2019牛客多校一 H. XOR (线性基)

大意: 给定序列, 求所有异或和为$0$的子序列大小之和.

先求出线性基, 假设大小为$r$.

对于一个数$x$, 假设它不在线性基内, 那么贡献为$2^{n-r-1}$

因为它与其余不在线性基内数的任意组合后均可以与线性基异或后变为$0$, 产生$1$的贡献.

所以问题就转化为求多少个数可以不在线性基内.

现任意求出一组线性基, 然后再暴力验证该组线性基内的数即可.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head
 
const int N = 1e5+10;
int n, vis[N];
ll a[N];
struct _ {
    ll a[66];
	_ () {memset(a,0,sizeof a);}
    inline bool ins(ll x) {
        REP(i,1,*a) x=min(x,a[i]^x);
        return x?a[++*a]=x:0;
    }
    inline _ operator + (const _ &rhs) const {
        _ r;
        REP(i,0,*a) r.a[i]=a[i];
        REP(i,1,rhs.a[0]) r.ins(rhs.a[i]);
        return r;
    }
	inline int chk(ll x) {
		REP(i,1,*a) x=min(x,a[i]^x);
		return !x;
	}
} pre[N], suf[N];
 
int main() {
	while (~scanf("%d", &n)) {
		REP(i,1,n) {
			scanf("%lld", a+i);
			if ((pre[i]=pre[i-1]).ins(a[i])) vis[i] = 1;
		}
		if (pre[n].a[0]==n) {
			puts("0");
			continue;
		}
		suf[n+1] = _();
		PER(i,1,n) (suf[i]=suf[i+1]).ins(a[i]);
		int sum = 0;
		REP(i,1,n) {
			if (!vis[i]) ++sum;
			else {
				vis[i] = 0;
				if ((pre[i-1]+suf[i+1]).chk(a[i])) ++sum;
			}
		}
		int ans = sum*qpow(2,n-pre[n].a[0]-1)%P;
		printf("%d\n", ans);
	}
}