HDU 1798 Tell me the area-优快云博客

本文链接：https://blog.youkuaiyun.com/WEYuLi/article/details/8718331

本文介绍了一道关于计算平面中两个圆交叠部分面积的编程题，通过数学方法求解两圆相交时的公共区域面积，并给出了使用C++实现的完整代码示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Tell me the area

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1361 Accepted Submission(s): 426

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

Sample Output

AC代码：

#include <iostream>
#include <cmath>
#include <cstdio>

using namespace std;


int main()
{
    double x1,y1,r1,x2,y2,r2,s,PI;
    PI=2*asin(1.0);                //PI不能写成3.1415926
    while(cin>>x1>>y1>>r1)
   {
      cin>>x2>>y2>>r2;

      double d=sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));
      if(d>=(r1+r2)||!r1||!r2)      //外离
      s=0;
      else if(d<=abs(r2-r1))        //内含
      {
        if(r2<r1)
        r1=r2;
        s=r1*r1*PI;
      } 
      else                                   //相交
      {
         double a1=acos((r1*r1+d*d-r2*r2)/(2.0*r1*d));       //a1,a2分别为两扇形圆心角的一半
         double a2=acos((r2*r2+d*d-r1*r1)/(2.0*r2*d));
         s=a1*r1*r1+a2*r2*r2-r1*r1*sin(a1)*cos(a1)-r2*r2*sin(a2)*cos(a2);        //相交面积等于两扇形面积之和减四边形

      }
       printf("%.3lf\n",s);
   }

return 0;
}