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
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 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 }

 

 
posted @ 2018-03-07 16:16  Z-Y-Y-S  阅读(309)  评论(1编辑  收藏  举报