// primef2.c  factor numbers up to 63 bits, need 64 bit computer, OS, compiler
//            factor up to 0.922*10^20   about 19 digits
//            the first one million primes are in the executable program 
#include <stdio.h>
#include <time.h>
#include "primef2.h" // primes and np come from here

int main(void)
{
  int i, j, nfact = 0;;
  long int rem, quo, prod, chk;
  long int fctr[100]; // actually could be 63, 2^63 has most
  double now, next;
  int rtn;

  printf("primef2.c running, enter a number, less than 64 bits, to factor\n");
  while(1)
  {
    rtn = scanf("%ld", &rem);
    if(rtn==0 || rtn==EOF) return 0;

    now = (double)clock()/(double)CLOCKS_PER_SEC;
    printf("factor %ld \n", rem);
    if(rem<2l)
    {
      printf("primef2.c finished \n");
      return 0;
    }
    j = 0;
    nfact = 0;
    while(j<np)
    {
      quo = rem/primes[j];
      prod = quo*primes[j];
      if(prod==rem)
      {
        printf("a factor is %ld \n", primes[j]);
        fctr[nfact] = primes[j];
        nfact++;
        rem = quo; 
      }
      else
      {
        j++;
      }
      if(rem==1) break;
    } // end while j<np
    if(j>=np)
    {
      fctr[nfact] = rem;
      nfact++;
    }
    if(rem != 1) printf("a factor is %ld \n", rem);
    chk = 1l; // check result
    for(i=0; i<nfact; i++) chk = chk * fctr[i];
    printf("check product of %d factors = %ld\n", nfact, chk);
    next = (double)clock()/(double)CLOCKS_PER_SEC;
    printf("factored and checked in %f seconds\n\n", next-now);
  } // end while, read next number
} // end primef2.c