HDU2509 Be the Winner

Be the Winner

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3062    Accepted Submission(s): 1709

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.

 

Output

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

 

Sample Input

2 2 2 1 3

 

Sample Output

No Yes

 

Source

ECJTU 2008 Autumn Contest

 

Recommend

lcy

太长不看版：

　　计算所有堆的异或和。若异或和不为0，那么一定有取法将其转化为0；若异或和为0，因为必须取，那么不得不将其转化成异或和不为0的状态。

　　最后面对着都取完的状态（0）的人输，按以上规律判定能不能占先机将异或和转化为0即可。

长也要看版：

反尼姆博弈(anti-nim)

该类问题的判定方法:

　　下面是从知乎找到的分析↓

奇异局势，所有堆的xor和==0.

假定S是非奇异局势，T是奇异局势。
一堆中石子数量>=2，表示充裕堆， =1表示孤单堆。

S0即非奇异局势下，充裕堆为0的状态
S1即非奇异局势下，充裕堆为1的状态
S2即非奇异局势下，充裕堆>=2的状态

T0即奇异局势下，充裕堆为0的状态
T2即奇异局势下，充裕堆>=2的状态

1.奇异局势的定义可知，S能转移到T，能转移到S， T只能转移到S

2.S0必败，T0必胜

3.S1必胜，因为S1只需要转移到S0即可。

4.S2必胜，T2必败。
1）T2只能转移到S1 和 S2
2）若T2转移到S1 则T2败，若T2转移到S2，S2只需要转回到T2即可。所以S2胜，T2败。

所以：
必胜态：T0,S1,S2
必败态：S0,T2

作者：公丕鑫
来源：知乎

另外还有一篇比较详细的证明分析：http://qianmacao.blog.163.com/blog/static/20339718020123555821140/

然后是代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        int i,j;
        int s=0;bool flag=0;
        for(i=1;i<=n;i++){
            scanf("%d",&j);
            s^=j;
            if(j>1)flag=1;
        }
        if(!flag){//充裕堆为0 
            if(n%2)//堆数为奇数，必败 
                printf("No
");
            else printf("Yes
");
        }
        else{//充裕堆>0 
            if(s==0)//奇异局势 
                printf("No
");
            else
                printf("Yes
");
        }
    }
    return 0;
}