Friday, September 16, 2016

Count Divisors of Factorial - GeeksforGeeks


Count Divisors of Factorial - GeeksforGeeks
Given a number n, count total number of divisors of n!.

A Simple Solution is to first compute factorial of given number, then count number divisors of the factorial. This solution is not efficient and may cause overflow due to factorial computation.
A better solution is based on Legendre’s formula . Below are the steps.
  1. Find all prime numbers less than or equal to n (input number). We can use Sieve Algorithm for this. Let n be 6. All prime numbers less than 6 are {2, 3, 5}.
  2. For each prime number p find the largest power of it that divides n!. We use below Legendre’s formula formula for this purpose.
    The value of largest power that divides p is ⌊n/p⌋ + ⌊n/(p2)⌋ + ⌊n/(p3)⌋ + ……
    Let these values be exp1, exp2, exp3,.. Using the above formula, we get below values for n = 6.
    • The largest power of 2 that divides 6!, exp1 = 4.
    • The largest power of 3 that divides 6!, exp2 = 2.
    • The largest power of 5 that divides 6!, exp3 = 1.
  3. The result is (exp1 + 1) * (exp2 + 1) * (exp3 + 1) … for all prime numbers, For n = 6, the values exp1, exp2, and exp3 are 4 2 and 1 respectively (computed in above step 2). So our result is (4 + 1)*(2 + 1) * (1 + 1) = 30
// allPrimes[] stores all prime numbers less
// than or equal to n.
vector<ull> allPrimes;
 
// Fills above vector allPrimes[] for a given n
void sieve(int n)
{
    // Create a boolean array "prime[0..n]" and
    // initialize all entries it as true. A value
    // in prime[i] will finally be false if i is
    // not a prime, else true.
    vector<bool> prime(n+1, true);
 
    // Loop to update prime[]
    for (int p=2; p*p<=n; p++)
    {
        // If prime[p] is not changed, then it
        // is a prime
        if (prime[p] == true)
        {
            // Update all multiples of p
            for (int i=p*2; i<=n; i += p)
                prime[i] = false;
        }
    }
 
    // Store primes in the vector allPrimes
    for (int p=2; p<=n; p++)
        if (prime[p])
            allPrimes.push_back(p);
}
 
// Function to find all result of factorial number
ull factorialDivisors(ull n)
{
    sieve(n);  // create sieve
 
    // Initialize result
    ull result = 1;
 
    // find exponents of all primes which divides n
    // and less than n
    for (int i=0; i < allPrimes.size(); i++)
    {
        // Current divisor
        ull p = allPrimes[i];
 
        // Find the highest power (stored in exp)'
        // of allPrimes[i] that divides n using
        // Legendre's formula.
        ull exp = 0;
        while (p <= n)
        {
            exp = exp + (n/p);
            p = p*allPrimes[i];
        }
 
        // Multiply exponents of all primes less
        // than n
        result = result*(exp+1);
    }
 
    // return total divisors
    return result;
}
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