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Digital Roots

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 80782 Accepted Submission(s): 25278

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

Source Greater New York 2000

Recommend We have carefully selected several similar problems for you: 1017 1021 1012 1048 1016 分析：最後得到的結果有九種可能性 1-9，當數n<10時，n%9余數為0的時候對應9,其余的余數和最終結果相同 如果n>=10 那麽 數根也會和呈現出1-9的循環，把余數0看作9的話，則余數呈現對應的1-9循環 那麽求出余數即可 代碼如下：

#include <cstdio>
#include <algorithm>
#include 
 
<cstring>
using namespace std;
char a[20100];
int main()
{
    int n,sum;
    while(scanf("%s",a)!=EOF)
    {
        if(strlen(a)==1&&a[0]==‘0‘)break;
       sum=0;
       for(int i=0;i<strlen(a);i++)
        sum+=a[i]-‘0‘;
        sum=sum%9;
       if(sum==0)puts("9");
       else printf("%d\n",sum);
    }
    return 0;
}

HDU 1013 Digital Roots(九余數定理）