SPOJ XMAX - XOR Maximization

XMAX - XOR Maximization

Given a set of integers S = { a1, a2, a3, ... a|S| }, we define a function X on S as follows:
X( S ) = a1 ^ a2 ^ a3 ^ ... ^ a|S|.
(^ stands for bitwise 'XOR' or 'exclusive or')

Given a set of N integers, compute the maximum of the X-function over all the subsets of the given starting set.

Input

The first line of input contains a single integer N, 1 <= N <= 105.
Each of the next N lines contain an integer ai, 1 <= ai <= 1018.

Output

To the first line of output print the solution.

Example

Input:

3
1
2
4
Output:

7

高斯消元类似。尽量变幻成上三角矩阵。每一位尽量留一个1

如矩阵：

010000

001000

000100

000010

就是一个理想的矩阵。

/* ***********************************************
Author        :guanjun
Created Time  :2016/9/8 15:30:01
File Name     :spoj_XMAX.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 100010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << ;
const double eps=1e-;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
    int x,y;
};
struct cmp{
    bool operator()(Node a,Node b){
        if(a.x==b.x) return a.y> b.y;
        return a.x>b.x;
    }
};

bool cmp(ll a,ll b){
    return a>b;
}
ll a[maxn];
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int n;
    while(cin>>n){
        for(int i=;i<=n;i++)cin>>a[i];
        sort(a+,a++n,cmp);
        int row=;
        for(int i=;i>=;i--){
            for(int j=row;j<=n;j++){
                if(a[j]&(1LL<<i)){
                    swap(a[row],a[j]);
                    for(int k=;k<=n;k++){
                        if((a[k]&(1LL<<i))&&(k!=row)){
                            a[k]=a[k]^a[row];
                        }
                        //puts("YES");
                    }
                    row++;
                }
            }
        }
        ll ans=0LL;
        for(int i=;i<=n;i++){
        //    cout<<a[i]<<endl;
            ans^=a[i];
        }
        cout<<ans<<endl;
    }
    return ;
}