HDOJ 4642 Fliping game

Fliping game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1770    Accepted Submission(s): 1157

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.

 

Sample Input

22 21 11 13 30 0 00 0 00 0 0

 

Sample Output

AliceBob

 

题意就是一局游戏，在n*m的棋盘里面有硬币，分为朝上或者朝下，每个人可以选择一个x,y，翻转x到n，y到m的硬币,求最后谁能获胜。

很少做博弈论，其实这个题很简单，每次翻转时都会把n，m的硬币翻转，要使最后nm==1；

如果输入nm为0那么就肯定需要奇数次，反之就需要偶数次。(我还是太年轻)

#include<stdio.h>#include<algorithm>#include<string>#include<string.h>#include<queue>#include<stack>#include<math.h>using namespace std;int main(){    int a[105][1005];    int n,m,t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++)        {            for(int j=1;j<=m;j++)                scanf("%d",&a[i][j]);        }        if(a[n][m]==1)            printf("Alice\n");        else            printf("Bob\n");    }return 0;}