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

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

 1 #include
 2 #include
 3 #include
 4 #include
 5 #include
 6 using namespace std;
 7 int ans,n;
 8 int gi()
 9 {
10   int x=0;char ch=getchar();
11   while (ch<'0'||ch>'9') ch=getchar();
12   while (ch>='0'&&ch<='9')
13     {
14       x=x*10+ch-'0';
15       ch=getchar();
16     }
17   return x;
18 }
19 int main()
20 {int i,T,TT,x;
21   cin>>T;
22   TT=T;
23   while (T--)
24     {
25       cin>>n;
26       ans=0;
27       for (i=1;i<=n;i++)
28     {
29       x=gi();
30       if (i%6==0||i%6==2||i%6==5)
31         ans^=x;
32     }
33       if (ans) printf("Case %d: Alice\n",TT-T);
34       else printf("Case %d: Bob\n",TT-T);
35     }
36 }

 

 

转载于:https://www.cnblogs.com/Y-E-T-I/p/8523094.html