HDU 1013_Digital Roots大数&模拟

Digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 47146    Accepted Submission(s): 14605

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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
  
  

 

Source

Greater New York 2000

不断把各个位上的数相加，直到得到一个个位数。
代码：

#include <iostream>
#include<cstdio>
using namespace std;
long long int check(long long int n)
{
    long long int a;
    long long int s=0;
    for(;;)
    {
        a=n%10;
        s=s+a;
        if(n/10==0)
            break;
        n=n/10;
    }
    return s;
}
int main()
{
    long long int i;
    long long int s;
    char c[1000];
    while(scanf("%s",c)!=EOF&&c[0]!='0')//字符串输入解决大数问题
    {
        s=0;
       for(i=0;c[i]!='\0';i++)
       {
           s=s+c[i]-'0';//变为整数
       }
        while(s/10!=0)//循环模拟过程，求解
        {
            s=check(s);
        }
        cout<<s<<endl;
    }
    return 0;
}