ok. got it Nitin. thanx :)
Sanju
:)
On Sun, Aug 21, 2011 at 10:06 AM, Nitin Nizhawan
<nitin.nizha...@gmail.com>wrote:
> I guess, earlier sarvan simply wanted to calculate
> result = SUM {0<=i < N } A[i]*(8^i)
>
> in which
> 8^i = 2^(3*i) which is equivalent to right shifting 3*i
>
> since each A[i] is octal it is just 3bits long , we need not add we can
> simply shift and do OR.
>
>
> On Sun, Aug 21, 2011 at 10:16 PM, Sanjay Rajpal <srn...@gmail.com> wrote:
>
>> @Nitin : could u explain ur logic ?
>>
>>
>> Sanju
>> :)
>>
>>
>>
>> On Sun, Aug 21, 2011 at 9:24 AM, Nitin Nizhawan <nitin.nizha...@gmail.com
>> > wrote:
>>
>>> @sanjay, oops, my intention was bitwise OR
>>>
>>>
>>> On Sun, Aug 21, 2011 at 4:25 PM, sarvesh saran <
>>> aquarian.thun...@gmail.com> wrote:
>>>
>>>> Hi Prakash,
>>>>
>>>> I'll paste the exact description of the problem:
>>>>
>>>> A non-empty array A of N elements contains octal representation of a
>>>> non-negative integer K, i.e. each element of A belongs to the interval [0;
>>>> 7] (both ends included).
>>>>
>>>> Write a function that returns the number of bits set to 1 in the binary
>>>> representation of K.
>>>>
>>>> thanks,
>>>>
>>>> Sarvesh
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>> i.e take any decimal number, convert to base 8 and then store each digit
>>>> of base 8 representation in an array.
>>>>
>>>> So the question is, given such an array get back the original number.
>>>>
>>>> thanks,
>>>> Sarvesh
>>>>
>>>>
>>>> On Sun, Aug 21, 2011 at 4:13 PM, Prakash D <cegprak...@gmail.com>wrote:
>>>>
>>>>> A[i]<<3*i
>>>>>
>>>>> why is it needed to convert from base 8 to base 10??
>>>>>
>>>>> On Sun, Aug 21, 2011 at 4:07 PM, Sanjay Rajpal <srn...@gmail.com>wrote:
>>>>>
>>>>>> Hi your intention was logical OR or BITWISE OR ?
>>>>>>
>>>>>> u did Logical.
>>>>>> Sanju
>>>>>> :)
>>>>>>
>>>>>>
>>>>>>
>>>>>> On Sun, Aug 21, 2011 at 3:30 AM, sarvesh saran <
>>>>>> aquarian.thun...@gmail.com> wrote:
>>>>>>
>>>>>>> Hi Nitin,
>>>>>>>
>>>>>>> thanks that makes sense. I will try that out.
>>>>>>>
>>>>>>> I have another question. Is there a really fast way of converting a
>>>>>>> hexadecimal string say "02F9A" to its decimal representation in C++?
>>>>>>>
>>>>>>> thanks,
>>>>>>> Sarvesh
>>>>>>>
>>>>>>> thanks,
>>>>>>> Sarvesh
>>>>>>>
>>>>>>>
>>>>>>> On Sun, Aug 21, 2011 at 3:41 PM, Nitin Nizhawan <
>>>>>>> nitin.nizha...@gmail.com> wrote:
>>>>>>>
>>>>>>>> int num = 0;
>>>>>>>> for(int i=0;i<A.size();i++){
>>>>>>>> num=num||(A[i]<<3*i);
>>>>>>>> }
>>>>>>>> printf("%d",num);
>>>>>>>>
>>>>>>>> I think this will do. Given the number is with in the range of
>>>>>>>> integer.
>>>>>>>>
>>>>>>>>
>>>>>>>> On Sun, Aug 21, 2011 at 3:40 PM, Nitin Nizhawan <
>>>>>>>> nitin.nizha...@gmail.com> wrote:
>>>>>>>>
>>>>>>>>> int num = 0;
>>>>>>>>> for(int i=0;i<A.size();i++){
>>>>>>>>> num=num||(A[i]<3*i);
>>>>>>>>> }
>>>>>>>>> printf("%d",num);
>>>>>>>>>
>>>>>>>>> I think this will do.
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> On Sun, Aug 21, 2011 at 2:25 PM, sarvesh saran <
>>>>>>>>> aquarian.thun...@gmail.com> wrote:
>>>>>>>>>
>>>>>>>>>> Hi,
>>>>>>>>>>
>>>>>>>>>> I have a vector<int> A or an array (for C guys) that contains the
>>>>>>>>>> octal representation of a number.
>>>>>>>>>>
>>>>>>>>>> So the array can be something like: [1,5,7] or [7,7,5,6,3,4,2] etc
>>>>>>>>>>
>>>>>>>>>> i.e no number in the array can be >= 8.
>>>>>>>>>>
>>>>>>>>>> Now given this array, I need to convert it its decimal
>>>>>>>>>> representation.
>>>>>>>>>>
>>>>>>>>>> The naive way to do it would be to scan array from left to right,
>>>>>>>>>> take each digit, multiply by 8 pow (x) where x is from 0 to ...n and
>>>>>>>>>> compute
>>>>>>>>>> sum.
>>>>>>>>>>
>>>>>>>>>> i.e something like:
>>>>>>>>>>
>>>>>>>>>> int oct = 1;
>>>>>>>>>> int num = 0;
>>>>>>>>>>
>>>>>>>>>> for(<array length>){
>>>>>>>>>> num+= oct * A[i];
>>>>>>>>>> oct = oct * 8;
>>>>>>>>>> }
>>>>>>>>>>
>>>>>>>>>> is there a faster way to do this? maybe using some STL container or
>>>>>>>>>> algorithm. ?
>>>>>>>>>>
>>>>>>>>>> thanks,
>>>>>>>>>> sarvesh
>>>>>>>>>>
>>>>>>>>>>
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