2013 多校联合4 1011 Fliping game （hdu 4642）

http://acm.hdu.edu.cn/showproblem.php?pid=4642

Fliping game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 140    Accepted Submission(s): 104


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.

思路：签到题，只要判断右下顶点是向上翻还是向下翻的即可，因为每一次选择的矩形必定包含该顶点，若该顶点为1，则必定要奇数次操作才可以将所有的点变为0，所以Alice必胜，否则Bob必胜。

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;

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