鉴于以下电网的数字（从1到36），我怎么能在网格内确定其坐标（X，Y）？

  | 0 1 2 3 4 5 6 7 8
------------------------------
0 | 1 2 3 4 5 6 7 8 9
1 | 10 11 12 13 14 15 16 17 18
2 | 19 20 21 22 23 24 25 26 27
3 | 28 29 30 31 32 33 34 35 36
 

即。我想获得如下：

  | 0 1 2 8
------------------------------
0 | （0,0）（1,0）（2,0）...（8,0）
1 | （0,1）（1,1）（2,1）...（8,1）
2 | （0,2）（1,2）（2,2）...（8,2）
3 | （0,3）（1,3）（2,3）...（8,3）
 

我曾尝试：

  X =编号％9-1;
Y =号/ 9;
 

其中适用于除那些在右边的最后一列的所有情况。

于是我想出了：

 如果（编号％9 == 0）{
    X = 8;
    Y =号/ 9-1;
}
其他{
    X =编号％9-1;
    Y =号/ 9;
}
 

我的问题是，是否有这样做的更聪明的方式？

  X =（数1）9％;
Y =（数1）/ 9;
 

Given the numbers in the following grid (from 1 to 36), how can I determine their coordinates (x,y) within the grid?

 |  0  1  2  3  4  5  6  7  8
------------------------------
0|  1  2  3  4  5  6  7  8  9
1| 10 11 12 13 14 15 16 17 18
2| 19 20 21 22 23 24 25 26 27
3| 28 29 30 31 32 33 34 35 36

i.e. what I want to obtain is the following:

 |     0     1     2        8
------------------------------
0| (0,0) (1,0) (2,0) ... (8,0)
1| (0,1) (1,1) (2,1) ... (8,1)
2| (0,2) (1,2) (2,2) ... (8,2)
3| (0,3) (1,3) (2,3) ... (8,3)

I have tried:

x = number%9-1;
y = number/9;

which works for all cases except the ones in the last column on the right.

So I came up with:

if (number%9==0) {
    x = 8;
    y = number/9-1;
}
else{
    x = number%9-1;
    y = number/9;
}

My question is, is there a smarter way of doing this?

x = (number-1)%9;
y = (number-1)/9;