SGU 275 To xor or not to xor 高斯消元求N个数中选择任意数XORmax

275. To xor or not to xor

 

The sequence of non-negative integers A1, A2, ..., AN is given. You are to find some subsequence Ai ₁, Ai ₂, ..., Ai _k (1 <= i ₁ < i ₂ < ... < i _k<= N) such, that Ai ₁ XOR Ai ₂ XOR ... XOR Ai _k has a maximum value.

Input

The first line of the input file contains the integer number N (1 <= N <= 100). The second line contains the sequence A1, A2, ..., AN (0 <= Ai <= 10^18). 

Output

Write to the output file a single integer number -- the maximum possible value of Ai ₁ XOR Ai ₂ XOR ... XOR Ai _k. 

Sample test(s)

Input

 

 

3 
11 9 5 

 

 

Output

 

 

14 

 题意：

　　从N个数中选择任意个数，求异或的最大值

题解：

　　从高位贪心

　　高斯消元判断能否构成1

　　复杂度 60*N*N

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18;
const double Pi = acos(-1.0);
const int N = 1e3+, M = 1e6, mod = 1e9+, inf = 2e9;

int n,a[][N];
int main() {
    scanf("%d",&n);
    for(int i = ; i < n; ++i) {
        LL x;int cnt = ;
        scanf("%I64d",&x);
        while(x) {
            a[cnt++][i] = x%;
            x/=;
        }
    }

    for(int  i = ; i < ; ++i) a[i][n] = ;
    LL ans = ;
    for(int  i = ; i >= ; --i) {
        int x = -;
        for(int j = ; j < n; ++j) {
            if(a[i][j]) {
                x = j;break;
            }
        }
        if(x == - && a[i][n] == ) {
            ans += 1LL<<i;
        } else if(x != -) {
            ans += 1LL<<i;
            for(int k = i - ; k >=; --k) {
                if(a[k][x]) {
                    for(int j = ; j <= n; ++j) a[k][j] ^= a[i][j];
                }
            }
        }
    }
    cout<<ans<<endl;
    return ;
}