HDU 5754 Life Winner Bo （博弈组合，思维）

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本文介绍了一种基于不同棋子类型的棋盘游戏博弈策略，包括国王、车、马和皇后四种棋子。每种棋子都有其独特的移动规则，文章探讨了在两人轮流移动棋子的情况下，如何通过最优策略决定游戏的胜负。

Life Winner Bo

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3009 Accepted Submission(s): 1093

Problem Description

Bo is a "Life Winner".He likes playing chessboard games with his girlfriend G.

The size of the chessboard is

N×M.The top left corner is numbered

(1,1) and the lower right corner is numberd

(N,M).

For each game,Bo and G take turns moving a chesspiece(Bo first).At first,the chesspiece is located at

(1,1).And the winner is the person who first moves the chesspiece to

(N,M).At one point,if the chess can't be moved and it isn't located at

(N,M),they end in a draw.

In general,the chesspiece can only be moved right or down.Formally,suppose it is located at

(x,y),it can be moved to the next point

(x′,y′) only if

x′≥x and

y′≥y.Also it can't be moved to the outside of chessboard.

Besides,There are four kinds of chess(They have movement rules respectively).

1.king.

2.rook(castle).

3.knight.

4.queen.

(The movement rule is as same as the chess.)

For each type of chess,you should find out that who will win the game if they both play in an optimal strategy.

Print the winner's name("B" or "G") or "D" if nobody wins the game.

Input

In the first line,there is a number

T as a case number.

In the next

T lines,there are three numbers type,

N and

M.

"type" means the kind of the chess.

T≤1000,2≤N,M≤1000,1≤type≤4

Output

For each question,print the answer.

Sample Input

41 5 52 5 53 5 54 5 5

Sample Output

GGDB

根据棋子的不同一共有四种博弈

前三种的np图

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <cmath>
#include<algorithm>
#define ans (sqrt(5)+1)/2
using namespace std;
int main(int argc, char *argv[])
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int s,m,n;
		scanf("%d %d %d",&s,&n,&m);
		if(s==1)
		{
			if(n%2==0||m%2==0)
			{
				printf("B\n");
			}
			else
			{
				printf("G\n");
			}
		}
		if(s==2)  
        {  
            if(n==m)  
                printf("G\n");  
            else  
                printf("B\n");  
        }  
		  if(s==3)
        {
           n--;
           m--;
           if(n>m)
           {
		   	swap(n,m);
		   }
		   
		   if(n==m&&n%3==0)
           {
		   	printf("G\n");
		   }
		   else if(m%3==2&&m-n==1)
		   {
		   	printf("B\n");
		   }
		   else
 			{
				printf("D\n");
			}
        }
		if(s==4)
		{
			m--;
			n--;
			if(n<m)
           	{
                   int temp=n;n=m;m=temp;
 	 		}
			int k=(n-m);
	  		n=(int)(((sqrt(5)+1)/2.0)*k);
			if(m==n)
			{
				printf("G\n");
			}
			else
			{
				printf("B\n");
			}
		}
	}
	return 0;
}