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：定義只能有1次 int n1 = 1, n2 = 1; for (int i = 3; i <= pos; i++) { elem = n2 + n1; n1 = n2; n2 = elem; } return true; } 主函數(shù)調用
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 改進后的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; } 主函數(shù)調用
const vector<int> *result=fib_new(5); cout << result->back(); const vector<int> *result=fib_new(5); cout << result->at(4)<< endl; //這個應該怎么多次調用返回，這個還沒明白，留個記號。
最后這個版本可以避免進行重復運算，使用了局部靜態(tài)對象。
