Happy 2004（快速幂+乘法逆元）

Happy 2004

问题描述 :

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.

输入:

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.

输出:

For each test case, in a separate line, please output the result of S modulo 29.

样例输入:

1
10000
0

样例输出:

6
10

设S(x)表示x的因子和。则题目求为:S(2004^X)mod 29
因子和S是积性函数,即满足性质1。

性质1 ：如果 gcd(a,b)=1  则 S(a*b)= S(a)*S(b)
2004^X=4^X * 3^X *167^X
S(2004^X)=S(2^(2X)) * S(3^X) * S(167^X)

性质2 ：如果 p 是素数 则 S(p^X)=1+p+p^2+…+p^X = (p^(X+1)-1)/(p-1)
因此：S(2004^X)=(2^(2X+1)-1) * (3^(X+1)-1)/2 * (167^(X+1)-1)/166
167%29 == 22
S(2004^X)=(2^(2X+1)-1) * (3^(X+1)-1)/2 * (22^(X+1)-1)/21

性质3 ：(a*b)/c %M= a%M * b%M * inv(c)
其中inv(c)即满足 (c*inv(c))%M=1的最小整数,这里M=29
则inv(1)=1，inv(2)=15，inv(22)=15

有上得：
S(2004^X)=(2^(2X+1)-1) * (3^(X+1)-1)/2 * (22^(X+1)-1)/21
=(2^(2X+1)-1) * (3^(X+1)-1)*15 * (22^(X+1)-1)*18

 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 #include <queue>
 #include <map>
 #include <set>
 using namespace std;
 #define LL long long
 const int INF=0x3f3f3f3f;
 const double eps=1e-;
 int p=;
 LL pow_mod(LL x,LL n)
 {
     LL res=;
     while(n>)
     {
         if(n&)    res=res*x%p;
         x=x*x%p;
         n>>=;
     }
     return res;
 }
 int main()
 {
     LL x,i;
     while(cin>>x&&x)
     {
         int a=pow_mod(,*x+);
         int b=pow_mod(,x+);
         int c=pow_mod(,x+);
         int s=((a-)*(b-)*(c-)**)%;
         cout<<s<<endl;
      }

 }