Question

try to print this in one loop itself.i have already done this in two loops(one nested into another).so please try to do in one loop itself.

1
2   4
3   6   9
4   8   12  16
5   10  15  20  25

Comments

can i use recursion?????

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yes you can.

Answer 1

#include <stdio.h>
int main(int argc, char * argv[])
{
        int i, j=1, k=1, n=atoi(argv[1]);
        for(i = 1; j <= n; k++)
        {
                printf("%d ",i);
                if(k == j)
                {
                        k = 0; i = ++j;
                        printf("\n");
                }
                else i += j;
        }
}

Comments

it says segmentation fault. probably you are trying to access out of your way.

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n=atoi(argv[1])

So, you must pass command line argument for the number of lines to be printed. Else replace this line with a scanf or n = 5.

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yes brother it's working.thanks a lot.

Answer 2

#include<stdio.h>
#include<conio.h>
int main()
{
    int j=1;
for(int i =1; i<=5;i++)
{
    if(i*j<=i*i)
    {
       printf("%d \t",i*j);
        i--;
        j=j+1;
    }
  else
    {
        printf("\n");
        j=1;
    }
}
   getch();
return 0;
}

Comments

Your logic is pretty good, just replace

 if(i*j<=i*i)

with

if( j <= i)

Also, try using a better compiler.

http://gatecse.in/wiki/Why_I_should_not_use_Turbo_C%3F

Answer 3

pattern(1,row*row); pattern(i,n)
{
 if(i*i>n)
    return;
 else
 {
  for(k=i;k<=i*i;k=k+i)
    print("%d",k);
  print("/n");
  pattern(i+1,n);
 }
}

Comments

I can try to print this pattern without any loop

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how @bhagirathi please can you give an example.

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Instead of the for loop ,call a recursive function again

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ya that is one more way to do the same ...