Game HDU - 3389 (Nim)

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

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

 题意：有n堆石子，下标1~n，两个人玩游戏，轮流操作，每轮玩家可以任意选择两个堆石子下标为a, b，并且满足b < a；a > 0；(a + b) % 2 = 1；(a + b) % 3 = 1。玩家可以将a中任意石头（大于0）拿放到b中，Alice先手，不能操作者失败。

题解：当不能操作时，所以石头必当移动到1，3，4中，当列出表后就会知道，满足：

(i % 3 == 1 || i & 3 == 2) && (i / 3) % 2 == 0  的堆合并到1中；

(i % 3 == 1 || i & 3 == 2) && (i / 3) % 2 == 1  的堆合并到4中；

i % 3 == 0  的堆合并到3中。

我们假设现在只有一颗石头，当这颗石头需要移动奇数才能到达终点时，则先手必胜。

现在存在很多石头就是nim博弈，求个异或和就行了。

所以现在只需要计算每颗石头移动到终点需要步数的奇偶性就行了。

简化后的代码如下：

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