HDU 1798 Tell me the area （数学）

题目链接

Problem Description

There are two circles in the plane (shown in the below picture), there is a common area between the two circles. The problem is easy that you just tell me the common area.

Input

There are many cases. In each case, there are two lines. Each line has three numbers: the coordinates (X and Y) of the centre of a circle, and the radius of the circle.

Output

For each case, you just print the common area which is rounded to three digits after the decimal point. For more details, just look at the sample.

Sample Input

0 0 2
2 2 1

Sample Output

0.108

分析：

两圆的位置关系有相离、相切和相交三种，相切又有外切和内切。

如果两圆的位置关系是相离或者外切的话，它们是没有公共部分的，及面积为0；

如果两圆内切，公共部分就为整个的小圆，面积也就是小圆的面积；

如果两圆相交的话，面积就应该是每个圆所对应的扇形减去所对应的三角形的面积和。

代码：

#include<stdio.h>
#include<iostream>
#include<math.h>
using namespace std;
int main()
{
    double q,w,m,n,a,b,c,x,y,z,PI;
    PI=2*asin(1.0);
    while(~scanf("%lf%lf%lf",&a,&b,&c))
    {
        scanf("%lf%lf%lf",&x,&y,&z);
        a=sqrt((a-x)*(a-x)+(b-y)*(b-y));///计算圆心距
        ///如果两圆相离、外切或至少一圆半径为0时，那么所求面积为0
        if(a>=c+z||!c||!z)x=0;
        ///如果两内切或内含，那么所求面积为小圆面积
        else if(a<=fabs(z-c))
        {
            if(z>c)z=c;
            x=z*z*PI;
        }
        ///如果两圆相交，面积求解如下
        else
        {
            ///由余弦定理求出公共弦在圆o1中对应的圆心角的一半
            b=acos((a*a+c*c-z*z)/(2*a*c));
            ///由余弦定理求出公共弦在圆o2中对应的圆心角的一半
            y=acos((a*a+z*z-c*c)/(2*a*z));
            ///计算圆o1中扇形面积
            m=b*c*c;
            ///计算圆o2中扇形面积
            n=y*z*z;
            ///计算圆o1中扇形所对应的三角形面积
            q=c*c*sin(b)*cos(b);
            ///计算圆o2中扇形所对应的三角形面积
            w=z*z*sin(y)*cos(y);
            ///q+w为图中四边形面积，两扇形面积之和与四边形面积之差即为
            ///所求面积。在图2中y为钝角，计算出的面积w为负值，这时q+w
            ///表示两三角面积之差，刚好还是四边形面积，因此对于图1和图
            ///2不必分情况讨论
            x=m+n-(q+w);
        }
        printf("%.3f\n",x);
    }
    return 0;
}