POJ2311 Cutting Game

题意

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Cutting Game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6007		Accepted: 2190

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)

分析

对于任何一个人，都不会先剪出1*n或者n*1，应该这样就必败了。

那我们考虑一个状态的后继中，最小的边也是2，这样就可以避免之前的问题，也不需要考虑类似ANTI-SG。

一旦出现2*2，2*3，3*2，这些都成了终止状态，不论怎么剪都会出现1*n,或者n*1

mex求出不属于集合的最小整数

纸片的SG值是后者的纸片的SG的异或值。

还是考察SG函数

时间复杂度(O(n^3))

代码

#include<iostream>
#include<cstring>
const int N=206;
int n,m,sg[N][N];
int SG(int x,int y){
	bool f[N];
	memset(f,0,sizeof f);
	if(sg[x][y]!=-1) return sg[x][y];
	for(int i=2;i<=x-i;++i) f[SG(i,y)^SG(x-i,y)]=1;
	for(int i=2;i<=y-i;++i) f[SG(x,i)^SG(x,y-i)]=1;
	int t=0;
	while(f[t]) ++t;
	return sg[x][y]=t;
}
int main(){
//	freopen(".in","r",stdin),freopen(".out","w",stdout);
	memset(sg,-1,sizeof sg);
	sg[2][2]=sg[2][3]=sg[3][2]=0;
	while(~scanf("%d%d",&n,&m)) puts(SG(n,m)?"WIN":"LOSE");
	return 0;
}