Problem Description Considera positive integer X,and let S be the sum of all positive integer divisors of2004^X. Your job is to determine S modulo 29 (the rest of the division of S by29).

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 29is equal to 6.     Input Theinput consists of several test cases. Each test case contains a line with theinteger X (1 <= X <= 10000000). 

A test case of X = 0 indicates the end of input, and should not be processed.     Output Foreach test case, in a separate line, please output the result of S modulo 29.     Sample Input 1 10000 0     Sample Output 6 10   /*************************** 解题思路： 参考大神的：http://www.cnblogs.com/372465774y/archive/2012/10/22/2733977.html 体重主要用到除数和函数。 除数和函数 ：F(n) 求n的约数的和 ( 约数大于等于1 小于n )  除数和函数是一个积性函数，满足性质 :当m , n 互质时， f(m*n) = f(m) * f(n) 如果 p 是一个素数，则 f(p^n) = 1 + p + p^2 +p^3 +p^4 + .... + p^(n-1) + p^n = (p^(n+1) -1)/p-1  （等比数列求和） 则题目中  f(2004^n) = f(2^(2*n)) * f(3^n) * f(167^n)     = (2^(2*n+1) -1) * (3^(n+1) -1)/2  *(167^(n+1) -1)/166     用到乘法逆元：(同余性质) a^k/d = a^k*(d-1)     d-1 即为d的逆元。   3的逆元为15     167 的逆元为18 具体参考:http://baike.baidu.com/link?url=pcN2WyxgeFP9isdQxd9bTobeiRH3MnXcrdIwHh7jCBsYkVyTfFhF5QiS-d8-HgNgslVb334pgqkClTiIp359Xa 然后还要用到 快速幂模：转换为位运算，这题要用这个，一般的会超时，具体看代码吧。 *************************/ Code: #include using namespace std; int Mod(int a,int b)// 快速幂模函数 { int t = 1; while(b) { if(b&1) t = t*a%29; b>>=1; a = a*a%29; } return t; } int main() { int n,a,b,c; while(scanf("%d",&n)&&n) { a=(Mod(2,2*n+1)-1); b=(Mod(3,n+1)-1)*15; c=(Mod(22,n+1)-1)*18; printf("%d ",a*b*c%29); } return 0; }