Cutting Game
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 2844   Accepted: 1036

Description

Urej loves to play various types of dull games. He usually asks other people to play with him. He says that playing those games can show his extraordinary wit. Recently Urej takes a great interest in a new game, and Erif Nezorf becomes the victim. To get away from suffering playing such a dull game, Erif Nezorf requests your help. The game uses a rectangular paper that consists of W*H grids. Two players cut the paper into two pieces of rectangular sections in turn. In each turn the player can cut either horizontally or vertically, keeping every grids unbroken. After N turns the paper will be broken into N+1 pieces, and in the later turn the players can choose any piece to cut. If one player cuts out a piece of paper with a single grid, he wins the game. If these two people are both quite clear, you should write a problem to tell whether the one who cut first can win or not.

Input

The input contains multiple test cases. Each test case contains only two integers W and H (2 <= W, H <= 200) in one line, which are the width and height of the original paper.

Output

For each test case, only one line should be printed. If the one who cut first can win the game, print "WIN", otherwise, print "LOSE".

Sample Input

2 2
3 2
4 2

Sample Output

LOSE
LOSE
WIN

Source

POJ Monthly,CHEN Shixi(xreborner)
 
sg函数
 
 #include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set> using namespace std; const int MAX_N = ;
int dp[MAX_N][MAX_N]; int grundy(int w, int h) {
if(dp[w][h] != -) return dp[w][h]; set<int> s;
for(int i = ; w - i >= ; i++) {
s.insert(grundy(i, h) ^ grundy(w - i,h));
}
for(int i = ; h - i >= ; ++i) {
s.insert(grundy(w, i) ^ grundy(w, h - i));
} int res = ;
while(s.count(res)) res++;
return dp[w][h] = res;
} int main()
{
//freopen("sw.in","r",stdin);
int w, h; for(int i = ; i <= ; ++i) {
for(int j = ; j <= ; ++j) dp[i][j] = -;
}
while(~scanf("%d%d",&w,&h)) {
printf("%s\n",grundy(w, h) ? "WIN" : "LOSE");
}
//cout << "Hello world!" << endl;
return ;
}

poj 2311的更多相关文章

  1. POJ 2311 Cutting Game (Multi-Nim)

    [题目链接] http://poj.org/problem?id=2311 [题目大意] 给出一张n*m的纸,每次可以在一张纸上面切一刀将其分为两半 谁先切出1*1的小纸片谁就赢了, [题解] 如果切 ...

  2. 【POJ 2311】 Cutting Game

    [题目链接] http://poj.org/problem?id=2311 [算法] 博弈论——SG函数 [代码] #include <algorithm> #include <bi ...

  3. POJ 2311 Cutting Game(二维SG+Multi-Nim)

    Cutting Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4798   Accepted: 1756 Desc ...

  4. POJ 2311 Cutting Game(Nim博弈-sg函数/记忆化搜索)

    Cutting Game 题意: 有一张被分成 w*h 的格子的长方形纸张,两人轮流沿着格子的边界水平或垂直切割,将纸张分割成两部分.切割了n次之后就得到了n+1张纸,每次都可以选择切得的某一张纸再进 ...

  5. poj 2311 Cutting Game 博弈论

    思路:求SG函数!! 代码如下: #include<iostream> #include<cstdio> #include<cmath> #include<c ...

  6. POJ 2311 Cutting Game(SG+记忆化)

    题目链接 #include<iostream> #include<cstdio> #include<cstring> using namespace std; ][ ...

  7. POJ 2311 Cutting Game [Multi-SG?]

    传送门 题意:n*m的纸片,一次切成两份,谁先切出1*1谁胜 Multi-SG? 不太一样啊 本题的要求是后继游戏中任意游戏获胜就可以了.... 这时候,如果游戏者发现某一单一游戏他必败他就不会再玩了 ...

  8. 4.1.7 Cutting Game(POJ 2311)

    Problem description: 两个人在玩如下游戏. 准备一张分成 w*h 的格子的长方形纸张,两人轮流切割纸张.要沿着格子的边界切割,水平或者垂直地将纸张切成两部分.切割了n次之后就得到了 ...

  9. POJ 2311 Cutting Game(SG函数)

    Cutting Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4806   Accepted: 1760 Desc ...

随机推荐

  1. C扩展 C++回顾到入门

    引言 C扩展也称C++, 是一个复(za)杂(ji)优(ken)秀(die)的语言. 本文通过开发中常用C++方式来了解和回顾C++这么语言. C++看了较多的书但还是觉得什么都不会. 只能说自己还付 ...

  2. C 几种异常机制简单讲述

    引言 这是关于C中如何使用异常机制的讨论.顺带讲一讲C中魔法函数的setjmp内部机制.通过它实现高级的异常try...catch. 允许我先扯一段面试题. 对于计算机面试题. 算法题等.觉得还是有意 ...

  3. RMAN - 备份异机恢复

    OS: Oracle Linux Server release 5.7 DB: Oracle Database 11g Enterprise Edition Release 11.2.0.3.0 - ...

  4. JVM学习总结五——性能监控及故障处理工具

    之前扯了四篇理论,这一篇终于可以动动手了.本篇我们将介绍JVM常用的一些工具,这些工具将是我们监控JVM状态.处理故障和调优分析的利器. 不过在开始之前,我还是要先车扯两句:工具终归只是帮助我们我们处 ...

  5. perl DBI 学习总结(转载)

    perl DBI 学习总结 源文地址:http://blog.csdn.net/like_zhz/article/details/5441946 DBI和DBD的不同关系模型: ########### ...

  6. Asp.Net Web API开发微信后台

    如果说用Asp.Net开发微信后台是非主流,那么Asp.Net Web API的微信后台绝对是不走寻常路. 需要说明的是,本人认为Asp.Net Web API在开发很多不同的请求方法的Restful ...

  7. json 读写 swift

    // // ViewController.swift // json读写 // // Created by mac on 15/7/14. // Copyright (c) 2015年 fangyuh ...

  8. 硬件相关-EMI & EMS & EMC

    EMI——Electro Magnetic Interference 电磁干扰 定义:是指电磁波与电子元件作用后而产生的干扰现象. 分类:有传导干扰和辐射干扰两种. 传导干扰: 是指通过导电介质把一个 ...

  9. activemq整合spring

  10. Daily Scrum 12.7

    摘要:本次会议主要是为了分配任务.我们对于各自将要进行的任务进行了讨论,并最终确定下了我们每个人Beta版本将要进行的任务.因为vs中任务的编写在此次会议之后,所以迭代时直接填写了已完成时间. Tas ...