Cutting Game

Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 4806   Accepted: 1760

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

分析

dfs求sg值。居然可以这样玩(⊙_⊙)

code

 #include<cstdio>
#include<set>
#include<cstring> using namespace std;
int sg[][]; int get_SG(int w,int h) {
if (sg[w][h] != -) return sg[w][h];
set<int>s;
for (int i=; w-i>=; ++i) //竖直切
s.insert(get_SG(i,h)^get_SG(w-i,h));
for (int i=; h-i>=; ++i) //水平切
s.insert(get_SG(w,i)^get_SG(w,h-i));
for (int j=; ; ++j)
if (!s.count(j)) {sg[w][h] = j;break;}
return sg[w][h]; }
int main () {
memset(sg,-,sizeof(sg));
int w,h;
while (~scanf("%d%d",&w,&h)) {
if (get_SG(w,h)==) puts("LOSE");
else puts("WIN");
}
return ;
}

POJ 2311 Cutting Game(SG函数)的更多相关文章

  1. POJ 2311 Cutting Game(SG函数)

    题目描述 意思就是说两个人轮流剪纸片,直到有一个人剪出1*1的方格就算这个人赢了.然后给出纸片的长和宽,求先手会赢还是会输 (1<=n,m<=200) 题解 看了一眼,这不是裸的SG吗 啪 ...

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

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

  3. 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 ...

  4. POJ 2960 S-Nim 博弈论 sg函数

    http://poj.org/problem?id=2960 sg函数几乎是模板题. 调试代码的最大障碍仍然是手残在循环里打错变量名,是时候换个hydra产的机械臂了[超想要.jpg] #includ ...

  5. HDU3544 Alice's Game && POJ 2960 S-Nim(SG函数)

    题意: 有一块xi*Yi的矩形巧克力,Alice只允许垂直分割巧克力,Bob只允许水平分割巧克力.具体来说,对于Alice,一块巧克力X i * Y i,只能分解成a * Y i和b * Y i其中a ...

  6. 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 ...

  7. poj 2960 S-Nim(SG函数)

    S-Nim Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 3694   Accepted: 1936 Description ...

  8. POJ 2311 Cutting Game (博弈)

    题意:给定一个长方形纸张,每次只能水平或者垂直切,如果切到1*1的方格就胜,问先手胜还是负. 析:根据Nim游戏可知,我们可以分别求出每个子游戏的和,就是答案,所以我们就枚举每一种切法,然后求出SG函 ...

  9. poj 2960 S-Nim【SG函数】

    预处理出SG函数,然后像普通nim一样做即可 #include<iostream> #include<cstdio> using namespace std; const in ...

随机推荐

  1. WPF动画的几种模式

    最近在用WPF做简单动画,以下是几点经验总结: 1. 使用DispatcherTimer做动画 VB6的年代大家就用Timer做动画了,不用多解释,这个DispatcherTimer和本身的Timer ...

  2. (转载)C#中的lock关键字

    lock 关键字可以用来确保代码块完成运行,而不会被其他线程中断.这是通过在代码块运行期间为给定对象获取互斥锁来实现的. 先来看看执行过程,代码示例如下: 假设线程A先执行,线程B稍微慢一点.线程A执 ...

  3. centos7.3.1611安装及初始配置

    安装前规划: 主机名称 网络配置 分区配置 分区配置 自定义分区,标准分区 /boot 200M (可选) swap 内存1.5倍到2倍(不大于8G) / 根分区(100G到200G) 其余的备用(数 ...

  4. 在vim中插入命令行的输出结果

    vim是linux中常见的编辑器,这里讲讲如何在vim中插入命令行的输出结果. 基本用法: 在指令模式下运行 :!command ,如!date将日期显示在vim底部,!ls列出当前目录 将命令结果插 ...

  5. badboy页面脚本发生错误,解决方案

    1.参考网址:https://jingyan.baidu.com/article/e9fb46e17537797520f76645.html?from=qqbrowser061108 本人亲自测试,方 ...

  6. (转载)office 2003 gaozhi.msi 缺失提示问题修复

    某些GHOST版win7,自带office 2003,每次启动word,它都会提示"稿纸没安装"云云,找不到那个文件.可是我搜遍了硬盘,确实没有那个文件.每次都要点取消,这个提示才 ...

  7. Nginx+Keepalived双主轮询负载均衡

    双主模式使用两个VIP,前段有2台服务器,互为主从,两台服务器同时工作,不存在资源浪费情况.同时在前端的DNS服务器对网站做多条A记录,实现了Nginx的负载均衡,当一台服务器故障时候,资源会转移到另 ...

  8. LeetCode Remove Duplicates from Sorted List 删除有序链表中的重复结点

    /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode ...

  9. ZOJ 2112 Dynamic Rankings(二分,树套树)

    动态区间询问kth,单点修改. 区间用线段树分解,线段树上每条线段存一颗平衡树. 不能直接得到kth,但是利用val和比val小的个数之间的单调性,二分值.log^3N. 修改则是一次logN*log ...

  10. 【CF1000C】Covered Points Count(离散化+差分)

    点此看题面 大致题意: 给出\(n\)条线段,分别求有多少点被覆盖\(1\)次.\(2\)次...\(n\)次. 正常的算法 好吧,这道题目确实有个很简单的贪心做法(只可惜我做的时候没有想到,结果想了 ...