Eddy‘s digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5113    Accepted Submission(s):
2851

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.

The Eddy‘s easy problem is that : give you the n,want
you to find the n^n‘s digital Roots.

Input

The input file will contain a list of positive integers
n, one per line. The end of the input will be indicated by an integer value of
zero. Notice:For each integer in the input n(n<10000).

Output

Output n^n‘s digital root on a separate line of the
output.

Sample Input

2

4

0

Sample Output

4

4

九余数定理：一个数的每位数字之和等于这个数对9取余，如果等于0就是9

 1 #include <iostream>
 2 #include <cstdio>
 3 using namespace std;
 4 int main()
 5 {
 6     int n,sum,i;
 7     while(cin>>n&&n)
 8     {
 9         sum=1;
10         i=n;
11         while(i--)
12             sum=(sum*n)%9;
13         if(sum==0)
14             cout<<9<<endl;
15         else
16             cout<<sum<<endl;
17     }
18     return 0;
19 }

Eddy's digital Roots（九余数定理）