[Ybtoj High-efficiency Advanced 1.1] [Recursion] Tiling scheme

[Ybtoj High-efficiency Advanced 1.1] [Recursion] Tiling scheme

topic

---

Problem-solving ideas

Let f[n] be the number of plans when the length is n
n=1

When n==2



When n>=3

• The last one is placed vertically 2*1, there are f[n-1] kinds of schemes
• The last one is 2*2, there are f[n-2] kinds of schemes
• The last one is to put 2 2*1 horizontally, there are f[n-2] kinds of schemes

---

Code

#include<iostream>
#include<cstdio>
using namespace std;
struct lzf{
    
    
	int l,a[120];
}f[300];
int n;
void gj(int x)
{
    
    
	 int t=0;
	 for (int i=1;i<=f[x-2].l;i++)
     {
    
    
     	 f[x].a[i]=f[x-2].a[i]*2+t;
     	 t=f[x].a[i]/10;
     	 f[x].a[i]=f[x].a[i]%10;
	 }
	 if (t>0)
	    f[x].l=f[x-2].l+1,f[x].a[f[x].l]=t;
	 t=0;
	 f[x].l=max(f[x].l,f[x-1].l);
	 for (int i=1;i<=f[x].l;i++) 
	 {
    
    
	 	 f[x].a[i]=f[x].a[i]+t+f[x-1].a[i];
	 	 t=f[x].a[i]/10;
	 	 f[x].a[i]=f[x].a[i]%10;
	 }
	 if (t)
	    f[x].l++,f[x].a[f[x].l]=t;
} 
int main()
{
    
    
	f[1].l=1;
	f[1].a[1]=1;
	f[2].l=1;
	f[2].a[1]=3;
	for (int i=3;i<=250;i++)
        gj(i);	
	while (scanf("%d",&n)>0)
	{
    
    
		  for (int i=f[n].l;i>0;i--)
		      printf("%d",f[n].a[i]);
		  printf("\n");
	}
	return 0;
}