九余数定理

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

24390

 

Sample Output

63

 

题意分析：数根的定义：将一个数的每位数相加得到一个新的数，如果该数为一位数，该新数即是数根；否则重复开头的步骤。例如：28945 -> 2+8+9+4+5=28 -> 2+8+10 -> 1+0 = 1;所以1即是28945的数根。

解题思路：当然，直接按题目的字面意思做也能AC。这里我只是想介绍一下九余数定理。

我么们知道，不管是什么数，数根一定是0~9其中的一个，因为不会有正数各位之和为0。

证明：
假设，数d的根为d%9( 暂时不取0，整除时取9）
当d < 10时，1~9这9个数肯定成立;
当d >= 10时，d的根为d%9 = (d-1)%9+1，即d的前一个数的数根加1.
得证.

源代码如下：

<strong>#include<cstdio>int main(void){while(1){    int num=0;        char ch;while(scanf("%c",&ch) && ch!='\n')  num += ch-'0';if(!num) return 0;else if(num%9 == 0) puts("9");else printf("%d\n",num%9);}return 0;}</strong>

类似的题：

Eddy's digital Roots 只不过这里变成求Nⁿ的数根

源代码如下：

<span style="font-size:18px;">#include<cstdio>int main(void){    int n,i,ans;    while(scanf("%d",&n),n)    {        ans = 1;        for(i = 0; i < n; i++)            ans = ans*n%9;        if(!ans) puts("9");        else printf("%d\n",ans);    }    return 0;}</span>