/* Stack of Boxes Challenge by John Wells Input a number N and create a stack of N layers. The top layer has a single 4x6 box formed like: +----+ | | | | +----+ Each lower layer has 4 boxes supporting the outer 4 corners of the layer above it. This is the pattern for N of 4. // +----+ +----+ // | | | | // | +----+ +----+ | // +--| | | |--+ // | +----++----+ | // +--| || |--+ // | +----+ | // +--| |--+ // +--| |--+ // | +----+ | // +--| || |--+ // | +----++----+ | // +--| | | |--+ // | +----+ +----+ | // | | | | // +----+ +----+ // // // //////// The pattern is symmetrical in both vertical and horizontal axis. Observing the pattern, you can see that it has a base pattern which is this: ///////////// // // // +-- // // | // // // ///////////// Using that part, you can create the bpattern by rotating or mirroring it. Unfortunately, I couldn't think of a way to do that. I'm not really sure if that method is feasible but if it is, I hope someone does it. My solution uses the horizontal symmetry property. The main process prints only the top half and saves it in a variable and then prints it in reverse. */ #include #include using namespace std; string t =""; void p(string q, int r=1){ for(int u=0; u>n; //Just edit the value. SL servers are problematic at times. n = 5; cout<<"N = "<< n<0; b+=0){ p(" ",v); if(b!=n)x==1?p("| "):p("+--"); p(s); p(" ",(b-2)*2); if(b!=1)p(s); if(b!=n)x==1?p(" |"):p("--+"); p(" ",v); p("\n"); if(x!=1){ b--; v++; } x = -(x); s = x==1? "+----+":"| |"; } //printing the top half of the pattern in reverse reverse(t.begin(),t.end()); cout<

Stack Of Boxes Challenge v1.1