POJ 2311 Cutting Game(二维SG+Multi-Nim)
纪向群,2017年世乒赛,315晚会播出时间
Cutting Game
Time Limit: 1000MS |
|
Memory Limit: 65536K |
Total Submissions: 4798 |
|
Accepted: 1756 |
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
,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;
}
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