hdu 1452 Happy 2004 (快速幂+取模乘法逆元）

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

   题意：找2014^x%29的值。

   快速幂和取模乘法逆元。乘法逆元：x=(1/b)%m. 求x的值。x*b=k*m+1,即k去最小正整数时，x也为整数。即b能被k*m+1整除。

  

#include<cstdio>
#include<cstring>
using namespace std;
int f1(int x,int y)
{
    int ans=1;
    while (y)
    {
        if (y&1) ans=ans*x%29;
        x=x*x%29;
        y>>=1;
    }
    return ans;
}
int f2(int x)
{
    int i=1;
    while (i)
    {
        if ((29*i+1)%x==0) break;
        i++;
    }
    return (29*i+1)/x;
}
int main()
{
    int x,a,b,c;
    while (~scanf("%d",&x))
    {
        if (!x) break;
        a=(f1(2,2*x+1)-1)%29;
        c=(((f1(3,x+1)-1)%29)*f2(2))%29;
        b=(((f1(22,x+1)-1)%29)*f2(21))%29;
        printf("%d
",(a*b*c)%29);
    }
    return 0;
}