尼姆博弈

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 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).

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

2

2 3

2 3 8

5 2 7

2 3

1 2 3

3 2 1

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：最基本的尼姆博弈

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