Highly Factorized Sieve of Eratosthenes - c++

Question

Just for fun, I'm trying to implement the pseudocode from this StackOverflow answer for the highly factorized Sieve of Eratosthenes in C++. I can't figure out why my code returns both prime and non-prime numbers. Am I implementing these for loops incorrectly? Should I be using while loops instead? I suspect that I'm not incrementing the for loops properly. Any help would be greatly appreciated. I've spent several hours trying to hunt down the flaw.
GordonBGood's pseudocode is inserted as comments, and I've used all the same variable names.
#include <iostream>
#include <vector>
#include <cmath>
const int limit = 1000000000;
const std::vector<int> r {23,29,31,37,41,43,47,53, 59,61,67,71,73,79,83, //positions + 19
89,97,101,103,107,109,113,121,127, 131,137,139,
143,149,151,157,163,167,169,173,179,181,187,191,193,
197,199,209,211,221,223,227,229};
int main()
{
// an array of length 11 times 13 times 17 times 19 = 46189 wheels initialized
// so that it doesn't contain multiples of the large wheel primes
// for n where n ← 210 × w + x where w ∈ {0,...46189}, x in r: // already
// if (n mod cp) not equal to 0 where cp ∈ {11,13,17,19}: // no 2,3,5,7
// mstr(n) ← true else mstr(n) ← false // factors
std::vector<bool> mstr(limit);
int n;
for (int w=0; w <= 46189; ++w) {
for (auto x = begin(r); x != end(r); ++x) {
n = 210*w + *x;
if (n % 11 != 0 && n % 13 != 0 && n % 17 != 0 && n % 19 != 0)
mstr[n]=true;
else
mstr[n]=false;
}
}
// Initialize the sieve as an array of the smaller wheels with
// enough wheels to include the representation for limit
// for n where n ← 210 × w + x, w ∈ {0,...(limit - 19) ÷ 210}, x in r:
// sieve(n) ← mstr(n mod (210 × 46189)) // init pre-culled primes.
std::vector<bool> sieve(limit+1000);
for (int w=0; w <= (limit-19)/210; ++w) {
for (auto x = begin(r); x != end(r); ++x) {
n = 210*w + *x;
sieve[n] = mstr[(n % (210*46189))];
}
}
// Eliminate composites by sieving, only for those occurrences on the
// wheel using wheel factorization version of the Sieve of Eratosthenes
// for n² ≤ limit when n ← 210 × k + x where k ∈ {0..}, x in r
// if sieve(n):
// // n is prime, cull its multiples
// s ← n² - n × (x - 23) // zero'th modulo cull start position
// while c0 ≤ limit when c0 ← s + n × m where m in r:
// c0d ← (c0 - 23) ÷ 210, cm ← (c0 - 23) mod 210 + 23 //means cm in r
// while c ≤ limit for c ← 210 × (c0d + n × j) + cm
// where j ∈ {0,...}:
// sieve(c) ← false // cull composites of this prime
int s, c, c0, c0d, cm, j;
for ( auto x = begin(r); x != end(r); ++x) {
for ( int k=0; (n=210*k + (*x)) <= sqrt(limit); ++k){
if (sieve[n]) {
s = n*n - n*((*x)-23);
for ( auto m = begin(r); (c0=s+n*(*m)) <= limit && m != end(r); ++m) {
c0d = (c0-23)/210;
cm = (c0-23)%210 + 23;
for ( int j=0; (c=210*(c0d+n*j)+cm) <= limit; ++j) {
sieve[c] = false;
}
}
}
}
}
// output 2, 3, 5, 7, 11, 13, 17, 19,
// for n ≤ limit when n ← 210 × k + x where k ∈ {0..}, x in r:
// if sieve(n): output n
std::cout << "2\n3\n5\n7\n11\n13\n17\n19\n";
for ( auto x = begin(r); x != end(r); ++x) {
for ( int k = 0; (n=210*k + (*x)) <= limit; ++k) {
if (sieve[n]);
std::cout << n << std::endl;
}
}
std::cout << std::endl;
return 0;
}