【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): 2355 Accepted Submission(s): 1348
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.
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
Author
eddy
Recommend
JGShining
1 // Project name : 1163 ( Eddy's digital Roots ) 2 // File name : main.cpp 3 // Author : Izumu 4 // Date & Time : Sun Jul 8 20:59:43 2012 5 6 7 #include <iostream> 8 using namespace std; 9 10 int main() 11 { 12 int n; 13 while (cin >> n && n) 14 { 15 int sum = 1; 16 for (int i = 0; i < n; i++) 17 { 18 sum = sum * n % 9; 19 } 20 21 if (sum == 0) 22 { 23 cout << "9" << endl; 24 } 25 else 26 { 27 cout << sum << endl; 28 } 29 } 30 return 0; 31 } 32 33 // end 34 // ism