hdu 1452 Happy 2004

Happy 2004

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 301    Accepted Submission(s): 220

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

1100000

 

 

Sample Output

610
/*************************************************************  计算 2004^X的因子和 s(2004^X)   mod M, M=29    s(2004^X)%29   因子和 s是积性函数,即 ：gcd(a,b)=1==> 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)    如果 p是素数 ==> s(p^X)=1+p+p^2++p^X = (p^(X+1)-1) /(p-1)    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    (a*b)/c %M= a%M* b%M  * inv(c)    c*inv(c)=1 %M    模为1的所有数  inv（c）为最小可以被c整除的      inv(2)=15,  inv(21)=18    2*15=1 mod 29, 18*21=1 mod 29    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 ***************************************************************/    //hdu 1452  #include <iostream>  #include <cstdio>  #include <cmath>    using namespace std;    int _pow( int a, int n )  {      int b = 1;      while( n > 1 )          if( n % 2 == 0 )          {              a = ( a * a ) % 29;              n /= 2;          }          else          {              b = b * a % 29;              n--;          }          return a * b % 29;  }  int main()  {      int X;      int a, b, c;      while( cin >> X, X )      {          a = _pow( 2, 2 * X + 1 );          b = _pow( 3, ( X + 1 ) );          c = _pow( 22, ( X + 1 ) );            cout << ( a - 1 ) * ( b - 1 ) * 15 * ( c - 1 ) * 18 % 29 << endl;      }      return 0;  }