Task: Long primes


C++: #include <iomanip>
#include <iostream>
#include <list>
using namespace std;
  
void sieve(int limit, list<int> &primes)
{
  bool *c = new bool[limit + 1];
  for (int i = 0; i <= limit; i++)
    c[i] = false;
  
  int p = 3, n = 0;
  int p2 = p * p;
  while (p2 <= limit)
  {
    for (int i = p2; i <= limit; i += 2 * p)
      c[i] = true;
    do
      p += 2;
    while (c[p]);
    p2 = p * p;
  }
  for (int i = 3; i <= limit; i += 2) 
    if (!c[i]) 
      primes.push_back(i);
  delete [] c;
}
 

int findPeriod(int n) 
{
  int r = 1, rr, period = 0;
  for (int i = 1; i <= n + 1; ++i)
    r = (10 * r) % n;
  rr = r;
  do
  {
    r = (10 * r) % n;
    period++;
  }
  while (r != rr);
  return period;
}
 
int main()
{
  int count = 0, index = 0;
  int numbers[] = {500, 1000, 2000, 4000, 8000, 16000, 32000, 64000};
  list<int> primes;
  list<int> longPrimes;
  int numberCount = sizeof(numbers) / sizeof(int);
  int *totals = new int[numberCount];
  cout << "Please wait." << endl << endl;
  sieve(64000, primes);  
  for (list<int>::iterator iterPrime = primes.begin();  
    iterPrime != primes.end(); 
    iterPrime++)
  {  
    if (findPeriod(*iterPrime) == *iterPrime - 1)
      longPrimes.push_back(*iterPrime);
  }    
  
  for (list<int>::iterator iterLongPrime = longPrimes.begin();
    iterLongPrime != longPrimes.end(); 
    iterLongPrime++)
  {
    if (*iterLongPrime > numbers[index])
      totals[index++] = count;
        ++count;
  }
  totals[numberCount - 1] = count;
  cout << "The long primes up to " << totals[0] << " are:" << endl;
  cout << "[";
  int i = 0;
  for (list<int>::iterator iterLongPrime = longPrimes.begin(); 
    iterLongPrime != longPrimes.end() && i < totals[0]; 
    iterLongPrime++, i++)
  {  
    cout << *iterLongPrime << " ";
  }  
  cout << "\b]" << endl;
  cout << endl << "The number of long primes up to:" << endl;
  for (int i = 0; i < 8; ++i)
    cout << "  " << setw(5) << numbers[i] << " is " << totals[i] << endl;
  delete [] totals;
}






Python: def sieve(limit):
    primes = []
    c = [False] * (limit + 1) 
    
    p = 3
    while True:
        p2 = p * p
        if p2 > limit: break
        for i in range(p2, limit, 2 * p): c[i] = True
        while True:
            p += 2
            if not c[p]: break

    for i in range(3, limit, 2):
        if not c[i]: primes.append(i)
    return primes


def findPeriod(n):
    r = 1
    for i in range(1, n): r = (10 * r) % n
    rr = r
    period = 0
    while True:
        r = (10 * r) % n
        period += 1
        if r == rr: break
    return period

primes = sieve(64000)
longPrimes = []
for prime in primes:
    if findPeriod(prime) == prime - 1:
        longPrimes.append(prime)
numbers = [500, 1000, 2000, 4000, 8000, 16000, 32000, 64000]
count = 0
index = 0
totals = [0] * len(numbers)
for longPrime in longPrimes:
    if longPrime > numbers[index]:
        totals[index] = count
        index += 1
    count += 1
totals[-1] = count
print('The long primes up to 500 are:')
print(str(longPrimes[:totals[0]]).replace(',', ''))
print('\nThe number of long primes up to:')
for (i, total) in enumerate(totals):
    print('  %5d is %d' % (numbers[i], total))