hdu 1452 Happy 2004 膜拜这推导过程

Happy 2004

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 920    Accepted Submission(s): 648

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

Source

ACM暑期集训队练习赛（六）

 

 /*
   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)
   (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)/166
   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(21)=18
   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<iostream>
 #include<stdio.h>
 #include<cstring>
 #include<cstdlib>
 using namespace std;
 typedef __int64 LL;

 const LL p = ;

 LL pow_mod(LL a,LL n)
 {
     LL ans=;
     while(n)
     {
         if(n&) ans=(ans*a)%p;
         n=n>>;
         a=(a*a)%p;
     }
     ans=ans-;
     if(ans<) ans=ans+p;
     return ans;
 }
 void solve(LL x)
 {
     LL sum=;
     sum=(sum*pow_mod(,x+)*)%p;
     sum=(sum*pow_mod(,*x+))%p;
     sum=(sum*pow_mod(,x+)*)%p;
     printf("%I64d\n",sum);
 }
 int main()
 {
     LL x;
     while(scanf("%I64d",&x)>)
     {
         if(x==)break;
         solve(x);
     }
     return ;
 }