Write a program to count the number of prime factors of a given positive integer and calculate the difference between the number and the lowest prime factor.

Sure, here is a Python program that does what you asked for:

def count_prime_factors_and_difference(n):
    # Initialize count of prime factors and lowest prime factor
    count = 0
    lowest_prime_factor = 0

    # Count the number of 2s that divide n
    while n % 2 == 0:
        n = n / 2
        count += 1
        if lowest_prime_factor == 0:
            lowest_prime_factor = 2

    # n must be odd at this point, thus skip the even numbers and iterate only for odd
    for i in range(3, int(n**0.5)+1, 2):
        while n % i == 0:
            n = n / i
            count += 1
            if lowest_prime_factor == 0:
                lowest_prime_factor = i

    # This condition is to handle the case when n is a prime number greater than 2
    if n > 2:
        count += 1
        if lowest_prime_factor == 0:
            lowest_prime_factor = n

    # Calculate the difference
    difference = n - lowest_prime_factor

    return count, difference

# Test the function
n = 315
count, difference = count_prime_factors_and_difference(n)
print("The number of prime factors is", count)
print("The difference between the number and the lowest prime factor is", difference)

This program first counts the number of 2s that divide the given number, then it counts the number of odd prime factors. If the given number is a prime number greater than 2, it also counts it as a prime factor. The lowest prime factor is also tracked during this process. Finally, it calculates the difference between the given number and the lowest prime factor.