COMP9021PrinciplesofProgrammingTerm2,2024CodingQuiz5Worth4marksanddueWeek8Thursday@9pmDescriptionYouareprovidedwithastubinwhichyouneedtoinsertyourcodewhereindicatedwithoutdoinganychangestotheexistingcode

ContentYouareclimbingastaircase.Ittakesnstepstoreachthetop.Eachtimeyoucaneitherclimb1or2steps.Inhowmanydistinctwayscanyouclimbtothetop? Example1:Input:n=2Output:2Explanation:Therearetwowaystoclimbtothet

Youareclimbingastaircase.Ittakesnstepstoreachthetop.Eachtimeyoucaneitherclimb1or2steps.Inhowmanydistinctwayscanyouclimbtothetop?Example1:Input:n=2Output:2Explanation:Therearetwowaystoclimbtothetop.1.1

DescriptionUsually,One-dimensionaldynamicplanningproblem,theparameterisalways$n$,theresultissimilartonumbersequence$a_n$,or$f(n)$($f$canbeviewedasfunctionorcorrespondingrelationship).Atthesametime,therewillbecertaincor

问题描述该题考察的是动态规划classSolution:defclimbStairs(self,n:int)->int:dp=[0]*nifn==1:return1dp[0]=1dp[1]=2foriinrange(2,n):dp[i]=dp[i-1]+dp[i-2]ret

第一次自己想到用dpclassSolution{public:intdp[100];intclimbStairs(intn){dp[0]=1;dp[1]=1;for(inti=2;i<100;++i)dp[i]=dp[i-1]

linkSolution虽然但是，这个sb题目真的很sb，不知道怎么评到3400的，也不知道为什么我又没有做出来

​​题目​​classSolution{public:intdp[10005];intclimbStairs(intn){dp[1]=1;dp[2]=2;for(inti=3;i<=n;i++)

题目描述70.爬楼梯解题思路本质上与斐波那契数是一样的：\(a_n=a_{n-1}+a_{n-2}\)构建for循环来遍历。代码classSolution{public:intclimbStairs(i