Area of the first white circle is pi* r*r
Area of first black circle that includes first white circle is pi * (r+1) *
(r+1)
So area of first black strip is pi * (r+1) * (r+1) - pi *r *r
If you continue doing this then area of second black strip is pi * (r+3) *
(r+3) - pi * (r+2) * (r+2)
So total area of n black strips is (pi * (r+1) * (r+1) - pi * r * r) + (pi
* (r+3) * (r+3) - pi * (r+2) * (r+2)) + .......
I will remove pi as pi is common for all. So above expression can be
written as
(r*r + 2*r + 1 - r *r) + (r*r + 6 * r + 9 - r * r + 4 * r + 4) + ......
( 2* r + 1) + ( 2*r + 5) + (2r + 9) + .....
2 * r * n + (1 + 5 + 9 + ....)
2* r * n + ( 1 + (1 + 4) + ( 1 + 8 ) + ....)
2 * r * n + ( 1 * n + ( 4 + 8 + 12 + ...))
2 * r * n + 1 * n + 4 * n * ( n-1) /2
2 * r * n + n + 2 * n * (n -1)
This is the formula everyone used to do the binary search
=
On Sat, Apr 27, 2013 at 1:04 PM, newbie007 <lescoutinh...@gmail.com> wrote:
> 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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