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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;
}
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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;
}
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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;
}
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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;
}
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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。
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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。
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