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]); ???}}解题笔记-CodeForces-908C New Year and Curling
原文地址：https://www.cnblogs.com/sigeryoung/p/solving-codeforces-908c.html