# POJ2234

Written by    10:00 July 29, 2014

POJ2234

Matches Game
 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9306 Accepted: 5372

Description

Here is a simple game. In this game, there are several piles of matches and two players. The two player play in turn. In each turn, one can choose a pile and take away arbitrary number of matches from the pile (Of course the number of matches, which is taken away, cannot be zero and cannot be larger than the number of matches in the chosen pile). If after a player’s turn, there is no match left, the player is the winner. Suppose that the two players are all very clear. Your job is to tell whether the player who plays first can win the game or not.

Input

The input consists of several lines, and in each line there is a test case. At the beginning of a line, there is an integer M (1 <= M <=20), which is the number of piles. Then comes M positive integers, which are not larger than 10000000. These M integers represent the number of matches in each pile.

Output

For each test case, output “Yes” in a single line, if the player who play first will win, otherwise output “No”.

Sample Input

Sample Output

Source

## 结论一

• 当XOR(M1, M2, …Mn) != 0 时，总是可以通过改变一个Mi的值，就可以让XOR(M1, M2, …Mn) = 0

## 结论二

• 当XOR(M1, M2, …Mn) = 0 时，任意一个Mi的值的改变，就可以让XOR(M1, M2, …Mn) != 0 也就是说如果一开始XOR(M1, M2, …Mn) = 0，那么先手者肯定会使XOR(M1, M2, …Mn) != 0，后手者无论如何都可以使XOR(M1, M2, …Mn) = 0，如此循环至最终XOR（0，0，0，0，0，0，0。。。） = 0，后手者必赢，所以XOR(M1, M2, …Mn) = 0就是必输局面，相反，如果一开始XOR(M1, M2, …Mn) != 0，那么就是必胜局面。

AC代码如下：

Category : acmstudy

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