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;}