static solution invx(solution solu2)
{	
	solution solu = solu2;
	float L1_U,L2_U;	

	if(solu.CACL_A > solu.CACL_B){
		L1_U = L1;
		L2_U = L2;
	}else{
		L1_U = L2;
		L2_U = L1;
	}

	float ax = solu.CACL_A-solu.CACL_B;

	float A = pow(H,2)+pow(L1_U,2)-pow(ax,2)-pow(L2_U,2);
	float B = 2*H*L1_U;
	float C = 2*ax*L2_U;

	float a = pow(B,2)*pow(L2_U,2)+pow(C,2)*pow(L1_U,2);
	float b = -2*L1_U*A*B*pow(L2_U,2)-2*H*pow(L1_U,2)*pow(C,2);
	float c = pow(A,2)*pow(L1_U,2)*pow(L2_U,2)-pow(C,2)*pow(L1_U,2)*pow(L2_U,2)+pow(C,2)*pow(L1_U,2)*pow(H,2);

	solution solu3 = caclduo(a,b,c);
	
	if(solu.CACL_A > solu.CACL_B){
		solu.solved.x = solu.CACL_B + sqrt(pow(L1,2)-pow(solu3.DeltaX,2));
		solu.solved.y = solu3.DeltaX;
	}else{
		solu.solved.x = solu.CACL_A + sqrt(pow(L2,2)-pow(solu3.DeltaX,2));
		solu.solved.y = H-solu3.DeltaX;
	}

	// cout << sqrt(pow(L2,2)-pow(solu.DeltaX,2)) << "," << sqrt(pow(L1,2)-pow(solu.DeltaX,2)) << "," << org_a << "," << org_b << "," << ax << endl;

	return solu;
}

static solution getx(float x,float y)   //计算
{
	solution solu = caclduo(1,-2*x,pow(x,2)+pow(y,2)-pow(L1,2));    //计算二元一次函数，并取较小的那个根

	if(solu.success)
	{
		x -= solu.DeltaX;   //在这一步我们得到了最小的那个根，也就是L(X0,0),然后把他转换到相对坐标轴上，把原点通过X轴平移到L点。
		float a1 = atan(y/x);   //请结合上图，理解下面的代码
		float sina2 = (H-L1*sin(a1))/L2;    //请结合上图，理解下面的代码 
		float cosa2 = sqrt(1-pow(sina2,2));      //请结合上图，理解下面的代码
		float ax = L2*cosa2;

	    // cout << "a1:" << a1 << "  sina1:" << sin(a1) << "  sina2:" << sina2 << "  cosa2:" << cosa2 << "  ax:" << ax <<"  DeltaX:" << solu.DeltaX << endl;

		solu.CACL_A = solu.DeltaX + (x-ax);     //解算完成，由于是在相对坐标系上计算了，需要把原点移回绝对坐标系
		solu.CACL_B = solu.DeltaX;
	}
	return solu;
}

//ax^2+bx+c=0,返回较小方程的根
static solution caclduo(float a,float b,float c)    //二元一次方程计算函数，输入a,b,c
{
	solution solu;
	float root = pow(b,2)-4*a*c;    //计算根
	solu.success = true;
	if(root>=0 || a!=0) //二次函数不存在
	{
		if(root == 0)   //二次函数只有一个解
		{
			solu.DeltaX = -b/2*a;   //返回解
			solu.success = true;
			goto end;
		}
		float x1 = (-b+sqrt(root))/(2*a);   //通过求根公式计算方程的解
		float x2 = (-b-sqrt(root))/(2*a);   

		// cout << "a:" << a << "  b:" << b << "  c:" << c << "  root:" << root << "  Dx1:" << x1 << "  Dx2:" << x2 << endl;

		if(x1>x2)
			solu.DeltaX = x2;          //由于安装方式，取最靠近原点的值作为增量值。
		else
			solu.DeltaX = x1;

	}
	else 
		solu.success = false;
	end:return solu;
}

...(约200行左右)
#define Ax 218
#define Ay 32
...

...(约160行左右)
const float L1 = 110.0;
const float L2 = 110.0;
const float H = 110.0;
...

g++ plot.c -lgdi32 -o plot.exe
./plot.exe