Editorial

Let $d(n)$ be the number of divisors of $n$. Obviously, $d(1)=1$.

Let $p$ be a prime integer. Then $p$ has two divisors: $1$ and $p$. Therefore, $d(p)=2$.

Let $n=p^k$ be a power of a prime number. Then $n$ has $k+1$ divisors: $1,p,p_2,p_3,...,p^k$. Hence, $d(p^k)=k+1$.

Let $n=p^k{q}^l$. Consider two sets: $P=1,p,p_2,p_3,...,p_k$ and $Q=1,q,q_2,q_3,...,q_l$

Any divisor $d$ of the number $p^k{q}^l$ can be represented in the form $x\cdot y$, where $x\in P,y\in Q$. A divisor $x$ from $P$ can be chosen in $k+1$ ways, and a divisor $y$ from $Q$ can be chosen in $l+1$ ways. Therefore, a divisor $d=x\cdot y$ can be formed in $(k+1)\cdot (l+1)$ ways.

Decompose the number $n$ into its prime factors: $n=p_1^{a_1}{p}_2^{a_2}...p_k^{a_k}$. The number of divisors of $n$ is equal to

Example

The number $n=18$ has $6$ divisors: $1,2,3,6,9,18$.

Let's decompose the number $n=18$ into its prime factors:

Therefore

Subtracting two divisors ($1$ and $18$), we get the answer: $4$ divisors.

Algorithm realization

The function CountDivisors decomposes the number $n$ into its prime factors and calculates the number of its divisors $d(n)$.

int CountDivisors(int n)
{
  int c, i, res = 1;
  for(i = 2; i * i <= n; i++)
  {
    if (n % i == 0)
    {
      c = 0;
      while(n % i == 0)
      {
        n /= i;
        c++;
      }
      res *= (c + 1);
    }
  }
  if (n > 1) res *= 2;
  return res;
}

The main part of the program. Read the value of $n$.

scanf("%d",&n);

Calculate and print the answer. From the total number of divisors $d(n)$ subtract two divisors: $1$ and the number $n$ itself.

printf("%d\n",CountDivisors(n)-2);

Python realization

import math

The function CountDivisors decomposes the number $n$ into its prime factors and calculates the number of its divisors $d(n)$.

def CountDivisors(n):
  res = 1;
  for i in range(2, int(math.sqrt(n)) + 1):
    if (n % i == 0):
      c = 0
      while (n % i == 0):
        n //= i
        c += 1
      res *= (c + 1);
  if (n > 1): res *= 2;
  return res;

The main part of the program. Read the value of $n$.

n = int(input())

Calculate and print the answer. From the total number of divisors $d(n)$ subtract two divisors: $1$ and the number $n$ itself.

print(CountDivisors(n) - 2)