Digital Roots
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36662    Accepted Submission(s): 11228

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
我的代码如下:

#include <stdio.h>

int main() {
    int n, sum, count, t, i, a[100];
    while(scanf("%d", &n) != EOF && n > 0) {
        count = 0; i = 0;
        t = sum = n;
        loop : while(t) {
            count++;
            a[i++] = t % 10;
            t /= 10;
        }
        while(sum >= 10) {
            t = 0;
            for(i = 0; i < count; i++) t += a[i];
            sum = t, count = i = 0;
            goto loop;
        }
        printf("%d\n", sum);
    }
    return 0;
}
//简单说下题意如
3 = 3
33 = 3 + 3 = 6
333 = 3 + 3 + 3 = 9
3333 = 3 + 3 + 3 + 3 = 12 = 1 + 2 = 3

或许数据太大，int装不下吧，这道题没必要存数据啊，边读边计算就行

坛里对此题有很全面的讨论，搜下

有么？给个传送看看....没找到

应该够大了吧?  我看到好多人都是边读边计算的，但是我偏偏没那么做，将一个很大很大的数拆开加起来，看了下好像不怎么大了...主要我想弄清楚我的问题在哪里?

#include <stdio.h>

int main() {
    int n, sum, t ;
    while(scanf("%d", &n) != EOF && n > 0)
    {
        t=sum=n;
        while(t>10)
        {   
            sum=0;
            while(t)
            {
              sum+= t % 10;
              t /= 10;
            }
            t=sum;
        }
        printf("%d\n", sum);
    }
    return 0;
}
参考一下这个程序

#include <stdio.h>
#include <string.h>
#define MAX 10000
char str[MAX];
long judge(long n)
{
    long sum=0;
    while(n)
    {
        sum+=(n%10);
        n/=10;
    }
    if(sum%10==sum) printf("%ld\n",sum);
    else judge(sum);
}
int main()
{
    int i;
    long sum;
    while(scanf("%s",str)!=EOF)
    {
        if(str[0]=='0') break;
        sum=0;
        for(i=0;i<strlen(str);i++) sum+=(str[i]-48);
        judge(sum);
    }
    return 0;
}

#include <stdio.h>
#include <string.h>
int main() {
    char s[1005];
    int a[1005], len, i, sum, n, t, b;
    while(scanf("%s", s) != EOF) {
        if(s[0] == '0') return 0;
        memset(a, 0, sizeof(a));
        sum = 0;
        len = strlen(s);
        for(i = 0; i < len; i++) {
            a[i] = s[i] - '0';
            sum += a[i];
        }
        n = sum;
        b = 0;
        loop : while(n) {
                    t = n % 10;
                    n /= 10;
                    b += t;
                    sum = b;
                }
        while(sum >= 10) {
            n = sum;
            b = 0;
            goto loop;
        }
        printf("%d\n", sum);
    }
    return 0;
}