500.cpp

#include <iostream>
#include <set>
#include <vector>
#include <cmath>

using namespace std;

typedef pair<long long, long long> pii;
set<pii> pq;

// limit <= 10^8
vector<int> prime_numbers (int limit) {
  const int SQRT_MAXN = 10000;
  const int S = 1000;

  vector<int> primes;
  vector<int> res_primes;
  bool nprime[SQRT_MAXN], bl[S];
  int cnt = 0;

  memset(nprime, 0, sizeof nprime);
  memset(bl, 0, sizeof bl);

  int nsqrt = (int) sqrt (limit + .0);
  for (int i=2; i<=nsqrt; ++i)
    if (!nprime[i]) {
      primes.push_back(i);
      if (i * 1ll * i <= nsqrt)
        for (int j=i*i; j<=nsqrt; j+=i)
          nprime[j] = true;
    }

  cnt = (int) primes.size();

  int result = 0;
  for (int k=0, maxk=limit/S; k<=maxk; ++k) {
    memset (bl, 0, sizeof bl);
    int start = k * S;
    for (int i=0; i<cnt; ++i) {
      int start_idx = (start + primes[i] - 1) / primes[i];
      int j = max(start_idx,2) * primes[i] - start;
      for (; j<S; j+=primes[i])
        bl[j] = true;
    }
    if (k == 0)
      bl[0] = bl[1] = true;
    for (int i=0; i<S && start+i<=limit; ++i)
      if (!bl[i])
        res_primes.push_back(start+i);
  }

  return res_primes;
}

int main () {
  vector<int> p = prime_numbers(10000000);

  for (int i = 0; i < (int)p.size(); i++)
    pq.insert(make_pair(p[i] * 1LL, p[i] * 1LL));

  long long num = 1;
  const long long MOD = 500500507LL;
  
  // for every exponent in 2^500500
  for (int two_exp = 1; two_exp <= 500500; two_exp++) {
    // take top of the heap
    pii top = *pq.begin();
    pq.erase(pq.begin());

    // multiply the result
    long long prime_exp = top.first;
    num = (num * prime_exp) % MOD;

    // put the new candidate back to the queue
    long long new_base = top.second;
    long long new_prime_exp = prime_exp * prime_exp;
    pq.insert(pii(new_prime_exp, new_base));
  }

  cout << num << endl;

  return 0;
}

Nice algorithm btw. Thanks for sharing 👍 Will study it & the maths behind it when done with the stuff at hand