I - Intersection HDU - 5120(圆环相交面积)
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.
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.OutputFor 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:求两个圆环相交的面积 思路:圆环相交面积=外圆1与外圆2相交面积-内圆1与外圆2相交面积-外圆1与内圆2相交面积+内圆1与内圆2相交面积
*/
#include <bits/stdc++.h> #define MAXK 3
#define pi acos(-1) using namespace std; int t;
double r,R;
double x[MAXK],y[MAXK]; double dis(double a1,double b1,double a2,double b2){
return sqrt((a1-a2)*(a1-a2)+(b1-b2)*(b1-b2));
} double cal(double a1,double b1,double R,double a2,double b2,double r){
double d=dis(a1,b1,a2,b2);
if(d>=(R+r)){
return 0.0;
}else if(d<=fabs(R-r)){
return min(pi*R*R,pi*r*r);
}else{
if(r>R){
swap(a1,a2);
swap(b1,b2);
swap(R,r);
}
double ok=sqrt(R*R-r*r);
double x=(R*R-r*r+d*d)/(*d);
x=sqrt(R*R-x*x);
double s1=acos(-(*x*x)/(*R*R));
double s2=acos(-(*x*x)/(*r*r));
if(d>=ok){//相交的很小
return (R*R*s1+r*r*s2-*d*x)/;
}else{//相交的很大
return pi*r*r-(r*r*s2-R*R*s1+*d*x)/;
}
} }
int main(){
//freopen("in.txt","r",stdin);
scanf("%d",&t);
for(int ca=;ca<=t;ca++){
printf("Case #%d: ",ca);
scanf("%lf%lf",&r,&R);
for(int i=;i<;i++){
scanf("%lf%lf",&x[i],&y[i]);
}
printf("%.6f\n",cal(x[],y[],R,x[],y[],R)-
cal(x[],y[],R,x[],y[],r)-
cal(x[],y[],r,x[],y[],R)+
cal(x[],y[],r,x[],y[],r));
}
return ;
}
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