Play a game

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2322 Accepted Submission(s): 1865

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

題意：給你一個n∗n的遊戲棋盤，一個石頭在任意的一個方格內，兩個人交替移動石子，石子只能被移動到相鄰且沒有移動過的格子內，誰不能移動誰輸.

思路：格子總會走完的，所以說根據格子數量判斷奇偶性就可以了.

ac程式碼：

/* ***********************************************
Author       : AnICoo1
Created Time : 2016-07-29-20.42 Friday
File Name    : D:\MyCode\2016-7月\2016-7-29.cpp
LANGUAGE     : C++
Copyright  2016 clh All Rights Reserved
************************************************ */
 

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-8
using namespace std;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}
//head
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF,n)
    {
        int x=n*n;
        if(n%2)
            printf("ailyanlu\n");
        else
            printf("8600\n");
    }
    return 0;
}