Niuke.com: minimum cost of climbing stairs

  This obviously needs to be done with dynamic programming , First , Let's set up an array of dynamic programming dp[n+1], among n Is the number of stairs ,dp[i] It means to arrive at the third i The minimum cost of a step . Set up a cost[n] To store the cost of climbing a certain stair . The price of the first stair is cost[n-1], And so on .

Next, the dynamic transfer equations are listed , Obviously , We either crossed two levels before , Or only one level has been crossed . If you have crossed two levels .

dp[i]=min(dp[i-2]+cost[i-2],dp[i-1]+cost[i-1])

The code is as follows ：

#include<bits/stdc++.h>
using namespace std;
int main(){
    int n;
    cin>>n;
    vector<int> cost(n);
    for(int i=0;i<n;++i){
        cin>>cost[i];
    }
    vector<int> dp(n+1);
    for(int i=2;i<=n;++i){
        dp[i]=min(dp[i-2]+cost[i-2], dp[i-1]+cost[i-1]);
    }
    cout<<dp[n]<<endl;
    return 0;
}