找到数字的总和，直到只剩下1位数 [英] Find the sum of the digits until only 1 digit remains

问题描述

大家好，



我在创建一个循环达到1位数时遇到了一些困难。例如，999992确实变为47和11，但不是2，因为它应该....



这是我的代码：

Hi guys,

I am having some difficulties creating a loop to get to 1 digit. E.g 999992 does change to 47 and 11, but not to 2 as it should....

Here's my code:

#include "stdafx.h"
#include <iostream>
using namespace std;

int main()
{
	int n, sum=0, total=0;
	cout << "Enter an integer." << endl;
	cin >> n;
	cout << "You have entered: " << n << endl;
	
	while (n > 0) {
	 sum += n % 10;
	 n /= 10;
	}

	cout << "The digitsum is: " << sum << endl;
	
	if (sum < 10)
	 cout << "The total digitsum is: "<< sum << endl;
	else 
	 do {
	  total += sum % 10;
	  sum /= 10;
	  cout << "The total digitsum is: " << total << endl;
	 } while (total < 10);
	
	system("pause");
	return 0;
}

任何人都可以帮我吗？

Can anyone help me out here?

推荐答案

请参阅我的评论功能。公平规则告诉使用：它足以找到第一个bug，然后你应该把它带入它帐户并修复它。



至少创建两个函数：一个计算数字之和，另一个代表外循环。您的查找数字总和的算法第一次正常工作。如果你把它作为一个单独的函数，变量 sum 将在每次调用时初始化为0。问题的其余部分将以微不足道的方式解决。现在轮到你了。

-SA

Please see my comments to the function. Fair rules tell use: it's enough to locate just first bug, then you should take it into it account and fix it.

Create at least two more functions: one calculating sum of digits, another one representing outer cycle. Your algorithm of finding sum of digits works correctly for the very first time. If you make it a separate function, the variable sum will be initialized to 0 on each call. The rest of the problem will be solved in a trivial way. It's your turn now.

—SA

这篇关于找到数字的总和，直到只剩下1位数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！