* 用拉格朗日插值法依据N个已知数据点即使函数值

* 输入: n--已知数据点的个数N-1

* x--已知数据点第一坐标的N维列向量

* y--已知数据点第二坐标的N维列向量

* xx-插值点第一坐标

* 输出: 函数返回值所求插值点的第二坐标

-* Use Lagrange interpolation method based on N-known data points even function * Input : n -- known data points the number N-1 * x -- known data points of the first N-dimensional coordinates listed Vector * y -- Known number stronghold of the second N-dimensional coordinates listed Vector * xx-point first interpolation * Coordinate output : function return values for the second interpolation point coordinates