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
題意
Nim變形，從某一堆必須拿掉一個，但其他的可以在放到其他堆裏面。
問給出的情況是否先手贏。
思路
如果n == 1：
先手直接拿掉 一定贏了。
如果n == 2
如果兩堆一樣多，先拿必輸，不一樣多，就按照nim博弈的方式操作。
如果n == 3
第三堆一定大於前兩隊的差值，所以走第三堆並且將前兩堆補成相等狀態，先拿必勝，這樣便可推廣到奇數的情況，即排序之後拿最大的一堆，並且一定可以補全前面的使它們成爲一對一對的，這樣先拿必勝
所以當n爲偶數時，如果是成對出現的，那麼先拿必輸，如果不是，