OOXX Game

Think:
这次组队赛的水题。。。。
判断 O 是奇数还是偶数
然后对应输出就可以了～～～

Problem Description

Fat brother and Maze are playing a kind of special (hentai) 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.
Input

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
Output

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.
Sample Input
3
1 4
OXXX
2 4
OOXX
OOXX
1 2
XX
Sample Output
Case 1: Maze
Case 2: Fat brother
Case 3: Fat brother

#include<stdio.h>#include<string.h>#include <algorithm>using namespace std;int main(){    int t;    int sum;    int n, m;    int Case = 1;    scanf("%d",&t);    while(t--)    {        scanf("%d%d", &n, &m);        char c;        sum = 0;        getchar();        for(int i = 0; i < n;i++)        {            for(int j = 0; j < m; j++)            {                scanf("%c", &c);                if(c == 'O')                sum++;            }            getchar();        }        if(sum % 2 == 0)            printf("Case %d: Fat brother\n", Case++);        else            printf("Case %d: Maze\n", Case++);    }}