Intersection
Time Limit: 4000/4000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 2754 Accepted Submission(s): 1025
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 ≤ 10 5), 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
2
2 3
0 0
0 0
2 3
0 0
5 0
Sample Output
Case #1: 15.707963
Case #2: 2.250778
Source 2014ACM/ICPC亞洲區北京站-重制賽（感謝北師和上交）
Recommend liuyiding
題意：兩圓環相交，求兩圓環相交的面積
對着模闆打了，錯了好多遍，最後知道是圓環相交的面積
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
#define Pi 3.14159265358979
struct node
{
double x,y,r;
}q[10];
double f(node a,node b)
{
double d=sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
if(a.r+b.r<=d)
{
return 0;
}
double r=a.r>b.r?b.r:a.r;
if(fabs(a.r-b.r)>=d)return Pi*r*r;
double k1=acos((a.r*a.r+d*d-b.r*b.r)/(2.0*a.r*d));
double k2=acos((b.r*b.r+d*d-a.r*a.r)/(2.0*b.r*d));
double s1=k1*a.r*a.r-a.r*sin(k1)*a.r*cos(k1);
double s2=k2*b.r*b.r-b.r*sin(k2)*b.r*cos(k2);
return s1+s2;
}
int main()
{
double r1,r2,x1,x2,y1,y2;
int t;
scanf("%d",&t);
for(int h=1;h<=t;h++)
{
scanf("%lf%lf",&r1,&r2);
scanf("%lf%lf",&q[0].x,&q[0].y);
scanf("%lf%lf",&q[1].x,&q[1].y);
q[1].r=r2;
q[0].r=r2;
double s1=f(q[0],q[1]);//dada
//printf("%.6lf\n",s1);
q[0].r=r1;
double s2=f(q[0],q[1]);//daxiao
//printf("%.6lf\n",s2);
q[1].r=r1;
double s3=f(q[0],q[1]);//xiaoxiao
printf("Case #%d: %.6lf\n",h,s1-2*s2+s3);
}
return 0;
}