HDU 3389 Game（博弈）

Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 500    Accepted Submission(s): 341

Problem Description

Bob and Alice are playing a new game. There are n boxes which have been numbered from 1 to n. Each box is either empty or contains several cards. Bob and Alice move the cards in turn. In each turn the corresponding player should choose a non-empty box A and choose another box B that B<A && (A+B)%2=1 && (A+B)%3=0. Then, take an arbitrary number (but not zero) of cards from box A to box B. The last one who can do a legal move wins. Alice is the first player. Please predict who will win the game.

 

Input

The first line contains an integer T (T<=100) indicating the number of test cases. The first line of each test case contains an integer n (1<=n<=10000). The second line has n integers which will not be bigger than 100. The i-th integer indicates the number of cards in the i-th box.

 

Output

For each test case, print the case number and the winner's name in a single line. Follow the format of the sample output.

 

Sample Input

  

   2
2
1 2
7
1 3 3 2 2 1 2
  

 

Sample Output

  

   Case 1: Alice
Case 2: Bob
  

 

Author

hanshuai@whu

 

Source

The 5th Guangting Cup Central China Invitational Programming Contest

 

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/*

参考: http://www.mamicode.com/info-detail-448117.html

*/


#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define bug printf("hihi\n")

#define eps 1e-8
typedef __int64 ll;

using namespace std;


#define INF 0x3f3f3f3f
#define N 10005

int main()
{
    int i,j,t,n,ca=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        int ans=0,x;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&x);
            if(i%6==0||i%6==2||i%6==5)
                ans^=x;
        }
       if(ans)
          printf("Case %d: Alice\n",++ca);
       else
          printf("Case %d: Bob\n",++ca);
    }
    return 0;
}