codeforces 505C C. Mr. Kitayuta, the Treasure Hunter(dp)

题目链接：

The Shuseki Islands are an archipelago of 30001 small islands in the Yutampo Sea. The islands are evenly spaced along a line, numbered from 0 to 30000 from the west to the east. These islands are known to contain many treasures. There are n gems in the Shuseki Islands in total, and the i-th gem is located on island pi.

Mr. Kitayuta has just arrived at island 0. With his great jumping ability, he will repeatedly perform jumps between islands to the east according to the following process:

First, he will jump from island 0 to island d.
After that, he will continue jumping according to the following rule. Let l be the length of the previous jump, that is, if his previous jump was from island prev to island cur, let l = cur - prev. He will perform a jump of length l - 1, l or l + 1 to the east. That is, he will jump to island (cur + l - 1), (cur + l) or (cur + l + 1) (if they exist). The length of a jump must be positive, that is, he cannot perform a jump of length 0 when l = 1. If there is no valid destination, he will stop jumping.

Mr. Kitayuta will collect the gems on the islands visited during the process. Find the maximum number of gems that he can collect.

Input

The first line of the input contains two space-separated integers n and d (1 ≤ n, d ≤ 30000), denoting the number of the gems in the Shuseki Islands and the length of the Mr. Kitayuta's first jump, respectively.

The next n lines describe the location of the gems. The i-th of them (1 ≤ i ≤ n) contains a integer pi (d ≤ p1 ≤ p2 ≤ ... ≤ pn ≤ 30000), denoting the number of the island that contains the i-th gem.

Output

Print the maximum number of gems that Mr. Kitayuta can collect.

Examples

input

output

input

output

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output

4

题意：

有30000个点,n个点放在pi上,第一次去d,每次走的距离只能为上次行走的距离l,l+1,l-1,但是一定要前进;问最后能得到的最大值是多少;


思路：

可以发现行走长度的变化范围不超过250;dp[i][j]表示走j长到达i能得到的最大值;转移方程看代码吧;

AC代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn=3e4+600;
const int N=3e4;
const int inf=1000000;
int n,d;
int p[maxn],num[maxn],dp[maxn][505];
int main()
{
    scanf("%d%d",&n,&d);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&p[i]);
        num[p[i]]++;
    }
    memset(dp,-inf,sizeof(dp));
    dp[d][250]=num[d];
    for(int i=1;i<=N;i++)
    {
        for(int j=0;j<=500;j++)
        {
            int x=j-250+d;
            if(i+x>i&&i+x<=N)dp[i+x][j]=max(dp[i+x][j],dp[i][j]+num[i+x]);
            if(i+x+1>i&&i+x+1<=N)dp[i+x+1][j+1]=max(dp[i+x+1][j+1],dp[i][j]+num[i+x+1]);
            if(i+x-1>i&&i+x-1<=N)dp[i+x-1][j-1]=max(dp[i+x-1][j-1],dp[i][j]+num[i+x-1]);
        }
    }
    int ans=0;
    for(int i=1;i<=N;i++)
    {
        for(int j=0;j<=500;j++)
        {
            ans=max(ans,dp[i][j]);
        }
    }
    cout<<ans<<"
";

}