GCD

[Former Member]

how to write program in abap to find GCD of three numbers

[Former Member]

how to write program in abap to find GCD of three numbers

[Former Member]

Looks as if I will have to ask.

What is GCD.

[Former Member]

martin its

greatest common divisor

[Former Member]

Hi,

What's the reason for asking this question: are you going to teach for some school children?

Try your self to find the GREATEST COMMON DIVISIOR of three numbers.

Take 3 varaibles: let say a1, a2. a3.

Do a1 times.

You start dividing continuously all these three numbers .

when ever you get the last number which devides these three numbers, that is the GCD.

endddo.

reward if useful

Regards,

Anji

[Azeemquadri]

*A crude way of doing GCD. Can use paramters on selection screen if reqd

data: n1 type i, " First Number

n2 type i, " Second Number

n3 type i, " Third Number

n4 type i, " Find reminder for first number

n5 type i, " Find reminder for Second Number

n6 type i, " Find reminder for third number

n7 type i, " Stores GCD

n type i. " Counter

n = 2.

n1 = 30.

n2 = 15.

n3 = 45.

do.

n4 = n1 mod n." Divide by counter

n5 = n2 mod n.

n6 = n3 mod n.

*Stop if the counter is greater than any given number.

if sy-index gt n1 or sy-index gt n2 or sy-index gt n3.

exit.

endif.

if n4 = 0 and n5 = 0 and n6 = 0.

n7 = n.

endif.

n = n + 1.

enddo.

write n7.

[Former Member]

Hi ,

Am not going to teach school children. Am myself learning ABAP so i want to practice some logics in ABAP so i tried for that but i cant able to get so only i asked.

[Former Member]

hi, thanks

[Former Member]

Hi ,

Execute the code .

REPORT zex13 .


PARAMETERS : a TYPE i ,
             b TYPE i ,
             c TYPE i .

DATA : d TYPE i,
       low TYPE i.
DATA : r1 TYPE i ,
       r2 TYPE i ,
       r3 TYPE i .



* find the least of the three numbers .

IF a < b.
  low = a.
ELSE .
  low = b.
ENDIF.

IF c < low .
  low = c.
ELSE .
  low = low.
ENDIF.


*find the gcd.

WHILE ( low >= 1 )   .

  r1 =  a MOD low.

  r2 =  b MOD low.

  r3 =  c MOD low.

  IF r1 = 0 AND r2 = 0 AND r3 = 0.
    WRITE :/ 'Gcd of', a, b , c , 'is' , low.
    EXIT.
   
  ENDIF.
  low = low - 1.

ENDWHILE.

Steps .

For ur query 


1. First find the least value of the given 3 numbers .

2. Divide the 3 numbers with this  low  value .

3.See the GCD cannot be greater than the low value at most it can be  equal to or less than the low value .    
       
4 . GCD of the 3 numbers is when u have a zero reminder for all the numbers .

      We will search this condition keeping in loop by approaching the Low value in the reverse till we get the condition true i.e R1 = 0 ,R2 = 0, R3 = 0.

5. Once a hit then exit out of the loop .
    that is ur gcd .

6. When this condition also is failing then GCD can be only 1 for the exception .


I hope this will solve ur query .

Still you can enhance the logic to the above code.

Revert back for further queries .

Regards,

Vijay.

:):)

[Former Member]

Hi vijay,

Thanks its working fine.