0 题面
C. New Year and Curling
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.
1 代码
#include <iostream> #include <cmath> #include <cstdio> #define SQR(x) (x)*(x) #define DIST(x, y) (SQR(x)+SQR(y)) #define ABS(x) ((x)>=0?x:(-x)) using namespace std; int x[1000]; double y[1000]; int main() { int n, r; cin >> n >> r; for(int i = 0; i < n; i++) { cin >> x[i]; } for(int i = 0; i < n; i++) { int max_j = -1; double max_y = -1; double offset = 0; for(int j = 0; j < i; j++) { if((SQR(x[j] - x[i]) <= 4 * r * r) && ((y[j] + sqrt((double)(4.0 * r * r) - (double)(SQR(x[j] - x[i])))) > max_y)) { max_j = j; max_y = (y[j] + sqrt((double)(4.0 * r * r) - (double)(SQR(x[j] - x[i])))); } } if(max_j == -1) y[i] = r; else y[i] = max_y; } for(int i = 0; i < n; i++) { printf("%.7f ", y[i]); } }
