用 dp[i] 表示从前 i 个数中任取出若干个数(不能取相邻的数)的最大值

对于第 i 个数，可以取也可以不取，如果取则 dp[i]=dp[i-2]+a[i] ,不取则 dp[i]=dp[i-1]，从二者中取最大值

所以递推公式 dp[i]=max(dp[i-2]+a[i],dp[i-1])

递推初始值为 dp[1]=a[i]

#include<bits/stdc++.h>
using namespace std;
#define ioio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define endl "\n"
#define debug(x) cout<<#x<<":"<<x<<endl;
#define L(k) k<<1
#define R(k) k<<1|1
#define P pair
#define P1 first
#define P2 second
#define u_map unordered_map
#define p_queue priority_queue
typedef long ll;
const double eps = 1e-6;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int INF2 = (1 << 31);
const int N = 1e2 + 7;
int dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};
/*-------------------------------------------------*/

int n;
int a[N];
int dp[N];

int main(){
	ioio
	cin>>n;
	for(int i=1;i<=n;i++)
		cin>>a[i];
	dp[1]=a[1];
	for(int i=2;i<=n;i++)
		dp[i]=max(dp[i-2]+a[i],dp[i-1]);
	cout<<dp[n]<<endl;
	return 0;
}

#include <bits/stdc++.h>
using namespace std;
#define accelerate ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define endl "\n";
#define mod 12345
#define ll long long
#define PII pair<int,int>
#define INF 0x3f3f3f3f
const int N=1e2+10;
ll n,m,k,x,y,T;
ll a[N];
ll pre[N][2];
int main(){
	accelerate;
	cin>>n;
	for(int i=1;i<=n;i++) cin>>a[i];
	for(int i=1;i<=n;i++){
		pre[i][0]+=max(pre[i-1][1],pre[i-1][0]);
		pre[i][1]+=pre[i-1][0]+a[i];
	}
	cout<<max(pre[n][0],pre[n][1]);
	return 0;
}