hdoj 1564 Play a game(博弈问题（寻找必胜点）)

Play a game

http://acm.hdu.edu.cn/showproblem.php?pid=1564

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1548    Accepted Submission(s): 1255

Problem Description

New Year is Coming!
ailyanlu is very happy today! and he is playing a chessboard game with 8600.
The size of the chessboard is n*n. A stone is placed in a corner square. They play alternatively with 8600 having the first move. Each time, player is allowed to move the stone to an unvisited neighbor square horizontally or vertically. The one who can't make a move will lose the game. If both play perfectly, who will win the game?

 

Input

The input is a sequence of positive integers each in a separate line.
The integers are between 1 and 10000, inclusive,(means 1 <= n <= 10000) indicating the size of the chessboard. The end of the input is indicated by a zero.

 

Output

Output the winner ("8600" or "ailyanlu") for each input line except the last zero.
No other characters should be inserted in the output.

 

Sample Input

2 0

 

Sample Output

8600

 

Author

ailyanlu

 

Source

Happy 2007

 

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/*

这道题赢家的关键就是最先抢夺过程中剩余可以走的个数为奇数的状态，总的个数为n*n刚开始可以走的个数为n*n-1，如果n为偶数n*n-1为奇数，则第一个走的人面临的就是必赢状态，否则第二走的人会赢。

*/ 

题目意思是每个人可以垂直方向走，也可以水平方向走，每回走一步，但是走过的地方就不能走了，无论怎么走都能看成转着圈走，走到中间就走不了了，奇数时走偶数步，偶数时走奇数步，所以规律就有了

#include

#include
#include
using namespace std;
int main()
{
int n;
while(scanf("%d",&n),n)
{
if(n&1)
 printf("ailyanlu\n");
 else
  printf("8600\n");
}
return 0;
}