HDOJ 4876 ZCC loves cards

枚举组合,在不考虑连续的情况下推断能否够覆盖L...R,对随机数据是一个非常大的减枝.

通过检測的暴力计算一遍

ZCC loves cards

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1346    Accepted Submission(s): 335

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 
 

 

Author

镇海中学

 

Source

field=problem&key=2014%20Multi-University%20Training%20Contest%202&source=1&searchmode=source" style="color:rgb(26,92,200); text-decoration:none">2014 Multi-University Training Contest 2

 

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#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
 
using namespace std;
 
int n,k,m,a[30],save[30],have[30],R,L;
bool vis[3000],cx[200];
 
void ckMax(int num,int sum)
{
	vis[sum]=true;
	if(num==k) return ;
	ckMax(num+1,sum^save[num]);
	ckMax(num+1,sum);
}
 
bool ck()
{
 	memset(vis,0,sizeof(vis));
 	ckMax(0,0);
 	for(int i=L;i<=R;i++)
 	{
 		if(vis[i]==false)	return false;
 	}
 	return true;
}
 
void CALU()
{
    if (!ck()) return;
    for(int i=0;i<k;i++)
    	have[i]=save[i];
    do
    {
    	memset(vis,0,sizeof(vis));
    	for(int i=0;i<k;i++)
    	{
    		int x=0;
    		for(int j=0;j<k;j++)
    		{
    			x^=have[(i+j)%k];
    			vis[x]=true;
    		}
    	}
    	for(int i=L;i<=L+k*k;i++)
    	{
    		if(vis[i]==false) break;
    		R=max(R,i);
    	}
    }while(next_permutation(have,have+k-1));
}      
 
void dfs(int num,int id)
{
	if(num==k)
	{
		CALU();
		return ;
	}
	for(int i=id;i<n;i++)
	{
		save[num]=a[i];
		dfs(num+1,i+1);
	}
}
 
int main()
{
	while(scanf("%d%d%d",&n,&k,&L)!=EOF)
	{
		R=L-1;
		for(int i=0;i<n;i++)
			scanf("%d",a+i);
		sort(a,a+n);
		dfs(0,0);
		if(R<L) printf("0\n");
		else printf("%d\n",R);
	}
	return 0;
}