View Code of Problem 60

#include<iostream>
using namespace std;
//1 0 1 2
//2 00 01 10 11 3
//3 000 001 010 011 100 101 110 111 5
//4 0000 0001 0010 0011 0100 0101 0110 0111
//  1000 1001 1010 1011 1100 1101 1110 1111 8
int main() {
	//显然根据规律可知为斐波那契数列
	int t, s, ss[41],count=1;
	cin>>t;
	ss[0]=0; 
	ss[1]=2;
	ss[2]=3;
	for(int i=3; i<=40; i++) ss[i]=ss[i-1]+ss[i-2];
	while(t--) {
		cin>>s;
		cout<<"Scenario #"<<count++<<":"<<endl;
		cout<<ss[s]<<endl;
		cout<<endl;
	} 
}