POJ2234 Matches Game（博弈&Nim博弈）

最新推荐文章于 2018-12-20 17:55:54 发布

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#POJ2234 #Nim博弈

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本文介绍了一种名为Nim的简单博弈游戏，并提供了一个用于判断玩家是否能够赢得游戏的算法实现。通过输入不同堆数及每堆火柴的数量，利用异或运算来确定先手玩家是否有获胜的可能性。

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

2 45 45
3 3 6 9

Sample Output

No
Yes

题意：Nim博弈模板

解题思路：Nim博弈公式：对于一个Nim游戏的局面(a1,a2,...,an)，它是P-position当且仅当a1^a2^...^an=0，其中^表示异或(xor)运算。（后手可保证必胜）

AC代码：

#include<stdio.h>

int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		int ans=0,x;
		for(int i=0;i<n;i++)
		{
			scanf("%d",&x);
			ans^=x;
		}
		if(ans) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}