HDU1452：Happy 2004（数论 & 唯一分解）

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

题意：给X，求2004的X次方的所有因子和mod 29。

思路：对于某个数的因子和，加入分解成p1^a1*p2^a2，因子和和(p1^0+p1*1+...p1^a1)*(p2^0+p2*1+...p2^a2)，可以化成等比数列求和，X次方就ai*X即可，最后要处理下逆元。

# include <iostream>
# include <cstdio>
# include <set>
# include <algorithm>
using namespace std;
const int mod = 29;
int a[6], b[6], cnt=0;
int qmod(int x, int y)
{
int ans = 1;
for(;y;y>>=1)
{
if(y&1) ans = ans*x%mod;
x = x*x%mod;
}
return ans;
}
int main()
{
int j=2004;
for(int i=2; i*i<=j; ++i)
{
if(j%i == 0)
{
a[cnt] = i;
while(j%i == 0) j /= i, ++b[cnt];
++cnt;
}
}
if(j) a[cnt] = j, b[cnt++] = 1;
int x;
while(~scanf("%d",&x),x)
{
int c[6], ans = 1, tmp;
for(int i=0; i<cnt; ++i) c[i] = b[i]*x;
for(int i=0; i<cnt; ++i)
{
tmp = (qmod(a[i], c[i]+1)-1+mod)%mod * qmod(a[i]-1, mod-2)%mod;
ans = ans*tmp%mod;
}
printf("%d\n",ans);
}
return 0;
}