HDU--1013Digital Roots

Digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 67997    Accepted Submission(s): 21258

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

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=1013

解题思路：这题的意思是求一个数的数根，比如24的数根是：2+4=6,39的数根是3+9=12，但是12大于10，所以1+2=3,3是39的数根

注意输入数的范围很大，故采用字符输入，把它的每一位都取出来，相加如果和大于等于的10的话，进入循环中，最后得出结果。

AC代码：

#include <stdio.h>#include <string.h>char str[100000000];int main(){    int leng;    while(scanf("%s",str)!=EOF)    {        leng=strlen(str);        if(leng==1 && str[0]=='0')        {            break;        }        int i,sum=0;        for(i=0; i<leng; i++)        {            sum=sum+(str[i]-'0');        }        int m,n;        while(sum>=10)        {            m=sum;            sum=0;            while(m)            {                n=m%10;                sum=sum+n;                m=m/10;            }            //printf("%d\n",sum);        }        printf("%d\n",sum);    }}