hdu-5754 Life Winner Bo(博弈)

题目链接：

Life Winner Bo

Time Limit: 2000/1000 MS (Java/Others)    

Memory Limit: 131072/131072 K (Java/Others)

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

4

1 5 5

2 5 5

3 5 5

4 5 5

 

Sample Output

G

G

D

B

 

题意：

 

四种棋子,king rook knight,queen;然后看谁先从(1,1)到(n,m);

 

思路：

 

前三种画图都可以找到位置的关系;可以把位置直接标出来;queen的是威佐夫博弈;

 

AC代码：

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
//#include <bits/stdc++.h>
#include <stack>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=2e6+10;
const int maxn=500+10;
const double eps=1e-8;

int flag;
void dfs(int x,int y)
{
   // cout<<x<<" "<<y<<endl;
    if(x==1&&y==1){flag=-1;return ;}
    if(x<1||y<1)return ;
    int fx,fy;
    fx=x-1,fy=y-2;
    if(fx==1&&fy==1){flag=1;return ;}
    fx=x-2;fy=y-1;
    if(fx==1&&fy==1){flag=1;return ;}
    fx=x-3;fy=y-3;
    dfs(fx,fy);
}

int main()
{
        //freopen("in.txt","r",stdin);
        int t;
        read(t);
        while(t--)
        {
            int type,n,m;
            read(type);read(n);read(m);
            if(type==1)
            {
                if((n-1)%2||(m-1)%2)printf("B\n");
                else printf("G\n");
            }
            else if(type==2)
            {
                if(n==m)printf("G\n");
                else printf("B\n");
            }
            else if(type==3)
            {
                    flag=0;
                    dfs(n,m);
                    //if(n==1&&m==1)flag=0;
                    if(flag==0)printf("D\n");
                    else if(flag==1)printf("B\n");
                    else printf("G\n");
            }
            else
            {
                     if(n<m) swap(n,m);
                     n--;m--;
                     int k=n-m;
                     n=(int)(k*(1+sqrt(5))/2.0);
                     if(n==m)   printf("G\n");
                     else   printf("B\n");
            }
        }

        return 0;
}