线性基 - 寻找异或第K大

XOR is a kind of bit operator, we define that as follow: for two binary base number A and B, let C=A XOR B, then for each bit of C, we can get its value by check the digit of corresponding position in A and B. And for each digit, 1 XOR 1 = 0, 1 XOR 0 = 1, 0 XOR 1 = 1, 0 XOR 0 = 0. And we simply write this operator as ^, like 3 ^ 1 = 2,4 ^ 3 = 7. XOR is an amazing operator and this is a question about XOR. We can choose several numbers and do XOR operatorion to them one by one, then we get another number. For example, if we choose 2,3 and 4, we can get 2^3^4=5. Now, you are given N numbers, and you can choose some of them(even a single number) to do XOR on them, and you can get many different numbers. Now I want you tell me which number is the K-th smallest number among them.

InputFirst line of the input is a single integer T(T<=30), indicates there are T test cases.

For each test case, the first line is an integer
N(1<=N<=10000), the number of numbers below. The second line
contains N integers (each number is between 1 and 10^18). The third line
is a number Q(1<=Q<=10000), the number of queries. The fourth
line contains Q numbers(each number is between 1 and 10^18)
K1,K2,......KQ.OutputFor each test case,first output Case #C: in a single line,C
means the number of the test case which is from 1 to T. Then for each
query, you should output a single line contains the Ki-th smallest
number in them, if there are less than Ki different numbers, output -1.Sample Input

2
2
1 2
4
1 2 3 4
3
1 2 3
5
1 2 3 4 5

Sample Output

Case #1:
1
2
3
-1
Case #2:
0
1
2
3
-1

Hint

If you choose a single number, the result you get is the number you choose.
Using long long instead of int because of the result may exceed 2^31-1.

 题意 ： 给你 n 个数 , q 个询问，每次询问第 K 异或小的值
 思路分析 ：线性基的板子题，构造出 n 个数的线性基，然后将每一位相互独立出来，然后看有多少个不唯一的，存在一个新的数组里，如果在新的数组中有K个元素，则异或出来的最终答案最多有 2^k , 
　　 当然这里面是包括 0 的，但是并不是所有的都可以异或出来 0 ，那要怎么确定这个能否异或出来 0 呢？如果构造出来的线性基有 k 位不为 1，那么说明每个数都提供了 1，则说明不会异或出 0 ，否则可以。
 代码示例：

#define ll long long

ll n;
ll a[65];

void insert(ll x){

    for(ll i = 60; i >= 0; i--){
        if ((1ll<<i)&x){
            if (!a[i]) {a[i] = x; return;}
            x ^= a[i];
        }
    }
}
ll cnt = 0;
ll p[65];

void rbuild(){
    for(ll i = 60; i >= 0; i--){
        for(ll j = i-1; j >= 0; j--){
            if ((1ll<<j)&a[i]) a[i] ^= a[j];
        }
    }

    for(ll i = 0; i <= 60; i++){
        if (a[i]) p[cnt++] = a[i];
    }
}

ll query(ll x){
    if (x >= (1ll<<cnt)) return -1;
    ll ans = 0;
    for(ll j = 60; j >= 0; j--){
        if ((1ll<<j)&x){
            ans ^= p[j];
        }
    }
    return ans;
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    ll t, q;
    ll x;
    ll kase = 1;

    cin >>t;
    while(t--){
        memset(a, 0, sizeof(a));
        memset(p, 0, sizeof(p));
        cnt = 0;
        cin >> n;
        for(ll i = 1; i <= n; i++){
            scanf("%lld", &x);
            insert(x);
        }
        rbuild();
        cin >> q;
        //printf("cnt = %d\n", cnt);
        printf("Case #%d:\n", kase++);
        for(ll i = 1; i <= q; i++){
            scanf("%lld", &x);
            if (cnt != n) x--;
            printf("%lld\n", query(x));
        }
    }
    return 0;
}