开发具有如下函数原型的递归与非递归两种函数equ，负责判断数组a与b的前n个元素值是否按下标对应完全相同，是则返回true，否则返回false。并编制主函数对它们进行调用，以验证其正确性。bool equ(int a[]， int b[]， int n) 提示：递归函数中可按如下方式来分解并处理问题，先判断最后一个元素是否相同，不同则返false；相同则看n是否等于1，是则返回true，否则进

-The development has the following function prototype recursive and two non-recursive function equ responsible for judging the array a and b of the first n element values ​ ​ corresponding to the press labeled exactly the same, is it returns true, otherwise it returns false. And the preparation of the main function to call them to verify their correctness. bool equ (int a [], int b [], int n) Note: according to the following ways to break down and deal with the problem, first determine the last element is the same, different then return false　same n is equal to the recursive function 1, it returns true, otherwise into the