,b2,…,bn}
He says that x∈span(A) (or span(B)) if and only if there exists a subset of A (or B) whose exclusive-or sum equals to x.
Bobo would like to know the number of x where x∈span(A) and x∈span(B) hold simultaneously.
输入描述：
The input contains zero or more test cases and is terminated by end-of-file. For each test case:
The first line contains an integer n.
The second line contains n integers a1,a2,…an
The third line contains n integers b1,b2,…,bn
*1≤n≤50

• 0 ≤ ai, bi ≤ 2^60
• The number of test cases does not exceed 5000.
  输出描述：
  For each case, output an integer which denotes the result.

题意：
给出A,B两个数组，定义x在Span（A）集合里为：A的一个子集的异或和为x（可以是空集），求有多少个x即在Span（A）也在Span（B）中。
题解：
构造A,B线性基，对A中的每个基在B中访问并存储，更新答案
code:

#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll a[61],b[61];
bool insert(ll x,ll *t)
{
    for(int i=60;i>=0;i--)
    {
        if(!((1ll<<i)&x))continue;
        if(!t[i])
        {
            t[i]=x;
            return false;
        }
        else x^=t[i];
        if(!x)return true;
    }
    return true;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        ll p;
        memset(a,0,sizeof a);
        memset(b,0,sizeof b);
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&p);
            insert(p,a);
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&p);
            insert(p,b);
        }
        ll ans=1;
        for(int i=60;i>=0;i--)
        {
            if(!b[i])continue;
            if(insert(b[i],a))ans<<=1;
        }
        printf("%lld\n",ans);
    }
    return 0;
}