HDU 1452 Happy 2004(约数和定理)

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本文探讨了一个特定的数论问题Happy2004，该问题要求计算2004的X次方的所有正除数之和S，并求S对29取模的结果。通过使用费马小定理求逆元的方法来高效解决此问题。

Happy 2004

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1904 Accepted Submission(s): 1394

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

Sample Output

  
   6
10

Source

ACM暑期集训队练习赛（六）

POINT：

了解一下约数和定理就行，用一下费马求逆元。

#include <iostream>
#include <stdio.h>
using namespace std;
#define  LL long long
const int p = 29;
int qkm(int base,int mi)
{
    base%=p;
    int ans=1;
    while(mi)
    {
        if(mi&1) (ans*=base)%=p;
        (base*=base)%=p;
        mi>>=1;
    }
    return ans;
}
int main()
{
    int x;
    while(~scanf("%d",&x)&&x)
    {
        int a=qkm(2,2*x+1)-1;
        int b=((qkm(3,x+1)-1)+p)%p*qkm(2,p-2);
        int c=((qkm(167,x+1)-1)+p)%p*qkm(166,p-2);
        printf("%d\n",a*b*c%p);
    }
    
}