杭电1163 Eddy's digital Roots

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

 

这道题我刚开始不知道怎么做，后来费了很大的力气想出了一个办法：我计算出了n^n的结果，这个结果我是用数组存储的！但是很不幸，这个程序提交的结果是超时！后来就开始在网上找这道题的相关解法。然后在找的过程中发现了“数根”“九余数”等相关概念！然后找到了解法！感觉这个题的解法特别的巧妙！我的最终AC的代码，几乎全部都是借鉴的别人的！很简洁！特别绝！

View Code

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 
 4 int main(int argc, char *argv[])
 5 {
 6     int n, mul, i;
 7     while( scanf( "%d", &n ) && n )
 8     {
 9            mul = 1;
10            for( i = 0; i < n; i++ )
11                 mul = (mul*n%9==0)?9:(mul*n%9);
12            printf( "%d\n", mul );
13     }
14   
15   //system("PAUSE");    
16   return 0;
17 }

 

 

转载于:https://www.cnblogs.com/yizhanhaha/archive/2013/04/15/3022018.html