Game

Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 50 Accepted Submission(s): 44

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

 

Recommend

notonlysuccess

 

/*
题意：1到n的盒子里，是空的或者装着卡片，每人轮流移动卡片，Alice先手，移动卡片的规则：每次选择一个A，然后选择一个B
    要求 B<A&&(A+B)%2==1&&(A+B)%3==0,从A中抽取任意一张卡片放到B中，谁不能移动了，就输了。

初步思路：这个就是先将能移动的盒子对找出来，然后统计能移动的牌

#错误：上面的想法不完善，只是一方面的认为卡片会从一上一级向下一级移动。

#补充：阶梯博弈，实际上卡片是在3的余数上进行转移的。

#感悟：手残用了数组，开小了，wa了两发才发现
*/
#include<bits/stdc++.h>
using namespace std;
int n;
int x;
int res=0;
int t;
void init(){
    res=0;
}
int main(){
    // freopen("in.txt","r",stdin);
    scanf("%d",&t);
    for(int ca=1;ca<=t;ca++){
        init();
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d",&x);
            if(i%6==0||i%6==2||i%6==5)
                res^=x;
        }
        if(res) printf("Case %d: Alice
",ca);
        else printf("Case %d: Bob
",ca);
    }
    return 0;
}