HDU4876:ZCC loves cards

Problem Description

ZCC loves playing cards. He has n magical cards and each has a number on it. He wants to choose k cards and place them around in any order to form a circle. He can choose any several  consecutive cards the number of which is m(1<=m<=k) to play a magic. The magic is simple that ZCC can get a number x=a1⊕a2...⊕am, which ai means the number on the ith card he chooses. He can play the magic infinite times, but  once he begin to play the magic, he can’t change anything in the card circle including the order.
ZCC has a lucky number L. ZCC want to obtain the number L~R by using one card circle. And if he can get other numbers which aren’t in the range [L,R], it doesn’t matter. Help him to find the maximal R.

 

Input

The input contains several test cases.The first line in each case contains three integers n, k and L(k≤n≤20,1≤k≤6,1≤L≤100). The next line contains n numbers means the numbers on the n cards. The ith number a[i] satisfies 1≤a[i]≤100.
You can assume that all the test case generated randomly.

 

Output

For each test case, output the maximal number R. And if L can’t be obtained, output 0.

 

Sample Input

   
   
   
   

    
    
    
    4 3 1
2 3 4 5
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    7


    
    
    
    
 
     
 
      Hint 
     
 ⊕ means xor 
    
    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    用全排列的方法来做，这我一开始还真没想到
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int n,k,l,r;
int vis[500],a[500],tem[500],s[500];

void set(int len,int sum)
{
    vis[sum] = 1;
    if(len == k)
        return ;
    set(len+1,sum^tem[len]);
    set(len+1,sum);
}

int check()
{
    memset(vis,0,sizeof(vis));
    set(0,0);
    for(int i = l; i<=r; i++)
        if(!vis[i])
            return 0;
    return 1;
}

void solve()
{
    if(!check()) return ;
    int i,j;
    for(i = 0; i<k; i++)
        s[i] = tem[i];
    do
    {
        memset(vis,0,sizeof(vis));
        for(i = 0; i<k; i++)
        {
            int ans = 0;
            for(j = i; j<k+i; j++)
            {
                ans^=s[(j%k)];
                vis[ans] = 1;
            }
        }
        for(i = l; i<=128; i++)//a[i]最大100，所以不会超过128
            if(!vis[i])
            {
                r = max(r,i-1);
                break;
            }

    }
    while(next_permutation(s+1,s+k));
}

void dfs(int now,int len)
{
    if(len == k)
    {
        solve();
        return ;
    }
    for(int i = now; i<n; i++)
    {
        tem[len] = a[i];
        dfs(i+1,len+1);
    }
}

int main()
{
    int i,j;
    while(~scanf("%d%d%d",&n,&k,&l))
    {
        for(i = 0; i<n; i++)
            scanf("%d",&a[i]);
        sort(a,a+n);//先排序，方便后面进行排列
        r = l-1;
        dfs(0,0);
        if(r<l)
            printf("0\n");
        else
            printf("%d\n",r);
    }

    return 0;
}