This happens to give the answer *ceil(log(ceil(log(P/L, C)), 2))*
On Sun, May 23, 2010 at 18:27, Bharath Raghavendran <[email protected]>wrote:
> Sorry .. I hit send too soon :P
>
> #include<iostream>
> #include<math.h>
> using namespace std;
>
> double log(double a, double b)
> {
> return log(a)/log(b);
> }
>
> int main()
> {
> double L,P,C;
> int T,t;
> cin>>T;
> for(t=1;t<=T;t++)
> {
> cin>>L>>P>>C;
>
> double a = log(log(P/L,C),2);
> int a_int = a;
> while(a_int < a) a_int++;
> a_int = max(a_int, 0);
> cout<<"Case #"<<t<<": "<<a_int<<endl;
>
> }
> }
>
> On 23 May 2010 18:24, Bharath Raghavendran <[email protected]> wrote:
>
>> Here is the mathematical formula :
>> double a = log(log(P/L,C),2);
>> int a_int = a;
>> while(a_int < a) a_int++;
>> a_int = max(a_int, 0);
>>
>>
>> On 23 May 2010 17:49, Bharath Raghavendran <[email protected]> wrote:
>>
>>> You are trying to find a value of a between L and P such that u r sure
>>> that a contestants can access at a time but C*a contestants cannot.
>>>
>>> Its basically a binary search for a satisfying element between L and P
>>> with a small modification that instead of taking (L+P)/2 each time, go for
>>> their geometric mean. This will maintain the ratio of first and last between
>>> the two halves and hence will be most optimal.
>>>
>>>
>>> On 23 May 2010 17:40, Saurabh Aggarwal <[email protected]> wrote:
>>>
>>>> Hi Guys,
>>>>
>>>> Could anyone please explain the logic behind the "Load Testing" problem
>>>> ?
>>>> From the solutions I gathered that they have found the number of bits in
>>>> X, where
>>>>
>>>> L * (C^X) > P
>>>>
>>>> But couldn't understand why this was done ?
>>>>
>>>> Thanks.
>>>>
>>>> --
>>>> Regards,
>>>> Saurabh Aggarwal
>>>>
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>>>>
>>>
>>>
>>
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Thanks & Regards,
Dhruva Sagar.
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