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=0;char ch=getchar();
  while (ch<'0'||ch>'9') ch=getchar();
  while (ch>='0'&&ch<='9')
    {
      x=x*10+ch-'0';
      ch=getchar();
    }
  return x;
}
int main()
{int i,T,TT,x;
  cin>>T;
  TT=T;
  while (T--)
    {
      cin>>n;
      ans=0;
      for (i=1;i<=n;i++)
    {
      x=gi();
      if (i%6==0||i%6==2||i%6==5)
        ans^=x;
    }
      if (ans) printf("Case %d: Alice
",TT-T);
      else printf("Case %d: Bob
",TT-T);
    }
}