杭电 1013

Digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 48103    Accepted Submission(s): 14939

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

24390

 

Sample Output

63

 

Source

Greater New York 2000

 

这道题有个小陷阱 需要特别注意 就是要考虑到大数的问题 通过一般的类型肯定要溢出

可以利用 字符型数组 进行运算 至于判断是否是两或多位数的时候 可以通过递归来判定

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int dr(int n)
{
    int t=0;
    while(n)
    {
            t+=n%10;
            n/=10;
            }//这个式子在不知道位数的时候也能将一个数的各个位数分离开来
            if(t/10)//此处是递归
            dr(t);
            else
            return t;
    }
int main()
{
    char a[1010];
    while(~scanf("%s",a))
    {
                         int sum=0,i;
                         if(a[0]=='0')
                         break;
                         else
                         {
                             for(i=0;i<strlen(a);i++)
                             sum+=a[i]-'0';
                             printf("%d\n",dr(sum));
                             }
                         }
    return 0;
    }