hdoj 1013Digital Roots



/*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*/

<span style="font-size:18px;">#include <stdio.h>
#include <string.h>
int main()
{
 int sum;
 char a[1000],n[1000];
 while(gets(n))
 {
  if (strcmp(n,"0")==0)
  {
   break;
  }
  else
  {
   int sum=0;
   for (int j=0;j<strlen(n);j++)
   {
                sum+=(n[j]-48);
   }
   while (sum>=10)
   {
    sprintf(a,"%d",sum);//把格式化的数据写入某个字符串缓冲区。
意思是把sum转化为字符串a。
 
    sum=0;
    for (int i=0;i<strlen(a);i++)
    {
     sum+=(a[i]-48);
    }
   }
     printf("%d
",sum);
  } 
 }
}</span>

这种方法不easy想到。

<span style="font-size:18px;">//9余数定理 
#include<stdio.h>
#include<string.h>
char a[10010];
int main()
{
 int n;
 while(~scanf("%s",a),strcmp(a,"0"))
 {
  int sum=0;
  int len=strlen(a);
  for(int i=0;i<len;++i)
  sum+=a[i]-'0';
  printf("%d
",(sum-1)%9+1);//为什么减一后再加一。是为了避免18这些数字。 
 }
 return 0;
}
</span>