Advanced Number Pattern - c

Question

Given a Number N, Print the following pattern.
Input Format
The input contains a number N
Constraints
1 < N < 100
Output Format
The required pattern
for input 5 is
1
2 9
3 8 10
4 7 11 14
5 6 12 13 15
and for input 3 is
1
2 5
3 4 6
this is the code i have tried .. but the results are not the same
#include<stdio.h>
void pattern(int n)
{
for(int i=1; i<=n; i++)
{
int k = i;
for(int j=1; j<=i; j++)
{
printf("%d ",k);
k = n - j + k;
}
printf("\n");
}
}
int main()
{
int n = 5;
pattern(n);
return 0;
}
this is the result of the above code
1
2 6
3 7 10
4 8 11 13
5 9 12 14 15
How should I modify the above code to get the Expected Output?

Answer 1

This is an interesting problem, and not an easy one. I'm not going to write a program to solve it (in part because I'm too lazy), but I can describe how I would solve it.
You already have an outer loop for(int i=1; i<=n; i++) which counts down the rows, and an inner loop for(int j=1; j<=i; j++) which counts across the columns. Those are both fine.
Inside the inner loop I would test if(j % 2 == 1). If j % 2 is 1 we're in an odd-numbered column, and we want to count down the column. But if j % 2 is 0, we're in an even column, and we have to do it the other way.
First I would have a variable which is the number that's supposed to be at the top of the column (1, 9, 10, 14, or 15 in the n=5 case). I'd have to compute that number two different ways, one for the "odd" columns and one for the "even".
And then I'd use that number as a base to count down the odd columns, and up the evens. Specifically: I'd add i to it in the odd columns, but subtract i in the even columns. But actually that's not quite right, because i is not 1 at the top of columns other than 1, so what I'd actually have to add or subtract would be some function of i and j. But I think you can work this out.

Answer 2

Here are my three cents.:)
#include <stdio.h>
int main(void)
{
while ( 1 )
{
const unsigned int UPPER_LIMIT = 100;
printf( "Enter a non-negative number no greater than %u (0 - exit): ",
UPPER_LIMIT );
unsigned int n;
if ( scanf( "%u", &n ) != 1 || n == 0 ) break;
if ( !( n < UPPER_LIMIT ) ) n = UPPER_LIMIT - 1;
putchar( '\n' );
for ( unsigned int i = 0; i < n; i++ )
{
for ( unsigned int j = 0; j < i + 1; j++ )
{
unsigned int value = j % 2 == 0
? i + 1 + j * n - j * ( j + 1 ) / 2
: ( j + 1 ) * n - j * ( j + 1 ) / 2 - i + j;
printf( "%2u ", value );
}
putchar( '\n' );
}
putchar( '\n' );
}
return 0;
}
The program output might look the following way
Enter a non-negative number no greater than 100 (0 - exit): 10
1
2 19
3 18 20
4 17 21 34
5 16 22 33 35
6 15 23 32 36 45
7 14 24 31 37 44 46
8 13 25 30 38 43 47 52
9 12 26 29 39 42 48 51 53
10 11 27 28 40 41 49 50 54 55
Enter a non-negative number no greater than 100 (0 - exit): 9
1
2 17
3 16 18
4 15 19 30
5 14 20 29 31
6 13 21 28 32 39
7 12 22 27 33 38 40
8 11 23 26 34 37 41 44
9 10 24 25 35 36 42 43 45
Enter a non-negative number no greater than 100 (0 - exit): 8
1
2 15
3 14 16
4 13 17 26
5 12 18 25 27
6 11 19 24 28 33
7 10 20 23 29 32 34
8 9 21 22 30 31 35 36
Enter a non-negative number no greater than 100 (0 - exit): 7
1
2 13
3 12 14
4 11 15 22
5 10 16 21 23
6 9 17 20 24 27
7 8 18 19 25 26 28
Enter a non-negative number no greater than 100 (0 - exit): 6
1
2 11
3 10 12
4 9 13 18
5 8 14 17 19
6 7 15 16 20 21
Enter a non-negative number no greater than 100 (0 - exit): 5
1
2 9
3 8 10
4 7 11 14
5 6 12 13 15
Enter a non-negative number no greater than 100 (0 - exit): 4
1
2 7
3 6 8
4 5 9 10
Enter a non-negative number no greater than 100 (0 - exit): 3
1
2 5
3 4 6
Enter a non-negative number no greater than 100 (0 - exit): 2
1
2 3
Enter a non-negative number no greater than 100 (0 - exit): 1
1
Enter a non-negative number no greater than 100 (0 - exit): 0
Or using the recursive approach of calculating the output value the program can look the following way.
#include <stdio.h>
int main(void)
{
while ( 1 )
{
const unsigned int UPPER_LIMIT = 100;
printf( "Enter a non-negative number no greater than %u (0 - exit): ",
UPPER_LIMIT );
unsigned int n;
if ( scanf( "%u", &n ) != 1 || n == 0 ) break;
if ( !( n < UPPER_LIMIT ) ) n = UPPER_LIMIT - 1;
putchar( '\n' );
for ( unsigned int i = 0; i < n; i++ )
{
unsigned int value = i + 1;
for ( unsigned int j = 0; j < i + 1; j++ )
{
printf( "%2u ", value );
value += j % 2 == 0 ? 2 * ( n - i ) - 1 : 2 * ( i - j );
}
putchar( '\n' );
}
putchar( '\n' );
}
return 0;
}
For example its output for the entered number equal to 10 looks like
Enter a non-negative number no greater than 100 (0 - exit): 10
1
2 19
3 18 20
4 17 21 34
5 16 22 33 35
6 15 23 32 36 45
7 14 24 31 37 44 46
8 13 25 30 38 43 47 52
9 12 26 29 39 42 48 51 53
10 11 27 28 40 41 49 50 54 55
Enter a non-negative number no greater than 100 (0 - exit): 0