hdu 3595 GG and MM （Every-SG游戏）

GG and MM

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 687    Accepted Submission(s): 303

Problem Description

GG and MM like playing a game since they are children. At the beginning of game, there are two piles of stones. MM chooses a pile of stones first, which has x stones, and then she can choose a positive number k and remove k*x stones out from the other pile of stones, which has y stones (I think all of you know that y>=k*x - -!). Then it comes the turn of GG, followed the rules above-mentioned as well. When someone can't remove any stone, then he/she loses the game, and this game is finished.
Many years later, GG and MM find this game is too simple, so they decided to play N games at one time for fun. MM plays first, as the same, and the one on his/her turn must play every unfinished game. Rules to remove are as same as above, and if someone cannot remove any stone (i.e., loses the last ending game), then he/she loses. Of course we can assume GG and MM are clever enough, and GG will not lose intentionally, O(∩_∩)O~

 

Input

The input file contains multiply test cases (no more than 100).
The first line of each test case is an integer N, N<=1000, which represents there are N games, then N lines following, each line has two numbers: p and q, standing for the number of the two piles of stones of each game, p, q<=1000(it seems that they are so leisure = =!), which represent the numbers of two piles of stones of every game.
The input will end with EOF.

 

Output

For each test case, output the name of the winner.

 

Sample Input

  
  

   
   3
1 1
1 1
1 1
1
3 2
  
  

 

Sample Output

  
  

   
   MM
GG
  
  

 

Author

alpc95

 

Source

2010 ACM-ICPC Multi-University Training Contest（16）——Host by NUDT 

题解：Every-SG游戏

这个游戏就是分成若干不相干的组，每次操作能移动的都要移动。这个游戏的特点不仅有必胜和必败，而且有时间长短的博弈。对于自己必胜的局面，希望步数越多越好，自己必败的局面，早点结束才有利。

必胜当且仅当所有的单一游戏步数最大的为奇数。最大的为奇数，当然是先手赢，其他的都已经提前结束。

其实判断的时候也可以比较一下必胜态与必败态那个持续的时间更长。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#define N 2003
using namespace std;
int sg[N][N],step[N][N];
int n,m;
int get(int x,int y)
{
	if (sg[x][y]!=-1) return sg[x][y];
    if (x>y) swap(x,y);
    int mx=0,mn=1000000000;
    for (int i=x;i<=y;i+=x) {
    	if (!get(x,y-i)) {
    		sg[x][y]=sg[y][x]=1;
    		mx=max(mx,step[x][y-i]);
		}
		else mn=min(mn,step[x][y-i]);
	}
	//cout<<x<<" "<<y<<" "<<mx<<" "<<mn<<" "<<sg[x][y]<<endl;
	if (sg[x][y]==1){
		step[x][y]=step[y][x]=mx+1;
		return sg[x][y];
	}
	step[x][y]=step[y][x]=mn+1;
	sg[x][y]=sg[y][x]=0;
	return sg[x][y];
}
int main()
{
	freopen("a.in","r",stdin);
	memset(sg,-1,sizeof(sg));
	for (int i=0;i<=1000;i++)
	 sg[i][0]=sg[0][i]=step[i][0]=step[0][i]=0;
	while (scanf("%d",&n)!=EOF) {
		int a=0,b=0;
		for (int i=1;i<=n;i++) {
			int x,y; scanf("%d%d",&x,&y);
			get(x,y);
			if (x==0||y==0) continue;
			if (sg[x][y]) a=max(a,step[x][y]);
			else b=max(b,step[x][y]);
		}
		if (a>b) puts("MM");
		else puts("GG");
	}
}