递归，公式：f(n)=(f(n-1)+1)*2,n>2;f(2)=4;
#include <iostream>
using namespace std;
int Fun(
int a){
int f, n = a;
if (n ==
2) f =
4;
else f = (Fun(n-
1) +
1) *
2;
return f;
}
int main(){
int n,S;
while (
cin >> n){
S = Fun(n);
cout << S << endl;
}
}
想复杂了，完全可以不用递归的- -！
#include <iostream>
using namespace std;
int main(){
int n,S=
1;
while (
cin >> n){
while (n >=
2){
S = (S +
1) *
2;
n--;
}
cout << S << endl;
S =
1;
}
}
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