Problem Description
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.
Input
The 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.
Output
For each test case, in a separate line, please output the result of S modulo 29.
Sample Input
1
10000
0
Sample Output
6
10
题意：找2014^x%29的值。
快速幂和取模乘法逆元。乘法逆元：x=(1/b)%m. 求x的值。x*b=k*m+1,即k去最小正整数时，x也为整数。即b能被k*m+1整除。
1 #include<cstdio>
2 #include<cstring>
3 using namespace std;
4 int f1(int x,int y)
5 {
6 int ans=1;
7 while (y)
8 {
9 if (y&1) ans=ans*x%29;
10 x=x*x%29;
11 y>>=1;
12 }
13 return ans;
14 }
15 int f2(int x)
16 {
17 int i=1;
18 while (i)
19 {
20 if ((29*i+1)%x==0) break;
21 i++;
22 }
23 return (29*i+1)/x;
24 }
25 int main()
26 {
27 int x,a,b,c;
28 while (~scanf("%d",&x))
29 {
30 if (!x) break;
31 a=(f1(2,2*x+1)-1)%29;
32 c=(((f1(3,x+1)-1)%29)*f2(2))%29;
33 b=(((f1(22,x+1)-1)%29)*f2(21))%29;
34 printf("%d\n",(a*b*c)%29);
35 }
36 return 0;
37 }