[Shebang]

Oh, I divided the odd number by 3. XD

Code:

#include <iostream> int main(){int n;std::cin >> n;while(n > 1){n%2==0?n/=2:n=n*3+1;std::cout<<n<<" ";}return 0;}

I wonder why n*=3+1 doesn't work, like n=n*3+1 ??? Anyway, it's 114 bytes now.

Code:

#include <iostream> int main(){int n;std::cin>>n;while(n>1){n%2==0?n/=2:n=n*3+1;std::cout<<n<<" ";}return 0;}

110 bytes after removing some spaces. :3

The reason you can't do n*=3+1 is because the *= assignment operator multiplies the left by everything on the right. Essentially, it's doing n=n*(3+1).

Speaking of shortening, you can change the ternary expression to be part of an assignment. So you can do n=n%2==0?n/2:3*n+1;

Furthermore, you can change the condition for the ternary to be just n%2. This is essentially the same logic I used. The reason for this is that n%2 will always return 0 or 1, where 0 is a "falsy" value and 1 is a "truthy" value. So, you can rearrange the statement to become n=n%2?3*n+1:n/2;

That's a good start, I think. You should also try to use command-line arguments so you don't need that long code to get it from STDIN.

[lux]

Had to make a slight change to join the 69 club.

Code:

for(i=process.argv[2];i>1;(i%2==0)?(i=i/2):(i=i*3+1),console.log(i))

---

The reason you can't do n*=3+1 is because the *= assignment operator multiplies the left by everything on the right. Essentially, it's doing n=n*(3+1).

Speaking of shortening, you can change the ternary expression to be part of an assignment. So you can do n=n%2==0?n/2:3*n+1;

Furthermore, you can change the condition for the ternary to be just n%2. This is essentially the same logic I used. The reason for this is that n%2 will always return 0 or 1, where 0 is a "falsy" value and 1 is a "truthy" value. So, you can rearrange the statement to become n=n%2?3*n+1:n/2;

That's a good start, I think. You should also try to use command-line arguments so you don't need that long code to get it from STDIN.

@OversouL read up on 'Operator Associativity'. Here's a link to get you started. http://www.maths.cam.ac.uk/undergrad/cat...ode38.html

[lux]

[Eclipse]

This is especially true if you're also a JS developer.

[OversouL]

The reason you can't do n*=3+1 is because the *= assignment operator multiplies the left by everything on the right. Essentially, it's doing n=n*(3+1).

Speaking of shortening, you can change the ternary expression to be part of an assignment. So you can do n=n%2==0?n/2:3*n+1;

Furthermore, you can change the condition for the ternary to be just n%2. This is essentially the same logic I used. The reason for this is that n%2 will always return 0 or 1, where 0 is a "falsy" value and 1 is a "truthy" value. So, you can rearrange the statement to become n=n%2?3*n+1:n/2;

That's a good start, I think. You should also try to use command-line arguments so you don't need that long code to get it from STDIN.

Oh, I see. Those are some neat tricks. Also what do you mean by command-line arguments? I googled it, but won't you still try to declare somethings in main before you can do it? Which just takes more bytes. Unless you're making something long, then I guess it will be better. Or I'm talking of something else? Far from what you meant?

Had to make a slight change to join the 69 club.

Code:

for(i=process.argv[2];i>1;(i%2==0)?(i=i/2):(i=i*3+1),console.log(i))

---

@OversouL read up on 'Operator Associativity'. Here's a link to get you started. http://www.maths.cam.ac.uk/undergrad/cat...ode38.html

Thanks for letting me know about it. I thought everything was read from left to right.

I should get involved in this more. I learn a lot from you guys.

[lux]

This is especially true if you're also a JS developer.

Okay, kid.

[OversouL]

I forgot to add that thanks to Shebang I'm now at 105 bytes.

Code:

#include <iostream> int main(){int n;std::cin>>n;while(n>1){n=n%2?3*n+1:n/2;std::cout<<n<<" ";}return 0;}

[Eclipse]

Okay, kid.

sorry man all hail gitblub

Spoiler:

jk ily rly

I forgot to add that thanks to Shebang I'm now at 105 bytes.

Code:

#include <iostream> int main(){int n;std::cin>>n;while(n>1){n=n%2?3*n+1:n/2;std::cout<<n<<" ";}return 0;}

Updated.