斐波那契数列的通项公式推导:https://blog.csdn.net/zhouchangyu1221/article/details/104974955。

class Solution {
public:
    int Fibonacci(int n) {
        if (n == 0) 
            return 0;
        int f1 = 1;
        int f2 = 1;
        if (n < 3) 
            return 1;
        int res;
        int i = 3;
        for(; i <= n ; i++)
        {
            res = f1 + f2;
            f1 = f2;
            f2 = res;
        }
        return res;
    }
};