Boredom, nice solution, good dp state

/**
 *    author:  compounding
 *    created: 2024-12-10 15:46:47
 **/
#include <bits/stdc++.h>
using namespace std;
mt19937_64 RNG(chrono::steady_clock::now().time_since_epoch().count());
#define NeedForSpeed                  \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);
#define int long long
#define all(x) (x).begin(), (x).end()
typedef vector<int> vi;
typedef vector<bool> vb;
typedef vector<vi> vvi;
typedef vector<pair<int, int>> vpi;
#define f first
#define s second
#define yes cout << "YES" << endl
#define no cout << "NO" << endl
#define endl "\n"
const int mod = 1000000007;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
const int N = (int)1e5 + 5;
int a[N], cnt[N];
void solve()
{
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        cnt[a[i]]++;
    }
    // dp[i] = maximum sum of elements we can get such that we allowed to delete elements <= i
    vi dp(N, 0);
    dp[1] = cnt[1]; // if we delete 1, then we get cnt[1]
    for (int i = 2; i < N; i++)
    {
        // if we delete i, then we get cnt[i] * i & allowed to delete elements <= i - 2 = f(i - 2)
        // if we delete i - 1, then we get f(i - 1)
        dp[i] = max(dp[i - 1], dp[i - 2] + cnt[i] * i);
    }
    cout << dp[N - 1] << endl;
    return;
}

signed main()
{
    NeedForSpeed;
    int t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}