HDOJ 4876 ZCC loves cards，hdojzcc

HDOJ 4876 ZCC loves cards，hdojzcc

枚舉組合,在不考慮連續的情況下判斷是否可以覆蓋L...R,對隨機資料是一個很大的減枝.

通過檢測的暴力計算一遍

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 DescriptionZCC 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. 
InputThe 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. 
OutputFor each test case, output the maximal number R. And if L can’t be obtained, output 0. 
Sample Input

4 3 12 3 4 5

 
Sample Output

7Hint ⊕ means xor  

 
Author鎮海中學 
Source2014 Multi-University Training Contest 2 
RecommendWe have carefully selected several similar problems for you:  4881 4880 4879 4878 4877  
Statistic | Submit | Discuss | Note

#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;}