LightOJ 1247 Matrix Game (尼姆博弈)

A - Matrix Game

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 containsn 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

题意：给定m行 每行n个数 每次选择一行取任意个数 谁取到最后一个谁赢。

题解：最基本常规的尼姆博弈 因为每一行可以任意取 所以每一行看成一堆即可

　　　将每一行进行异或 所得结果ans如果不为0先手赢 否则后手赢

#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
    int t,cas=1;
    cin>>t;
    while(t--)
    {
        int m,n,data,ans=0;
        cin>>m>>n;
        for(int i=0;i<m;i++)
        {
            int sum=0;
            for(int j=0;j<n;j++)
            {
                cin>>data;
                sum+=data;
            }
            ans^=sum;
        }
        if(ans)
        printf("Case %d: Alice
",cas++);
        else
        printf("Case %d: Bob
",cas++);
    }

    return 0;
}