C++ program to print spiral pattern of numbers

Displaying numbers in different formats is one of the problems of learning basic coding.

Different coding concepts like conditional statements and loop statements. have In different programs we use special characters like asterisks to print triangles or square. In this article, we will print numbers in a spiral, just like the squares in C.

We take the number of rows n as input, and then start from the upper left corner Move to the right, then down, then left, then up, then right again, and so on etc.

Spiral pattern and numbers

1   2   3   4   5   6   7
24  25  26  27  28  29  8
23  40  41  42  43  30  9
22  39  48  49  44  31  10
21  38  47  46  45  32  11
20  37  36  35  34  33  12
19  18  17  16  15  14  13

To solve this problem, we will use a two-dimensional matrix of size n x n, in this case, we take n = 7. Then fill the matrix in a spiral starting from the upper left corner. final printout the entire matrix. Here we are printing from 1 to 7 on the first line and then the process is changing it's direction, move toward the bottom to 13, then to the left to 19, and finally Go up until 24, then right, and so on. Let's look at a better algorithm understand.

algorithm

• Enter s as the number of lines
• Create a s x s matrix and initialize them with 0
• Number: = 1
• Initialize i, j, m to 0
• Initialize n := s - 1, p := 0 and q := s - 1
• When num does not exceed s * s, execute
  • For the range of j from p to q, execute
  • mat[ m, j ] := num
  • Number := Number 1
  • Finish
  • meter:=meter 1
  • For i from m to n, execute
  • mat[ i, q ] := num
  • Number := Number 1
  • Finish
  • q := q - 1
  • For the range of j from q to p, decrease it by 1 each time and execute
  • mat[ n, j ] := num
  • Number := Number 1
  • Finish
  • n := n - 1
  • For i from n to m, decrease i by 1 and execute
  • mat[ i, p ] := num
  • Number := Number 1
  • Finish
  • p := p 1
• Finish
• For i from 0 to s-1, perform the following operations
  • For j ranging from 0 to s - 1, execute
    • Display mat[ i, j ]
  • Finish
  • Move the cursor to the next line
• Finish

The Chinese translation of

Example

is:

Example

#include <iostream>
using namespace std;
void solve( int s ){
   int mat[ s ][ s ] = {0};
   int i, j, m, n, p, q, num;
   num = 1; // start count from 1
   i = 0;
   j = 0;
   m = 0; // row index lower limit
   n = s - 1; // row index upper limit
   p = 0; // column index lower limit
   q = s - 1; // column index upper limit
   while ( num <= s * s ) {
   
      // place numbers horizontally left to right
      for ( j = p; j <= q; j++ ) {
         mat[ m ][ j ] = num;
         num = num + 1;
      }
      m = m + 1;
   
      // fill vertically from top to bottom
      for ( i = m; i <= n; i++ ) {
         mat[ i ][ q ] = num;
         num = num + 1;
      }
      q = q - 1;

      // fill horizontally from right to left
      for ( j = q; j >= p; j-- ) {
         mat[ n ][ j ] = num;
         num = num + 1;
      }
      n = n - 1;

      // fill vertically from bottom to top
      for ( i = n; i >= m; i-- ) {
         mat[ i ][ p ] = num;
         num++;
      }
      p = p + 1;
   }

   // display the mat
   for ( i = 0; i < s; i++ ) {
      for ( j = 0; j < s; j++ ) {
         printf("%d\t", mat[i][j]);
      }
      printf("\n");
   }
}
int main(){
   int n = 5;
   cout << "Spiral numbers for " << n << " lines." << endl;
   solve( n );
}

Output

Spiral numbers for 5 lines.
1	2	3	4	5	
16	17	18	19	6	
15	24	25	20	7	
14	23	22	21	8	
13	12	11	10	9

Output results (when n = 12)

Spiral numbers for 12 lines.
1	2	3	4	5	6	7	8	9	10	11	12	
44	45	46	47	48	49	50	51	52	53	54	13	
43	80	81	82	83	84	85	86	87	88	55	14	
42	79	108	109	110	111	112	113	114	89	56	15	
41	78	107	128	129	130	131	132	115	90	57	16	
40	77	106	127	140	141	142	133	116	91	58	17	
39	76	105	126	139	144	143	134	117	92	59	18	
38	75	104	125	138	137	136	135	118	93	60	19	
37	74	103	124	123	122	121	120	119	94	61	20	
36	73	102	101	100	99	98	97	96	95	62	21	
35	72	71	70	69	68	67	66	65	64	63	22	
34	33	32	31	30	29	28	27	26	25	24	23

in conclusion

Displaying number patterns is a fairly common problem when learning programming language. In this article we learned how to display numbers in the square in which the element is located Print in a spiral form in C, starting from the top left corner and moving downwards At the end of column n, we move down, then at the end of row n, we move left, then After reaching the first row, move up until row 2^nd, then repeat this process over and over again until... Complete the entire square. Unlike other numeric pattern problems, it requires a 2D array Solve this problem effectively.

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