Thanks for all the input guys,
Here is the code... Could anyone spot any problem with this?
#include <iostream>
#include <fstream>
#include <ctime>
#include <time.h>
#include <iomanip>
#define CPS 2
using namespace std;
int main(int argc, const char * argv[])
{
ifstream infile("input2.in");//I renamed GCJ input file for simplicity...
ofstream outfile("output.out");
int nCases;
infile >> nCases;
cout<<"# Cases: "<<nCases<<endl;
for (int ci = 0; ci < nCases; ci++) {
double C, F, X;
infile >> C;
infile >> F;
infile >> X;
int n = (X/C) - (CPS/F);
if (n < 0) {
n = 0;
}
double tn = 0;//Time Taken to by nFarms
for (int i=1; i <=n; i++) {
tn += C/(CPS+((i-1)*F));
}
double tw = tn + X/(CPS+n*F);
cout<<"Case #"<<std::fixed<< std::setprecision(7)<<ci+1<<": "<<tw<<endl;
outfile<<"Case #"<<ci+1<<": "<<std::fixed<<
std::setprecision(7)<<tw<<endl;
}
infile.close();
outfile.close();
return 0;
}
Regards
On Monday, April 14, 2014 9:22:14 AM UTC-4, Sivasathivel Kandasamy wrote:
> I attempted the second problem, which took all the time irrespective of my
> confidence in my logic. I mean I tested my algorithm with the samples given
> in the problem page and was able to get answers correct to the last decimal.
> However, when I used their input and produced the output, Google Code Jam
> didn't accept my outputs. I'm totally at loss where the problem was... And I
> posted the problem in Stack Exchange, where many seems to agree with my
> logic.
>
>
>
> Losing the qualification round is painful, but I would be worth it if I could
> know where I went wrong...
>
>
>
> Here is the logic I came with... (with some algebra...):
>
>
>
> Let us assume, n as the minimum number of Farms to buy to win in the shortest
> time. n is a non-negative integer value.
>
> a is the number of cookies one gets per second, a=2;
>
>
>
>
>
> n is the minimum number of farms to buy to win, if and only if the following
> equation is satisfied...
>
> i.e time taken to buy n−1 farms + time to reach goal with (n−1) Farms > time
> taken to buy n farms + time to reach goal with n Farms
>
>
>
> However, time taken to buy n Farms =∑C/(a+(i−1)F); i = 1,...,n
>
>
>
> Therefore LHS becomes:
>
> (∑C/(a+(i−1)F))+(X/(a+(n−1)F))
>
>
>
> RHS becomes
>
> (∑C/(a+(i−1)F))+(X/(a+nF))+C/(a+(n−1)F)
>
>
>
> by rearranging the equations,
>
> (X/(a+(n−1)F))>(X/(a+nF))+(C/(a+(n−1)F))(eq.1)
>
> (X/(a+(n−1)F))−(X/(a+nF))>(C/(a+(n−1)F))
>
>
>
> After some rearrangements, it becomes:
>
>
>
> XF/(a+nF)>C
>
> XF>C(a+nF)
>
> XF/C>a+nF
>
> (XF/C)−a>nF
>
> (1/F)((XF/C)−a)>n
>
> (X/C)−(a/F)>n
>
>
>
> or
>
>
>
> n <(X/C)−(a/F)
>
> --- by ensuring that n is an non-negative integer, this equation would give
> the minimum number of farms,n, to buy. Assuming it is not possible to buy
> fraction of farms
>
>
>
> With n Farms, the total time taken to reach the goal would be given by RHS of
> (eq.1)
>
>
>
> Could anyone tell me, where I went wrong?
>
> As I sample, I tried with the samples and produced results correct to the
> last decimal...!
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