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 ycoordinate 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.
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.
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.
6 2 5 5 6 8 3 12
2 6.0 9.87298334621 13.3370849613 12.5187346573 13.3370849613
The final positions of the disks will look as follows:
In particular, note the position of the last disk.
想麻烦了，其实很简单。
问题一：如何判断两个圆是否接触。利用勾股定理判断。
问题二：如果解除了，那么他比接触的还要高多少？还是勾股定理，求出高度差。
问题三：暴力，枚举第i个之前的所有的园即可
#include<bits/stdc++.h>
using namespace std;
double x[1010],y[1010];
int main()
{
double n,r;
while(cin>>n>>r)
{
for(int i=0; i<n; i++)
{
cin>>x[i];
y[i]=r;
for(int j=0; j<i; j++)
{
double l=4*r*r-(x[i]-x[j])*(x[i]-x[j]);
y[i]=max(y[i],y[j]+sqrt(l));
}
printf("%.10f ",y[i]);
}
puts("\n");
}
return 0;
}