HDU 1564 Play a game (找规律博弈)
Play a game
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1030 Accepted Submission(s): 838
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?
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.
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.
No other characters should be inserted in the output.
Sample Input
2
0
Sample Output
8600
Author
ailyanlu
Source
Recommend
8600
从一个n*n的角落出发,每次移动到相邻的,而且没有经过的格子上。谁不能操作了谁输。
结论就是n为偶数,先手赢,奇数,后手赢。
可以看下图
S表示起点。
如果n为偶数,那么所有格子可以被2*1的砖块覆盖掉。
这样先手每次都移动到当前1*2的另外一块。先手必赢。
如果n为奇数。出了起始那个店,其余点都可以被覆盖。
所有后手赢
#include <string.h> #include <stdio.h> #include <algorithm> #include <iostream> using namespace std; int main() { int n; while(scanf("%d",&n)==1 &&n) { if(n%2==0)printf("8600\n"); else printf("ailyanlu\n"); } return 0; }
人一我百!人十我万!永不放弃~~~怀着自信的心,去追逐梦想
分类:
ACM(HOJ)题目解题报告
, ACM博弈
【推荐】国内首个AI IDE,深度理解中文开发场景,立即下载体验Trae
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步
· AI与.NET技术实操系列(二):开始使用ML.NET
· 记一次.NET内存居高不下排查解决与启示
· 探究高空视频全景AR技术的实现原理
· 理解Rust引用及其生命周期标识(上)
· 浏览器原生「磁吸」效果!Anchor Positioning 锚点定位神器解析
· DeepSeek 开源周回顾「GitHub 热点速览」
· 物流快递公司核心技术能力-地址解析分单基础技术分享
· .NET 10首个预览版发布:重大改进与新特性概览!
· AI与.NET技术实操系列(二):开始使用ML.NET
· .NET10 - 预览版1新功能体验(一)
2011-07-22 乘积最大
2011-07-22 盒子放球
2011-07-22 升降交替数列
2011-07-22 数字和
2011-07-22 无连续整数的子集数问题
2011-07-22 无相邻1问题
2011-07-22 ACM POJ 2245Lotto解题报告