HDU3949 XOR(线性基第k小)

最新推荐文章于 2020-07-17 17:00:13 发布

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最新推荐文章于 2020-07-17 17:00:13 发布

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原文链接：http://www.cnblogs.com/zwfymqz/p/9192926.html

本文介绍了一个关于XOR运算的问题，通过构建线性基并利用高斯消元法来解决特定查询问题。文章详细阐述了如何通过二进制分解求解第K小的XOR结果，并给出了完整的代码实现。

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 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.

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 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.

Author

elfness

Source

题目大意：给出$n$个数，问两两异或后第$k$小的数是多少

看了很多篇博客，发现都是在围绕着高斯消元解xor方程组来的。

然后我惊讶的发现，原来高斯消元解xor解方程组其实就是求出线性基然后再消元

通过消元保证线性基内有元素的每一列只有一个$1$

然后把$k$二进制分解，如果第$i$是$1$就异或上第$i$个有解的线性基

同时要特判$0$的情况，若线性基的大小与元素的大小相同则不能异或为$0$(线性无关)，否则可以异或为零，这时我们只要求出第$k-1$小就可以了

这里把$k$二进制分解后的$0/1$实际对应了线性基中元素选/不选，可以证明这样一定是对的

#include<cstdio>
#include<cstring>
#include<algorithm>
#define int  long long 
using namespace std;
const int MAXN = 1e5 + 10,  B = 31;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int  P[MAXN];
void Insert(int  x) {
    for(int i = B; i >= 0; i--) {
        if((x >> i) & 1) {
            if(P[i]) x = x ^ P[i];
            else {P[i] = x; return ;}
        }
    }
}
void Debug(int *a, int N) {
    for(int i = 0; i <= N; i++) {
        for(int j = 0; j <= B; j++) 
            printf("%d ", (P[i] >> j) & 1);
        puts("");
    }
    puts("********");
}
main() { 
    int QwQ = read();
    for(int test = 1; test <= QwQ; test++) {
        printf("Case #%I64d:\n", test);
        memset(P, 0, sizeof(P));
        int  N = read();
        for(int i = 1; i <= N; i++) 
            Insert(read());
        for(int i = B; i >= 0; i--) {
            if(P[i]) {
                for(int j = i + 1; j <= B; j++)
                    if((P[j] >> i) & 1) P[j] ^= P[i];
            }
        }
        int now = 0;
        for(int i = 0; i <= B; i++)
            if(P[i])
                P[now++] = P[i];
        int Q = read();
        while(Q--) {
            int  K = read(), ans = 0;
            if(now != N) K--;
            if(K >= (1ll  << now)) {puts("-1"); continue;}
            for(int i = 0; i <= B; i++)
                if((K >> i) & 1)
                    ans ^= P[i];
            printf("%I64d\n", ans);
        }
    }
}