(08-22-2015, 09:55 AM)OversouL Wrote: 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

(08-22-2015, 12:29 PM)Shebang Wrote: 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.

(08-22-2015, 03:58 PM)Lux Wrote:

(08-22-2015, 12:29 PM)Shebang Wrote: 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.

(08-22-2015, 02:48 PM)Lux Wrote: 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))

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@OversouL read up on 'Operator Associativity'. Here's a link to get you started. http://www.maths.cam.ac.uk/undergrad/catam/ccatsl/manual/node38.html

(08-22-2015, 04:07 PM)eclipse Wrote: This is especially true if you're also a JS developer.

(08-22-2015, 04:18 PM)Lux Wrote: Okay, kid.

(08-22-2015, 04:46 PM)OversouL Wrote: 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;}