棋盘填充

/*问题描述*/

/*在一个n*n的棋盘中，有一个地方是 不同的，需要填充 用'L'型的方块填充。

*/

/*用递归的方式，分成四个部分，根据该块的位置，选择一个点填充，另外的也填充。。

*/

int a[4][4];

void chess(int line,int cols,int p_line,int p_cols,int size){

printf("begin\n");

if(size==1)

return ;

size /=2;

printf("size=%d\n",size);

/*divide it into 4,judge them if the particular one in it,then take action*/

if(p_line

/* it means that the element in this unit ,and at the left top of the big unit */

chess(line,cols,p_line,p_cols,size);

}

else{

/*not in this unit ,so we should value it's right below number */

a[line+size-1][cols+size-1]=1;

printf("a[%d][%d]=1",line+size-1,cols+size-1);

chess(line,cols,line+size-1,cols+size-1,size);

}

/*since it have four directorys,so the process will repeated*/

/*right top*/

if(p_line =cols+size){

chess(line,cols+size,p_line,p_cols,size);

}

else{

printf("a[%d][%d]=1",line+size-1,cols+size);

a[line+size-1][cols+size]=1;

chess(line,cols+size,line+size-1,cols+size,size);

}

/*left below */

if(p_line >=line+size && p_cols

chess(line+size,cols,p_line,p_cols,size);

}

else{

printf("a[%d][%d]=1",line+size,cols+size-1);

a[line+size][cols+size-1]=1;

chess(line+size,cols,line+size,cols+size-1,size);

}

/*right below*/

if(p_line >= line+size && p_cols >= cols+size){

chess(line+size,cols+size,p_line,p_cols,size);

}

else{

printf("a[%d][%d]=1",line+size,cols+size);

a[line+size][cols+size]=1;

chess(line+size,cols+size,line+size,cols+size,size);

}

}

int main(){

int i=0,j=0;

for(;i<4;++i)

for(j=0;j<4;++j)

a[i][j]=0;

a[1][1]=1;

for(i=0;i<4;++i){

for(j=0;j<4;++j)

printf("%d",a[i][j]);

printf("\n");

}

printf("\n\n");

chess(0,0,1,1,4);

for(i=0;i<4;++i){

for(j=0;j<4;++j)

printf("%d",a[i][j]);

printf("\n");

}

exit(0);

}

/*analysize */

/*according this program ,we should have found that 

 * this skill have advantages in the solution of such problem

 * put a big problem into many little steps

 * follow the steps above we will reach the practicularly element ,and draw the element arount it*/

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