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======拉格朗日插值====== 已知$n+1$个点，可确定一个n次的线性多项式，可用拉格朗日插值求解该多项式的某一点的取值。 由公式可确定 $$f(k)=\sum_{i=1}^ny[i]\prod_{i\ne j}\frac{k-x[j]}{x[i]-x[j]}$$ 由上述公式可确定该多项式的$x=k$的时候的取值$f(k)$