Total number of divisors for a given number using prime Factorization

#include <bits/stdc++.h>

using namespace std;

int divCount(int n)

{

bool mark[n+1];

memset(mark, true, sizeof(mark));

for(int i=2; i*i<=n; i++)

{

if(mark[i]==true)

{

for(int j=i*2; j<n; j+=i)

{

mark[j]=false;

}

}

}

int total=1;

for(int i=2; i<=n; i++)

{

if(mark[i])

{

int cnt=0;

if(n%i==0)

{

while(n%i==0)

{

n/=i;

cnt++;

}

total=total*(cnt+1);

}

}

}

return total;

}

int main ()

{

int n;

cin >> n;

cout << divCount(n) << endl;

return 0;

}

another example:

//this is my work__ finding number of divisor
#include <bits/stdc++.h>
using namespace std;

bool isprime(int x)
{
for(int i=2; i*i<=x; i++)
{
if(x%i==0){
return false;
}
}
return false;
}
int divcount(int n)
{
int total=1;
for(int i=2; i<=n; i++)
{
if(isprime(i)==false)
{
int cnt=0;
if(n%i==0)
{
while(n%i==0)
{
  n/=i;
  cnt++;
}
total=total*(cnt+1);
}
}
}
return total;
}
int main ()
{
int n;
cin >> n;
cout << divcount(n) << endl;
return 0;
}