n^n的数字根

Eddy's digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2569    Accepted Submission(s): 1478

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
  

 

题意：n<10000，数字根是各个位相加之和是一位数就是数字根，否则继续重复，继续求加和的数字根，直到为一位数，这个数就是数字根。。。

性质：一个数的数字根为该数字除以9的余数，两个数之和的数字根为
两数分别得数字根之和除以9的余数。

用快速幂取模解决即可，，但是要考虑到取余为0时，数字根就是9，因为（9*k）%9==0，总是9的倍数，，

ps：至于这个性质怎么证明我就不知道了，这是我在做多校的时候，看题解知道的，希望路过的大牛指点。。。。

Accepted

1163

0MS

480K

478 B

我的代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#define ll long long
using namespace std;
ll quickpow(ll n)
{
    ll a=n%9==0?9:n%9,b=1;
    while(n>0)
    {
        if(n&1)
        {
            b=(b*a)%9;
            if(b==0)
            return 9;
        }
        a=(a*a)%9;
        a=a==0?9:a;
        n>>=1;
    }
    return b;
}
int main()
{
    ll n;
    while(scanf("%I64d",&n)&&n)
    {
      printf("%I64d\n",quickpow(n));
    }
}