HDU

Fliping game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2129    Accepted Submission(s): 1427

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 (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.

 

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

 

题目链接 

 

题意 

　　A与B在玩一个游戏，规则如下，在一个n*m的方格中，每一个格子都可为1或为0，每个人都能选定一个点，将以这个点为左上角，(n，m)为右下角的矩形内每个格子的数值翻转（0变1，1变0），其中选定的点只能作1到0的翻转，无路可走的人则输（面对全为0的矩阵）。A先手。给出一个棋盘，问谁获胜？

 

分析 

　　这种题，第一眼感觉就是有某种规律或者必胜的局面。果不其然，当右下角的格子为1时，A获胜，反之，B获胜。因为不管谁操作，右下角的方格都是会翻转的。

 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include <queue>
#include <vector>
#include<bitset>
using namespace std;
typedef long long LL;
const int maxn = 400;
const int mod = 772002+233;
typedef pair<int,int> pii;
#define X first
#define Y second
#define pb push_back
//#define mp make_pair
#define ms(a,b) memset(a,b,sizeof(a))

const int inf = 0x3f3f3f3f;

int main(){
    int t;
    cin>>t;
    while(t--){
        int n,m;
        cin>>n>>m;
        int f=0;
        int x;
        for(int i=0;i<n;i++)
        for(int j=0;j<m;j++){
            scanf("%d",&x);
            if(i==n-1&&j==m-1&&x) f=1;
        }
        if(f) puts("Alice");
        else puts("Bob");
    }
}