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
Recommend
lcy
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
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
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：
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 #include<cmath>
6 using namespace std;
7 int main(){
8 int n;
9 while(scanf("%d",&n)!=EOF){
10 int i,j;
11 int s=0;bool flag=0;
12 for(i=1;i<=n;i++){
13 scanf("%d",&j);
14 s^=j;
15 if(j>1)flag=1;
16 }
17 if(!flag){//充裕堆为0
18 if(n%2)//堆数为奇数，必败
19 printf("No\n");
20 else printf("Yes\n");
21 }
22 else{//充裕堆>0
23 if(s==0)//奇异局势
24 printf("No\n");
25 else
26 printf("Yes\n");
27 }
28 }
29 return 0;
30 }