/* ----> link in the comments <---- The below pattern will form the basis of this challenge. /\ \/ /\ / \ / \ /\/ \/\ \/\ /\/ \ / \ / \/ /\ \/ Given an input width and height, each >=1, output the above ASCII art pattern repeated that many times, joining (and overlapping) at the small diamonds. * * * * * * * * * For example, here is an input with width = 2 and height = 1: /\ /\ \/ \/ /\ /\ / \ / \ / \ / \ /\/ \/\/ \/\ \/\ /\/\ /\/ \ / \ / \ / \ / \/ \/ /\ /\ \/ \/ Here is an input width = 2 and height = 2: /\ /\ \/ \/ /\ /\ / \ / \ / \ / \ /\/ \/\/ \/\ \/\ /\/\ /\/ \ / \ / \ / \ / \/ \/ /\ /\ \/ \/ /\ /\ / \ / \ / \ / \ /\/ \/\/ \/\ \/\ /\/\ /\/ \ / \ / \ / \ / \/ \/ /\ /\ \/ \/ Rules and I/O Input and output can be given by any convenient method. You can print it to STDOUT or return it as a function result. Either a full program or a function are acceptable. Any amount of extraneous whitespace is acceptable, so long as the characters line up appropriately. Standard loopholes are forbidden. This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins */ /* SOLUTION _______________________________ 098765432109 9 \/ . 8 /\ . 7 / \ . 6 / \ . 5\/ \/ 4/\ /\ 3 \ / 2 \ / 1 \/ 0 /\ 9 * * * * \ * * * * factor =y%10 - x10 if factor = -4 || 4 --> \ 47 * * * * / * * * * factor =x%10+y%10; if factor = 5 || 13 --> / 92 the first version void p(int x,int y){ int f,g,t;char c,n=10; ++x*=n;++y*=n; do{t=x; while(88); } The final version below after some tricks to lower the length of the code. */ #include #include auto p(int x,int y){ int n=10,t=x=++x*n; for(++y*=n;y>8;) t>7?putchar(t<9?y--,n:t%n-y%n+4&7?t%n+y%n-5&7?32:47:92),t--:t=x; } // using C(clang) std::string code = "n,t;p(x,y){n=10,t=x=++x*n;++y;for(y*=n;y>8;)t>7?putchar(t<9?y--,n:t%n-y%n+4&7?t%n+y%n-5&7?32:47:92),t--:(t=x);}";// 111 bytes int main (){ std::cout<<"length: " <

pattern_pro