hdu 1564 Play a game（博弈——找规律）

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 m
4000
ove. 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

题意：有一个n*n的棋盘，棋盘的角落有一个石子，两个人轮流移动石子（8600先来），规定每次石子只能移动到当前位置的相邻位（且这个位置未被移动到过），最后谁不能移动石子则谁输，问最后谁赢

结论：当x=n*n为奇数时，后手赢，否则先手赢

证明：当x为偶数时，有y=x-1个位置可走，且y为奇数

又，若可走的步数为奇数时，先手必赢

且奇数一定可以表示为一个奇数加一个偶数的和

也就是说，以先手为主导走的路，最后一定能走到奇数步的路上，因此先手必胜

同理，若y为偶数，偶数可以表示为两个奇数的和或者两个偶数的和

只有当偶数变为奇数，也就是变为以后手为主导时，才一定可以走到奇数步的路上，因此后手必胜

详细证明博客：Boapath

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

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