HDU 1143 Tri Tiling recursion problem

Fills a 3*n rectangle with a 1*2 rectangle, which must not be filled when n is odd,n*3%2==1

The next thing we want to do with this is to start with a simple scenario where recursion is about reducing the size of the problem and solving it by solving the problem. If it's 3*2, there are only 3 possible cases.

So let's consider whether some of the 3*4 rectangles are a combination of these two 3*2 rectangles, and of course include a special case where the stitching method is shown

Put 4, 6, 8 .... As a whole piece, there are two cases (positive, inverted)

Then we can divide the various cases into the following two major categories

A rectangle formed by itself (i.e. not divisible)

The case can be divided, namely 2*f (n-2i) (2i>2)/3*f (n-2i) (at this time 2i==2)

The code is as follows:

1#include <iostream>2#include <cstdio>3 using namespacestd;4 intTillintN)5 {6     ints,i;7     if(n!=2)8     {9s=2;Ten     } One     Else A         return 3; -      for(i=2;i<N;) -     { the         if(i==2) -s=s+3*till (ni); -         Else -s=s+2*till (ni); +i=i+2; -     } +     returns; A } at intMain () - { -     intN; -      while(cin>>N) -     { -         if(n==-1) in              Break; -         if(n%2==1) tocout<<0<<Endl; +         Else if(n==0) -cout<<1<<Endl; the         Else *Cout<<till (n) <<Endl; $     }Panax Notoginseng     return 0; -}

HDU 1143 Tri Tiling recursion problem