* Use Lagrange interpolation method based on N

* 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

拉格朗日插值法-C代码.doc