求两圆环相交部分的面积

Matt is a big fan of logo design. Recently he falls in love with logo made up by rings. The following figures are some famous examples you may know.



A ring is a 2-D figure bounded by two circles sharing the common center. The radius for these circles are denoted by r and R (r < R). For more details, refer to the gray part in the illustration below.



Matt just designed a new logo consisting of two rings with the same size in the 2-D plane. For his interests, Matt would like to know the area of the intersection of these two rings.

Input

The first line contains only one integer T (T ≤ 10
5), which indicates the number of test cases. For each test case, the first line contains two integers r, R (0 ≤ r < R ≤ 10).

Each of the following two lines contains two integers x i, y i (0 ≤ x
i, y i ≤ 20) indicating the coordinates of the center of each ring.

Output

For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1) and y is the area of intersection rounded to 6 decimal places.

Sample Input

2
2 3
0 0
0 0
2 3
0 0
5 0

Sample Output

Case #1: 15.707963
Case #2: 2.250778

这道题就是告诉两个圆环，阴影部分为圆环的实体，要求两个圆环实体部分相交的面积，画出图后发现（用A,a记第一个圆环， B,b记第二个圆环， area函数为求两圆的

相交面积），area（A, B） - area（A, b）- area（B, a） + area（a, b）（因为多减了两小圆相交的面积，所以后面要再加上），第一次提交时用的c++ 的输入输出流输

入数据，结果超时，改为scanf()函数后就过了。。

#include <cstdlib>
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>

using namespace std;

const double PI = acos(-1);

// 圆的结构
struct circle {
double x; // 圆心横坐标
double y; // 圆心纵坐标
double r; // 半径
};

// 计算圆心距
double dist(circle a, circle b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

// 计算两相交圆的面积
double area(circle a, circle b) {

if((dist(a, b)+min(a.r,b.r))<=max(a.r,b.r)) // 内含或重合
{
if(a.r<b.r)
return PI*a.r*a.r;
else
return PI*b.r*b.r;
}
else if(dist(a, b)>=(a.r+b.r)) // 相离或相切
return 0.0;
else // 相交
{
double length=dist(a, b);
// 利用三角形余弦定理求圆心角
double d1=2*acos((a.r*a.r+length*length-b.r*b.r)/(2*a.r*length));
double d2=2*acos((b.r*b.r+length*length-a.r*a.r)/(2*b.r*length));
// 利用圆心角求得扇形面积再减去三角形面积后两部分相加就是相交部分面积
double area1=a.r*a.r*d1/2-a.r*a.r*sin(d1)/2;
double area2=b.r*b.r*d2/2-b.r*b.r*sin(d2)/2;
double area=area1+area2;
return area;
}
}
int main()
{
int T;
circle a, b, A, B;
double r, R;
scanf("%d", &T) ;
for(int i=1; i<=T; i++)
{
scanf("%lf%lf", &r, &R);
a.r = b.r = r;
A.r = B.r = R;
scanf("%lf%lf", &a.x, &a.y);
A.x = a.x;
A.y = a.y;
scanf("%lf%lf", &b.x, &b.y);
B.x = b.x;
B.y = b.y;
double sec = area(A, B) - area(A, b) - area(B, a) + area(a, b);
printf("Case #%d: %.6lf\n", i, sec);

}

return 0;
}