类似汉诺塔的题
大佬推导
#include<bits/stdc++.h>
using namespace std;
int n;
int hnt(int n)
{
if(n == 1) return 2;
else return 3 * hnt(n-1)+ 2 ;
}
int main()
{
while(~scanf("%d", &n))
{
cout << hnt(n) << endl;
cout << pow(3,n) - 1 << endl;
}
return 0;
}