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. arm 基本

    ARMr0-r4 传递参数与返回值r7 帧指针 指向母函数被调用子函数在栈看中的交界栈帧指针(Frame Pointer).指向前一个保存的栈帧(stack frame)和链接寄存器(link reg ...

  2. PBOC规范(2.0->3.0)对照表

    1    数据方面 TAG                                               PBOC2.0                                 ...

  3. hdu 4585 Shaolin

    原题链接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=46093 #include<algorithm> #in ...

  4. XCode6之后预编译文件的创建

    首先,在你的项目创建一个.pch预编译头文件(一直点Next)

  5. SharePoint 2010 设置宽度1024px

    在模板页中找到 s4-workspace,设置class=”s4-nosetwidth“,然后再设置宽度为1024px:如果要居中,设置style=“margin:0 auto” 这样也会有一个问题: ...

  6. Mono for Android (1) 之布局

    最近和同事交接工作,首次接触mono for android, 结果画view时少了layout,页面没办法出来,各种冥思,各种找问题,最后把关于布局的一些共享出来(同事写的,哈哈):   Andro ...

  7. Oracle Insert 多行(转)

    1.一般的insert 操作. 使用语法insert into table_name[(column[,column...])] values (value[,value…])的insert语句,每条 ...

  8. 浅谈ERP系统实施后如何完善企业内部控制制度建设

    ERP与企业内部控制制度,前者提升企业的管理水平,后者为企业发展保驾护航,两项工作都是企业各项工作的重中之重. ERP是企业资源规划Enterprise Resource Planning的缩写.企业 ...

  9. Notes of the scrum meeting(11/2)

    meeting time:13:00~13:30p.m.,November 2nd,2013 meeting place:3号公寓楼一层 attendees: 顾育豪                  ...

  10. 闹钟类app构想

    NABC--闹钟app N:我们打算针对那些易健忘的人来制作一款闹钟功能的记事本,具来说服务的对象有很多:有健忘的人,还有情侣,北漂的人及其父母(...),常年见不到亲人(双方),后期我们若提前完成基 ...