C++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。