# Eddy‘s digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5783    Accepted Submission(s): 3180

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

1.任何数加上9的数字根等于它本身的数字根。
2.任何数乘以9的数字根等于9。
3.多个数之和的数字根等于这多个数的数字根的和的数字根。
4.多个数之积的数字根等于这多个数的数字根的积的数字根。

```//计算x的y次幂(快速)
int quickpow(int x,int y)
{
int ret = 1;
while(y){
if(y&1)
ret *= x;
x *= x;
y >>= 1;
}
return ret;
}```
```//计算x的y次幂对mod取模(快速)
int quickpowmod(int x,int y,int mod)
{
int ret = 1;
x %= mod;
while(y){
if(y&1)
ret = ret*x%mod;
x = x*x%mod;
y >>= 1;
}
return ret;
}```

``` 1 #include <cstdio>
2
3 int quickpowmod(int x,int y,int mod)
4 {
5     int ret = 1;
6     x %= mod;
7     while(y){
8         if(y&1)
9             ret = ret*x%mod;
10         x = x*x%mod;;
11         y >>= 1;
12     }
13     return ret;
14 }
15
16 int main()
17 {
18     int n;
19     while(scanf("%d",&n), n){
20         int ans = quickpowmod(n,n,9);
21         printf("%d\n",ans == 0 ? 9 : ans);
22     }
23     return 0;
24 }```

HDU 1163 Eddy's digital Roots

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