Digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 79180    Accepted Submission(s):
24760

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

 1 #include <stdio.h>
 2 #include<string.h>
 3 int main()
 4 {
 5     char a[1500];
 6     int sum;
 7     while(scanf("%s",&a)!=EOF)
 8     {
 9         if(strlen(a)==1&&a[0]==‘0‘)
10             break;
11         sum=0;
12         for(int i=0;i<strlen(a);i++)
13         {
14             sum=sum+a[i]-‘0‘;
15             if(sum>=10)
16                 sum=sum/10+sum%10;
17         }
18         printf("%d\n",sum);
19     }
20     return 0;
21 }

 1 /*因为输入的数字可能很长，而一个数的digital root和该数每一位的和的digital root相等，所以可以直接算出所输入的数字的和（不会超过范围），然后根据常规方法求digital root。
 2 */
 3 #include <stdio.h>
 4 #include <string.h>
 5
 6 int root(int x)
 7 {
 8     int r=0;
 9     while (x)
10     {
11         r+=x%10;
12         x/=10;
13     }
14     if (r<10)
15         return r;
16     else
17         return root(r);
18 }
19
20 int main()
21 {
22     char c;
23     int r=0;
24     while (c=getchar(),1)
25     {
26         if (c==‘\n‘)
27         {
28             if (r==0)
29                 break;
30             printf("%d\n",root(r));
31             r=0;
32         }
33         else
34         {
35             r+=c-48;
36         }
37     }
38 }

 1 //9余数定理 ！！
 2 #include<stdio.h>
 3 #include<string.h>
 4 char a[10010];
 5 int main()
 6 {
 7     int n;
 8     while(~scanf("%s",a),strcmp(a,"0"))
 9     {
10         int sum=0;
11         int len=strlen(a);
12         for(int i=0;i<len;++i)
13         sum+=a[i]-‘0‘;//字符化为数字
14         printf("%d\n",(sum-1)%9+1);
15     }
16     return 0;
17 }