[keeper962]

I need to generate #1-1000 on the screen and then the computer counts which of those numbers are divisible by 3.

I am able to computer odd numbers I know how to do that but I can't get the rest of the numbers so 1-1000 show.

I have an idea about how to count which are dividable by 3:

if (n%3==0 && n<=1000),// n is the number 
b=b+1;//this would be the count system
cout<<b<<endl;

I am not sure if that works so can someone help me?

Here is the code I have so far:

#include <iostream>
#include <cmath>
#include <iomanip>
int divisible (int&n);

int n;

using namespace std;
 
#include <stdio.h>

int main()
{
    int InputNumber = 0;
    int i, j, k = 0;
    int PreviousNumber = 0;

    cout<<"Input number greater than two"<<endl;
    cin>>InputNumber;

 
    for(PreviousNumber = 0; PreviousNumber <= InputNumber; PreviousNumber++)
    {

      
            if(PreviousNumber % 3 == 0 && PreviousNumber<=1000)
            {
                k++;
            }
        
        if(k == 0)
        {
        cout<<PreviousNumber<<" ";
        }
        k=0;
    }
     if (n%3==0 && n<=1000),// n is the number 
       b=b+1;//this would be the count system
       cout<<b<<endl;
   
    return 0;
}

So Can anyone help??

This post has been edited by keeper962: 27 March 2010 - 09:06 AM

[keeper962]

CAN ANYONE HELP???????

[eker676]

What are you trying to do? Your problem isn't very clear.

Are you trying to check which numbers between 1 and 1000 are divisible by 3?

All you need here is a for loop and an if statement.

int nrDivByThree = 0;
for(int i = 1; i <= 1000; i++)
{
  if(i % 3 == 0)
  {
    cout << i << endl;
    nrDivByThree++;
  }
}

cout << "# of numbers divisible by three: " << nrDivByThree << endl;

Is that what you are looking for?

This post has been edited by eker676: 27 March 2010 - 10:05 AM

[keeper962]

well I need to display the numbers 1-1000 on the screen and the computer counts which of the 1000 numbers are divisiable by 3 and displays that number. So when you start the program it displays 1, 2, 3, 4..1000 and also displays the number of (out of 1000) are divisable by 3 so if 300 are divisable by 3 out of the 1000 300 is displayed.

[eker676]

Well that requires one for loop and an if statement. I don't know what all those variables are for but all you need is one variable plus the counting variable.

int divis = 0;

for(int i = 1; i <= 1000; i++)
{
  cout << i << endl;

  if(i % 3 == 0)
  {
    divis++;
  }
}

cout << "Numbers divisible by three = " << divis << endl;

[sarmanu]

Change the return type of "divisible" to "void", since your function does not return anything. And "nrDivByThree" should be a local variable for your function. You don't use it outside your function.

[eker676]

Using global variables, variables that can be accessed throughout the program, is bad practice when a local variable will suffice. A local variable goes out of scope or deletes itself when it comes to the end of a function/loop/etc.

On to your program,

Look at the comments:

#include <iostream>
#include <cmath>
#include <iomanip>

// Change your function prototype and def.
void divisible ();

// int nrDivByThree=0; <--- You don't need a global variable

using namespace std;

int main()
{
 //divisible(nrDivByThree);
 divisible();
 return 0;
}

void divisible ()
{
  int nrDivByThree = 0; // <---- ADD VARIABLE HERE

for(int i = 1; i <= 1000; i++)
{
  cout << i << " ";
  if(i % 3 == 0)
  {
    nrDivByThree++;
  }
}

cout <<endl<< "# of numbers divisible by three: " << nrDivByThree << endl;
}

This post has been edited by eker676: 27 March 2010 - 08:02 PM

[keeper962]

Thank You for cleaning up my program and for helping me I really appreciate it.