求Fib数列

1. 第一版本程序

int fib(int pos)
    {
        int elem = 1;
        int n1 = 1, n2 = 1;
        for (int i = 3; i <= pos; i++)
        {
            elem = n2 + n1;
            n1 = n2;
            n2 = elem;
        }
        return elem;
    }

2. 第二版本

bool fib(int pos, int &elem)
    {
        if(pos < 0 || pos > 1024)
        {
            elem = 0;
            return false;
        }
        elem = 1; //注意：定义只能有1次
        int n1 = 1, n2 = 1;
        for (int i = 3; i <= pos; i++)
        {
            elem = n2 + n1;
            n1 = n2;
            n2 = elem;
        }
        return true;
    }

主函数调用

int main()
    {    
            int pos;
        cout <<"Please enter a position: ";
        cin >> pos;
    
        int elem;
        if(fib(pos, elem))
        {
            cout << "element # " << pos
                 << " is " << elem << endl;
        }
        else
            cout << "Sorry. Couldn't calculate element #"
                 << pos <<endl;
    }

3. 第三版本 改进后的fib

const vector<int>* fib_new(int size)
    {
        const int max_size = 1024;
        static vector<int> elems;
    
        if(size <= 0 || size >= max_size)
        {
            cerr << "fib_new(): oops:invalid size:"
                 << size << "-- can't fulfill request.\n";
            return 0;
        }
        for(int ix = elems.size(); ix < size; ix++)
        {
            if (ix == 0 || ix == 1)
                elems.push_back(1);
            else
                elems.push_back(elems[ix - 1] + elems[ix - 2]);
        }
        return &elems;
    }

主函数调用

        const vector<int> *result=fib_new(5);
        cout << result->back();
        
        const vector<int> *result=fib_new(5);
        cout << result->at(4)<< endl;
        //这个应该怎么多次调用返回，这个还没明白，留个记号。

最后这个版本可以避免进行重复运算，使用了局部静态对象。