斐波那契数列

#include <iostream>
using namespace std;

long long fib(int n)
{
    int fb[] = {0,1};
    if(n<2)
         return fb[n];
     long long  one = 0,two = 1;
     long long save = 0;
     for(int i=2;i<=n;++i)
     {
         save = two;
         two = one+two;
         one = save;
     }
     return two;
}
int main()
{
    long n;
    while(cin>>n)
    {
        cout<<fib(n)<<endl;
    }
    return 0;
}