LIGHT OJ 1331 - Agent J 【余弦定理+海伦公式】

1331 - Agent J

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Time Limit: 1 second(s)

Memory Limit: 32 MB

Agent J is preparing to steal an antique diamond piece from a museum. As it is fully guarded and they are guarding it using high technologies, it's not easy to steal the piece. There are three circular laser scanners in the museum which are the main headache for Agent J. The scanners are centered in a certain position, and they keep rotating maintaining a certain radius. And they are placed such that their coverage areas touch each other as shown in the picture below:

Here R₁, R₂ and R₃ are the radii of the coverage areas of the three laser scanners. The diamond is placed in the place blue shaded region as in the picture. Now your task is to find the area of this region for Agent J, as he needs to know where he should land to steal the diamond.

Input

Input starts with an integer T (≤ 1000), denoting the number of test cases.

Each case starts with a line containing three real numbers denoting R₁, R₂ and R₃ (0 < R₁, R₂, R₃ ≤ 100). And no number contains more than two digits after the decimal point.

Output

For each case, print the case number and the area of the place where the diamond piece is located. Error less than 10^-6 will be ignored.

Sample Input	Output for Sample Input
3 1.0 1.0 1.0 2 2 2 3 3 3	Case 1: 0.16125448 Case 2: 0.645017923 Case 3: 1.4512903270

 

题意：求上图蓝色部分面积；

思路：连接三个圆心，构成三角形，蓝色面积=三角形面积-三个圆弧面积，三角形面积用海伦公式，圆弧面积必须知道圆弧所对的圆心角大小，问题就是知道三角形的三条边求三个角的大小，用余弦定理解决；

失误：刚开始没有想到余弦定理（其实是忘记了），只想到正玄定理，求了好长时间求不出来，浪费时间不少，以后这种判定是数学题的就把他转化成数学题来解，不然数学中的东西不容易想到。

AC代码：

#include
#include

#define PI acos(-1.0)

double Get(double a,double b,double c)
{
	double t=(a*a+b*b-c*c)/(2*a*b);
	return acos(t)/2.0;
}

int main()
{
	int T,Kase=0;
	double R1,R2,R3,S1,S2,p,a,b,c;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%lf %lf %lf",&R1,&R2,&R3);
		a=R1+R2; b=R2+R3; c=R3+R1;
		p=(a+b+c)/2.0;
		S1=sqrt(p*(p-a)*(p-b)*(p-c));
		S2=Get(a,b,c)*R2*R2+Get(a,c,b)*R1*R1+Get(b,c,a)*R3*R3;
		printf("Case %d: %lf\n",++Kase,S1-S2);
	}
	return 0;
}