# euler 47 distinct prime factors

May 4, 2015

This one turned out to be very similar to the previous one (except for the small detail of iterating from the smallest prime upwards, instead of the other way round).

``````class PrimeIterator {
public:
PrimeIterator() : prime_index_(-1) {}

int getNextPrime() {
assert((prime_index_ + 1) < primes_.size());
return primes_[++prime_index_];
}

bool hasMorePrimes() {
assert(prime_index_ >= -1 && (prime_index_ + 1) <= primes_.size());
return (prime_index_ + 1) < primes_.size();
}

void fillPrimes(int max) {
while (primes_.back() <= max) {
}
// Reset to before the first element
prime_index_ = -1;
}

private:
assert(primes_.size() > 0);
int n = primes_.back();
while (true) {
++n;
bool is_prime = true;
for (const int& p : primes_) {
if (n % p == 0) {
is_prime = false;
break;  // Try next number
}
}
if (is_prime) {
primes_.push_back(n);
return;
}
}
}

static vector<int> primes_;

int prime_index_;
};

vector<int> PrimeIterator::primes_ = {2};

int getNumPrimeFactors(int n) {
int numPrimeFactors = 0;
PrimeIterator pit;
pit.fillPrimes(n);
while (n > 1 && pit.hasMorePrimes()) {
int p = pit.getNextPrime();
if (n % p == 0) {
while (n % p == 0) {
n /= p;
}
++numPrimeFactors;
}
}
return numPrimeFactors;
}

int main() {
cout << "Euler #47 ... \n";

int runLength = 0;
for (int n = 2; ; ++n) {
if (getNumPrimeFactors(n) == 4) {
++runLength;
} else {
runLength = 0;
}
if (runLength == 4) {
cout << "Found a sequence of four numbers with four prime "
<< "factors starting at : " << n - 3 << endl;
return 0;
}
}
}
``````

Runs as follows:

``````\$ time ./Test
Euler #47 ...
Found a sequence of four numbers with four prime factors starting at : <redacted>

real    0m3.814s
user    0m3.812s
sys     0m0.000s
``````

(Yep, a noticeable delay, but anything under 5 seconds is good for me …)