1195C. Basketball Exercise

C. Basketball Exercise：题目

经典简单dp，考虑前两天的就行

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
vector<int> a((int)6e5);
vector<int> b((int)6e5);
ll dp[(int)5e5][3];
int main()
{int n;cin>>n;for (int i=2;i<=n+1;i++) cin>>a[i];for (int i=2;i<=n+1;i++) cin>>b[i];for (int i=2;i<=n+1;i++){dp[i][1] = max(dp[i-1][2],dp[i-2][2])+a[i];dp[i][2] = max(dp[i-1][1],dp[i-2][1])+b[i];}cout<<max(dp[n+1][1],dp[n+1][2])<<endl;
}

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