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

原题：

Three circles C_{a}C​a​​, C_{b}C​b​​, and C_{c}C​c​​, all with radius $R$and tangent to each other, are located in two-dimensional space as shown in Figure $1$. A smaller circle C_{1}C​1​​ with radius R_{1}R​1​​ (R_{1}R​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​​ ($2\le k\le N$), 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 $R$ and $k$, please write a program to calculate the value of R_{k}R​k​​, i.e., the radius of the $k-th$ 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​​ = $1259.8998$ for some $k$, then the answer you should report is $1259$.

Another example, if R_{k}R​k​​ = $39.1029$ for some $k$, then the answer you should report is $39$.

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

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

Input Format

Contains $l+3$ lines.

Line $1$: $l$ ----------------- the number of test cases, $l$ is an integer.

Line $2$: $R$ ---------------- $R$ is a an integer followed by a decimal point,then followed by a digit.

Line $3$: $k$ ---------------- test case #$1$, $k$ is an integer.

$\dots$

Line $i+2$: $k$ ----------------- test case # $i$.

$\dots$

Line $l+2$: $k$ ------------ test case #$l$.

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

Output Format

Contains $l$ lines.

Line $1$: $k$ R_{k}R​k​​ ----------------output for the value of $k$ and R_{k}R​k​​ at the test case #$1$, each of which should be separated by a blank.

$\dots$

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

Line $l$: $k$ R_{k}R​k​​ ----------------output for $k$ and the value ofR_{k}R​k​​ at the test case # $l$, 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 
#include 
#include 
#include 
#include 
using namespace std;
const int PI = 3.14159;
const double sqrt3 = sqrt(3.0);
int main() {
    int L; scanf("%d", &L);
    double R; scanf("%lf", &R);
    for (int i = 1; i <= L; i++) {
        int K; scanf("%d", &K);
        double r0 = R, y0 = 0;
        for (int j = 1; j <= K; j++) {
            double y = (1.0*(y0+r0-R)*(y0+r0-R) - 4.0*R*R) / (2.0*(y0+r0-R) - 2.0*sqrt3*R);
            double r = y - y0 - r0;
            r0 = r, y0 = y;
        }
        int ans = r0;
        printf("%d %d\n", K, ans);
    }
    int temp; scanf("%d", &temp);
    return 0;
}