POJ2311 Cutting Game SG函数

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题目链接：POJ2311

Cutting Game

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3850 Accepted: 1434

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 23 24 2

Sample Output

LOSELOSEWIN

题意：一个游戏，每次把一块布剪成2半（整数），剪到1*1就算赢，问给出布的长宽，先手会赢还是输。

题目分析：典型的博弈论题目，首先考虑SG函数，对于一块w*h的布有2种剪法，剪w或者h，设剪成i的长度，那么有i<w-i或h-i，把所有的剪到的sg值异或起来就是sg[w][h]的值。但是考虑到求sg函数时间复杂度太高O(N^3),因此可以先打表。发现异或和为0的只有那么几个数，在sg中把这些点标记出来，其他不用管就好了。

////  main.cpp//  POJ2311////  Created by teddywang on 2016/8/11.//  Copyright © 2016年 teddywang. All rights reserved.//#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int sg[205][205];int  get_sg(int n,int m)//打表函数，没有用到{    if(sg[n][m]!=-1) return sg[n][m];    int vis[1000];    memset(vis,0,sizeof(vis));    for(int i=2;i<=n-i;i++)    {        vis[get_sg(i, m)^get_sg(n-i,m)]=1;    }    for(int i=2;i<=m-i;i++)    {        vis[get_sg(n, i)^get_sg(n,m-i)]=1;    }    for(int i=0;;i++)    {        if(!vis[i]) return i;    }    }void init(){    int s[100]={2,3,7, 11, 17, 23, 27, 31, 37, 41, 45, 57, 61 ,65, 75, 79, 91, 95, 99, 109, 113, 125, 129, 133, 143, 147, 159, 163, 167, 177, 181, 193, 197 };    for(int i=0;s[i];i++)    {        for(int j=0;s[j];j++)        {            sg[s[i]][s[j]]=0;        }    }}int main(){    memset(sg,-1,sizeof(sg));    init();    sg[1][1]=0;   /* for(int i=2;i<=200;i++)    {        for(int j=2;j<=200;j++)        {            sg[i][j]=get_sg(i,j);            if(sg[i][j]==0) printf("%d ",j);        }        printf("\n");    }*/    int w,h;    while(cin>>w>>h)    {        if(sg[w][h]==0) printf("LOSE\n");        else printf("WIN\n");    }}

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