题源：leetcode
链接：https://leetcode-cn.com/problems/climbing-stairs/
这也是道简单的动态规划题，状态转移方程为：f(x) = f(x-1)+f(x-2)
1 class Solution { 2 public: 3 int climbStairs(int n) { 4 int p = 0; 5 int q = 0; 6 int r = 1; 7 8 for(int i = 0; i < n; i++){ 9 p = q; 10 q = r; 11 r = p+q; 12 } 13 return r; 14 } 15 };
原文：https://www.cnblogs.com/hosheazhang/p/15060532.html