代码中有条件地划分N的问题

所以,我正在努力解决 following 问题陈述

对于整数N,求出可以执行以下操作的次数:

Select an integer z which satisfies the 3 conditions:

1. Z can be expressed as p^e where p is prime and e > 0
2. z is a factor of N
3. z is different from all previous zs

Now replace N with N / z

这是我尝试的代码:

#include <algorithm>
#include <iostream>
#include <numeric>
#include <vector>

const bool isPrime(long long num) {
    for (long long div = 2; div <= sqrt(num); div++) {
        if (num % div == 0) {
            return false;
        }
    }
    return true;
}

std::vector<long long> getFactors(const long long number) {
    std::vector<long long> vec_of_divisors;
    for (long long i = 1; i <= number / 2; i++) {
        if (number % i == 0) {
            vec_of_divisors.push_back(i);
        }
    }
    vec_of_divisors.push_back(number);
    return vec_of_divisors;
}

bool hasPrimeFactors(long long num) {
    std::vector<long long> divisors = getFactors(num);
    std::vector<long long> prime_vec;
    for (const auto& divisor : divisors) {
        if (isPrime(divisor)) {
            return true;
        }
    }
    return false;
}

int main() {
    long long N = 997764507000;
    //std::cin >> N;
    long count = 0;
    std::vector<long long> usedFactors;

    auto notUsed = [&](long long num) {
        for (auto& elem : usedFactors) {
            if (elem == num) {
                return false;
            }
        }
        return true;
    };
    long long temp;
    do {
        temp = N;
        std::vector<long long> factors = getFactors(N);
        for (auto& factor : factors) {
            if ((factor != 1) && hasPrimeFactors(factor) && notUsed(factor)) {
                usedFactors.push_back(factor);
                N = N / factor;
                count++;
                break;
            }
        }
    } while (temp != N);
    std::cout << count << std::endl;
    return 0;
}

它超过了时间限制 2secs

看起来代码真的很低效。 我错过了什么?

我试过调试,但调试完成后需要很长时间 10^11