题解 | #不相邻取数#

动态规划的简单运用

#include<iostream>
using namespace std;
int num[200001] = {0},dp[200001];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,ans = 0;
    cin >> n;
    for(int i = 0;i < n;++i){
        cin >> num[i];
        dp[i] = num[i];
    }
    for(int i = 2;i < n;++i){
        if(i - 3 < 0)
            dp[i] = dp[i - 2] + num[i];
        else
            dp[i] = max(dp[i - 2],dp[i - 3]) + num[i];
    }
    for(int i = 0;i < n;++i)
        ans = max(ans,dp[i]);
    cout << ans;
}