hdu 4642 Fliping game

Description

Alice and Bob 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 one in each grid and every coin may be upward or downward freely. Then they take turns to choose a rectangle (x  ₁, y  ₁)-(n, m) (1 ≤ x  ₁≤n, 1≤y  ₁≤m) and flips all the coins (upward to downward, downward to upward) in it (i.e. flip all positions (x, y) where x  ₁≤x≤n, y  ₁≤y≤m)). The only restriction is that the top-left corner (i.e. (x  ₁, y  ₁)) must be changing from upward to downward. The game ends when all coins are downward, and the one who cannot play in his (her) turns loses the game. Here's the problem: Who will win the game if both use the best strategy? You can assume that Alice 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 starts with two integers N and M indicate the size of the board. Then goes N line, each line with M integers shows the state of each coin, 1<=N,M<=100. 0 means that this coin is downward in the initial, 1 means that this coin is upward in the initial.

 

Output

For each case, output the winner’s name, either Alice or Bob.

 

Sample Input

        
        
        
        

         
         
         
          2
2 2
1 1
1 1
3 3
0 0 0
0 0 0
0 0 0 
        
        
        
        

 

Sample Output

        
        
        
        

         
         
         
          Alice
Bob 
        
        
        
        

 

题目大意：在一个N * M恩木板上由N * M个硬币，给出正反面，然后两个二B来玩游戏，没次都是爱丽丝先，选一个坐标（x0 y0 )，硬币为正， 将满足（ x0 < x  && y0 < y）的坐标点上的硬币正反调换， 问， 最后没有硬币反导致输了这个2B游戏的人是谁。

解题思路：这个是个很2B的游戏，因为每次不管玩家选哪个位置翻转，都会翻转n , m这个位置，所以这要判断n,m这个位置硬币的状态就可以了。

#include <stdio.h>
int main(){
    int n, m, t, a;
    scanf("%d", &t);
    while (t--){
	scanf("%d%d", &n, &m);
	for (int i = 0; i < n; i++)
	    for (int j = 0; j < m; j++)
		scanf("%d", &a);
	printf("%s\n", a?"Alice":"Bob");
    }
return 0;
}