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2013
12-11

HDU 1452 Happy 2004-数论-[解题报告] C++

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

 AC code:

#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;
 }

解题报告转自:http://www.cnblogs.com/gaigai/archive/2012/04/24/2468453.html


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