zoj1115-----------Digital Roots

最新推荐文章于 2017-06-29 16:33:11 发布

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本文介绍了一种计算数字根的方法，即通过不断累加整数的各位数字直至得到一位数的过程。文章提供了一个C++实现的例子，包括如何读取输入、进行累加计算，并通过循环进一步简化数字直到达到数字根。

Background

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.

Example

Input

24
39
0

Output

6
3

题目很容易理解，

因为数字很大，所以将其视为一个字符串，用gets()输入数组s[ ]，

然后根据题意将数组中的数字相加

这时所得数字位数必然比较低，根据余数继续运算

代码：

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;

int main()
{
   char s[1000];
   int i,j,sum,temp;
   while(gets(s)&&strcmp(s,"0")!=0)  //strcmp是字符串比较函数 需要头文件string.h
   {
      int sum=0;
      for(i=0;s[i]!='\0';i++)
         sum=sum+s[i]-'0';           //减去ASCII码才是其数值大小
      while(sum>=10)                 //根据余数求较小的这个数
      {
         j=sum;
         temp=0;
         while(j>0)
         {
            temp=temp+j%10;
            j=j/10;
         }
         sum=temp;
      }     
      cout<<sum<<endl;         
   }
   return 0;
}