HDU3389 Game

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

根据staircase nim游戏的结论

假设移动的终点为0，那么只要考虑奇数位的Nim和就行了

如果不是按位移动的，那么就变成距离终点步数为奇数的Nim和

画图归纳，发现当i%6等于0，2，5时，到终点步数为奇数

取那些位的Nim和

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 using namespace std;
 int ans,n;
 int gi()
 {
   int x=;char ch=getchar();
   while (ch<''||ch>'') ch=getchar();
   while (ch>=''&&ch<='')
     {
       x=x*+ch-'';
       ch=getchar();
     }
   return x;
 }
 int main()
 {int i,T,TT,x;
   cin>>T;
   TT=T;
   while (T--)
     {
       cin>>n;
       ans=;
       for (i=;i<=n;i++)
     {
       x=gi();
       if (i%==||i%==||i%==)
         ans^=x;
     }
       if (ans) printf("Case %d: Alice\n",TT-T);
       else printf("Case %d: Bob\n",TT-T);
     }
 }