UC知道c++编程，关于回数和质数。
来源：百度知道 编辑：UC知道 时间：2024/06/28 05:10:25
Problem Description
Palindromes are words or numbers that are read the same forwards and backwards. For example: 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121 are palindrome numbers.
Prime numbers are the numbers that are dividable by only one and by themselves. For example: 1, 2, 3, 5, 7, 11, 13, 17, 19, 23 are prime numbers.
Develop an algorithm and then translate it to a C++ program to prompt user for an integer number, N. Then it prints out all prime palindrome numbers less that N.
For example: 1, 2, 3, 5, 7, 11, 101 are all palindrome prime numbers less than 102.
Use top-down design, and divide the problem into appropriate sub-task like we did in the class. Then translate them into C++ functions.
Note 1: You should develop at least four functions. The body of the main program should be very simple, and it should call appropriate functions to do the task.
Note 2: We have not studied Strings and arrays yet, s
Palindromes are words or numbers that are read the same forwards and backwards. For example: 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121 are palindrome numbers.
Prime numbers are the numbers that are dividable by only one and by themselves. For example: 1, 2, 3, 5, 7, 11, 13, 17, 19, 23 are prime numbers.
Develop an algorithm and then translate it to a C++ program to prompt user for an integer number, N. Then it prints out all prime palindrome numbers less that N.
For example: 1, 2, 3, 5, 7, 11, 101 are all palindrome prime numbers less than 102.
Use top-down design, and divide the problem into appropriate sub-task like we did in the class. Then translate them into C++ functions.
Note 1: You should develop at least four functions. The body of the main program should be very simple, and it should call appropriate functions to do the task.
Note 2: We have not studied Strings and arrays yet, s
不是什么难事吧
它只不过要求 不用数组，不用字符串
必须用函数，函数放在头文件里面，要求调用
不用数组确实麻烦。。。
建立头文件，在里面做两个函数，两个函数都只有一个参数，返回值可以是0/1，也可以是别的 ，你自己看着办
函数1实现：检查一个数是不是回数，用求余的方法算，看首位跟末位是不是一样，一样就是，不一样就不是，函数返回一个数，确认输入的这个参数到底是不是回数
函数2实现：检查一个数是不是素数，这个就不用我说了吧
最后建立 主函数 调用原来那个头文件
主函数里面一个循环， 1--N
用两个函数一个一个查看，都符合条件的就是所需要的数
完成。。。。。要是用数组 或许思路清晰一点
c+课本上有，你自己查去
给个链接