『算法-ACM 竞赛-数学-数论』Hdu 1452 Happy 2004（积性函数性质+和函数公式+快速模幂+乘法逆元）

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

设$F(x)=2004^x的因子和$ 因为这是个积性函数，则有$f(N)=\prod_{i=1}^nf(q_i ^{q_i}) 其中 N可以表示为\prod_{i=1}^nq_i ^{q_i}$
$f（2004 ^ n）= f(2 ^{(2 * n)})* f（3 ^ n)* f（167 ^ n）
=（2 ^{（2 * n + 1）}-1）*（3 ^{（n + 1）}-1）/ 2 *（167 ^{（n + 1）}-1）/ 166$

用到乘法逆元：(同余性质）

a ^ k / d = a ^ k *（d-1）d-1 即为 d 的逆元。3 的逆元为 15 167 的逆元为 18

JAVA C++ 没区别，最近在 JAVA 要考试了额，熟悉一下。

import java.util.Scanner;

public class Main {
	public static long Q_pow( long a, long p, long mod)
	{
		long ans = 1%mod;
	     while(p>0)  {
	         if(p%2==1)  ans = ans*a%mod;  //防止在对P取模前溢出
	         a = a*a%mod;
	         p >>=1;  //比除法快多了
	     }
	     return ans;
	}
	public static void main(String[] args) {
		int n;
		Scanner in = new Scanner(System.in);
		while(in.hasNext())
		{
			n=in.nextInt();
			if(n==0) break;
			long ans=((Q_pow(167,n+1,29)-1)*18)%29;
			ans=(ans*((Q_pow(3,n+1,29)-1)*15)%29)%29;
			ans=(ans*(Q_pow(2,2*n+1,29)-1))%29;
			System.out.println(ans);
		}
		in.close();
	}

}