HDU - 1564(32/600)

最新推荐文章于 2021-01-21 00:23:24 发布

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本文介绍了一个关于棋盘游戏的胜负判断问题，玩家在n*n的棋盘上交替移动棋子，目标是让对手无法行动。文章通过分析游戏规则得出结论：当棋盘边长为奇数时，后手玩家获胜；为偶数时，先手玩家获胜。

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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

怎么看都没看懂
后来就看的题解…

是这样的这个东西同一个格子只能有一个人走过
不管路径怎么样走的人一定是固定的

而且从原点开始想要赢就只能走奇数长的路

奇数=奇数+偶数
偶数=奇数+奇数=偶数+偶数

只有先手对奇数路的时候才可能赢

当n现在是奇数的时候
有偶数个格子
先手走了一步就是奇数个格子
后走就是偶数…

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
    int n;
    while (cin >> n)
    {
        if (!n)break;
        if (n % 2 == 0)cout << "8600" << endl;
        else cout << "ailyanlu" << endl;
    }
}