POJ 2311 Cutting Game(二维SG+Multi-Nim)

2018-02-26 08:15:01来源:cnblogs.com作者:自为风月马前卒人点击

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


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.


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.


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



POJ Monthly,CHEN Shixi(xreborner)  比较有意思的一道题目,如果你知道什么叫Multi-Nim,那么这道题就比较简单了最关键的地方就是一个游戏的SG函数为后继状态异或和的mex注意边长需要从2枚举,否则2*2这个状态会挂掉 
#include<cstdio>#include<cstring>using namespace std;const int MAXN=233;int read(){    char c=getchar();int x=0,f=1;    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}    return x*f;}int SG[MAXN][MAXN];//当前剩余i行 j列的SG函数 int S[MAXN];int main(){    #ifdef WIN32    freopen("a.in","r",stdin);    #else    #endif    int N=201,M=201;    for(int i=2;i<=N;i++)    {        for(int j=2;j<=N;j++)        {            memset(S,0,sizeof(S));            for(int k=2;k<=i-2;k++) S[ SG[k][j]^SG[i-k][j] ]=1;                for(int k=2;k<=j-2;k++) S[ SG[i][k]^SG[i][j-k] ]=1;            for(int k=0;;k++) if(!S[k]) {SG[i][j]=k;break;}        }    }    while(scanf("%d%d",&N,&M)!=EOF)        puts(SG[N][M]?"WIN":"LOSE");    return 0;}