HDU 3389 Game(博弈 Nim 找规律)

Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 610    Accepted Submission(s): 426

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

 

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：

游戏规则就是把卡片移动到更靠左的盒子中，A B 两个盒子满足（A+B) % 3 == 0 并且满足A、B奇偶性质不同。

思路：

将每个盒子看作分开的石子堆，考察每一类盒子向左侧放卡片的策略。 发现最终卡片会被放到1、3、4三个盒子中，即最终结果。故经过奇数步骤到达这三个盒子中的必然是胜点，经过偶数步骤到达这些盒子的是必败情况。必败情况的SG函数值是0，对异或不会产生影响。

分情况  奇数：  %3 == 0 偶数步骤后到达盒子3

%3 == 1 偶数步骤后到达盒子1

%3 == 2 ji步骤后到达盒子2

偶数：  %3 == 0 奇数步骤后到达盒子3

%3 == 1 偶数步骤后到达盒子1

%3 == 2 奇数步骤后到达盒子4

偶数步骤到达最终结果的情况均可以忽略不计，剩下的情况可以想象成N多组尼姆游戏，不断异或即可得到答案。

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