题目意思

Solution

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define pai pair<int, int>
#define ms(__x__,__val__) memset(__x__, __val__, sizeof(__x__))
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void print(ll x, int op = 10) { if (!x) { putchar('0'); if (op)    putchar(op); return; }    char F[40]; ll tmp = x > 0 ? x : -x;    if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]);    if (op)    putchar(op); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }
const int dir[][2] = { {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;

const int N = 3e4 + 7;
const int M = 500 + 7;
int dp[N][M], vis[N];

int main() {
    // vis[i] 第i座岛上的gems数量
    // dp[i][j] 用走d+j步的办法从前一个岛跳到第i个岛收集到的最大gems数量
    // 1 + 2 + 3 +……+ n = n*(n+1)/2 <= 30000
    // 最大增加的步长 n < 250
    int n = read(), d = read();
    for (int i = 1; i <= n; ++i) {
        int x = read();
        ++vis[x];
    }
    ms(dp, -1);
    dp[d][250] = vis[d]; //把负数补正，250代表0步
    int ans = vis[d];
    for (int i = d; i <= 30000; ++i) {
        for (int j = 1; j <= 500; ++j) {
            if (dp[i][j] == -1)    continue; //这个步长下无法到达起点直接跳过
            int ed = i + d + j - 250;
            if (ed <= 30000) {
                dp[ed][j] = max(dp[ed][j], dp[i][j] + vis[ed]);
                ans = max(ans, dp[ed][j]);
            }
            if (ed + 1 <= 30000) {
                dp[ed + 1][j + 1] = max(dp[ed + 1][j + 1], dp[i][j] + vis[ed + 1]);
                ans = max(ans, dp[ed + 1][j + 1]);
            }
            if (ed - 1 <= 30000 and ed - 1 > i) { //终点是ed-1,不能小于起点i
                dp[ed - 1][j - 1] = max(dp[ed - 1][j - 1], dp[i][j] + vis[ed - 1]);
                ans = max(ans, dp[ed - 1][j - 1]);
            }
        }
    }
    print(ans);
    return 0;
}