H电-Problem Archive-5120-Intersection

Intersection

Time Limit: 4000/4000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 177    Accepted Submission(s): 76

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 ≤ 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 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

22 30 00 02 30 05 0

 

Sample Output

Case #1: 15.707963Case #2: 2.250778

 

Source

2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交）

S：

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>using namespace std;double w = 1e-8;double pi = acos(-1.0);struct node{    double x, y;}p, q;double f(node a,double r,node b,double R){    double d = sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));    if(d >= r + R + w)  return 0;    if(fabs(R - r) + w >= d){        int rr = min(r, R);        return pi * rr * rr;    }    double x = (r * r + d * d - R * R) / (2 * d);    double g1 = acos(x / r);    double g2 = acos((d - x) / R);    return r * r * g1 + R * R * g2 - d * r * sin(g1);}int main(){    int T, k = 1;    double ans;    scanf("%d", &T);    while(T--){        double x1, x2, y1, y2, r1, r2;        scanf("%lf%lf%lf%lf%lf%lf", &r1, &r2, &x1, &y1, &x2, &y2);        q.x = x1, q.y = y1;        p.x = x2, p.y = y2;        /*        if(x1 == x2 && y1 == y2){            printf("Case #%d: %.6lf\n", k++, ans);            continue;        }        */        double ans = f(q, r2, p, r2) - f(q, r1, p, r2) - f(q, r2, p, r1) + f(q, r1, p, r1);        printf("Case #%d: %.6lf\n", k++, ans);    }    return 0;}