ProjectEuler: Amicable numbers

21.c

#include<stdio.h>
static long get_factor_num(long n) {
    int factor;
    int exp_count = 1;
    int ans = 1;
    for (factor = 2; n > 1; factor++) {
        exp_count = 1;
        while (n % factor == 0) {
            n /= factor;
            exp_count *= factor;
        }
        ans *= (exp_count * factor - 1) / (factor - 1);
    }
    return ans;
}
int main() {
    int i, sum = 0;
    for (i=2; i<10000; i++) {
        int n = get_factor_num(i) - i;
        if (n == i)
            continue;
        if (get_factor_num(n) - n == i)
            sum += i;
    }
    printf("%d\n", sum);
}