hdoj Intersection 5120 (数学计算几何) 求两个相交圆的面积
2015-11-07 13:08
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Intersection
Time Limit: 4000/4000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 1643 Accepted Submission(s): 620
#include<stdio.h> #include<string.h> #include<math.h> #define P acos(-1.0)//定义π #define E 1e-6 #define min(a,b) (a>b?b:a) double dis,r,R; struct zz { double x,y; }a,b; double ll(double x1,double y1,double x2,double y2) { return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); } double fun(double r1,double r2) { if(dis>=(r1+r2)) return 0; double s1,s2,ss1,ss2; if(dis<=abs(r1-r2)) { r1=min(r1,r2); return P*r1*r1; } s1=(r1*r1+dis*dis-r2*r2)/(2*r1*dis); s2=(r2*r2+dis*dis-r1*r1)/(2*r2*dis); ss1=acos(s1); ss2=acos(s2); return ss1*r1*r1+ss2*r2*r2-r1*dis*sin(ss1); } int main() { int t; int T=1; scanf("%d",&t); while(t--) { scanf("%lf%lf",&r,&R); scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y); dis=ll(a.x,a.y,b.x,b.y); if(abs(dis)<E) { printf("Case #%d: %.6lf\n",T++,P*(R*R-r*r)); continue; } double ss=0; ss=fun(R,R); ss-=2*fun(R,r)-fun(r,r); printf("Case #%d: %.6lf\n",T++,ss); } return 0; }
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
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.
[align=left]Sample Input[/align]
2
2 3
0 0
0 0
2 3
0 0
5 0
[align=left]Sample Output[/align]
Case #1: 15.707963
Case #2: 2.250778//题意:给你两个圆的半径,r,R;再给你这两个圆的圆心点,让你求出这两个圆的不相交面积。
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