HDU 1013 Digital Roots(九余数定理）

Posted 2020-10-06 hinata_hajime

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

 

 

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分析：最后得到的结果有九种可能性 1-9，当数n<10时，n%9余数为0的时候对应9,其余的余数和最终结果相同

  如果n>=10 那么 数根也会和呈现出1-9的循环，把余数0看作9的话，则余数呈现对应的1-9循环  那么求出余数即可

代码如下：

#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;
}