Intersection
Time Limit: 4000/4000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 3178 Accepted Submission(s): 1203
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
题意：给你圆环的内半径和外半径，再给你两个圆环(圆环相同)的中点坐标，求两个圆环重合的面积
思路：先画个图，由图可知黑色部分为相交的面积，将一个圆环分解成一个小圆和一个大圆，用area表示两个圆相交的面积，那么黑色部分面积就是area(A,B)-area(A,b)-area(B,a)+area(a,b)
只用写一个两个圆相交面积的函数即可
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define PI 3.14159265357
struct circle
{
double x,y;
double r;
};
double dis(circle a,circle b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double solve(circle a, circle b) {
double d = dis(a, b);
if (d >= a.r + b.r)
return 0;
if (d <= fabs(a.r - b.r)) {
double r = a.r < b.r ? a.r : b.r;
return PI * r * r;
}
double ang1 = acos((a.r * a.r + d * d - b.r * b.r) / 2. / a.r / d);
double ang2 = acos((b.r * b.r + d * d - a.r * a.r) / 2. / b.r / d);
double ret = ang1 * a.r * a.r + ang2 * b.r * b.r - d * a.r * sin(ang1);
return ret;
}
int main()
{
circle a,b,A,B;
double r,R;
int t,cas=1;
scanf("%d",&t);
while(t--)
{
scanf("%lf%lf",&r,&R);
a.r=b.r=r;
A.r=B.r=R;
scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y);
A.x=a.x;
A.y=a.y;
B.x=b.x;
B.y=b.y;
printf("Case #%d: ",cas++);
double area=solve(A,B)-solve(A,b)-solve(B,a)+solve(a,b);
printf("%lf\n",area);
}
return 0;
}
