Number of Divisors Using Prime Factorization

Question: Given a positive integer n, we must find the total number of divisors for n.

Let d( n ) be the number of divisors for the natural number, n. We begin by writing the number as a product of prime factors: n = p^a* q^b* r ^c … then the number of divisors, d( n ) = ( a +1)( b +1)( c +1)…

int NOD ( ll n )
{
    int nod = 1;

    for ( auto p : primes ) {
        if ( 1LL * p * p > n ) break;

        if ( n % p == 0 ) {
            int cnt = 0;

            while ( n % p == 0 ) {
                n /= p;
                cnt++;
            }

            cnt++;
            nod *= cnt;
        }
    }

    if ( n > 1 ) {
        nod *= 2;
    }

    return nod;
}

Mathematical Proof: How many divisors does a number have?)...

If the limits are not big, then You can use this code:

int nod[1123];
int lim = 1e3;

    for ( int i = 1; i <= lim; i++ ) {
        for ( int j = i; j <= lim; j += i ) {
            nod[j]++; 
        }
    }

Practice Questions:

583 - Prime Factors

11466 - Largest Prime Divisor

SPOJ.com - Problem MAIN12B