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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>>>>>
>>>>>
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