New Year and Curling CodeForces - 908C

New Year and Curling

CodeForces - 908C

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.

Example

Input

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.

题意： 已知一些圆在(xi,10100)
d25b
处，然后依次落下，下落时只要碰到某个圆，就会停止，求这些圆的最后的y值，也即是水平高度

分析： 首先看了下n的范围很小，可以到O(n2)的复杂度，直接暴力枚举，可以画图分析下，见下图



可以看到高度h = y1 + sqrt(2r*2r - (x2-x1)(x2-x1));枚举个最高的高度即可，因为一旦碰到就不会落下了

code:
#include <iostream>
#include <cstdio>
#include<cmath>
using namespace std;
int x[1005];
double y[1005];
int main(){
int n,r;
cin >> n >> r;
int i;
for(i = 0; i < n; i++){
cin >> x[i];
}
for(i = 0; i < n; i++){
y[i] = r;//初始高度为r
int j;
for(j = 0; j < i; j++){//枚举每个圆形
if(abs(x[j]-x[i])<=2*r){//如果可以相邻
y[i] = max(y[i],y[j]+sqrt(4*r*r-(x[j]-x[i])*(x[j]-x[i])));
//高度等于r和前一个圆形的高度加上计算出的值中取最大
}
}
}
for(i = 0; i < n; i++){
printf("%.12f ",y[i]);
}
return 0;
}