thanks to all

On Tue, Aug 9, 2011 at 8:45 PM, shady <[email protected]> wrote:

> he is questioning the complexity and not the algorithm... btw, you are
> right
>
>
> On Tue, Aug 9, 2011 at 8:41 PM, Don <[email protected]> wrote:
>
>> I don't think that this function is doing what you want it to do. If
>> you ask for a^b, it returns a^1 in most cases.
>>
>> Try this instead.
>>
>> int power(int a, int b)
>> {
>>  int result = 1;
>>  if (b == 1) result = a;
>>  else if (b>1)
>>  {
>>    result = power(a,b/2);
>>    result *= result;
>>    if (b%2) result *= a;
>>  }
>>  return result;
>> }
>>
>> On Aug 8, 10:37 pm, rohit <[email protected]> wrote:
>> >  int get_power(int a, int b)
>> > {
>> > if(!b)
>> > return 1;
>> > if(b%2)
>> >  return a * get_power(a, b/2);
>> >  return get_power(a, b/2);
>> >  }
>> >
>> > int func(int p)
>> > { int sum = 0;
>> >  for(int i = 1; i <= p; ++i)
>> > {
>> > sum += get_power(i, 5);}
>> >
>> > return sum;
>> >
>> > }
>>
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>>
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