CF505C Mr. Kitayuta, the Treasure Hunter(DP)

题目链接

题意：





题解：












AC代码

/*
    Author : zzugzx
    Lang : C++
    Blog : blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;

#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define endl '\n'
#define SZ(x) (int)x.size()
#define mem(a, b) memset(a, b, sizeof(a))

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod = 1e9 + 7;
//const int mod = 998244353;

const double eps = 1e-6;
const double pi = acos(-1.0);
const int maxn = 1e6 + 10;
const int N = 3e4 + 5;
const ll inf = 0x3f3f3f3f;
const int dir[][2]={{0, 1}, {1, 0}, {0, -1}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
int dp[N][510], base = 250;
int a[N];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
//  freopen("in.txt", "r", stdin);
//  freopen("out.txt", "w", stdout);
    int n, d;
    cin >> n >> d;
    for (int i = 1, x; i <= n; i++)
        cin >> x, a[x]++;
    mem(dp, -inf);
    dp[d][base] = a[d];
    int ans = 0;
    for (int i = 1; i <= 30000; i++)
        for (int j = 1; j <= 500; j++) {
            int last = i - (d + j - base);
            if (last < 0 || last >= i) continue;
            for (int k = -1; k <= 1; k++)
                if (j + k > 0) dp[i][j] = max(dp[i][j], dp[last][j + k] + a[i]);
            ans = max(ans, dp[i][j]);
        }
    cout << ans;
    return 0;
}