HDU 1013 Digital Roots

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
   
   
   
   

 

思路：因为能输入的数字可以很大，所以需要用字符串来写，（这个有点坑，当初可把我坑惨了。）明白了这一点就好办多了，题意：给定一个数，把这个数各个位上的数字相加，如果大于10，就循环，直到小于10位置，输出最终结果。

代码如下：

#include<stdio.h>
#include<string.h>
char dight[1000]; int main() {
    memset(dight,0,sizeof(dight)); int sum; while(scanf("%s",dight)&&strcmp(dight,"0")!=0) {
        sum=0; int len; int i;
        len=strlen(dight); for(i=0;i<len;i++) {
            sum=sum+dight[i]-'0'; //字符换为数字来相加。 if(sum>=10) {
                sum=sum/10+sum%10; //大于10的话，各个位数相加。 } }

            printf("%d\n",sum); } return 0; }