IntersectionTime Limit: 4000/4000 MS (Java/Others)Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 1036Accepted Submission(s): 407
Problem Description
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 ≤ 105), 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 xi, yi (0 ≤ xi, yi ≤ 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
22 30 00 02 30 05 0
Sample Output
Case #1: 15.707963Case #2: 2.250778
题意：
求两圆环相交的面积。
参考代码：
#include<stdio.h> #include<math.h> #define PI acos(-1.0) double sum(double a1,double b1,double r1,double a2,double b2,double r2) {double A1,A2,s1,s2,s,d;d=sqrt((a2-a1)*(a2-a1)+(b2-b1)*(b2-b1));if(d>=r1+r2)return 0.000;else if(d<=fabs(r1-r2)){if(r1>r2)return PI*r2*r2;elsereturn PI*r1*r1;}else{A1=2*acos((d*d+r1*r1-r2*r2)/(2*d*r1));A2=2*acos((d*d+r2*r2-r1*r1)/(2*d*r2));s1=0.5*r1*r1*sin(A1)+0.5*r2*r2*sin(A2);s2=A1/2*r1*r1+A2/2*r2*r2;s=s2-s1;return s;} } int main(){int t;scanf("%d",&t);for(int q=1;q<=t;q++){double r1,r2,a1,a2,b1,b2;scanf("%lf%lf%lf%lf%lf%lf",&r1,&r2,&a1,&b1,&a2,&b2);printf("Case #%d: ",q);printf("%.6lf\n",sum(a1,b1,r2,a2,b2,r2)-2*sum(a1,b1,r2,a2,b2,r1)+sum(a1,b1,r1,a2,b2,r1));}return 0;}
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，其实每一朵花，都有它自己的生命，
