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
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int ans,n;
int gi()
{
int x=;char ch=getchar();
while (ch<''||ch>'') ch=getchar();
while (ch>=''&&ch<='')
{
x=x*+ch-'';
ch=getchar();
}
return x;
}
int main()
{int i,T,TT,x;
cin>>T;
TT=T;
while (T--)
{
cin>>n;
ans=;
for (i=;i<=n;i++)
{
x=gi();
if (i%==||i%==||i%==)
ans^=x;
}
if (ans) printf("Case %d: Alice\n",TT-T);
else printf("Case %d: Bob\n",TT-T);
}
}