For the first circle, you need pi*(r+1)²-pi*r², which is pi*((r+1)² - r²) cm²,
since 1 mililiter covers pi cm², you need (r+1)²-r² paint.
following this logic, for each circle you need:
1st: (r+1)² - (r-0)²
2nd: (r+3)² - (r-2)²
3rd: (r+5)² - (r-4)²
4th: (r+7)² - (r-6)²
and so on...
note the relation between each one: 1 0, 3 2, 5 4, 7 6 ...
generalizing, the "numbers" for the i-th circle are 2i-1 and 2i-2
So, the general formula for the amount of paint you need for the i-th circle is:
(r+(2i-1))² - (r+(2i-2))²
which, simplifying, leads to: c
The amount of paint you need to paint all circles from i to n, are:
Summation(4i+2r-3) from i=1 to n
The formula for this summation is: n(2n + 2r - 1)
In the algorithm, he uses (2 * r + 1) * n + 2.0 * n * (n - 1), which is
equivalent to 2n²+2nr-n = n(2n+2r-1).
So, basically, what the algorithm does is a binary search to search for "n"
(the number of circles) where the amount of paint needed will be smaller than t
(amount of paint you have).
Hope this helps you nderstand.
(Now, why the comparison to 1.5t is used, is also a mystery to me :P).
> Hi,
>
>
>
> I understand that they're using binary search, but I don't know how can it
> get to the solution.
>
> Could someone be very nice and explain the code below, please?
>
> This is from coder "wata":
>
>
>
> void solve() {
>
> long left = 0, right = 1L << 40;
>
> while (right - left > 1) {
>
> long n = (left + right) / 2;
>
> if ((double)(2 * r + 1) * n + 2.0 * n * (n - 1) > 1.5 * t) {
>
> right = n;
>
> } else if ((2 * r + 1) * n + 2 * n * (n - 1) > t) {
>
> right = n;
>
> } else {
>
> left = n;
>
> }
>
> }
>
> System.out.println(left);
>
> }
>
>
>
> Why those values, why 1.5? Man, I don't understand this code :(
>
>
>
> Em sábado, 27 de abril de 2013 05h42min58s UTC-3, Vaibhav Tulsyan escreveu:
>
> > I was seeing the solutions of the top 10 contestants for the large input of
> > Bull's Eye. They all seem to have used some method involving variables like
> > beginning,end and mid. Can anybody explain to me what method they've
> > applied exactly?
>
> > I just used basic maths to solve it. They seem to have used some better
> > algorithm.
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