Answer / j.n.abhishek

void main()
{
int n,i,j;
clrscr();
printf("ENTER THE LIMIT:");
scanf("%d",&n);
for(i=0;i<=(2*n);i++)
{
if(i!=n)
{
for(j=0;j<(2*n);j++)
{
printf(" ");
}
printf("+\n");
}
else
{
for(j=0;j<=(2*n);j++)
{
printf("+ ");
}
printf("\n");
}
}
getch();
}

Answer / senthil

more generic answer considering for any value n (odd or even)
void printPlus(int n)
{
int i, j;

for(i=0; i<=n; i++)
{
if(i == n/2)
{
for(j=0; j<n; j++)
{
printf("* ");
}
printf("\n");


// to ignore last step of drawing for odd cnt
if(n%2 != 0)
{
i++;
}
}
else
{
for(j=0; j<n/2; j++)
{
printf(" ");
}

if(n%2 == 0)
{
// even
printf("\b*\n");
}
else
{
// odd
printf("*\n");
}
}
}
}

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