Clockwise filling of two-dimensional arrays (implemented in C)

As shown in: For any given n, print the following data clockwise:

3

1 2 3

6 4

5

----------------------------------

4

1 2 3 4

9 10 5

8 6

7

----------------------------------

5

1 2 3 4 5

12 13 14 6

11 15 7

10 8

9

...

...

...

For such data filling, the filling principle is similar to the clockwise filling problem of a full two-dimensional array. Next we will analyze the data filling problem of such an array.

Based on the clockwise filling of the question, we can divide the problem into clockwise solving:

<1> think of this array as an equi triangle and separate it from the outer self. For example, when n = 5,

1 2 3 4 5 and

12 6 13 14

11 7 15

10 8

9

These two equi triangles are used for analysis.

<2> Fill each equi-edge triangle in clockwise direction.

1, the elements of the first line are a [k] [J] = A [k] [J-1] + 1;

2, the elements on the diagonal of the pair are a [I] [S + 2-k-i] = A [I-1] [S + 3-k-i] + 1;

3. Fill in a [I-1] [k] = A [I] [k] + 1 from the last line in sequence for the elements in the first column;

The most complicated issue is the filling of sub-diagonal elements. In fact, it is not difficult to find that the subscript of the element on the diagonal is S + 2-k;

(S is the number of elements in array a [] [], n is the number of elements in the first line of the K equi triangle, and K is the K equi triangle currently being filled)

 

I will post my own program for you to share. If anything is inappropriate, you are welcome to criticize and advise.

 

# Include <stdio. h> <br/> # define N 21 <br/> int s; <br/> int a [n] [N]; </P> <p> void fun (int n, int K) <br/> {<br/> int I, j; <br/> If (k> 1) <br/> A [k] [k] = A [k] [k-1] + 1; <br/> else <br/> A [k] [k] = 1; <br/> // fill in rows <br/> for (j = k + 1; j <= N + k-1; j + +) <br/> A [k] [J] = A [k] [J-1] + 1; <br/> // sub-diagonal padding <br/> for (I = k + 1; I <= N + k-1; I ++) <br/> A [I] [S + 2-k-i] = A [I-1] [S + 3-k-i] + 1; <br/> // column filling <br/> for (I = N + k-1; I> K + 1; I --) <br/> A [I-1] [k] = A [I] [k] + 1; </P> <p >}< br/> // output data <br/> void print (int n) <br/>{< br/> int I, J; <br/> for (I = N; I> 0; I --) <br/> {<br/> for (j = 1; j <= I; j ++) <br/> printf ("% 4D", a [n-I + 1] [J]); <br/> printf ("/N "); <br/>}< br/> void main () <br/>{< br/> int X, I; <br/> I = 1; <br/> scanf ("% d", & S); <br/> X = s; <br/> DO <br/> {<br/> fun (X, I); <br/> X = X-3; <br/> I ++; <br/>}while (x> 0); <br/> Print (s); <br/>}</P> <p>