已知仿射变换的系数，求仿射变换的反向系数

对于如下从（xx，yy）到（X，Y）的仿射变换：

X = xx + ax * xx + bx * yy + cx
Y = yy + ay * xx + by * yy + cy

已知（ax，bx，cx），（ay，by，cy）为其仿射变换系数，则（X，Y）到（xx，yy）必然也存在仿射变换关系：

xx = X + axp* X + bxp * Y + cxp
yy = Y + ayp * X + byp * Y + cyp

求其对应的反向系数（axp，bxp，cxp），（ayp，byp，cyp）。

int main()
{	
	double ax = -0.00012457722524773310000000000;
	double bx = -0.00026952578840466643000000000;
	double cx = -40.53236981248609500000000000000;

	double ay = 0.00088450690917435935000000000;
	double by = -0.00006127575130507561900000000;
	double cy = -210.50907546352857000000000000000;

	double tt = ax * by + ax + by + 1 - bx * ay;

	double axp = (by+1) / tt - 1;
	double bxp = -bx / tt;
	double cxp = (bx * cy - cx - by * cx) / tt;
	double ayp = -ay / tt;
	double byp = (ax + 1) / tt - 1;
	double cyp = (ay * cx - cy - ax * cy) / tt;

	printf("%.10lf	%.10lf	%.10lf
", axp, bxp, cxp);
	printf("%.10lf	%.10lf	%.10lf
", ayp, byp, cyp);

	double xx = 500;
	double yy = 600;

	double X = xx + ax* xx + bx* yy + cx;
	double Y = yy + ay * xx + by * yy + cy;

	printf("原始：%.10lf	%.10lf
", xx, yy);
	printf("正运算：%.10lf	%.10lf
", X, Y);
	
	//axp = 0.00012480028850991864000000000;
	//bxp = 0.00026947569230899039000000000;
	//cxp = 40.47053059568900600000000000000;

	//ayp = -0.00088529760825697967000000000;
	//byp = 0.00006131347116844487700000000;
	//cyp = 210.73126518749581000000000000000;

	xx = X + axp* X + bxp * Y + cxp;
	yy = Y + ayp * X + byp * Y + cyp;
	printf("逆运算：%.10lf	%.10lf
", xx, yy);	

    return 0;
}