根据给定的数字查找其因数的高效C++代码

根据给定的数字查找其因数的高效C++代码

//Algorithm derived from the method described here:
//http://mathschallenge.net/index.php?section=faq&ref=number/number_of_divisors
int mathTools::numFactors(unsigned long long number){
    if(number < 2){
        return 1;
    }
    //Answer starts at one because all numbers can be divided by 1
    int answer = 1;
    int counter = 0;
	//Adds up the number of times divisible by 2
    while(number % 2 == 0){
        number = number / 2;
        counter++;
    }
    answer = answer * (counter + 1);
    counter = 0;
	//Gets all other multiples of the number except possibly the last
    for(unsigned long long i = 3; i < ceil(sqrt(number)) + 1; i++){
        while(number % i == 0){
            number = number / i;
            counter++;
        }
        answer = answer * (counter + 1);
        counter = 0;
    }
	//If the last number is not a square this will account for it.
    if(ceil(sqrt(number) != floor(sqrt(number)))){
            counter++;
    }
    answer = answer * (counter + 1);
    return answer;
}