**
题干要求
其中n是多项式的阶数，a[]中存储系数，x是给定点。函数须返回多项式f(x)的值。 这里主要讲解自定义函数的原理：
一、涉及的数学知识
- 用到多项式的合并与展开
简单 例子：3x^2+2x+1=x(3x+2)+1
- 本题假设我们输入 :n=3 ; x =2.0;
a[0]=1 ; a[1]=1;a[2]=2 ;a[3]=3;
则f(x)=a[0] * (x^0) + a[1] * (x^1)+ a[2] * (x ^2)+a[3] * (x^3)=
x(x(a[3])x+a[2])+a[1])+a[0] ;
用循环计算 x(x(a[3])x+a[2])+a[1])+a[0]； 然后依照先算括号里面的原则计算。
用到的循环语句是 add = add * x + a[i];
add 等于0是为了第一次计算引出a[3];
二、代码实现
#include <stdio.h>
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