WGS84转GCJ-02坐标系容易y=x+x2+x3+x4+...，但是反过来就难了。。原因在于高次代数方程求根。(https://www.bilibili.com/read/cv2121106)(一元n次方程的解法) 牛顿法和一元n次方程
要实现高精度转换，需要进行迭代，而这比较耗时。如果只取前4位，多项式则容易求解，但是精度不够。
有没有较快的方法实现高精度逆变换。
C++：https://www.cnblogs.com/2008nmj/p/14198389.html
以lonGCJ-02,latGCJ-02为观测值，而lonWGS84,latWGS84为未知数，列误差方程并线性化，则线性化后的误差方程为：
v=Ax-l
其中，
，
，
而，系数矩阵A展开为：
要使得误差v=0，那么需要什么条件：
/** * \brief Covert geodetic coordinate in GCJ-02 coordinate system to geodetic coordinate * in WGS84 coordinate system * * \param [in] gcj02lon: longitude in GCJ-02 coordinate system [unit:degree] * \param [in] gcj02lat: latitude in GCJ-02 coordinate system [unit:degree] * \return Returns geodetic coordinate in WGS84 coordinate system * \remark The encryption formula is known,and use the analytical derivation method to * solve the problem with high precision. * \detail Assuming the encryption formula is * * gcj02lon = Wgs2Gcj(wgs84lon, wgs84lat) * gcj02lat = Wgs2Gcj(wgs84lon, wgs84lat) * * In the rectification process, (wgs84lon, wgs84lat) are unknown items. Obviously, * this is a system of nonlinear equations. * * The linear formed error functions of forward intersection come from * consideration of a Taylor series expansion. * V = AX - b * here: * V: The residuals of the observed values * A: The jacobian matrix * X: The modification of the unknown items * b: The constant terms of the error functions * * Then the error functions written in vector form are: * | V_lon | = | dlongcj_dlonwgs dlongcj_dlatwgs | | d_lonwgs | - | l_lon | * | V_lat | = | dlatgcj_dlonwgs dlatgcj_dlatwgs | | d_latwgs | - | l_lat | * here: * l_lon = longcj - longcj' // the modification of longcj * l_lat = latgcj - latgcj' // the modification of latgcj * longcj : the observed longitude in GCJ-02 * latgcj : the observed latitude in GCJ-02 * longcj' = Wgs2Gcj(wgs84lon',wgs84lat') // estimated longitude in GCJ-02 * latgcj' = Wgs2Gcj(wgs84lon',wgs84lat') // estimated latitude in GCJ-02 * wgs84lon' : estimated longitude in WGS84 * wgs84lat' : estimated latitude in WGS84 * d_lonwgs : unknown items * d_latwgs : unknown items * wgs84lon = wgs84lon' + d_lonwgs // update * wgs84lat = wgs84lat' + d_latwgs * * let V = [V_lon V_lat]T = 0, then * d_latwgs = (l_lon * dlatgcj_dlonwgs - l_lat * dlongcj_dlonwgs) / * (dlongcj_dlatwgs * dlatgcj_dlonwgs - dlatgcj_dlatwgs * dlongcj_dlonwgs) * d_lonwgs = (l_lon - dlongcj_dlatwgs * d_latwgs) / dlongcj_dlonwgs * * This iterative procedure is repeated until X= [d_lonwgs d_latwgs]T are * sufficiently small. * \time 01:54:46 2020/06/13 */
