OOXX Game

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总时间限制: 1000ms 内存限制: 32768kB

描述

  Fat brother and Maze are playing a kind of special game on an N*M board (N rows, M columns). At the beginning, there are N*M coins in this board with two symbol “O” or “X”. Then they take turns to choose a grid with symbol “O” and change it into “X”. The
game ends when all the symbols in the board are “X”, and the one who cannot play in his (her) turns loses the game.Fat brother and Maze like this kind of OOXX game very much and play it day and night. They don’t even need a little rest after each game!

Here's the problem: Who will win the game if both use the best strategy? You can assume that Maze always goes first.

输入

The first line of the date is an integer T, which is the number of the text cases.

Then T cases follow, each case contains two integers N and M indicate the size of the board. Then goes N line, each line with M character shows the state of the board.

1 <= T <=100, 1 <= n <=100, 1 <= m <=100

输出

For each case, output the case number first, and then output the winner’s name, either Fat brother or Maze. See the sample input and output for more details.

样例输入

3 

1  4 

OXXX 

2  4

OOXX 

OOXX 

1  2 

XX

样例输出

Case 1: Maze 

Case 2: Fat brother 

Case 3: Fat brother

来源
第四届福建省大学生程序设计竞赛

#include<stdio.h>
#include<iostream>

using namespace std;

int main()
{
int nc , i_nc , j_nc;
//cin >> nc;
scanf("%d",&nc);
for(int k = 0 ; k < nc ; k++)
{
int count = 0;
char ch;
scanf("%d %d", &i_nc , &j_nc);
for(int i = 0 ; i < i_nc ; i++)
{
for(int j = 0 ; j < j_nc ; j++)
{
cin >> ch;
if(ch == 'O')
count++;
}
}

if(count%2)
printf("Case %d: Maze \n",k + 1);
else
printf("Case %d: Fat brother \n",k + 1);
}
}