HDU 5120 Intersection

IntersectionTime limit:4000/4000 MS (java/others) Memory limit:512000/512000 K (java/others)
Total submission (s): 967 Accepted Submission (s): 371


Problem Descriptionmatt is a big fan of logo design. Recently he falls in love with logo made up by rings. The following figures is some famous examples.


A ring is a 2-d figure bounded by the circles sharing the common center. The radius for these circles is 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 the rings with the same size in the 2-d plane. For he interests, Matt would like to know the area of the intersection of these.
Inputthe first line contains only one integer T (t≤105), which indicates the number of test cases. For each test case, the first line contains the integers R, R (0≤r < r≤10).

Each of the following, lines contains, integers xi, yi (0≤xi, yi≤20) indicating the coordinates of the center of Each ring.
Outputfor each test case, output a single line "Case #x: Y", where x was the case number (starting from 1) and Y-is the being A 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

Source

2014ACM/ICPC Asia Beijing Station-Replay (thanks to the North Division and handing in)

Geometric calculations,,, a little trouble

 #include <cstdio> #include <cmath> #include <algorithm>using namespace Std;int a1,a2,b1,b2;double d,p;double area;double fun (int a,int b) {double sa,sb;d=sqrt ((A1-A2) * (A1-A2) + (b1-b       2) * (B1-B2));//center distance double Rr=min (A, B), if (D<=abs (a))//internal or inner-cut Area=acos ( -1.0) *rr*rr;else if (d>=a+b)//tangent or disjoint    area=0.0;else{//Intersection for Area p= (a+b+d)/2.0;//Helen formula inside the P-sa=acos ((a*a+d*d-b*b)/(2.0*a*d));//c^2=a^2+b^2-2*a*b*cos; Find Central angle Sb=acos ((b*b+d*d-a*a)/(2.0*b*d)); Ibid. area=sa*a*a+sb*b*b-2*sqrt (p* (p-a) * (p-b) * (p-d)),//two sector area and the area of the triangle minus the surface, the area of the intersection} return areas;} int main () {int t,k=0;int r,r;double s1,s2,s3,sum;while (scanf ("%d", &t)!=eof) {while (t--) {k++; scanf ("%d%d%d%d%d%d ", &R,&R,&A1,&B1,&A2,&B2); S1=fun (R,R); S2=fun (R,R); S3=fun (r,r);//The ring can be seen as two circles, you can see the law, the area of two rings is equal to//two large circle intersection area minus two times the large circle and the small Circle intersection area, plus two small circle intersection area if (r>r) sum=s1-2*s3+s2; else sum=s2-2*s3+s1; printf ("Case #%d:%.6lf\n", k,sum);}} return 0;} 

HDU 5120 Intersection