@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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