圆相切半径4园

Three circles CaC_{a}C​a​​, CbC_{b}C​b​​, and CcC_{c}C​c​​, all with radius RRR and tangent to each other, are located in two-dimensional space as shown in Figure 111. A smaller circle C1C_{1}C​1​​ with radius R1R_{1}R​1​​ (R1<RR_{1}<RR​1​​<R) is then inserted into the blank area bounded by CaC_{a}C​a​​, CbC_{b}C​b​​, and CcC_{c}C​c​​ so that C1C_{1}C​1​​ is tangent to the three outer circles, CaC_{a}C​a​​, CbC_{b}C​b​​, and CcC_{c}C​c​​. Now, we keep inserting a number of smaller and smaller circles Ck (2≤k≤N)C_{k} (2 leq k leq N)C​k​​ (2≤k≤N) with the corresponding radius RkR_{k}R​k​​ into the blank area bounded by CaC_{a}C​a​​, CcC_{c}C​c​​ and Ck−1C_{k-1}C​k−1​​ (2≤k≤N)(2 leq k leq N)(2≤k≤N), so that every time when the insertion occurs, the inserted circle CkC_{k}C​k​​ is always tangent to the three outer circles CaC_{a}C​a​​, CcC_{c}C​c​​ and Ck−1C_{k-1}C​k−1​​, as shown in Figure 111

Figure 1.

(Left) Inserting a smaller circle C1C_{1}C​1​​ into a blank area bounded by the circle CaC_{a}C​a​​, CbC_{b}C​b​​ and CcC_{c}C​c​​.

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

Now, given the parameters RRR and kkk, please write a program to calculate the value of RkR_{k}R​k​​, i.e., the radius of the k−thk-thk−th inserted circle. Please note that since the value of RkR_kR​k​​ may not be an integer, you only need to report the integer part of RkR_{k}R​k​​. For example, if you find that RkR_{k}R​k​​ = 1259.89981259.89981259.8998 for some kkk, then the answer you should report is 125912591259.

Another example, if RkR_{k}R​k​​ = 39.102939.102939.1029 for some kkk, then the answer you should report is 393939.

Assume that the total number of the inserted circles is no more than 101010, i.e., N≤10N leq 10N≤10. Furthermore, you may assume π=3.14159pi = 3.14159π=3.14159. The range of each parameter is as below:

1≤k≤N1 leq k leq N1≤k≤N, and 104≤R≤10710^{4} leq R leq 10^{7}10​4​​≤R≤10​7​​.

Input Format

Contains l+3l + 3l+3 lines.

Line 111: lll ----------------- the number of test cases, lll is an integer.

Line 222: RRR ---------------- RRR is a an integer followed by a decimal point,then followed by a digit.

Line 333: kkk ---------------- test case #111, kkk is an integer.

…ldots…

Line i+2i+2i+2: kkk ----------------- test case # iii.

…ldots…

Line l+2l +2l+2: kkk ------------ test case #lll.

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

Output Format

Contains lll lines.

Line 111: kkk RkR_{k}R​k​​ ----------------output for the value of kkk and RkR_{k}R​k​​ at the test case #111, each of which should be separated by a blank.

…ldots…

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

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

样例输入

1
152973.6
1
-1

样例输出

1 23665

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int main(){
    int T,n;
    double R;
    while(scanf("%d",&T),T!=-1){
        scanf("%lf",&R);
        while(T--){
            scanf("%d",&n);
            double k1=1/R,k2=1/R,k3=1/R;
            double ans;
            for(int i=0;i<n;++i) {
                double B=-2*(k1+k2+k3);
                double C=-(k1+k2+k3)*(k1+k2+k3)+2*(k1*k1+k2*k2+k3*k3);
                double D=B*B-4*C;
                double k4=(-B+sqrt(D))/2;
                ans=1/k4;
                k3=k4;
            }
            printf("%d %d
",n,floor(ans));
        }
    }
}