Could you explain it with an example ?

I hope you are considering that n^n may be well beyond any datatype range.


On Mon, Nov 23, 2009 at 3:24 PM, Bharath <[email protected]> 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 <[email protected]> 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 <
>> [email protected]> 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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