博弈——Play a game

最新推荐文章于 2020-10-29 23:04:12 发布

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本文介绍了一个简单的棋盘游戏，玩家需轮流移动棋子至未访问过的相邻格，无法行动者判负。通过分析游戏规则，得出奇数边长棋盘Ailyanlu胜，偶数则8600胜的结论。

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

题意：ailyanlu和8600玩游戏，有一个边长为n的正方形棋盘，棋子最开始在棋盘的某一个角落，玩家轮流移动这颗棋子，每回合可以移动到这颗棋子的前后左右且之前没移动到的格子，如果某个玩家不能移动棋子，则为输。8600先移动，输出赢家。

思路：

8600的最小必败态是n=1，n=1又可以转移到n=2，同样n=2可以转移到n=3，这样推下去，就是n是奇数是ailyanlu赢，否则8600赢。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

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