How to solve this peculiar nested for loops exercise?

Question

I've been studying C programming for a while and I've stumbled on this exercise I can't seem to solve:

• Write nests of loops that cause the following output to be displayed:

  0
  01
  012
  0123
  01234
  012345
  01234
  0123
  012
  01
  0
  

Until now I've managed to get to the sixth row, but I can't finish the exercise. This is the code I've written to obtain half of the solution to this problem:

#include <stdio.h>

int
main(void)
{

int i, j;

for(i = 1; i <= 6; ++i) {
for(j = 0; j < i; ++j)

    printf("%d", j);
    printf("\n");

 }

return(0);

}

The answer had some reasearch effort, was clear; maybe useful for someone studying the same subject, so it was downvoted for no reason.

Answer 1

You can actually do this with a single nested loop:

#include <stdio.h>

int getLength(int i) {
    /* Since this is homework, I'll leave this for you to complete. */
    if (i < ?) return ?;
    else return ?;
}

int main(void) {
    for(int i = 0; i < 11; ++i) {
        int length = getLength(i);
        for(int j = 0; j < length; ++j) {
             printf("%d", j);
        }
        printf("\n");
    }

    return 0;
}

Answer 2

There's a clue in the question: "Write nests of loops". Count up, then count down.

Answer 3

Since this is an exercise I will only hint at a solution, as to learn you really should do this yourself.

1. for loops can go in reverse.
2. You can have more than one nested loop.

Answer 4

for(int i=0;i<=10;i++)
{
    if(i<=5)
    {
        for(int k=0;k<=i;k++)

        printf("%d",k);
    }
    else if(i>5)
    {
        for(int j=0;j<=(10-i);j++)
            printf("%d",j);
    }
    printf("\n");
}

i hope that will help

this is the defualt answer

Answer 5

I thought user should tell till which number to print. so my code is a bit complex.

include

include

int main (void) {

   int rows;
   int counter;
   int backward;
   int forward;

printf("Till which number you want to print??");
scanf("%d",&rows);
for(counter=0;counter<=rows;counter++)
    {for(forward=0;forward<=counter;forward++)
        printf("%d",forward);
            printf("\n");
        }

for(counter=rows;counter>=0;counter--)
            {for(backward=0;backward<=counter;backward++)
        printf("%d",backward);
                printf("\n");    
        }

getch();
return 0;

}

Thank u !