cpp several may between seve int first xor script

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.

        
 題意 ： 給你 n 個數 , q 個詢問，每次詢問第 K 異或小的值
 思路分析 ：線性基的板子題，構造出 n 個數的線性基，然後將每一位相互獨立出來，然後看有多少個不唯一的，存在一個新的數組裏，如果在新的數組中有K個元素，則異或出來的最終答案最多有 2^k , 
　　 當然這裏面是包括 0 的，但是並不是所有的都可以異或出來 0 ，那要怎麽確定這個能否異或出來 0 呢？如果構造出來的線性基有 k 位不為 1，那麽說明每個數都提供了 1，則說明不會異或出 0 ，否則可以。
 代碼示例：

#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;
}

線性基 - 尋找異或第K大