counting steps

ok heres my problem..
1. if x is even, divide x by 2.
2. if x is odd, multiply x by 3 and add 1, and count 1 step.
3. Repeat steps 1 and 2 until you arrive at the number 1.

For example, if we begin with 13, the sequence we obtain is:
13, 40, 20, 10, 5, 16, 8, 4, 2, 1
with 2 steps (from 13 to 40 and from 5 to 16).

this is my code so far... cant seem to get the loop working but i got the even/odd going...

Code:

---

#include <cstdlib>
#include <iostream>

using namespace std;
int steps(int a);

int main()
{
    int month;
    cout<<"enter month: "<<endl;
    cin>>month;
   
    cout<<steps(month)<<endl;
   
    system("PAUSE");
    return EXIT_SUCCESS;
}

int steps(int a)
{
    int s = 1;
    int count = 0;
    int answer;
   
    for (int x = 0; x>1; x++) <---- i have a feeling something is wrong here...
    {
      if (a%2==0)
      {
      answer = a/2;
     
      }
      else
      {
      answer = (a * 3) + 1;
      count++;
      cout<<answer<<endl;
      }
    }
   
    return count;
}

---

thanks in advanced!!

use a do-while loop because you dont know how many iterations your going through, but with a for loop you do.

Code:

---

do{
      if (a%2==0)
      {answer = a/2;}
      else
      {
          answer = (a * 3) + 1;
          count++;
          cout<<answer<<endl;
      }
     
}
while(a!=1);

---

Quote:

---

Originally Posted by I BLcK I

use a do-while loop because you dont know how many iterations your going through, but with a for loop you do.

Code:

---

do{
      if (a%2==0)
      {answer = a/2;}
      else
      {
          answer = (a * 3) + 1;
          count++;
          cout<<answer<<endl;
      }
     
}
while(a!=1);

---

---

can u do it without the DO while loop ?

a while loop is probably better, so that it works for a = 1

also, inside the loop, you probably want to assign the things back to "a", like "a = a/2" and "a = a * 3 + 1"

Code:

---

int steps(int a)
{
    int count = 0;
   
    while (a > 1)
    {
      if (a%2==0)
      {
      a = a/2;
     
      }
      else
      {
      a = (a * 3) + 1;
      count++;
      }
    }
   
    return count;
}

---

Quote:

---

Originally Posted by spoon!

a while loop is probably better, so that it works for a = 1

also, inside the loop, you probably want to assign the things back to "a", like "a = a/2" and "a = a * 3 + 1"

Code:

---

int steps(int a)
{
    int count = 0;
   
    while (a > 1)
    {
      if (a%2==0)
      {
      a = a/2;
     
      }
      else
      {
      a = (a * 3) + 1;
      count++;
      }
    }
   
    return count;
}

---

---

thanks spoon, it was those small changes that made it work!!
appreciate it :)

Its much more simplier and logically reasonable to use a do while loop, but i dont see why you would want to use a for loop. To answer your question, yes I can do it with a do, and for loop, but its much more comprehendable to use a do-while loop if someone else where to read your code.

Quote:

---

Originally Posted by I BLcK I

Its much more simplier and logically reasonable to use a do while loop, but i dont see why you would want to use a for loop. To answer your question, yes I can do it with a do, and for loop, but its much more comprehendable to use a do-while loop if someone else where to read your code.

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ok thanks mate! much appreciated :)