Codeforces Good Bye 2017 C. New Year and Curling（模拟水题）

拼搏现实的明天。 2022-06-02 02:16 140阅读 0赞

C. New Year and Curling

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard 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 (*x**i*, 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 *x*1, *x*2, ..., *x**n* (1 ≤ *x**i* ≤ 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 ![c62ea64d4651240724c5ac4779b671c741edec24.png][] 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:

![1d99f96dd950feea0dec51525aa5c4f94dafd5b5.png][]

In particular, note the position of the last disk.

题解：

昨天本来想打这个比赛的，结果11点的时候很困了就直接去睡了，第二天老刘说昨天只要a3题就可以加90多分。。然后我就直接做了第三题。。感觉很水。。感觉总是败给了时差

思路：

因为才1000个圆，直接枚举之前已经放入的圆满足横坐标范围内\[x-r,x+r\]，然后用勾股定理，圆心距离作为斜边为2r,直角边为abs(a\[i\]-a\[j\]),算出另一个直角边加上当前扫到符合条件的节点的纵坐标b\[j\]，然后取其中最大值就是答案

代码：

#include<stdio.h>
    #include<string.h>
    #include<stdlib.h>
    #include<queue>
    #include<stack>
    #include<math.h>
    #include<vector>
    #include<map>
    #include<set>
    #include<stdlib.h>
    #include<cmath>
    #include<string>
    #include<algorithm>
    #include<iostream>
    #include<stdio.h>
    using namespace std;
    #define ll long long
    double a[1005],b[1005];
    int main()
    {
        int i,j,k,n;
        double r;
        scanf("%d%lf",&n,&r);
        for(i=0;i<n;i++)
        {
            scanf("%lf",&a[i]);
            double y=r;
            for(j=0;j<i;j++)
            {
                if(abs(a[i]-a[j])<=2*r)
                {
                    y=max(y,b[j]+sqrt(4.0*r*r-abs(a[i]-a[j])*abs(a[i]-a[j])));
                }
            }
            b[i]=y;
        }
        for(i=0;i<n;i++)
        {
            printf("%.10lf\n",b[i]);
        }
        return 0;
    }

[c62ea64d4651240724c5ac4779b671c741edec24.png]: /images/20220602/2462eee2fd284932b821dda0513ea12b.png
[1d99f96dd950feea0dec51525aa5c4f94dafd5b5.png]: /images/20220602/a8713f39d2bb4499823e4b9f573c1c71.png