Hdu-1452-Happy 2004-费马小定理推除法逆元+同余定理+积性函数

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. 

InputThe 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. 
OutputFor each test case, in a separate line, please output the result of S modulo 29. 

Sample Input

1
10000
0

Sample Output

6
10



同余定理:a*b%c=((a%c)*(b%c))%c 或者可以=a%c*b%c
　　　　(a-b)%c=a%c-b%c,但是需要注意的是，如果计算出来为负数，需要加上出来c*1

#include<stdio.h>
typedef long long ll;

const int mod=29;
int ksm(int x,int n)
{
    int res=1;
    while(n>0)
    {
        if(n&1)
            res=res*x%mod;
        x=x*x%mod;
        n>>=1;
    }
    return res;
}

int main()
{
    int x;
    int k2=ksm(2,27);
    int k166=ksm(166,27);

    while(~scanf("%d",&x)&&x)
    {
        int k3=ksm(3,x+1);
        if(k3-1<0)
            k3=k3-1+29;
        else
            k3--;
            
        int k167=ksm(167,x+1);
        if(k167-1<0)
            k167=k167-1+29;
        else
            k167--;

        int s2=ksm(2,2*x+1);
        if(s2-1<0)
            s2=s2-1+29;
        else
            s2--;
            
        int s3=k2*k3%29;
        int s167=k167*k166%29;
        int sum=s2*s3*s167;
        printf("%d\n",sum%29);
    }
    return 0;
}