Tell me the area
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1801 Accepted Submission(s): 542
Problem Description
There are two circles in the plane (shown in the below picture), there is a common area between the two circles. The problem is easy that you just tell me the common area.
Input
There are many cases. In each case, there are two lines. Each line has three numbers: the coordinates (X and Y) of the centre of a circle, and the radius of the circle.
Output
For each case, you just print the common area which is rounded to three digits after the decimal point. For more details, just look at the sample.
Sample Input
0 0 2
2 2 1
Sample Output
0.108
Author
wangye
Source
Recommend
利用余弦公式计算出角度，然后通过扇形和三角形的关系求出答案即可
#include<cmath> #include<cstdio> #include<cstring> #include<stdlib.h> #include<algorithm> using namespace std; double distance(double x1,double y1,double x2,double y2) { return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); } int main() { //freopen("in.txt","r",stdin); double PI=2*asin(1.0); double x1,y1,r1; double x2,y2,r2; double dis,r,cosa,cosb; double a,b,c,d,ans; while(scanf("%lf %lf %lf %lf %lf %lf",&x1,&y1,&r1,&x2,&y2,&r2)!=EOF) { dis=distance(x1,y1,x2,y2); if(dis>=r1+r2||r1==0||r2==0)//两圆相离，外切时，相交面积为0 ans=0; else if(dis<=fabs(r1-r2))//两圆内含，内切时，相交面积为较小圆的面积 { r=r1<r2?r1:r2; ans=r*r*PI; } else//两圆相交时 { cosa=acos((r1*r1+dis*dis-r2*r2)/(2*r1*dis)); cosb=acos((r2*r2+dis*dis-r1*r1)/(2*r2*dis));//余弦公式求出两圆连线和半径的夹角 a=r1*r1*cosa; b=r2*r2*cosb;//扇形面积 c=r1*r1*cos(cosa)*sin(cosa); d=r2*r2*cos(cosb)*sin(cosb);//三角形面积 // s1=0.5*r1*r1*(cos1-sin(cos1)); // s2=0.5*r2*r2*(cos2-sin(cos2)); ans=(a+b)-(c+d);//这里用两个扇形的面积之和减去三角形之和，是因为相交的情况分为两个圆的圆心是否在同一个圆里两种情况 } printf("%.3lf ",ans); } return 0; }