HDU 4642 Fliping game

最新推荐文章于 2019-10-01 10:52:38 发布

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最新推荐文章于 2019-10-01 10:52:38 发布

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分类专栏： ---博弈入门 ---简单数学题 HDUOJ

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本文介绍了一种基于特殊棋盘的游戏胜负判断方法，通过分析棋盘上特定位置的初始状态来预测最终胜者。当最右下角的棋子为特定状态时，先手玩家获胜，反之则后手玩家获胜。

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

Sample Output

  
   Alice
Bob

Source

2013 Multi-University Training Contest 4

解题思路：一看就是多校的签到题，很水的数学题，证明过程下次下个详细的，就是说，按顺序输入的过程中，如果最右下角的那个点是1的话，那么就需要奇数步

才能搞定，所以A赢，其他情况下B赢。

#include <iostream>
# include<iostream>
# include<cstdio>

using namespace std;

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