313 Super Ugly Number

Write a program to find the nth super ugly number.
Super ugly numbers are positive numbers whose all prime factors are in the given prime list primes of size k. For example, [1, 2, 4, 7, 8, 13, 14, 16, 19, 26, 28, 32] is the sequence of the first 12 super ugly numbers given primes = [2, 7, 13, 19] of size 4.
Note:
(1) 1 is a super ugly number for any given primes.
(2) The given numbers in primes are in ascending order.
(3) 0 < k ≤ 100, 0 < n ≤ 106, 0 < primes[i] < 1000.
bool existsInPrimeList(vector<int> &primes, int n) {
for (int i = 0; i < primes.size(); i++) {
if (n == primes.at(i)) return true;
}
return false;
}
// primes: all numbers the evenly divide n except for 1 and itself
vector<int> generatePrimeFactors(int n) {
vector<int> primes;
int z = 2;
for (int i = 2; i < n; i++) {
// evenly divides
if (n % i == 0) {
primes.push_back(i);
i = 2;
n /= i;
for (int j = 2; j <= n; i++) {
if (n % j == 0) {
primes.push_back(j);
j = 2;
n /= j;
}
}
break;
}
}
return primes;
}
int nthSuperUglyNumber(int n, vector<int>& primes) {
if (n == 1) { return 1; }
// all prime factors are from the prime list
vector<int> factors;
int count = 1;
int current_n = 2;
bool exists;
while (count + 1 < n) {
exists = true;
factors = generatePrimeFactors(current_n);
// make sure factors are all from the prime list
for (auto i : factors) {
if (!existsInPrimeList(primes, i)) {
exists = false;
break;
}
}
if (exists) {
count++;
}
current_n++;
factors.empty();
}
return current_n + 1;
}
int main()
{
//vector<int> test = { 2,3,5 };
//cout << nthSuperUglyNumber(1, test) << endl;
//test = { 2 };
//cout << nthSuperUglyNumber(3, test) << endl;
vector<int> test = { 2,3,5 };
cout << nthSuperUglyNumber(50, test) << endl;
}