描述
There are two circles on the plane. Now you must to calculate the area which they cover the plane. For example, in Figure 1, the area of the red region is the answer of this problem.
输入
The input contains multiple test cases. The first line contains an integer T describing the number of test cases. Each case contains two lines. One line describes one circle. For each line has three integers x, y, r, indicating the coordinate of the centre and radius. All the numbers are separated by spaces. All the input integers are within the range of [-1000, 1000].
输出
For each test case, output one line containing a number with 3 digits after decimal point representing the answer describing above.
样例输入
2
2 2 2
1 4 3
2 2 1
-2 -2 1
样例输出
32.462
6.283
只能说是高中数学知识，模拟一下就完了。但是我好像把学得还回去了，结果想了好久好久。。。。。。
#include<stdio.h> #include<math.h> double min(double a,double b) { if(a>b) a=b; return a; } double max(double a,double b) { if(a<b) a=b; return a; } int main() { double x1,x2,y1,y2,r1,r2,S; int n; scanf("%d",&n); while(n--) { scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&r1,&x2,&y2,&r2); double d=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); double s1=acos(-1)*r1*r1; double s2=acos(-1)*r2*r2; if(d>=r1+r2) S=s1+s2; else if(d<=max(r1,r2)-min(r1,r2)) S=max(s1,s2); else { double A=2*acos((r1*r1+d*d-r2*r2)/(2*r1*d)); double shan1=A*r1*r1/2; double sanjiao1=sin(A)*r1*r1/2; double B=2*acos((r2*r2+d*d-r1*r1)/(2*r2*d)); double shan2=B*r2*r2/2; double sanjiao2=sin(B)*r2*r2/2; S=s1+s2-(shan1+shan2-sanjiao1-sanjiao2); } printf("%.3lf\n",S); } }
原文：https://www.cnblogs.com/mayouyou/p/9030693.html