There

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


打表偷笑,谁有好方法自由提供!!


#include <iostream>
#include<memory.h>
using namespace std;

int main()
{
    int n;
    while(cin>>n&&n!=0)
    {
        switch(n%28)
        {
            case 1:cout<<6<<endl;break;
            case 2:cout<<16<<endl;break;
            case 3:cout<<8<<endl;break;
            case 4:cout<<10<<endl;break;
            case 5:cout<<25<<endl;break;
            case 6:cout<<7<<endl;break;
            case 7:cout<<14<<endl;break;
            case 8:cout<<3<<endl;break;
            case 9:cout<<23<<endl;break;
            case 10:cout<<17<<endl;break;
            case 11:cout<<3<<endl;break;
            case 12:cout<<17<<endl;break;
            case 13:cout<<0<<endl;break;
            case 14:cout<<27<<endl;break;
            case 15:cout<<7<<endl;break;
            case 16:cout<<14<<endl;break;
            case 17:cout<<15<<endl;break;
            case 18:cout<<17<<endl;break;
            case 19:cout<<26<<endl;break;
            case 20:cout<<26<<endl;break;
            case 21:cout<<20<<endl;break;
            case 22:cout<<17<<endl;break;
            case 23:cout<<9<<endl;break;
            case 24:cout<<22<<endl;break;
            case 25:cout<<22<<endl;break;
            case 26:cout<<23<<endl;break;
            case 27:cout<<0<<endl;break;
            case 0:cout<<1<<endl;break;
        }
    }
    return 0;
}

posted @ 2015-08-02 09:57  martinue  阅读(146)  评论(0)    收藏  举报