Discussion
Alice and Bob decide to play a new stone game.At the beginning of the game they pick n(1<=n<=10) piles of stones in a line. Alice and Bob move the stones in turn.
At each step of the game,the player choose a pile,remove at least one stones,then freely move stones from this pile to any other pile that still has stones.
For example:n=4 and the piles have (3,1,4,2) stones.If the player chose the first pile and remove one.Then it can reach the follow states.
2 1 4 2
1 2 4 2（move one stone to Pile 2）
1 1 5 2（move one stone to Pile 3）
1 1 4 3（move one stone to Pile 4）
0 2 5 2（move one stone to Pile 2 and another one to Pile 3）
0 2 4 3（move one stone to Pile 2 and another one to Pile 4）
0 1 5 3（move one stone to Pile 3 and another one to Pile 4）
0 3 4 2（move two stones to Pile 2）
0 1 6 2（move two stones to Pile 3）
0 1 4 4（move two stones to Pile 4）
Alice always moves first. Suppose that both Alice and Bob do their best in the game.
You are to write a program to determine who will finally win the game.
Input
The input contains several test cases. The first line of each test case contains an integer number n, denoting the number of piles. The following n integers describe the number of stones in each pile at the beginning of the game, you may assume the number of stones in each pile will not exceed 100.
The last test case is followed by one zero.
Output
For each test case, if Alice win the game,output 1,otherwise output 0.
Sample Input
3
2 1 3
2
1 1
0
Sample Output
1
0
题意
Nim变形，从某一堆必须拿掉一个，但其他的可以在放到其他堆里面。
问给出的情况是否先手赢。
思路
如果n == 1：
先手直接拿掉 一定赢了。
如果n == 2
如果两堆一样多，先拿必输，不一样多，就按照nim博弈的方式操作。
如果n == 3
第三堆一定大于前两队的差值，所以走第三堆并且将前两堆补成相等状态，先拿必胜，这样便可推广到奇数的情况，即排序之后拿最大的一堆，并且一定可以补全前面的使它们成为一对一对的，这样先拿必胜
所以当n为偶数时，如果是成对出现的，那么先拿必输，如果不是，