HDU 2501 Tiling_easy version

递推式：f[n]=2*f[n-2]+f[n-1]

#include <cstdio>
#include <iostream>
using namespace std;
long long f[51];
int main()
{
    int i,n;
    f[1]=1; f[2]=3;
    for(int i=3; i<51; i++)
    f[i]=f[i-1]+2*f[i-2];
    cin>>n;
    while(cin>>n) cout<<f[n]<<endl;
    return 0;
}