POJ 2311 Cutting Game（SG函数）

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[500][500];


int get_SG(int w,int h) {
    if (sg[w][h] != -1) return sg[w][h];
    set<int>s;
    for (int i=2; w-i>=2; ++i) //竖直切
        s.insert(get_SG(i,h)^get_SG(w-i,h));
    for (int i=2; h-i>=2; ++i) //水平切
        s.insert(get_SG(w,i)^get_SG(w,h-i));
    for (int j=0; ; ++j) 
        if (!s.count(j)) {sg[w][h] = j;break;}
    return sg[w][h];

}
int main () {
    memset(sg,-1,sizeof(sg));
    int w,h;
    while (~scanf("%d%d",&w,&h)) {
        if (get_SG(w,h)==0) puts("LOSE");
        else puts("WIN");
    }
    return 0;
}