On Mon, Nov 23, 2009 at 3:57 PM, Saurabh Aggarwal <saurabh...@gmail.com>wrote:
> Could you explain it with an example ?
>
> I hope you are considering that n^n may be well beyond any datatype range.
Yes I am considering n to be well beyond any datatype range.
>
>
>
> On Mon, Nov 23, 2009 at 3:24 PM, Bharath <bharath1...@gmail.com> wrote:
>
>> For first k digits:
>>
>> (unsigned long)floor(pow(10.0, modf(n*log10((double)n), &dummy) + k - 1))
>>
>>
>> On Mon, Nov 23, 2009 at 3:24 PM, Bharath <bharath1...@gmail.com> wrote:
>>
>>> For last k digits
>>>
>>> int foo(int n, int k)
>>> {
>>> int m=1;
>>> for(; k > 0; k--) m*=10;
>>>
>>> int r=1, t=n % m;
>>> while(n)
>>> {
>>> if (n % 2)
>>> r = r * t % m;
>>> t = t * t % m;
>>> n >>= 1;
>>> }
>>>
>>> return r;
>>> }
>>>
>>> On Sat, Nov 21, 2009 at 9:44 AM, Siddharth Prakash Singh <
>>> sps...@gmail.com> wrote:
>>>
>>>> Can anybody suggest an algorithm to find first and last k digits of
>>>> n^n.
>>>>
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>>>>
>>>
>>>
>>> --
>>> <<Bharath>>
>>>
>>>
>>
>>
>> --
>> <<Bharath>>
>>
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>
>
>
> --
> Regards,
> Saurabh Aggarwal
>
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--
Siddharth Prakash Singh
http://www.spsneo.com
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