G.Finding the Radius for an Inserted Circle 2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

地址：https://nanti.jisuanke.com/t/17314

题目：

Three circles C_{a}C​a​​, C_{b}C​b​​, and C_{c}C​c​​, all with radius RR and tangent to each other, are located in two-dimensional space as shown in Figure 11. A smaller circle C_{1}C​1​​ with radius R_{1}R​1​​ (R_{1}<RR​1​​<R) is then inserted into the blank area bounded by C_{a}C​a​​, C_{b}C​b​​, and C_{c}C​c​​ so that C_{1}C​1​​ is tangent to the three outer circles, C_{a}C​a​​, C_{b}C​b​​, and C_{c}C​c​​. Now, we keep inserting a number of smaller and smaller circles C_{k} (2 leq k leq N)C​k​​ (2≤k≤N) with the corresponding radius R_{k}R​k​​ into the blank area bounded by C_{a}C​a​​, C_{c}C​c​​ and C_{k-1}C​k−1​​ (2 leq k leq N)(2≤k≤N), so that every time when the insertion occurs, the inserted circle C_{k}C​k​​ is always tangent to the three outer circles C_{a}C​a​​, C_{c}C​c​​ and C_{k-1}C​k−1​​, as shown in Figure 11

Figure 1.

(Left) Inserting a smaller circle C_{1}C​1​​ into a blank area bounded by the circle C_{a}C​a​​, C_{b}C​b​​ and C_{c}C​c​​.

(Right) An enlarged view of inserting a smaller and smaller circle C_{k}C​k​​ into a blank area bounded by C_{a}C​a​​, C_{c}C​c​​ and C_{k-1}C​k−1​​ (2 leq k leq N2≤k≤N), so that the inserted circle C_{k}C​k​​ is always tangent to the three outer circles, C_{a}C​a​​, C_{c}C​c​​, and C_{k-1}C​k−1​​.

Now, given the parameters RR and kk, please write a program to calculate the value of R_{k}R​k​​, i.e., the radius of the k-thk−th inserted circle. Please note that since the value of R_kR​k​​ may not be an integer, you only need to report the integer part of R_{k}R​k​​. For example, if you find that R_{k}R​k​​ = 1259.89981259.8998 for some kk, then the answer you should report is 12591259.

Another example, if R_{k}R​k​​ = 39.102939.1029 for some kk, then the answer you should report is 3939.

Assume that the total number of the inserted circles is no more than 1010, i.e., N leq 10N≤10. Furthermore, you may assume pi = 3.14159π=3.14159. The range of each parameter is as below:

1 leq k leq N1≤k≤N, and 10^{4} leq R leq 10^{7}10​4​​≤R≤10​7​​.

Input Format

Contains l + 3l+3 lines.

Line 11: ll ----------------- the number of test cases, ll is an integer.

Line 22: RR ---------------- RR is a an integer followed by a decimal point,then followed by a digit.

Line 33: kk ---------------- test case #11, kk is an integer.

ldots…

Line i+2i+2: kk ----------------- test case # ii.

ldots…

Line l +2l+2: kk ------------ test case #ll.

Line l + 3l+3: -1−1 ---------- a constant -1−1 representing the end of the input file.

Output Format

Contains ll lines.

Line 11: kk R_{k}R​k​​ ----------------output for the value of kk and R_{k}R​k​​ at the test case #11, each of which should be separated by a blank.

ldots…

Line ii: kk R_{k}R​k​​ ----------------output for kk and the value of R_{k}R​k​​ at the test case # ii, each of which should be separated by a blank.

Line ll: kk R_{k}R​k​​ ----------------output for kk and the value ofR_{k}R​k​​ at the test case # ll, each of which should be separated by a blank.

样例输入

1
152973.6
1
-1

样例输出

1 23665

题目来源

2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

 

思路：

　　圆的反演。

　　很容易想到把上面两大圆的切点作为反演中心，这样会得到下图。

　　绿色的是反演前的圆，黄色的是反演后的图形，两个大圆成了平行直线，下面的大圆成了直线间的小圆，后面添加的圆都在这个小圆的下面。

　　所以求出小圆的圆心的y即可，然后反演回去可以得到半径。

#include <bits/stdc++.h>

using namespace std;

#define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int,int> PII;
const double eps=1e-8;
const double PI=acos(-1.0);
const int K=1e6+7;
const int mod=1e9+7;


int main(void)
{
    double r,x,y,ls,dis,ans[11];
    int t;
    cin>>t>>r;
    x=0.5*r,ls=-0.5*sqrt(3.0)*r;
    dis=x*x+ls*ls;
    ls=ls/dis;
    r=1.0/(2*r);
    for(int i=1;i<=10;i++)
    {
        y=ls-r*2;
        ans[i]=0.5*(1.0/(y-r)-1.0/(y+r));
        ls=y;
    }
    for(int i=1,k;i<=t;i++)
        scanf("%d",&k),printf("%d %d
",k,(int)ans[k]);
    return 0;
}