Code Golf -- Print N Squared :: Kickass Programmers

GRUP STEAM

Kickass Programmers quprogs

Minta untuk Bergabung

GRUP STEAM

Kickass Programmers quprogs

23
ANGGOTA

SEDANG BERMAIN

ONLINE

Dibentuk

6 November 2015

Bahasa

Bhs. Inggris

Ringkasan Pengumuman Diskusi Event Anggota Komentar Kurator

Semua Diskusi > General Discussion > Rincian Topik

makka 3 Apr 2016 @ 11:45pm

Code Golf -- Print N Squared

Write a function that takes in a non-negative integer N and prints out or returns a hollow ASCII-art square whose sides are made of N copies of the number N. Link to full challenge, details and unit tests here[codegolf.stackexchange.com]

An example for if N were 10, the output would be:

10101010101010101010 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10101010101010101010

Discuss and leave solutions in the thread.

Terakhir diedit oleh makka; 3 Apr 2016 @ 11:46pm

< >

Menampilkan 1-4 dari 4 komentar

makka 4 Apr 2016 @ 12:00am

C++14, 156 chars

#define f for(i=0;i++<n;c<<t); [](string t){auto&c=cout;int n=stoi(t),i;f c<<'\n';for(i=0;++i<n-1;c<<t,c.width(~-n*size(t)+1),c.fill(0),c<<t+'\n');if(n-1)f}

This one is pretty cool though it took a long time of fidgeting to get it to work. Unpacked:

#define f for ( i = 0; i++ < n; c << t ); [](string t) { auto& c = cout; int n = stoi(t), i; f // print first row c << '\n'; // kind of annoying but no way to get rid of (yes I tried // c << '\n'+t instead of c << t+'\n') for ( i = 0; ++i < n - 1; ) { c << t; // output the number // then we we get the width of necessary spaces c.width(~-n*size(t)+1); // Equivalent to (n-1)*size(t) + 1, but we save // two characters because ~- takes precedence over // the multiplication c.fill(0); // fill with spaces, ' ' == 0 c << t+'\n'; } if ( n-1 ) f // This if statement is dissapointing }

And like always... to run it you need to use [](string t) { ... }("10");

C++14, 162 chars

The above was my first attempt at the solution and took about 20 minutes. This next one took close to an hour, but unfortunately was not as good as my first attempt. It's a lot more sane, but 6 more chars, if it were not for edge cases 1 & 0 this could match my first solution. I figured I'd post it anyways since I spent so much time on it.

#define f for(i=-2;i++<n;c<<t);c<<'\n'; [](string t){auto&c=cout;int n=stoi(t)-2,i;auto g=t;g.append(abs(n)*size(t),' ');g+=t+'\n';f;for(i=0;i++<n;c<<g);if(n+1)f;}

and unpacked:

[](string t) { auto&c = cout; int n = stoi(t)-2,i; auto g = t; // creating "num ... num" // this abs(n) is very unfortunate, blame edge cases 1 & 0 g.append(abs(n)*size(t),' '); g += t+'\n'; f; for ( i = 0; i++ < n; c << g); if ( n+1 ) f; }

Terakhir diedit oleh makka; 4 Apr 2016 @ 12:03am

birb

4 Apr 2016 @ 10:56am

Rust, 141 137 chars

|i|{let f=||{for _ in 0..i{print!("{}",i)}println!("")};f();if i>1{for _ in 0..i-2{println!("{}{0:1$}",i,i.to_string().len()*(i-1))}f()}}

Unpacked:

|i| { let f = || { for _ in 0..i { print!("{}",i) } println!("") }; f(); if i>1 { for _ in 0..i-2 { println!("{}{0:1$}",i,i.to_string().len()*(i-1)) } f() } }

Terakhir diedit oleh birb; 4 Apr 2016 @ 11:15am

mingmingrr 5 Apr 2016 @ 11:37am

Python 3, 102 chars

s=str(n);print(''.join('\r'+s*n if n==0 or i%n==0 else '\n'+s+' '*len(s)*(n-2)+s for i in range(n+1)))

Unpacked and refactored:

s = str(n) x = '' for i in range(n + 1): if n == 0 or i % n == 0: # carriage return here clears the current line x += '\r' + s * n else: x += '\n' + s + ' ' * len(s) * (n - 2) + s print(x)

Python 3, 60 chars

I found out later someone had this code almost letter for letter, grrr...
Does not support edge cases 1 or 0.

s=str(n);m=n-2;print(s*n+('\n'+s+' '*len(s)*m+s)*m+'\n'+s*n)

Unpacked:

s = str(n) m = n - 2 print(s * n + ('\n' + s + ' ' * len(s) * m + s) * m + '\n' + s * n)