hdu 4876 ZCC loves cards 2014 Multi-University Training Contest 2

ZCC loves cards

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

Total Submission(s): 1234 Accepted Submission(s): 301



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

2014 Multi-University Training Contest 2

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官方思路：



说难不难，说水不水的题，做的很无语。照着官方思路敲就可以了，过是没有问题的，有大神优化到90+ms，有兴趣的可以搞搞，个人认为没必要做成那样。

敲代码的时候自己做了组优化，结果直接导致我查了一下午错 。自己坑自己能怪谁呢

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int p[25],a[25];
int res[333];
int v[333];
int ans,n,k,l;
void work2()
{
do{
memset(v,0,sizeof(v));
for(int i=0;i<k;i++){
int t=0;
int s=0;
while(t<k){
int q=t+i;
q%=k;
s^=a[p[q]];
v[s]++;
t++;
}
for(int j=l;;j++){
if(v[j])continue;
else{
ans=max(ans,j-1);
break;
}
}
}
}while(next_permutation(p+1,p+k));
}
void work(int cnt,int s)
{
if(cnt==k) return;
res[s]=1;
int d=s^a[p[cnt]];
res[d]=1;
work(cnt+1,s);
work(cnt+1,d);
}
void dfs(int cnt)
{
if(cnt==k){
memset(res,0,sizeof(res));
work(0,0);
for(int i=l;;i++){
if(res[i]) continue;
else if(i-1>ans) work2();
return;
}
}
int s=cnt ?p[cnt-1]+1:0;
for(int i=s;i<n;i++){
p[cnt]=i;
dfs(cnt+1);
}
}
int main()
{
while(scanf("%d%d%d",&n,&k,&l)!=EOF){
for(int i=0;i<n;i++){
scanf("%d",&a[i]);
memset(p,0,sizeof(p));
}
ans=0;
dfs(0);
if(ans<l) printf("0\n");
else printf("%d\n",ans);
}
return 0;
}