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自然万物都趋向从有序变得无序

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HDU 3949 XOR （线性基）

发表于 更新于 分类于 Valine：

题目链接

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Problem 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.

Sample intput

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

题意:

在n个数中选出任意个数，求第k小异或和。

思路:

利用线性基求解，线性基大概是用来解决给定一个数数列，我们可以从数列中任意选出多个数进行异或，求第k大或者第k小的异或和。

求解线性基是利用类似于高斯消元的办法，把所有数转换成二进制，那么对于二进制的每一位，如果当前位已经有了一个1，那么就需要将其他的1都消掉，已达到每一位都只有一个1的目的，如果我们最后得到k个1，那么我们将可以表示共2^k个不同的数。

有一个坑点就是0的表示，能表示0表示我们二进制下一定有一些位全都是0.

代码:

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/*************************************************************************
    > File Name: HDU-3949.cpp
    > Author: wood
    > Mail: cbcruel@gmail.com
    > Created Time: 2018年08月07日 星期二 14时18分46秒
 ************************************************************************/

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<string>

using namespace std;
typedef long long LL;
#define sqr(x) (x)*(x)
const double eps = 1e-8;
const double C = 0.57721566490153286060651209;
const double pi = acos(-1.0);
const LL mod = 1e9 + 7;
const int maxn = 1e4 + 10;
const int MAX_BASE = 62;

LL a[maxn], b[66];
vector<LL> qq;
bool zero;

void solve(int  n) {
    int num = 0;
    memset(b, 0, sizeof(b));
    for (int i = 0; i < n; ++i)
        for (int j = MAX_BASE; j >= 0; --j)
            if (a[i] >> j & 1) {
                if (b[j]) a[i] ^= b[j];
                else {
                    b[j] = a[i];
                    num++;
                    for (int k = j - 1; k >= 0; --k)
                        if (b[k] && (b[j] >> k & 1))
                            b[j] ^= b[k];
                    for (int k = j + 1; k <= MAX_BASE; ++k)
                        if (b[k] >> j & 1)
                            b[k] ^= b[j];
                    break;
                }
            }
    if(num != n) zero = true;
    else zero = false;
    qq.clear();
    for(int i = 0; i <= MAX_BASE; ++i)
        if(b[i]) qq.emplace_back(b[i]);
}

LL gao(LL k){
    if(zero) k--;
    if(k >= (1LL<<qq.size())) return -1LL;
    LL ans = 0;
    for(int i = 0; i < qq.size(); ++i)
        if((k >> i) &1) ans ^= qq[i];
    return ans;
}

int main(){
    int t, kase = 0;
    //ios::sync_with_stdio(0);
    cin >> t;
    while(t--){
        int n,q;
        cin >> n;
        for(int i = 0; i < n; ++i) cin >> a[i];
        cin >> q;
        cout << "Case #" << ++kase << ":" << endl;
        solve(n);
        while(q--){
            LL k;
            cin >> k;
            cout << gao(k) << endl;
        }
    }
    return 0;
}