构造线性基后将其消至对任意位至多只有一个元素该位为1。于是就可以贪心了，将k拆成二进制就好。注意check一下是否能异或出0。

#include
     
       #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           using namespace std;#define N 100010#define ll long longll read(){ ll x=0,f=1;char c=getchar(); while (c<'0'||c>'9') { if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f;}bool flag=0;ll n,m,base[51],s;int main(){ freopen("kxor.in","r",stdin); freopen("kxor.out","w",stdout); n=read(); for (int i=1;i<=n;i++) { ll x=read(); for (int j=50;~j;j--) if (x&(1ll<
           
            =s) printf("-1\n"); else { ll ans=0; for (int j=0;j<=50;j++) if (base[j]) { ans^=(k&1)*base[j]; k>>=1; } printf("%lld\n",ans); } } fclose(stdin);fclose(stdout); return 0;}