Task: Long primes
C: #include <stdio.h>
#include <stdlib.h>
 
#define TRUE 1
#define FALSE 0
 
typedef int bool;
 
void sieve(int limit, int primes[], int *count) {
    bool *c = calloc(limit + 1, sizeof(bool)); 
    
    int i, p = 3, p2, n = 0;
    p2 = p * p;
    while (p2 <= limit) {
        for (i = p2; i <= limit; i += 2 * p) 
            c[i] = TRUE;
        do {
            p += 2;
        } while (c[p]);
        p2 = p * p;
    }
    for (i = 3; i <= limit; i += 2) {
        if (!c[i]) primes[n++] = i;
    }
    *count = n;
    free(c);
}
 

int findPeriod(int n) {
    int i, r = 1, rr, period = 0;
    for (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 i, prime, count = 0, index = 0, primeCount, longCount = 0, numberCount;
    int *primes, *longPrimes, *totals;
    int numbers[] = {500, 1000, 2000, 4000, 8000, 16000, 32000, 64000};
     
    primes = calloc(6500, sizeof(int));
    numberCount = sizeof(numbers) / sizeof(int);
    totals = calloc(numberCount, sizeof(int));
    sieve(64000, primes, &primeCount);
    longPrimes = calloc(primeCount, sizeof(int)); 
    
    for (i = 0; i < primeCount; ++i) {
        prime = primes[i];
        if (findPeriod(prime) == prime - 1) {
            longPrimes[longCount++] = prime;
        }
    }
    for (i = 0; i < longCount; ++i, ++count) {
        if (longPrimes[i] > numbers[index]) {
            totals[index++] = count;
        }
    }
    totals[numberCount - 1] = count;
    printf("The long primes up to %d are:\n", numbers[0]);
    printf("[");
    for (i = 0; i < totals[0]; ++i) {
        printf("%d ", longPrimes[i]);
    }
    printf("\b]\n");
 
    printf("\nThe number of long primes up to:\n");
    for (i = 0; i < 8; ++i) {
        printf("  %5d is %d\n", numbers[i], totals[i]);
    }
    free(totals);
    free(longPrimes);
    free(primes);
    return 0;
}

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;
}