[Al]演算法:有n級階梯,每次走1步或2步,最多有多少種走法
阿新 • • 發佈:2018-12-06
@Filename : floor.c * @Author : Mr.Zhong * @Date : 2018-11-02 * @Description: n級階梯,每次走一步或2步,最多有多少種走法 * @Analysis : 斐波那契數列 -> 遞迴思維 ***************************************************************/ #include <iostream> using namespace std; int getStepNum(int n) { if(n < 0) return 0; if(n == 0 || n ==1 || n ==2) return n; if(n > 2) return getStepNum(n-1)+getStepNum(n-2); } int main() { int levels; cout<<"Please enter how many levels of stairs :"<<endl; cin>>levels; int method = getStepNum(levels); cout<<"There are "<<method<<" methods in total"<<endl; return 1; } --------------------- 作者:Tobiu 在這裡插入程式碼片 來源:CSDN 原文:https://blog.csdn.net/localhostcom/article/details/83662968 版權宣告:本文為博主原創文章,轉載請附上博文連結!