Loop

[Bolong Yu]

How to create a loop that prints the numbers 1 to 20 but skips 3, 11, and 16.

[Zaita]

http://www.cplusplus.com/articles/jEN36Up4/

Let's be honest here, you haven't event attempted to solve this yourself. It's not difficult.

[IceThatJaw]

If you are creative with Math you might be able to find a slick way to achieve this without using a brute force method.

It there a pattern to these numbers or is just these 3 specific ones? At first glance, I don't see any patterns but I am not too great at the Maths.

If you want to brute force it just write out the logic in pseudocode.

X = 0
Begin LOOP Till X = 20
IF X != 3 AND X != 11 AND X != 16
Print X
END IF
END LOOP

[jumper007]

Yeah well, the basic C++ code for that is:

#include <iostream>

using namespace std;

int main (int argc, char **argv)
{
     for (int i = 1; i <= 20; i++)
          if (i != 3 && i != 11 && i != 16)
               cout << i << “ “;
     return 0;
}

Hope I could help,
~ Raul ~

[newbieg]

For loops are probably a better option since we're dealing with an iterator ( i ), but there is also the while loop as an option.
Jumper007's code uses the and (&&) signs in his if statement, I used or (||) signs.

They both work in this instance because jumper's code also uses the not equal sign (!=). Basically his code is saying if i doesn't equal 3 and i doesn't equal 11 and i doesn't equal 16 then print out the number.
My version says if i does equal 3 or 11 or 16 then tell them a number is skipped, otherwise print out the number.

int i=0;
while(i<=20)
{
    if(i==3 || i==11 || i==16)
    {
        cout<<"A number was skipped"<<endl;
    }
    else
    {
        cout<<i<<endl;
    }
}

This is also an example of difference in syntax, mine isn't exactly wrong, but the the extra brackets I used aren't necessary after the if and else functions since there is only one action line after both of them. I prefer using brackets because it helps me organize my code when I'm skimming it.