Nowadays, little haha got a problem from his teacher.His teacher wants to design a big logo for the campus with some circles tangent with each other. And now, here comes the problem. The teacher want to draw the logo on a big plane. You could see the example of the graph in the Figure1
At first, haha's teacher gives him two big
circles, which are tangent with each other. And, then, he wants to add
more small circles in the area where is outside of the small circle, but
on the other hand, inside the bigger one (you may understand this
easily if you look carefully at the Figure1.
Each small circles are added by the following principles.
* you should add the small circles in the order like Figure1.
* every time you add a small circle, you should make sure that it is tangented with the other circles (2 or 3 circles) like Figure1.
The teacher wants to know the total amount of pigment he would use when he creates his master piece.haha doesn't know how to answer the question, so he comes to you.
Task
add in the figure. You are supposed to write a program to calculate the
total area of all the small circles.
InputThe first line contains a integer t(1≤t≤1200), which means the number of the test cases. For each test case, the first line insist of two integers R1 and R2 separated by a space (1≤R≤100),
which are the radius of the two big circles. You could assume that the
two circles are internally tangented. The second line have a simple
integer N (1≤N≤10 000 000), which is the number of small circles the teacher want to add.
OutputFor each test case:
Contains a number in a single line, which shows the total area of
the small circles. You should out put your answer with exactly 5 digits
after the decimal point (NO SPJ).
Sample Input
2
5 4
1
4 5
1
Sample Output
3.14159
3.14159
笛卡尔定理
；
。 -------百度百科
中，两根x₁、x₂有如下关系：
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cmath>
using namespace std;
typedef long long ll;
const double eps = 1e-;
const double PI = acos(-1.0); void Debug()
{
puts("");
cout<<"+++++++++++++++++++++++++++分界线++++++++++++++++++++++++++++++"<<endl;
for(int i=; i<; i++)
{
for(int j=; j<; j++)
{
cout<<<<" ";
}
cout<<endl;
}
cout<<"+++++++++++++++++++++++++++分界线++++++++++++++++++++++++++++++"<<endl;
puts("");
} int n;
double r1, r2, r3,r4;
double ans;
void sovel()
{
if(r1 < r2) swap(r1,r2);
r3 = r1-r2;
double k1 = -1.0/r1, k2 = 1.0/r2, k3 = 1.0/r3, k4 = k1+k2+k3;
ans = r3*r3;
for(int i = ; i <= n; i++)
{
r4 = 1.0/k4;
ans += r4*r4;
if(r4*r4 < eps) break;
if(i+ <= n) ans +=r4*r4, i++;
double k5 = *(k1+k2+k4) - k3;
k3 = k4;
k4 = k5;
}
printf("%.5f\n", ans*PI); } int main()
{
std::ios::sync_with_stdio(false);
std::cin.tie();
int T;
cin >> T;
while(T--)
{
cin >> r1 >>r2 >> n;
sovel(); }
return ;
}
圆的反演 ：