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Description

In an extended version of the game Lights Out, is a puzzle with 5 rows of 6 buttons each (the actual puzzle has 5 rows of 5 buttons each). Each button has a light. When a button is pressed, that button and each of its (up to four) neighbors above, below, right and left, has the state of its light reversed. (If on, the light is turned off; if off, the light is turned on.) Buttons in the corners change the state of 3 buttons; buttons on an edge change the state of 4 buttons and other buttons change the state of 5. For example, if the buttons marked X on the left below were to be pressed,the display would change to the image on the right. 

The aim of the game is, starting from any initial set of lights on in the display, to press buttons to get the display to a state where all lights are off. When adjacent buttons are pressed, the action of one button can undo the effect of another. For instance, in the display below, pressing buttons marked X in the left display results in the right display.Note that the buttons in row 2 column 3 and row 2 column 5 both change the state of the button in row 2 column 4,so that, in the end, its state is unchanged. 

Note: 

1. It does not matter what order the buttons are pressed. 

2. If a button is pressed a second time, it exactly cancels the effect of the first press, so no button ever need be pressed more than once. 

3. As illustrated in the second diagram, all the lights in the first row may be turned off, by pressing the corresponding buttons in the second row. By repeating this process in each row, all the lights in the first 

four rows may be turned out. Similarly, by pressing buttons in columns 2, 3 ?, all lights in the first 5 columns may be turned off. 

Write a program to solve the puzzle.

Input

Output

Sample Input

20 1 1 0 1 01 0 0 1 1 10 0 1 0 0 11 0 0 1 0 10 1 1 1 0 00 0 1 0 1 01 0 1 0 1 10 0 1 0 1 11 0 1 1 0 00 1 0 1 0 0

Sample Output

PUZZLE #11 0 1 0 0 11 1 0 1 0 10 0 1 0 1 11 0 0 1 0 00 1 0 0 0 0PUZZLE #21 0 0 1 1 11 1 0 0 0 00 0 0 1 0 01 1 0 1 0 11 0 1 1 0 1

分析来源：http://www.xuebuyuan.com/1394020.html

题意：有一个5*6的矩阵，每个位置都表示按钮和灯，1表示亮，0表示灭。每当按下一个位置的按钮，它和它周围灯的状态全部翻转，问在这样的一个方阵中按下哪些按钮可以把整个方阵都变成灭的，这时1表示按了，0表示没按。

以下分析部分转自：http://blog.csdn.net/shiren_Bod/article/details/5766907

这个游戏有一些技巧： 

这个问题可以转化成数学问题。 

我们记L为待求解的原始布局矩阵。A(i,j)表示按下第i行第j列的按钮时的作用范围矩阵。在上述例子中， 

A(1,1)= 

A(2,2)= 

假设x(i,j)表示：想要使得L回到全灭状态，第i行第j列的按钮是否需要按下。0表示不按，1表示按下。那么，这个游戏就转化为如下方程的求解： 

其中x(i,j)是未知数。方程右边的0表示零矩阵，表示全灭的状态。直观的理解就是：原来的L状态，经过了若干个A(i,j)的变换，最终变成0:全灭状态。 

这是一个矩阵方程。两个矩阵相等，充要条件是矩阵中每个元素都相等。将上述方程展开，便转化成了一个9元1次方程组： 

这个方程有唯一解： 

也就是说，按下第一行的3个按钮，和右下角的按钮，就

1个开关最多控制5个灯，在构造的矩阵中，a[i][j]=1表示第i个开关可以影响到j号灯

构造出的矩阵如图：

#include 
   
    #include 
    
     using namespace std;int map[32][32];int ans[32];void Guass(){	for (int i=0;i<30;i++){	//控制行		if (map[i][i]==0){				for (int j=i+1;j<30;j++){	//找到不为0的那一行，然后进行交换				if (map[j][i]!=0){					for (int k=i;k<31;k++){						swap(map[j][k],map[i][k]);					}					break;				}			}		}				for (int j=0;j<30;j++){			if (i!=j&&map[j][i]){				for (int k=i;k<31;k++){					map[j][k]=map[i][k]^map[j][k];				}			}		}	}	for (int i=0;i<30;i++){		ans[i]=map[i][30];	}}int main(){	int t,kn,km,kx,ky;	cin>>t;	for (int cas=1;cas<=t;cas++){		for (int i=0;i<30;i++)			cin>>map[i][30];		for(int i=0;i<30;i++){      //构造30个方程                kn=i/6;              km=i%6;              for(int j=0;j<30;j++){                  kx=j/6;                   ky=j%6;                  if(abs(kx-kn)+abs(ky-km)<=1)                      map[i][j]=1;                  else                      map[i][j]=0;              }          }        Guass();        cout<<"PUZZLE #"<
     
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