thanx mukul for explanation.... checked it randomly on 123.. it works
fine...
On Sat, Aug 6, 2011 at 11:27 PM, Ashish Sachdeva <ashish.asachd...@gmail.com
> wrote:
> mukul, pls explain my following doubts:
> 1. whats the need of subtracting the numbers, like u subtracted 91 then 52
> etc...
> 2. the ones digit 7 is ok... didnt get how did u reach 4??
>
>
> On Sat, Aug 6, 2011 at 3:22 PM, Mukul Gupta <mukul.gupta...@gmail.com>wrote:
>
>> Vaibhav,
>> Hmm, fine that. I am more interested in the 2nd approach. The first
>> approach is simple brute force.
>>
>> On Sat, Aug 6, 2011 at 3:11 PM, <vaibhavmitta...@gmail.com> wrote:
>>
>>> Mukul, in first approach instead of sending the string again and again u
>>> can use the formula
>>> (a*b)%m = ((a%m)*(b%m))%m
>>> this way u can do sumthin like dis
>>>
>>> int count = 0, a = 1;
>>> while(a != 0) {
>>> count++;
>>> a = ((a*10)%n + 1) %n;
>>> }
>>>
>>> n later output a string consisting of count one's..
>>>
>>> Regards
>>> VM
>>>
>>> 3rd Year, Computer Engineering,
>>> Netaji Subhas Institute of Technology.
>>>
>>> On , Mukul Gupta <mukul.gupta...@gmail.com> wrote:
>>> > Manee, Nice Question.
>>> > I have thought of two algorithms. I wanted to know how one judges them.
>>> Both have similar time complexity but the 2nd one is slightly complex and
>>> much more logical.
>>> >
>>> > 1. Keeping on adding 1 as a string of 1's and apply it to this modulo
>>> function to check when it becomes 0.
>>> >
>>> >
>>> > long long modulo(char b[],long long a)
>>> > {long long d=0,len,i,j,k;
>>> > len=strlen(b);
>>> > for (k=0;k {d*=10;
>>>
>>> > d+=b[k]-48;
>>> > d=d%a;
>>> > }
>>> >
>>> > return d;
>>> >
>>> > }
>>> >
>>> >
>>> > 2. Any number ending in 3 will have the last digit as 1 if it is
>>> multiplied by 7.
>>> > Consider a case 13 ...let the required answer have 11.....111. as
>>> its representation.....13 x 7 = 91.....
>>> > So subtracting the 3 digit of of 111..1111 by 91...we get
>>> 111...11020....Now we know that the ones digit of the required number is
>>> 7...
>>> >
>>> > Similarly, if the last digit of a ten's digit has to be '2'...The
>>> number has to be multiplied by 4.....So we subtract 13 x 4 = 52 from.....
>>> > 11111.111102 to get 11...050...So we get the ten's digit as 4....
>>> >
>>> > Similarly, now for a number to end in 5...it has to be multiplied
>>> by 5....we subtract...65 from 111...105....to get 111..1040...
>>> > Hundred's digit is 5
>>> > Similarly, now for a number to end in 4...it has to be multiplied
>>> by 8 ... we subtract 104 from 111...104....to get 111...000. and thus we end
>>> the process as we have got the remainder as 0.
>>> >
>>> > Thus, our required answer is 13 x 8547 = 111111
>>> >
>>> > Now I want to know...that both the methods have similar complexity ie.
>>> O(k) where k is the number of 1's. However, 2nd is much more logical and
>>> complex. What does the company look for?
>>> >
>>> > Suggest some better methods or make ammends.
>>> >
>>> > Regards,
>>> >
>>> > Mukul Gupta
>>> > 3rd Year, Computer Engineering,
>>> > Netaji Subhas Institute of Technology.
>>> >
>>> >
>>> >
>>> >
>>> >
>>> >
>>> >
>>> >
>>> >
>>> > On Sat, Aug 6, 2011 at 9:51 AM, sahil gujral gujralsa...@gmail.com>
>>> wrote:
>>> >
>>> > yes u r wrong..111111111 is nt divisible by 23
>>> >
>>> >
>>> > On Sat, Aug 6, 2011 at 9:15 AM, sumit sumitispar...@gmail.com> wrote:
>>> >
>>> >
>>> > This looks quite simple.
>>> >
>>> > Every number ending in 3 follows a pattern.eg-
>>> >
>>> > 3 - 111
>>> >
>>> > 13 - 111111
>>> >
>>> > 23 - 111111111 etc
>>> >
>>> > we can find the reauired no. by :
>>> >
>>> > suppose input no. is 33
>>> >
>>> > In every case leave the no at 1's place(least significant) i.e. 3, In
>>> >
>>> > 33 you will be left with 3(after removal of 3 at first place).
>>> >
>>> > Now ,3 *(rest of nos +1 ) is your answer (in case of 33 it is 3*(3+1)
>>> >
>>> > = 12 i.e 111111111111).
>>> >
>>> > for 103 it is 3*(10+1) = 33 1's.
>>> >
>>> >
>>> >
>>> > Correct if I am wrong.
>>> >
>>> >
>>> >
>>> >
>>> >
>>> >
>>> > On Aug 5, 4:33 pm, Manee mani.ma...@gmail.com> wrote:
>>> >
>>> > > ADOBE asks the very basic C/C++ questions
>>> >
>>> > >
>>> >
>>> > > one of their toughest however was :
>>> >
>>> > >
>>> >
>>> > > every number ending in 3 has a multiple of the form "111...111"
>>> >
>>> > >
>>> >
>>> > > e.g 3 has 111
>>> >
>>> > > 13 has 111111
>>> >
>>> > > so on..
>>> >
>>> > >
>>> >
>>> > > find the algo for finding the number for an input number ending in 3.
>>> >
>>> > >
>>> >
>>> > > On Aug 5, 2:33 pm, Agyat jalsa.n.sa...@gmail.com> wrote:
>>> >
>>> > >
>>> >
>>> > >
>>> >
>>> > >
>>> >
>>> > >
>>> >
>>> > >
>>> >
>>> > >
>>> >
>>> > >
>>> >
>>> > > > hey, guys adobe is visiting our campus. So those who know questions
>>> >
>>> > > > that adobe asked in written or interview, please post here as it
>>> will
>>> >
>>> > > > be of great help (as adobe has visited some colleges already).
>>> >
>>> > > > Thank you in advance.
>>> >
>>> >
>>> >
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