[2019牛客网多校训练第1场]XOR

链接：https://ac.nowcoder.com/acm/contest/881/H
来源：牛客网

题目描述

Bobo has a set A of $n$ integers $a_1,a_2,\dots ,a_n$.
He wants to know the sum of sizes for all subsets of A whose xor sum is zero modulo $(10^9+7)$.
Formally, find $(\sum S\subseteq A,\oplus x\in Sx=0|S|)mod(10^9+7)$. Note that $\oplus$ denotes the exclusive-or (XOR).

输入描述:

The input consists of several test cases and is terminated by end-of-file.
The first line of each test case contains an integer $n$.
The second line contains n integers $a_1,a_2,\dots ,a_n$.

• $1\le n\le 10^5$
• $0\le a_i\le 10^{18}$
• The number of test cases does not exceed $100$.
• The sum of $n$ does not exceed $2\times 10^6$.

输出描述:

For each test case, print an integer which denotes the result.

输入

1
0
3
1 2 3
2
1000000000000000000 1000000000000000000

输出

1
3
2

题意：
给定一个序列S，问这个序列的所有子序列中，子序列所有的元素的异或和为0的子序列长度和为多少？

题解：
线性基
利用期望的线性性，等价于计算每个数在集合中的方案数。
先消一次元，设会得到 $r$ 个基，那么对于剩下 $n-r$ 个数，方案数都是 $2^{n–r–1}$
再对除了$r$ 个基外的 $n-r$ 个数消个元，得到另一组基 B，枚举 $r$ 个基中的一个，用另外的 $r-1+|B|$ 个数消元，会被消去的基底对答案的贡献也是$2^{n–r–1}$

#include
#define ll long long
#define MOD 1000000007LL
using namespace std;

int n;
ll a[100004];
ll fp(ll x,ll y){
    if(y==0)return 1LL;
    ll temp=fp(x,y>>1);
    if(y&1){
        return (((temp*temp)%MOD)*x)%MOD;
    }
    else {
        return (temp*temp)%MOD;
    }
}
struct LinearBasis{
    int sz;
    ll bs[74];
    set<int>num;
    LinearBasis(int Sz=61){
        sz=Sz;
        memset(bs,0,sizeof(bs));
        num.clear();
    }
    bool ins(ll x,int sit){
        for(ll i=sz-1,bin=(1LL<<(sz-1));i>=0;i--,bin>>=1){
            if(x&bin){
                if(!bs[i]){
                    bs[i]=x;
                    num.insert(sit);
                    break;
                }
                x^=bs[i];
            }
        }
        return x!=0;
    }
}A,B,C;

int w33ha(){
    A=LinearBasis();
    B=LinearBasis();
    C=LinearBasis();
    ll ans=0;
    for(int i=1;i<=n;i++){
        scanf("%lld",&a[i]);
        A.ins(a[i],i);
    }
    ll delta=fp(2LL,n-A.num.size()-1);
    ans=delta%MOD;
    ans=(ans*(n-A.num.size()))%MOD;
    for(int i=1;i<=n;i++){
        if(!A.num.count(i)){
            B.ins(a[i],i);
        }
    }
    for(int x:A.num){
        C=B;
        for(int y:A.num){
            if(x==y)continue;
            C.ins(a[y],y);
        }
        if(!C.ins(a[x],x)){
            ans=(ans+delta)%MOD;
        }
    }
    printf("%lld\n",ans);
    return 0;
}
int main(){
    while(scanf("%d",&n)!=EOF)w33ha();
    return 0;
}