杭电1013——Digital Roots（简单题）

最新推荐文章于 2020-03-25 22:10:54 发布

flying_fish_233

最新推荐文章于 2020-03-25 22:10:54 发布

阅读量483

点赞数

CC 4.0 BY-SA版权

分类专栏：杭电acm

本文链接：https://blog.youkuaiyun.com/qianchangdiyin/article/details/49359681

杭电acm 专栏收录该内容

30 篇文章

订阅专栏

本文详细介绍了数字根的概念，即通过不断将数字相加直至得到一位数的过程，并提供了有效的算法实现。针对输入的正整数，文章阐述了如何通过编程解决此问题，包括从使用int型数据到正确使用string类的转变，以应对大数值输入的挑战。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Problem Description
The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value contains two or more digits, those digits are summed and the process is repeated. This is continued as long as necessary to obtain a single digit.

For example, consider the positive integer 24. Adding the 2 and the 4 yields a value of 6. Since 6 is a single digit, 6 is the digital root of 24. Now consider the positive integer 39. Adding the 3 and the 9 yields 12. Since 12 is not a single digit, the process must be repeated. Adding the 1 and the 2 yeilds 3, a single digit and also the digital root of 39.

Input
The input file will contain a list of positive integers, one per line. The end of the input will be indicated by an integer value of zero.

Output
For each integer in the input, output its digital root on a separate line of the output.

Sample Input
24
39
0

Sample Output
6
3

一开始用int型数据保存输入，结果WA。后来发现，输入的n，可以很大，所以应该用string类保存输入。

#include<iostream>
#include<string>
#include<cstdio>
using namespace std;

int main()
{
    string num="";  
    while(cin>>num&&num.compare("0")!=0)
    {
        while(num.length()>1)
        {
            int tnum=0,i;
            for(i=0;i<num.length();i++)
            {
                tnum+=num[i]-'0';
            }
            char *temp=new char[num.length()+1];
            sprintf(temp,"%d",tnum);
            num=string(temp);   
        }
        cout<<num<<"\n";
    }
    return 0;
}