C++求Fib数列
发布时间:2020-12-16 05:36:46 所属栏目:百科 来源:网络整理
导读: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 lt; 0 || pos 1024) { elem = 0; return false; } elem =
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。 (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |