HDU4642:Fliping game

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 (x₁, y₁)-(n, m) (1 ≤ x₁≤n, 1≤y₁≤m) and flips all the coins (upward to downward, downward to upward) in it (i.e. flip all positions (x, y) where x₁≤x≤n, y₁≤y≤m)). The only restriction is that the top-left corner (i.e. (x₁, y₁)) 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.

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.

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

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

Alice
Bob


题意：给出一个矩阵，每个格子有一个硬币，初始状态1代表面朝上，0代表面朝下，两人轮流翻硬币，每次选一个面朝上的硬币，这个硬币为左上角，n,m的硬币为右下角，然后将这个矩形内的硬币都翻转一次，使得全部硬币变成面朝下的时候，轮到谁的时候不能再翻转硬币的时候，那个人输

思路：我们可以发现，无论是谁翻硬币，n,m位置的硬币必然会被翻转，那么其实我们只需要考虑n,m处的硬币，只要是1，那么必然要无论其他位置怎么翻转，这个位置必须要奇数次才能变成0，所以必然是奇数次，A胜，反之如果是0，必然是偶数次，B胜

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;

#define ls 2*i
#define rs 2*i+1
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 205
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define rank rank1
const int mod = 10007;

int t,n,m,i,j,x;

int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(i = 0; i<n; i++)
        {
            for(j = 0; j<m; j++)
                scanf("%d",&x);
        }
        if(x)
            printf("Alice\n");
        else
            printf("Bob\n");
    }

    return 0;
}