how to get this output

[Loveleen Saroya]

  1
        121
        12321 
        1234321
        123454321
 

can anybody help me how to get this output
thanks in advance

[John Dell'Oso]

Sounds like a homework question. We would like to help, but you have to demonstrate that you have made an attempt at a solution yourself. Do you have any code so far that you can show us?

Regards,
JD

[Svend Rost]

Ok..

Let's look at the output.

    
        1
        121
        12321 
        1234321
        123454321

The output tells us, that:

1) You need a loop.
2) Something is being added

Does this help?

/Svend Rost

Edit: Was to "friendly"...
[ February 22, 2006: Message edited by: Svend Rost ]

[Stan James]

Are you familiar with loop syntax? This "for" syntax is very common. If you don't know this bit, look up "for" in a language reference.

for(int i = 0; i < 10; i++ )
{
   System.out.println(i);
}

Try this and see what it prints.

Computer geeks often count from zero like this, because in assembly language we often count the distance from the start of a list. The first entry is zero steps away from the begining.

See if you can change that loop to print 1 through 5 instead of 0 through 9. Show us what you make! Sharing is the fun part. Then think about how that's going to help with your assignment.
[ February 22, 2006: Message edited by: Stan James ]

[Loveleen Saroya]

i got the output but was looking if there is any other way to do it.. like making a part string and then reverse and concat.

  
class PrintTest { 
public static void main(String[] args) { 
int i;
for(int j=1;j<10;j++) 
{
for(i=1;i>0;i++)
{ 
System.out.print(i); 
System.out.print(" "); 
if(i==j) { break; }
} 
i--;
while(i>0) 
{ System.out.print(i--); 
                  System.out.print(" "); 
                  if(i==0) { break; } 
                 } 
                             System.out.println();
} } } 
  

[Joni Salonen]

First note that 11*11 = 121, 111*111=12321, etcetera.

Then that 10^n - 1 is a number that consists of n 9's, so (10^n - 1)/9 consists of n 1's.

So therefore we get:

int powerOfTen = 1;
for (int len = 0; len < 5; len++)
{
    powerOfTen = powerOfTen*10;
    int ones = (powerOfTen-1)/9;
    System.out.println(ones*ones);
}

[Loveleen Saroya]

Thanks joni

[Loveleen Saroya]

Thanks Joni

[Christophe Verré]

First note that 11*11 = 121, 111*111=12321, etcetera.

That's clever
But that may not be the purpose of the assignment.
The person asking for this kind of output probably expects his students to use loops, not one liner mathematical trics (That may come as a shock if he did not realize this himself)

[Stan James]

Heh heh, I'd bet the instructor is looking for loops and turn in both solutions for sure. Was it Bernoulli who got in trouble as a child ... A teacher told the class to add all the numbers from 1 to 100 just to kill time, and one bright student realized it's 1+100 + 2+99 ... or 101*50 pairs. Blew the afternoon nap.

Oh, you were way ahead of me on loops. I might try to build both ends of the string at the same time in a single inner loop:

for( some loop control or other )
{
   left = left + i;
   right = i + right;
}
println left + i + right

[ February 23, 2006: Message edited by: Stan James ]

[Jan Groth]

it was Leibnitz who found the sum-formula... :-)