hdu2509Be the Winner(反nim博弈)

Be the Winner

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4939    Accepted Submission(s): 2724

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

题意：有n堆苹果，每堆有数个。两个人轮流取，每次至少取一个，取走最后一个的算输。

题解：是反nim博弈。先手必胜的两个结论：（1）每堆苹果都只有一个，而且有偶数堆（异或和为0），那么先手必胜。（2）至少有一堆苹果中个数超过一个，那么异或和不为0，先手必胜。

https://www.cnblogs.com/SilverNebula/p/5658629.html      顶

 #include<bits/stdc++.h>
 using namespace std;
 int main() {
     int n;
     while(~scanf("%d",&n))
     {
         int ans=,ai,num=;
         for(int i=;i<n;i++)
         {
             scanf("%d",&ai);
             ans^=ai;
             if(ai>)num=;
         }
         if(num)
         {
             if(ans==)printf("No\n");
             else printf("Yes\n");
         }
         else
         {
             if(ans==)printf("Yes\n");
             else printf("No\n");
         }
     }
     return ;
 }