设两点为(x1, y1), (x2, y2)那可用f(x)=g(x)(x-x1)(x-x2)+y1*(x-x2)/(x1-x2)+y2*(x-x1)/(x2-x1) 这里g(x)可选取任意函数则上面的f(x)满足过这两点f(x1)=y1, f(x2)=y2
