1408141504-hd-Digital Roots.cpp

                                       Digital Roots

                                      Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                       Total Submission(s): 48740    Accepted Submission(s): 15160

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

 

题目大意

         要求一个数的数根。如果结果是一位数那么这就是数根。如果结果包含两个或两个以上的数字，这些数字相加，并重复该过程，以获得一个单一的数字。

 

错误原因

       没有考虑到给出的正整数有没有超过int型。要仔细阅读题意，若题目没有标明所给数据在int范围之内，就要考虑int范围之外的数据。

 

解题思路

       首先要考虑int范围之外的数据的处理，那就要输入的格式为字符串，然后再转换为整形。因为就算是这个数据为1000位，他所有位数相加最大也不过9*1000,在int范围内。然后再用循环按照题意求解。

 

代码

#include<stdio.h>#include<string.h>char s[1100];int main(){    int sum,i,n,len;    while(scanf("%s",s)&&strcmp(s,"0")!=0)    {        len=strlen(s);        sum=0;        for(i=0;i<len;i++)            sum+=s[i]-'0';        for(i=0;;i++)        {            n=0;            while(sum)            {                n+=sum%10;                sum/=10;            }            if(n/10==0)                break;            else                sum=n;        }        printf("%d\n",n);    }    return 0;}