813D - Two Melodies - CodeForces Solution

#include <bits/stdc++.h>
#define F first
#define S second
#define pp pop_back
#define pb push_back
#define ll long long
#define ld long double
#define llu unsigned long long
#define all(x) (x).begin(), (x).end()
#define FOR(i, j, n) for (int i = j; i < n; i++)
#define FOR1(i, j, n) for (int i = j; i <= n; i++)
#define FOR_K(i, j, k, n) for (int i = j; i < n; i += k)
#define FOR_K1(i, j, k, n) for (int i = j; i <= n; i += k)
using namespace std;

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<pii, int> ppi;
typedef pair<int, pii> pip;
typedef pair<pll, ll> ppl;
typedef pair<ll, pll> plp;

const int N = 5e3 + 10;
const int M = 1e5 + 10;

int n, a[N];
int dp[N][N], mx[7], mn[M], ans;

int main() {
	ios::sync_with_stdio(0), cin.tie(0);
	cin >> n;
	FOR1 (i, 1, n)
		cin >> a[i];
	FOR1 (i, 0, n) {
		FOR1 (j, 1, n) mn[a[j]] = 0;
		FOR (j, 0, 7) mx[j] = 0;
		FOR1 (j, 1, n) {
			if (j > i)
				dp[i][j] = max(max(mn[a[j] - 1] + 1, mn[a[j] + 1] + 1), max(mx[a[j] % 7] + 1, dp[0][i] + 1)), ans = max(ans, dp[i][j]);
			mn[a[j]] = max(mn[a[j]], dp[min(i, j)][max(i, j)]);
			mx[a[j] % 7] = max(mx[a[j] % 7], dp[min(i, j)][max(i, j)]);
		}
	}
	cout << ans << endl;
	return 0;
}