HDU 1013Digital Roots 吃完饭，水个题。

Digital Roots

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 53073 Accepted Submission(s): 16565

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

Sample Output

九余数定理，弃九法，“弃九法”也叫做弃九验算法，利用这种方法可以验算加、减计算的结果是否错误。把一个数的各位数字相加，直到和是一个一位数（和是9，要减去9得0），这个数就叫做原来数的弃九数。可以用来检验加减乘除算式的正确性

这题是大数，要用字符串。

#include <iostream>#include <stdio.h>#include <string>#include <cstring>#include <algorithm>#define N 100009using namespace std;char s[N];int main(){    while(~scanf("%s",s))    {        if(s[0]=='0')break;        int ans=0;      int len=strlen(s);      if(len==1)      {          cout<<s<<endl;          continue;      }      for(int i=0;i<len;i++)      {          ans+=(s[i]-'0');      }      ans=ans%9;      if(ans==0)      cout<<9<<endl;      else        cout<<ans<<endl;    }    return 0;}