HDU3949

Description

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 2³4=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.

Input

First 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.

Output

For 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.

思路

这题用到了构造线性基的算法。
由于线性基里面每个元素最高位各不相同，所以异或出来的数的二进制中第i位的值与最高位为i的那一个元素有关（如果存在）。这样就可以直接构造出第k大的数。
注意0是没办法由线性基里面的数构造出来的，所以要特判原集合的数可不可以异或出0。方法就是如果原集合含有0或线性基个数不等于原集合个数就说明可以构造出0。

#include<bits/stdc++.h>
typedef long long ll;
const int N = 2e5;
const ll M = 998244353;
const int nN = 100;
using namespace std;
#define endl '
'

ll num[nN];
ll p[nN];
int rk[nN];


void insert(ll x) {
    for(int i = 61; i + 1; i--) {
        if(!(x >> i)) 
            continue;
        if(!num[i]) {
            num[i] = x;
            break;
        }
        x ^= num[i];
    }
}


int main() {
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    int cas = 0;
    while(t--) {
        cas++;
        bool ex = false;
        int cnt = 0;
        memset(num, 0, sizeof num);
        memset(p, 0, sizeof p);
        memset(rk, 0, sizeof rk);
        int n;
        cin >> n;
        for(int i = 0; i < n; i++) {
            ll v;
            cin >> v;
            if(v == 0) ex = true;
            insert(v);
        }
        for(int i = 0; i <= 61; i++) {
            if(num[i]) cnt++;
            int wid = 0;
            while(num[i] >> wid) wid++;
            p[wid] = num[i];
        }
        if(cnt != n) ex = true;
        cnt = 0;
        for(int i = 1; i <= 61; i++)  {
            if(p[i]) rk[cnt++] = i;  
            
        }
        
        int q;
        cin >> q;
        cout << "Case #" << cas << ":" << endl;
        while(q--)  {
            ll k;
            cin >> k;
            if(ex) k--;
            if(k == 0) {cout << "0" << endl; continue;}
            int wid = 0;
            while(k >> wid) wid++;
            if(wid > cnt) cout << "-1" << endl;
            else {
                ll res = 0;
                for(int i = wid-1; i >= 0; i--) {
                    if((k >> i) & 1) {
                        if(!((res >> (rk[i] - 1)) & 1)) res ^= p[rk[i]];
                    } else {
                        if(((res >> (rk[i] - 1)) & 1)) res ^= p[rk[i]];
                    }
                }
                cout << res << endl;
            }
            
        }
    }
}