thanks anurag :)

On 12 June 2010 20:07, Anurag Sharma <anuragvic...@gmail.com> wrote:

> Since we are given numerator 'n' and denominator 'd' separately already.
> and considering n and d as integers and d!=0 we can safely assume n/d as
> either a terminating fraction or a non terminating but recurring fraction,
> in which case we have to find the recurring digits of the fraction.
>
> Now what I suggested was almost same as Ravi's approach.
> take a Set 'S' keeping tuples (R,Q) where R is the current remainder and Q
> is the factor such that d*Q is subtracted from the number to get R.
> In other words. if at an intermediate step of division we have 'a' as the
> divident left then Q=floor(a/d) and R=a%d
>
> Keep dividing 'n' by 'd' like it is done manually. After every division
> check-
> 1. If the current remainder is not present in 'S' then add current
> remainder 'R' and corresponding quotient 'Q' in the set
> 2. If R is found in the set S, then all the following entries in the set
> until end will constitute the recurring digits.
> taking Ravi's example:-
>
> Example:
>   7) 9 (1.*285714*28                            S=[]
>        7
>        --
>         20                                           S=[(2,2)]
>          14
>          ---
>            60                                        S=[(2,2), (6,8)]
>             56
>              ---
>               40                                     S=[(2,2), (6,8),
> (4,5)]
>               35
>               ---
>                  50                                  S=[(2,2), (6,8),
> (4,5), (5,7)]
>                   49
>                    ---
>                      10                              S=[(2,2), (6,8),
> (4,5), (5,7), (1,1)]
>                         7
>                      ----
>                          30                          S=[(2,2), (6,8),
> (4,5), (5,7), (1,1), (3,4)]
>                           28                               ^
>                           ----                               |
>                               20                     2 is found in S here,
> so recurring digits are "285714"
>                                14
>
>   ----
>                                    60
>                                     56
>                                          repeats
>
>
> hope its clear
>
>
> Anurag Sharma
>
>
> On Sat, Jun 12, 2010 at 4:02 PM, divya jain <sweetdivya....@gmail.com>wrote:
>
>> @anurag
>>
>> i dint get ur approach..which numerator n denominator u r talking
>> about..plz explain.. thanks in advance
>>
>> On 11 June 2010 08:57, Anurag Sharma <anuragvic...@gmail.com> wrote:
>>
>>> Please note that the fractional repeating part is recurring. and so that
>>> 4th temporary variable assignment will be this way->
>>> temp=x*10000 - x= 233456.34563456...  - 23.34563456.... = 233433.0  (
>>> mark the fractional part is 0 now since the infinitely repeating 3456...
>>> will get cancelled)
>>> In this  case you can say that 4 places are repeating. But yes its
>>> according to the maths and in any programming language whenever you divide
>>> the numerator and denominator you wont get this infinitely recurring decimal
>>> places.
>>>
>>> @divya, also your approach wont work if the recurring fractional digits
>>> start after few places from the decimal like in the case of
>>> 23.123345634563456....  (note here after the decimal place 123 is not
>>> repeating while 3456.. after this 123 is repeating.)
>>>
>>> What I suggest in this case is keep dividing the numerator by denominator
>>> and at every step keep inserting the tupple (remainder, quotient) of that
>>> division step in a set. and before inserting in the set check whether it
>>> already exists. If yes then the all the quotients following from that point
>>> (including the point) will be recurring.
>>>
>>> Regards,
>>>
>>> Anurag Sharma
>>>
>>>
>>>
>>> On Thu, Jun 10, 2010 at 8:25 AM, Veer Sharma 
>>> <thisisv...@rediffmail.com>wrote:
>>>
>>>> Seems it wont work...
>>>> x=23.34563456
>>>>
>>>> temp = x*100 -x = 233.4563456 - 23.34563456 = 210.11071104
>>>> temp = x*100 -x = 2334.563456 - 23.34563456 = 2311.21782144
>>>> temp = x*1000 -x =  23345.63456 - 23.34563456 = 23322.28892544
>>>> temp = x*10000 -x =  233456.3456 - 23.34563456 = 233432.99996544
>>>> temp = x*100000 -x = 2334563.456 - 23.34563456 = 2334540.11036544
>>>>
>>>> ...
>>>>
>>>> On Jun 9, 11:24 pm, Anurag Sharma <anuragvic...@gmail.com> wrote:
>>>> > multiply the original number x=23.34563456
>>>> >
>>>> > Anurag Sharma
>>>> >
>>>> > On Wed, Jun 9, 2010 at 10:36 PM, Veer Sharma <
>>>> thisisv...@rediffmail.com>wrote:
>>>> >
>>>> >
>>>> >
>>>> > > One question:
>>>> >
>>>> > > No x = 23.34563456
>>>> > > temp = x X 10 = 233.4563456
>>>> > > temp = temp - x = 210.11071104
>>>> > > decimal part zero? No.
>>>> > > Now multiply the no. with 100. Which no? original x (= 23.34563456)
>>>> or
>>>> > > new no. temp (=210.11071104)?
>>>> >
>>>> > > On Jun 9, 8:12 pm, divya jain <sweetdivya....@gmail.com> wrote:
>>>> > > > multiply the no. with 10 nd store in temp. now subtract no from
>>>> temp.
>>>> > > check
>>>> > > > if the decimal part is zero if yes.  then 1st digit after decimal
>>>> is
>>>> > > > recurring. if no. multiply the no with 100 and repeat . if this
>>>> time
>>>> > > decimal
>>>> > > > part is zero then 2 digits after decimal r recurring nd so on..
>>>> >
>>>> > > > On 8 June 2010 21:45, Veer Sharma <thisisv...@rediffmail.com>
>>>> wrote:
>>>> >
>>>> > > > > You have a Numerator and Denominator. After division you might
>>>> get a
>>>> > > > > recurring decimal points float as the answer.
>>>> >
>>>> > > > > Problem is: You need to identify the recurring part for a given
>>>> > > > > decimal no?
>>>> > > > > For example 23.34563456 ...
>>>> > > > > return 3456
>>>> >
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