Hdu-1452-Happy 2004-费马小定理推除法逆元+同余定理+积性函数

Consider a positive integer X,and let S be the sum of all positive integer divisors of 2004^X. Your job is to determine S modulo 29 (the rest of the division of S by 29).

Take X = 1 for an example. The positive integer divisors of 2004^1 are 1, 2, 3, 4, 6, 12, 167, 334, 501, 668, 1002 and 2004. Therefore S = 4704 and S modulo 29 is equal to 6.

InputThe input consists of several test cases. Each test case contains a line with the integer X (1 <= X <= 10000000).

A test case of X = 0 indicates the end of input, and should not be processed. 
OutputFor each test case, in a separate line, please output the result of S modulo 29.

Sample Input

1
10000
0

Sample Output

6
10



同余定理:a*b%c=((a%c)*(b%c))%c 或者可以=a%c*b%c
　　　　(a-b)%c=a%c-b%c,但是需要注意的是，如果计算出来为负数，需要加上出来c*1

 #include<stdio.h>
 typedef long long ll;

 const int mod=;
 int ksm(int x,int n)
 {
     int res=;
     while(n>)
     {
         if(n&)
             res=res*x%mod;
         x=x*x%mod;
         n>>=;
     }
     return res;
 }

 int main()
 {
     int x;
     int k2=ksm(,);
     int k166=ksm(,);

     while(~scanf("%d",&x)&&x)
     {
         int k3=ksm(,x+);
         if(k3-<)
             k3=k3-+;
         else
             k3--;

         int k167=ksm(,x+);
         if(k167-<)
             k167=k167-+;
         else
             k167--;

         int s2=ksm(,*x+);
         if(s2-<)
             s2=s2-+;
         else
             s2--;

         int s3=k2*k3%;
         int s167=k167*k166%;
         int sum=s2*s3*s167;
         printf("%d\n",sum%);
     }
     return ;
 }