HDU 1163 Eddy's digital Roots

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

 

 

 

解析： 考察数字根。

数字根定义：对于一个正整数，如果它是个位数，它的数字根就是它本身。如果它不是个位数，则将其各位上的数字相加，所得的和如果是个位数，那么这个数就是它的数字根；否则将这个数各位上的数字相加，直至所得的和为个位数为止。

数n的数字根:digital root = (n-1)%9+1。为了方便计算，我们也可以写为digital root = (n%9 == 0 ? 9 : n%9)。

数字根性质：

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。

根据以上规律，结合快速幂及其取模的方法，可以得到较快速的算法。

 

//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;
}

#include <cstdio>

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;
}

int main()
{
    int n;
    while(scanf("%d",&n), n){
        int ans = quickpowmod(n,n,9);
        printf("%d
",ans == 0 ? 9 : ans);
    }
    return 0;
}