hdu 4876 ZCC loves cards

ZCC loves cards

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1929    Accepted Submission(s): 503

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 12 3 4 5

 

Sample Output

7
Hint
 ⊕ means xor 
 

 

Author

镇海中学

 

Source

2014 Multi-University Training Contest 2

 

暴力+剪枝

#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <cmath>#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#include <bitset>#include <functional>#include <sstream>#include <iomanip>#include <cmath>#include <cstdlib>#include <ctime>//#pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;#define INF 1e9#define MAXN 21#define maxn 30#define mod 1000000007#define eps 1e-7#define pi 3.1415926535897932384626433#define rep(i,n) for(int i=0;i<n;i++)#define rep1(i,n) for(int i=1;i<=n;i++)#define scan(n) scanf("%d",&n)#define scanll(n) scanf("%I64d",&n)#define scan2(n,m) scanf("%d%d",&n,&m)#define scans(s) scanf("%s",s);#define ini(a) memset(a,0,sizeof(a))#define out(n) printf("%d\n",n)ll gcd(ll a,ll b) {return b==0?a:gcd(b,a%b);}using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1int l,r,k,n;int a[maxn], tmp[maxn], save[maxn];int vis[200];void calmax(int cur,int sum) // 2 ^ k = 64枚举{vis[sum] = 1;if(cur == k) return ;calmax(cur + 1,sum ^ save[cur]);calmax(cur + 1,sum);}bool check(){ini(vis);calmax(0,0);for(int i = l;i <= r; i++) //当随意取也达不到当前r时，说明这组选法不满足，不用进入全排列{if(!vis[i]) return false;}return true;}void solve(){if(!check()) return;rep(i,k) tmp[i] = save[i];do //全排列计算，一次复杂度A(5,5) * (6 * 6 + 128){ini(vis);for(int i = 0;i < k; i++){int ans = 0;for(int j = i; j < k + i; j++){ans ^= tmp[j % k];vis[ans] = 1;}}for(int i = l;i <= 128; i++){if(!vis[i]){r = max(r,i-1);break;}}}while(next_permutation(tmp + 1, tmp + k));}void dfs(int cur,int num) //C(20,6)取数{if(num == k){solve();return;}for(int i = cur;i < n; i++){save[num] = a[i];dfs(i + 1,num + 1);}}int main(){#ifndef ONLINE_JUDGEfreopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);#endif  while(~scanf("%d%d%d",&n,&k,&l)){rep(i,n) scan(a[i]);sort(a,a+n);r = l - 1;dfs(0, 0);if(r < l) puts("0");else out(r);}return 0;}