@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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