edaBit-coding-challenges/The-Centrifuge-Problem/code.cpp

/*  A centrifuge, as you probably know, is a laboratory device used to 
 *  separate fluids based on density. The separation is achieved through 
 *  centripetal force by spinning a collection of test tubes at high speeds. 
 *  This means, the configuration needs to be in balance.

 *  Create a function that takes two numbers as arguments n and k and 
 *  returns true if the configuration is balanced and false if it's not. 
 *  To check out the formula, look at the resources tab.
 *  Here are the valid configurations for n = 6, k = 2, 3, 4 and 6.
 *  One test tube (k = 1) is never in balance.
 *  One hole (n = 1) is never in balance, even empty.
 */
std::vector<int> factors(int n){
    int z = 2;
		std::vector<int> output; 
    while (z * z <= n){
        if (n % z == 0){
            output.push_back(z);
            n /= z;
				}else{z++;}
		}            
    if(n > 1){output.push_back(n);}
	return output;
}
//  	It(test3){Assert::That(cFuge(12, 7), Equals(true));}
bool cFuge(int n, int k) {
	int Sum = 0;
	//  if k or n is one, return false
	if(1 == k || 1 == n){return false;}
	if(0 == k){return true;}
	if(n == k){return true;}
	//  if k and n are even, return true
	if ((k%2 == 0) && (n%2 == 0)){return true;}
	std::vector<int> nfactors = factors(n);
	//  find the prime factorization of a number and see if the sum of those numbers is equal to k
	for(int i = 0;i<nfactors.size();i++){Sum += nfactors[i];}
	if(Sum == k){return true;}
	std::vector<int> kfactors = factors(k);
	//  if n & k share a common factor, return true.
	for(int i = 0;i<nfactors.size();i++){
		for(int j = 0;j<kfactors.size();j++){
			if(nfactors[i] == kfactors[j]){return true;}	
		}
	}
	return false;
}