Finding total number of unique factorization - c++

Question

I want to find total factors of any number.
In number theory, factorization is the breaking down of a composite number into smaller non-trivial divisors, which when multiplied together equal the original integer. Your job is to calculate number of unique factorization(containing at least two positive integers greater than one) of a number.
For example: 12 has 3 unique factorizations: 2*2*3, 2*6, 3*4 . Note:
3*4 and 4*3 are not considered different.
I have attempted to find that but not getting exact for all.
Here is my code :
#include<iostream>
using namespace std;
int count=0;
void factor(int n,int c,int n1)
{
for(int i=n1; i<n ; i++)
{
if(c*i==n)
{count++;
return;}
else
if(c*i>n)
return;
else
factor(n,c*i,i+1);
}
return;
}
int main()
{
int num,n;
cin>>num;
for(int i=0 ; i<num ; i++)
{
cin>>n;
count=0;
factor(n,1,1);
cout<<count<<endl;
}
return 0;
}
Input is number of test cases followed by test-cases(Numbers).
Example : Input: 3 12 36 3150
Output: 3 8 91

Answer 1

I think you are looking for number of factorizations of a number which are unique.
For this I think you need to find the count of number of prime factor of that number. Say for
12 = 2, 2, 3
Total count = 3;
For 2, 2, 3 we need
(2*2)*3 ~ 4*3
2*(2*3) ~ 2*6
2*2*3 ~ 2*2*3
To solve this we have idea found in Grimaldi, discrete and combinatorial mathematics.
To find number of ways of adding to a number(n) is 2^(n-1) -1. For 3 we have...
3 =
1+1+1
2+1
1+2
Total count = 2^(3-1) -1 = 4-1 = 3
We can use analogy to see that
1+1+1 is equivalent to 2*2*3
1+2 is equivalent to 2*(2*3)
2+1 is equivalent to (2*2)*3
Say number of prime factors = n
So we have number of factorizations = 2^(n-1)-1
The code:
#include <stdio.h>
int power(int x, int y)
{
int prod =1, i ;
for(i=1; i<=y;i++) prod *= x;
return prod;
}
int main()
{
int number,div;
int count = 0, ti, t;
printf("Input: ");
scanf("%d",&t);
for(ti=1; ti<=t;ti++)
{
scanf("%d", &number);
div = 2;count = 0;
while(number != 0)
{
if(number%div!=0) div = div + 1;
else
{
number = number / div;
//printf("%d ",div);
count++;
if(number==1) break;
}
}
printf("%d ", power(2,count-1)-1);
}
return 0;
}

Answer 2

Using mod is really useful in attempting to factor:
for(int i = 1; i <= fnum; ++i){ //where fnum is the number you wish to factor
if(!(fnum % i)) ++count;
}
return count;
Of cross this is the number of factors, not unique factors, if you want the number of unique factors, you have to do some additional work.

Answer 3

The solution is to realize that of all permutations, precisely one is sorted. 2 * 4 * 7 * 3 gives the same result as 2 * 3 * 4 * 7. That means that when you've found one factor, you should not check the remainder for lower factors. However, you should check if the same factor appears again: 12 = 2 * 2 * 3. The sequence 2 2 3 is also sorted.
BTW, you should give your variables clearer names, or at least add some comments describing them.