hdoj5120Intersection【求圆相交的面积】

Intersection

Time Limit: 4000/4000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 1254    Accepted Submission(s): 488

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 x_i, y_i (0 ≤ x_i, y_i ≤ 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

 

给出两个同心圆的圆心坐标和同心圆半径求如图所示阴影面积

#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#define PI acos(-1.0)#define eps 1e-8using namespace std;int sgn(double n){if(fabs(n)<eps)return 0;if(n<0)return -1;return 1;}double MIN(double a,double b){return a<b?a:b;}double dist(double x1,double y1,double x2,double y2){return sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));}double getArea(double x1,double y1,double x2,double y2,double r1,double r2){double d=dist(x1,y1,x2,y2);if(sgn(d-r1-r2)>=0)return 0;//不相交if(sgn(d-fabs(r1-r2))<=0){//包含 double r=MIN(r1,r2);return PI*r*r;}       double cos_r1=(d*d+r1*r1-r2*r2)/(2.0*d*r1);//利用余弦定理算角度 double angle1=acos(cos_r1);double cos_r2=(d*d+r2*r2-r1*r1)/(2.0*d*r2);double angle2=acos(cos_r2);return fabs(r1*r1*angle1-0.5*r1*r1*sin(angle1*2.0)+r2*r2*angle2-0.5*r2*r2*sin(2.0*angle2));//算相交的面积 }int main(){int t,n,i,j,k=1;double r,R,x1,x2,y1,y2;scanf("%d",&t);while(t--){scanf("%lf%lf",&r,&R);scanf("%lf%lf",&x1,&y1);scanf("%lf%lf",&x2,&y2);double big_circleinter=getArea(x1,y1,x2,y2,R,R);//两个大圆相交面积 double small_circleinter=getArea(x1,y1,x2,y2,r,r);//两个小圆相交面积 double small_big_circleinter=getArea(x1,y1,x2,y2,r,R);//小圆和大圆相交面积 printf("Case #%d: %.6lf\n",k++,big_circleinter-2.0*small_big_circleinter+small_circleinter);}return 0;}