Fliping game(博弈找规律）

Problem 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

   22 21 11 13 30 0 00 0 00 0 0  

Sample Output

   AliceBob  

Problem Description

Input

Output

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

Sample Input

    22 21 11 13 30 0 00 0 00 0 0   

Sample Output

    AliceBob   

大致题意：在一个矩阵中，每格都放着任意向上或向下的硬币，选择一个点与最右下的点组成的矩形，所包含的硬币翻过来，直到所有的硬币都向下。

通过手推几个例子不难发现当最右下的点为0时，后手胜，为1 时先手胜。

不要用 cin , cout !!!!过不去

#include <stdio.h>
#include <iostream>
using namespace std;
int main()
{
	int T , n , m;
	int x;
	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" , &x);
			}
		}
		if(x == 1)
		{
			printf("Alice\n");
		}
		else
		{
			printf("Bob\n");
		}
	}
	return 0;
 }