Matrix Game （Nim博弈）

 Matrix Game （Nim博弈）：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=112620#problem/A 传送门：nefu

题面描述：

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit  Status

Description

Given an m x n matrix, where m denotes the number of rows and n denotes the number of columns and in each cell a pile of stones is given. For example, let there be a 2 x 3 matrix, and the piles are

2 3 8

5 2 7

That means that in cell(1, 1) there is a pile with 2 stones, in cell(1, 2) there is a pile with 3 stones and so on.

Now Alice and Bob are playing a strange game in this matrix. Alice starts first and they alternate turns. In each turn a player selects a row, and can draw any number of stones from any number of cells in that row. But he/she must draw at least one stone. For example, if Alice chooses the 2^nd row in the given matrix, she can pick 2 stones from cell(2, 1), 0 stones from cell (2, 2), 7 stones from cell(2, 3). Or she can pick 5 stones from cell(2, 1), 1 stone from cell(2, 2), 4 stones from cell(2, 3). There are many other ways but she must pick at least one stone from all piles. The player who can't take any stones loses.

Now if both play optimally who will win?

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a line containing two integers: m and n (1 ≤ m, n ≤ 50). Each of the next m lines contains n space separated integers that form the matrix. All the integers will be between 0 and 10⁹ (inclusive).

Output

For each case, print the case number and 'Alice' if Alice wins, or 'Bob' otherwise.

Sample Input

2

2 3

2 3 8

5 2 7

2 3

1 2 3

3 2 1

Sample Output

Case 1: Alice

Case 2: Bob

题目大意：

共有m行n列的石子，Alice和Bob依次进行取石子游戏，每一次都必须要在某一行中进行取石子，且取石子的个数要大于等于1个，先取完者先赢。

题目分析：

由于是在每一行中取石子，所以要把每一行的石子的个数加起来，再构造成一个新的堆，这样就一共会有m个堆，然后再按照正常的NIm博弈去求解问题，先取完先赢的状态，详细解释见Nim博弈的转载博客。

代码实现：

#include <iostream>

using namespace std;

int data[55];

int main()
{
    int t,casenum=0;
    int m,n;
    long long sum;
    long long ans;
    cin>>t;
    while(t--)
    {
        cin>>m>>n;
        sum=0;
        ans=0;
        for(int i=0;i<m;i++)
        {
            sum=0;
            for(int j=0;j<n;j++)
            {
                cin>>data[j];
                sum+=data[j];
            }
            ans^=sum;
        }
        if(ans)
        cout<<"Case "<<++casenum<<": Alice"<<endl;
        else
        cout<<"Case "<<++casenum<<": Bob"<<endl;
    }

    return 0;
}