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

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原题：

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

Figure 1.

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

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

Now, given the parameters $R$ and $k$ , please write a program to calculate the value of $R_{k}$ , i.e., the radius of the $k - t h$ inserted circle. Please note that since the value of $R_k$ may not be an integer, you only need to report the integer part of $R_{k}$ . For example, if you find that $R_{k}$ = $1259.8998$ for some $k$ , then the answer you should report is $1259$ .

Another example, if $R_{k}$ = $39.1029$ 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., $\leq 10$ . Furthermore, you may assume $\pi = 3.14159$ . The range of each parameter is as below:

$\leq k \leq N$ , and $10^{4} \leq R \leq 10^{7}$ .

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.

$\ldots$

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

$\ldots$

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

$\ldots$

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

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

样例输入

1152973.61-1

样例输出

1 23665

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>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;}

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