HDU-1564-Play a game 【找规律】

本文介绍了一个简单的棋盘游戏胜负判断算法。游戏在一个n*n的棋盘上进行,两名玩家交替移动角落开始的棋子,只能移动到未访问过的相邻格子。当一方无法移动时即为输家。通过判断棋盘大小的奇偶性可以得出获胜者。


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



Play a game




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
 


从一个n*n棋盘的一角出发,每次移动到相邻的,而且没有经过的格子上。谁不能操作了谁输。  还没走过的格子数为奇数时,先手胜。所以n为偶数,先手胜。

[cpp]  view plain  copy
 print ?
  1. #include<iostream>  
  2. using namespace std;  
  3. int main()  
  4. {  
  5.     int n;  
  6.     while(cin>>n)  
  7.     {  
  8.         if(n==0)  
  9.         break;  
  10.         if(n%2)  
  11.         cout<<"ailyanlu"<<endl;  
  12.         else  
  13.         cout<<"8600"<<endl;   
  14.     }  
  15.     return 0;  
  16.  }   


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值