Prime factorization logic

[Tej Samra]

Can someone explain to me the logic of this program?

#include <cstdlib>
#include <iostream>

using namespace std;

int main()
{
  int n;
  int i; 

cout <<"Enter a Number:";
cin>>n;   // lets say n=140
cout<<"Prime Factors:"; 
for(i=2;i<=n;i++) 
{ 
if(n%i==0) 
{ 
cout << i <<","; // this should show all factors of 140 even if they're not prime

n=n/i; //dont understand this. would n become half if i starts at 2?
i--; //dont understand either would i count down from 140 and does this affect n=n/i?
if(n==1) 
break; 
} 


} 

where in this program does it filter prime numbers from the rest and how does it know how many to display?

  
    system("PAUSE");
    return (0);
}

[ne555]

n=n/i; //dont understand this. would n become half if i starts at 2?
Yep, but only if n is even

"Your" code is equivalent to

for( int K=2; K<=n; ++K)
  while( n%K == 0 ){ //if the number is a divisor
    n /= K; //then divide it
    //put K in a divisor container, show it, etc.
  }

_ ¿Why it only show prime factors?
_ Because you start with K=2.
In every step you kill every all the 'K' factors from n, so later n%composite would fail.

_ ¿Why i--?
_ Bad style. It's a substitute for the while loop.
You are always doing i++, so in order to kill all the factors, you decrement i and then increment it (do nothing)