【BZOJ 4269】再见Xor

zky学长提供的线性基求法：

for(int i=1;i<=n;i++)
for(int j=64;j>=1;j--)
{
    if(a[i]>>(j-1)&1)
    {
        if(!lb[j]){lb[j]=a[i];break;}
        else a[i]^=lb[j];
    }
}

Gauss消元求线性基的方法：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define read(x) x=getint()
using namespace std;
const int N = 100003;
int getint() {
	int k = 0, fh = 1; char c = getchar();
	for(; c < ‘0‘ || c > ‘9‘; c = getchar())
		if (c == ‘-‘) fh = -1;
	for(; c >= ‘0‘ && c <= ‘9‘; c = getchar())
		k = (k << 1) + (k << 3) + c - ‘0‘;
	return k * fh;
}
int n, a[N];
void Gauss() {
	int tmp = 0, i;
	for(int j = 1 << 30; j; j >>= 1) {
		for(i = tmp + 1; i <= n; ++i)
			if (a[i] & j)
				break;
		if (i > n) continue;
		swap(a[++tmp], a[i]);
		for(i = 1; i <= n; ++i)
			if (i != tmp && a[i] & j)
				a[i] ^= a[tmp];
	}
	n = tmp;
}
int main() {
	read(n);
	for(int i = 1; i <= n; ++i)
		read(a[i]);
	Gauss();
	int ans = 0;
	for(int i = 1; i <= n ;++i)
		ans ^= a[i];
	printf("%d %d\n", ans, ans ^ a[n]);
	return 0;
}

没了