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Computer Science

Cool goIITian

Joined:	22 Mar 2007
Post:	63

12 Apr 2009 22:42:26 IST

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340

a small C++ program

None

hey can someone just tell me the logic for this

to print the following pattern

*****1#####1

****2####2

***3###3

**4##4

*5#5

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Comments (3)

RyuAmakusa

Blazing goIITian

Joined: 21 Mar 2008

Posts: 776

14 Apr 2009 13:32:02 IST

Like 1 people liked this

see the pattern ....
INV:1<=i<=5 and u have printed
****1#####1
****2####2
****(6-i)times,i,####....(6-i)times,i
loop
{
}
assert: i = 6 and u have printed
*****1#####1
****2####2
***3###3
**4##4
*5#5
INV ...it is called the ivariant it is a condition which is always satisfied at that point at every iteration
assert:......it is what u can assertain that would have happened whaen the loop end and the control is transfered to that point
and the INV is basicaly the weakening of assert: ......did u understand it........it is not any language in comp sci but it is
the language of the algorithm...........the C++ or java or python...what ever.......are just a matter of syntax.......
what u really have to learn is this, once u have done this.....then u can say that u have learnt programming

Swapnika.... finally into a branded college....NIT m cuming..!!!

Blazing goIITian

Joined: 7 Aug 2008

Posts: 1047

14 Apr 2009 13:35:45 IST

Like 0 people liked this

u can take the positions as 1st row 5th column,,,,n use the for loop with cout <<(the symbol u want to display)...u have to use a nested if else loop also...this program is nt easy....

Vikram Saxena

Forum Expert

Joined: 1 Apr 2007

Posts: 3732

24 Apr 2009 08:36:01 IST

Like 0 people liked this

for(i=1;i<=n;i++)
{
for(j=n;j>=i;j--)
printf("*");
printf("%d",i);
for(j=n;j>=i;j--)
printf("#");
}

-all the best for your future

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