@snehal  : YUP

On Fri, Jan 21, 2011 at 5:57 PM, abhijith reddy <[email protected]>wrote:

> @snehal .. misread it .. my apologies.
>
>
> On Fri, Jan 21, 2011 at 5:56 PM, abhijith reddy 
> <[email protected]>wrote:
>
>> O(lg(n)*lg(n)) is the complexity of the solution ! Not the solution.
>> M=3^x
>> We binary search on x and then compute 3^x in log(x) time using
>> exponentiation. Hence the complexity.
>>
>>
>>
>> On Fri, Jan 21, 2011 at 5:50 PM, snehal jain <[email protected]>wrote:
>>
>>> @juvir++
>>> it was mentioned in question not to use log or power. isnt there any
>>> approach using bitwise operators
>>>
>>>
>>> On Fri, Jan 21, 2011 at 5:24 PM, Manmeet Singh <[email protected]>wrote:
>>>
>>>> this will be O(log(n) * log(n)) solution
>>>>
>>>>
>>>> On Fri, Jan 21, 2011 at 4:29 PM, abhijith reddy <
>>>> [email protected]> wrote:
>>>>
>>>>> Below is code for modular exponentation in general
>>>>>
>>>>> // (a^b)%c
>>>>> int modexp(int a,int b,int c)
>>>>> {
>>>>>    int ans=1;
>>>>>    while(b)
>>>>>    {
>>>>>      if(b&1) ans=(ans*a)%c;
>>>>>      a=(a*a)%c;
>>>>>      b>>=1;
>>>>>    }
>>>>>    return ans;
>>>>>
>>>>> }
>>>>>
>>>>>
>>>>> On Fri, Jan 21, 2011 at 3:27 PM, juver++ <[email protected]>wrote:
>>>>>
>>>>>> int l = 0, r = ...;
>>>>>> while (l < r) {
>>>>>>   int m = (l + r) / 2;
>>>>>>   int p = power(3, m);
>>>>>>   if (p > M) r = m - 1;
>>>>>>   else if (p < M) l = m + 1;
>>>>>>   else print 3^m = M;
>>>>>> }
>>>>>>
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