hdu 1163 九余数定理

Eddy's digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5928    Accepted Submission(s): 3270

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

 

Author

eddy

思路：求n^n对9取模，特判0的情况；

 对于那个定理的证明；

设那个数为abcd；这个数=1000*a+100*b+10*c+d；

　　　　　　　　a+b+c+d=这个数-9999a-99b-99c；

　　　　　　　　由于9999a-99b-9c被九整除；得证；

 #include<iostream>
 #include<cstdio>
 #include<cmath>
 #include<string>
 #include<queue>
 #include<algorithm>
 #include<stack>
 #include<cstring>
 #include<vector>
 #include<list>
 #include<set>
 #include<map>
 using namespace std;
 #define ll long long
 #define mod 1000000007
 #define inf 999999999
 #define esp 0.00000000001
 //#pragma comment(linker, "/STACK:102400000,102400000")
 int quickpow(int a,int b,int c)
 {
     int ans=;
     while(b)
     {
         if(b&)
         {
             ans*=a;
             ans%=c;
         }
         b>>=;
         a*=a;
         a%=c;
     }
     return ans==?:ans;
 }
 int main()
 {
     int x,y,z,i,t;
     while(~scanf("%d",&x))
     {
         if(!x)break;
         printf("%d\n",quickpow(x,x,));
     }
     return ;
 }