HDU 2509 Be the Winner（尼姆模板与hdu 1907类似的题目）

本文链接：https://blog.youkuaiyun.com/YJX_xx/article/details/24935927

本文解析了BetheWinner博弈问题，玩家需从分成若干组的苹果中取走连续的一段，目标是避免取走最后一个苹果。通过分析苹果堆数量及每堆苹果数量，确定是否有必胜策略。

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原题：http://acm.hdu.edu.cn/showproblem.php?pid=2509

尼姆博弈详解：http://blog.youkuaiyun.com/yjx_xx/article/details/24914861

hdu 1907：http://blog.youkuaiyun.com/yjx_xx/article/details/24914861

Be the Winner

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1759 Accepted Submission(s): 951

Problem Description

Let's consider m apples divided into n groups. Each group contains no more than 100 apples, arranged in a line. You can take any number of consecutive apples at one time.
For example "@@@" can be turned into "@@" or "@" or "@ @"(two piles). two people get apples one after another and the one who takes the last is
the loser. Fra wants to know in which situations he can win by playing strategies (that is, no matter what action the rival takes, fra will win).

Input

You will be given several cases. Each test case begins with a single number n (1 <= n <= 100), followed by a line with n numbers, the number of apples in each pile. There is a blank line between cases. （------>这句害人不浅呐~~无奈了）

Output

If a winning strategies can be found, print a single line with "Yes", otherwise print "No".

Sample Input

Sample Output

  
   No
Yes

题意与hdu1907差不多，代码也差不多，就不解释了~~

AC代码如下：

#include <stdio.h>
/*
author:YangSir
time:2014/5/3
*/
int main()
{
	int n,i,a,sum,flag,b;
	flag=0;
	while(~scanf("%d",&n))
	{
		//if(flag)
		//	printf("\n");
	//	flag=1;又被题意误解，坑死了
		b=0;
		sum=0;
		for(i=0;i<n;i++)
		{
			scanf("%d",&a);
			if(a>1)
				b=1;
			sum^=a;
		}
		if((sum&&b==0)||(sum==0&&b))
			printf("No\n");
		else
			printf("Yes\n");
	}
	return 0;
}