G - Happy 2004------(HDU 1452)

传送门

Happy 2004

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1346    Accepted Submission(s): 977

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
   
   
   
   

 

题目大意：

求 2004^X的所有因子的和 模上29的值；

解题思路：

首先根据算术基本定理（1），

若 n 的标准素因子分解表达式为 n = p1^e1 * p2^e2 * …… * pk^ek

设f(n) 为n 的所有因子之和，则有:

f(n) = (p1^(e1+1)-1)/(p1-1) * (p2^(e2+1)-1)/(p2-1) * ... * (pk^(ek+1)-1)/(pk-1)

因为2004 = 2^2 * 3 * 167，所以2004^x也肯定是2^(2*x) * 3^x * 167^x;

然后根据算术基本定理：

167 Mod 29 == 22 所以直接用22就行

((2^(2*x+1)-1)/(2-1) * (3^(x+1)-1)/(3-1) *  (22^(x+1)-1)/(22-1)) Mod 29 

设a = (2^(2*x+1)-1)/1;

设b = (3^(x+1)-1)/2 ;

设c = (22^(x+1)-1)/21);

我们只要分别求出 a Mod 29, b Mod 29, c Mod 29的值就行啦，因为他们带着除数所以 我们进行求解逆元,分别是

1Mod29的逆元 2Mod29的逆元 21Mod29的逆元，这个可以根据扩展欧几里得算法得到，求出之后，我们就进行快速幂 分别求出2^(2x+1)Mod 29 , 3^(x+1)Mod 29,  22^(x+1)Mod 29 的值，然后减一 与 前面所求的逆元进行相乘就ok了

最后输出的就是 a*b*c Mod 29，下面是详细代码：

#include <iostream>

using namespace std;
typedef long long LL;
LL quick_mod(LL a, LL b, LL m)
{
    LL ans = 1;
    while(b)
    {
        if(b & 1)
            ans = (ans*a)%m;
        b>>=1;
        a = (a*a)%m;
    }
    return ans;
}
void exgcd(LL a, LL b, LL &x, LL &y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return;
    }
    LL x1, y1;
    exgcd(b, a%b, x1, y1);
    x = y1;
    y = x1 - (a/b)*y1;
}
int main()
{
    LL n, a, b, c, y, _a, _b, _c;
    while(cin>>n,n)
    {
        exgcd(1,29,_a,y);
        exgcd(2,29,_b,y);
        exgcd(21,29,_c,y);
        _a = (_a+29)%29;
        _b = (_b+29)%29;
        _c = (_c+29)%29;
        ///cout<<167%29<<endl;
        ///cout<<_a<<" "<<_b<<" "<<_c<<endl;
        a=(quick_mod(2,2*n+1,29)-1)*_a;
        b=(quick_mod(3,n+1,29)-1)*_b;
        c=(quick_mod(22,n+1,29)-1)*_c;
        cout<<(a*b*c%29)<<endl;
    }
    return 0;
}