Hi Sarvesh,
For this problem this is my solution.
A=[1,5,6]
n=len(A)
Ans=[0,1,1,2,1,2,2,3]
count=0
for i in range(0,n):
count=count+Ans[A[i]]
print count
Is there any flaw?
Thanks & Regards,
Anantha Krishnan
On Sun, Aug 21, 2011 at 4:25 PM, sarvesh saran
<[email protected]>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 <[email protected]> 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 <[email protected]> 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 <
>>> [email protected]> 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 <
>>>> [email protected]> 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 <
>>>>> [email protected]> 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 <
>>>>>> [email protected]> 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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