<h1 id="heading">题目</h1>
<blockquote>
<h2 id="description">Description</h2>
<blockquote>
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 &lt; R). For more details, refer to the gray part in the illustration below. 
<div class='image'><a href='http://acm.hdu.edu.cn/data/images/C556-1009-2.jpg' target='_blank' rel='noopener noreferrer'><img src="http://acm.hdu.edu.cn/data/images/C556-1009-2.jpg" /></a></div> 
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.
 
</blockquote>
<h2 id="input">Input</h2>
<blockquote>
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 &lt; 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.
</blockquote>
<h2 id="output">Output</h2>
<blockquote>
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.
</blockquote>
<h2 id="sample-input">Sample Input</h2>
<blockquote>
2 
2 3 
0 0 
0 0 
2 3 
0 0 
5 0
</blockquote>
<h2 id="sample-output">Sample Output</h2>
<blockquote>
Case #1: 15.707963 
Case #2: 2.250778
</blockquote>
</blockquote>
<h1 id="heading-1">题解</h1>
纯数学公式推导题 
由于牵扯的小数,需要使用 <code>double</code> 进行运算 
由于会涉及浮点误差 
因此在比较大小的时候需要注意把较大数加上一个偏移量
貌似用海伦公式会出现问题,最好用余弦定理
总共有三种情况
<ul>
<li>完全相离,面积为 <code>0</code></li>
<li>中心重合,面积就是大圆减小圆</li>
<li>其他情况,需要4部分相加减计算得出</li>
</ul>
<h1 id="heading-2">代码</h1>
<pre style="background-color:#fff">/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/
#include &lt;cstdio&gt;
#include &lt;algorithm&gt;
#include &lt;cstring&gt;
#include &lt;cmath&gt;
#include &lt;string&gt;
#include &lt;iostream&gt;
#include &lt;vector&gt;
#include &lt;list&gt;
#include &lt;queue&gt;
#include &lt;stack&gt;
#include &lt;map&gt;
#include &lt;set&gt;
#include &lt;functional&gt;
#include &lt;bitset&gt;
using namespace std;

const int maxn = 45;
const int maxm = 1000005;
const double pi = 3.141592653589;
int a[maxn];
int dp[maxn][maxm];

struct O {
 double x;
 double y;
 double r;
 O(double x,double y,double r) {
 this-&gt;x = x;
 this-&gt;y = y;
 this-&gt;r = r;
 }
};
inline double Ha(double a,double b,double c) {
 double p = (a + b + c) / 2;
 double ans = sqrt(p*(p - a)*(p - b)*(p - c));
 return ans;
}
inline double distance(double x1,double y1,double x2,double y2) {
 double ans = sqrt((x1 - x2)*(x1 - x2) + (y1 - y2)*(y1 - y2));
 return ans;
}
inline double Sshanxing(double r,double h) {
 double th = asin(h / r);
 return th * r *r / 2;
}
inline double Ssanjiao(double h,double r) {
 return sqrt(r*r - h*h)*h / 2;
}
inline double Syuan(double r) {
 return pi*r*r;
}


inline double S(O o1,O o2) {
 double dis = distance(o1.x,o1.y,o2.x,o2.y);
 if(dis + 1e-8 &gt; o2.r + o1.r)
 return 0;
 if(dis &lt; fabs(o2.r - o1.r) + 1e-8) {
 double r = min(o1.r,o2.r);
 return pi * r * r;
 }


 double costh1 = (o1.r*o1.r + dis*dis - o2.r*o2.r) / (2 * o1.r*dis);
 double costh2 = (o2.r*o2.r + dis*dis - o1.r*o1.r) / (2 * o2.r*dis);
 double th1 = acos(costh1);
 double th2 = acos(costh2);
 return th1*o1.r*o1.r + th2*o2.r*o2.r - o1.r*dis*sin(th1);

 /*double h = Ha(o1.r,o2.r,dis) / dis * 2;
 double S1 = Sshanxing(o1.r,h) - Ssanjiao(h,o1.r);
 double S2 = Sshanxing(o2.r,h) - Ssanjiao(h,o2.r);
 double s = (S1 + S2) * 2;
 return s;*/
}

void Do() {
 double r,R;
 double x1,x2,y1,y2;
 scanf(&#34;%lf%lf&#34;,&amp;r,&amp;R);
 scanf(&#34;%lf%lf&#34;,&amp;x1,&amp;y1);
 scanf(&#34;%lf%lf&#34;,&amp;x2,&amp;y2);

 O o1x(x1,y1,r);
 O o1d(x1,y1,R);
 O o2x(x2,y2,r);
 O o2d(x2,y2,R);

 double ans = S(o1d,o2d) - S(o1x,o2d) - S(o1d,o2x) + S(o1x,o2x);

 printf(&#34;%.6f\n&#34;,ans);
}

int main() {
 int T;
 scanf(&#34;%d&#34;,&amp;T);
 for(int i = 1;i &lt;= T;i++) {
 printf(&#34;Case #%d: &#34;,i);
 Do();
 }
 return 0;
}
</pre>

HDU 5120.Intersection