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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

 

 

Source

 代码如下：

/*此题是求2004^x%29的因子和。。。首先:任何一个数可以化成几个素因子幂的乘积的形式。。2004=2^2*3*167;这里167这个数字可以化简下,167%29=22;令g(p, e) = (p^(e+1) - 1) / (p-1), 其中p为素因子，e为素因子的幂则s(n) = g(p1, e1) * g(p2, e2) * ... * g(pk, ek)。因此：因子和可化为：s(2004^n)=[(2^(2n+1)-1)/1]*[(3^(n+1)-1)/2]*[22^(n+1)/21]%29;这里可以用逆元进行运算1的逆元为1;2的逆元为15;21的逆元为18;所以式子进一步简化为:s(2004^n)=(2^(2n+1)-1)*(3^(n+1)-1)*15*(22^(n+1))*18%29;15*28又能进一步取余所以最终式子为：s(2004^n)=(2^(2n+1)-1)*(3^(n+1)-1)*(22^(n+1))*9%29;*/#include 
   
    #include 
    
     #include 
     
      #include 
      
       using namespace std;int n;int fastpow (int a,int b){    int sum=1;    while (b>0)    {        if(b%2)        {            sum=sum*a%29;        }        a=a*a%29;        b>>=1;    }    return sum;}int main(){    while (scanf("%d",&n)!=EOF&&n)    {        int a,b,c;        a=fastpow(2,(n<<1)+1)-1;        b=fastpow(3,n+1)-1;        c=fastpow(22,n+1)-1;        printf("%d\n",9*a*b*c%29);    }    return 0;}
      
     
    
   

 

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