1564 Play a game

最新推荐文章于 2019-09-23 21:40:17 发布

WNNSCNS

最新推荐文章于 2019-09-23 21:40:17 发布

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分类专栏：题解

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本文介绍了一种棋盘游戏的胜负判断策略，玩家在n*n的棋盘上交替移动角落的石子，不能移动者输。通过寻找规律，发现奇数尺寸棋盘首胜，偶数则后手胜。

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Play a game

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1907 Accepted Submission(s): 1533

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

这题看起来很难的样子，不知道行走的方式以及起点，而且数据量又是那么的大，如果是开一个二维数组老实的推算肯定是不行的，这时候就要想到找规律，找到规律就能解决了。

代码如下：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <vector>
#include <algorithm>
#include <stdlib.h>
#include <set>
#include <vector>
using namespace std;
#define M 1005
#define inf 0xfffffff
int main(){
 int n;
 while(scanf("%d",&n)!=EOF){
  if(n == 0)
   break;
  if(n%2){
   printf("ailyanlu\n");
  }
  else
   printf("8600\n");
 }
 return 0;
}