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

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 severalconsecutivecards 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, butonce 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

7Hint⊕ means xor

Author 镇海中学

Source 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;}