Eddy's digital Roots
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5183 Accepted Submission(s): 2897
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
Sample Output
4
4
题目大意：将一个数n的各位数字加起来，如果得到的是一个一位数，那么这个数就叫n的数根，如果是两位数或多位数，则重复这个过程，直到得到的数字是一位数。现在给出n，求n^n的数根。
题目解析：最终的答案是小于10的，相当于各位数的和模10，可以不转化为对9取模，当余数为0时，则相当于模10余9。这样，便转化为了求各位数的和模9。这样便能运用九余数定理了。九余数定理的内容是这样的，一个数的各位之和除以9的余数等于这个数除以9的余数。
代码如下：
# include<iostream>
# include<cstdio>
# include<cstring>
# include<algorithm>
using namespace std;
int main()
{
int n;
while(scanf("%d",&n)&&n)
{
int ans=;
for(int i=;i<=n;++i)
ans=ans*n%;
if(ans==)
printf("9\n");
else
printf("%d\n",ans);
}
return ;
}