找出一个数字的因素。 没有得到准确的结果

Question

Can someone help correct my algorithm? I've tested it on a few numbers, and it doesn't output the complete factorization. For numbers with a large number of factors, it just completely fails.

int num = 20;

for(int i = 2; i <= num; i++)
{
    if(num%i == 0)
    {
        cout << i << endl;
        cout << num << endl;
        num = num/i;    
    }
}

EDIT: The two answers provided did not work, still not getting full results.

EDIT2: Divisors VS Factors

Answer 1

Judging from you comment to @ Luchian Grigore , you're confusing divisors with (prime) factorization . Divisors of a number are all numbers for which num % i == 0 is true. Factorization means getting a representation of num by a product of smaller numbers. If you want uniqueness of factorization, you usually use prime factorization.

To get all the divisors your code should be

for ( int i = 1; i <= num; ++i ) // note that 1 and num are both trivially divisors of num
{
    if ( num % i == 0 ) // only check for divisibility
    {
        std::cout << i << std::endl;
    }
}

to get the (prime) factorization, it's

for ( int i = 2; i <= num; ++i )
{
    while ( num % i == 0 ) // check for divisibility
    {
        num /= i;
        std::cout << i << std::endl;
    }
    // at this point, i cannot be a divisor of the (possibly modified) num.
}

Answer 2

The problem is that you're increasing i even if it is a divisor, and you shouldn't unless you find all its occurences.

So, for 4, you'd have 2 twice. But after the first 2 you encounter, you exit the loop because i is incremented to 3 and num became 2.

The following should work:

for(int i = 2; i <= num; )
{
    if(num%i == 0)
    {
        cout << i << endl;
        cout << num << endl;
        num = num/i;    
    }
    else
    {
        i++;
    }
}

Answer 3

for(int i = 2; i <= num; i++)
{
    if(num%i == 0)
    {
        cout << i << endl;
        cout << num << endl;
        num = num/i;    
        i--; // Add this to account for multiple divisors
    }
}

for(int i = 2; i <= num; i++)
{
    if(num%i == 0)
    {
        cout << i << endl;
        cout << num << endl;
    }
}

Answer 4

This should work. Note, should use c++11 for move constructor, otherwise you are going to want to pass in a std::list& instead.

std::list<int64_t> factor(int64_t f)
{
    std::list<int64_t> factors;
    for(int64_t ii = 2; ii<=f; ii++) {
        while(f % ii == 0) {
            f = f/ii;
            factors.push_back(ii);
        }
    }

    return factors;
}