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;

}