happy2004

 

Happy 2004

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Time Limit:1s	Memory limit:32M
Accepted Submit:142	Total Submit:279

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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 2004¹ 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

Output for the Sample Input

6
10

Original: 2004 ACM/ICPC Beijing Practice Problem

 

解题：

       运用到欧拉定理，费马小定理。要求某个数A的所有因子和，首先将该数分解：

#include <iostream>
using namespace std;

int happy(int n)
{
    int r,f,t,d;
    f=t=d=1;
    for (r=0;r<=n;++r)
    {
        f=f*(r==n?2:4)%29;        //f表示2^(2n+1)的值
        t=t*3%29;                //t表示3^(n+1)的值
        d=d*22%29;                //d表示167^(n+1)的值
    }
    r=(f-1)*(t-1)*(d-1)%29;        //S=r/332,r的值
    return r*9%29;                //S%29=r*9%29
}
int main()
{
    int n;
    while (cin>>n && n!=0)
    {
        cout<<happy(n%28)<<endl;
    }
    return 0;
}