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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