枚舉組合,在不考慮連續(xù)的情況下推斷能否夠覆蓋L...R,對隨機數(shù)據(jù)是一個非常大的減枝.

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 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 鎮(zhèn)海中學
Source 2014 Multi-University Training Contest 2
Recommend We 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; }

轉(zhuǎn)載于:https://www.cnblogs.com/bhlsheji/p/5382429.html

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