HDU 3389 Game 阶梯博弈-CFANZ编程社区

Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 391 Accepted Submission(s): 262

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

/*
HDU 3389 阶梯博弈 
多的拿到少的上面去，满足一定条件
 
阶梯博弈:博弈在一列阶梯上进行,每个阶梯上放着自然数个点,
两个人进行阶梯博弈，每一步则是将一个集体上的若干个点
(>=1 )移到前面去,最后没有点可以移动的人输。

(A+B)=6k+3(k >= 0)，对于两个盒子A,B可行的转移也可以想出来，
B%6==0的盒子可以转移到B%6==3的盒子里， 所以整体不变 
B%6==1的盒子可以转移到B%6==2的盒子
B%6==4的盒子可以转移到B%6==5的盒子
0 -> 3 -> 0 -> 3
2 -> 1 -> 2 -> 1
5 -> 4 -> 5 -> 4
最后的1, 3, 4盒子都是不能继续转移的盒子，
首先对于每一个转移公式都是互不影响的，对于第一个转移公式我们可以发现如果当前决策者处于不利形势，
他想要挽回形式而移动x%6==3的盒子到x%6==0的盒子是完全没用的，
因为对手总是可以继续把你转移的盒子再转移回x%6==3的盒子里使得你还是处于同样的劣势，

1,3,4 的状态不能转移到其他状态; 其他每个状态皆可转移;
且位置特定, 如2->1,5->4, 6->3, 7->2, 8->1 9->6

其本质我们有N级阶梯,现在要在%3的余数间转移, 0->0, 1->2, 2->1; 
其最后的结果为1, 3, 4; 那么他们的转移的步数的奇偶性也会确定，
我们只要选择步数为奇数的位置做nim博弈就行了;
而可以通过打表归纳证明得出模6为0、2、5的位置移动步数为奇，其余为偶。
*/
#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;

int main()
{
    int T,cas,ans,i,x;
    int n;
    cas=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        ans=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&x);
            if(i%6==0||i%6==2||i%6==5)
                ans^=x;
        }
        printf("Case %d: ",cas++);
        if(ans) 
            puts("Alice");
        else 
            puts("Bob");
    }
    return 0;
}