Good Bye 2017-C. New Year and Curling

2018-01-02 12:51:08来源:CSDN作者:qq_34531807人点击

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C. New Year and Curlingtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

Carol is currently curling.

She has n disks each with radius r on the 2D plane.

Initially she has all these disks above the line y = 10100.

She then will slide the disks towards the line y = 0 one by one in order from 1 to n.

When she slides the i-th disk, she will place its center at the point (xi, 10100). She will then push it so the disk’s y coordinate continuously decreases, and x coordinate stays constant. The disk stops once it touches the line y = 0 or it touches any previous disk. Note that once a disk stops moving, it will not move again, even if hit by another disk.

Compute the y-coordinates of centers of all the disks after all disks have been pushed.

Input

The first line will contain two integers n and r (1 ≤ n, r ≤ 1 000), the number of disks, and the radius of the disks, respectively.

The next line will contain n integers x1, x2, ..., xn (1 ≤ xi ≤ 1 000) — the x-coordinates of the disks.

Output

Print a single line with n numbers. The i-th number denotes the y-coordinate of the center of the i-th disk. The output will be accepted if it has absolute or relative error at most 10 - 6.

Namely, let's assume that your answer for a particular value of a coordinate is a and the answer of the jury is b. The checker program will consider your answer correct if  for all coordinates.

Exampleinput
6 2
5 5 6 8 3 12
output
2 6.0 9.87298334621 13.3370849613 12.5187346573 13.3370849613
Note

The final positions of the disks will look as follows:


In particular, note the position of the last disk.

这题感觉比第二题简单,完全考数学.用Θ(n²)解决。

Code:

var  n,r,i,j:longint;  x:array[0..1005] of longint;  o:array[0..1005] of double;function max(a,b:double):double;begin  if a>b then exit(a) else exit(b);end;begin  readln(n,r);  for i:=1 to n do  begin    read(x[i]);    o[i]:=r;    for j:=1 to i-1 do      if abs(x[i]-x[j])<=2*r then        o[i]:=max(o[i],sqrt(sqr(2*r)-sqr(x[i]-x[j]))+o[j]);  end;  for i:=1 to n do write(o[i]:0:10,' ');end.

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