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CF 505C(Mr. Kitayuta, the Treasure Hunter

編輯：關於C++

C. Mr. Kitayuta, the Treasure Hunter time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

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.
Sample test(s) input
```
4 10
10
21
27
27
```
output
```
3
```
input
```
8 8
9
19
28
36
45
55
66
78
```
output
```
6
```
input
```
13 7
8
8
9
16
17
17
18
21
23
24
24
26
30
```
output
```
4
```
Note
In the first sample, the optimal route is 0 ?→? 10 (+1 gem) ?→? 19 ?→? 27 (+2 gems) ?→?...

In the second sample, the optimal route is 0 ?→? 8 ?→? 15 ?→? 21?→? 28 (+1 gem) ?→? 36 (+1 gem) ?→? 45 (+1 gem) ?→? 55 (+1 gem) ?→? 66 (+1 gem) ?→? 78 (+1 gem) ?→?...

In the third sample, the optimal route is 0 ?→? 7 ?→? 13 ?→? 18 (+1 gem) ?→? 24 (+2 gems) ?→? 30 (+1 gem) ?→?...

這題直接dp很簡單，但是d的范圍很大

但是可以證明最終用到的d的取值最多不超過500

截取一段explanation:

Below is the explanation from yosupo, translated by me.

[From here]

Let m be the number of the islands (that is, 30001). First, let us describe a solution with time and memory complexity of O(m2).

We will apply Dynamic Programming. let dp[i][j] be the number of the gems that Mr. Kitayuta can collect after he jumps to island i, when the length of his previous jump is j (let us assume that he have not collect the gems on island i). Then, you can calculate the values of the table dp by the following:
- dp[i][j]?=?0, if i?≥?m
  (actually these islands do not exist, but we can suppose that they exist and when Mr. Kitayuta jumps to these islands, he stops jumping)
- dp[i][j]?=? (the number of the gems on island i) ?+?max(dp[i?+?j][j],?dp[i?+?j?+?1][j?+?1]), if i?m,?j?=?1
  (he cannot perform a jump of length 0)
- dp[i][j]?=? (the number of the gems on island i) ?+?max(dp[i?+?j?-?1][j?-?1],?dp[i?+?j][j],?dp[i?+?j?+?1][j?+?1]), if i?m,?j?≥?2
  This solution is unfeasible in terms of both time and memory. However, the following observation makes it an Accepted solution: there are only 491 values of j that we have to consider, which are d?-?245,?d?-?244,?d?-?243,?...,?d?+?244 and d?+?245.
  
  Why? First, let us find the upper bound of j. Suppose Mr. Kitayuta always performs the "l?+?1" jump (l: the length of the previous jump). Then, he will reach the end of the islands before he performs a jump of length d?+?246, because
  d?+?(d?+?1)?+?(d?+?2)?+?...?+?(d?+?245)?≥?1?+?2?+?...?+?245?=?245·(245?+?1)?/?2?=?30135?>?30000. Thus, he will never be able to perform a jump of length d?+?246 or longer.
  
  Next, let us consider the lower bound of j in a similar way. If d?≤?246, then obviously he will not be able to perform a jump of length d?-?246 or shorter, because the length of a jump must be positive. Suppose Mr. Kitayuta always performs the "l?-?1" jump, where d?≥?247. Then, again he will reach the end of the islands before he performs a jump of length d?-?246, because
  d?+?(d?-?1)?+?(d?-?2)?+?...?+?(d?-?245)?≥?245?+?244?+?...?+?1?=?245·(245?+?1)?/?2?=?30135?>?30000. Thus, he will never be able to perform a jump of length d?-?246 or shorter.
  
  Therefore, we have obtained a working solution: similar to the O(m2) one, but we will only consider the value of j between d?-?245 andd?+?245. The time and memory complexity of this solution will be O(m1.5), since the value "245" is slightly larger than $\$ .<喎?/kf/ware/vc/" target="_blank" class="keylink">vcD4KPHA+ClRoaXMgc29sdXRpb24gY2FuIGJlIGltcGxlbWVudGVkIGJ5LCBmb3IgZXhhbXBsZSwgdXNpbmcgYSA="normal" two dimensional array with a offset like this: dp[i][j - offset]. The time limit is set tight in order to fail most of naive solutions with search using std::map or something, so using hash maps (unordered_map) will be risky although the complexity will be the same as the described solution.
  
  [End]
```
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (30000+10)
#define MAXD (30000+10)
#define M (30001)
#define MP(a,b) make_pair(a,b) 
#define MAX_d_change (250+10)
#define C (250)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int n,d,a[MAXN]={0},s[MAXN]={0},f[MAXN][MAX_d_change*2]={0};
int main()
{
//	freopen("Treasure.in","r",stdin);
//	freopen(".out","w",stdout);
	cin>>n>>d;
	For(i,n)
	{
		int p;
		scanf("%d",&p);
		a[p]++;
	}
	For(i,M) s[i]=s[i-1]+a[i];
	
	int ans=0;
	memset(f,-1,sizeof(f));
	ans=f[d][C]=a[d];
	
	Fork(i,d,M)
	{
		Rep(j,2*C+1)
			if (f[i][j]>=0)
			{
				int dis=j-C+d;
				if (dis>0&&i+dis<=M) {f[i+dis][j]=max(f[i+dis][j],f[i][j]+a[i+dis]);ans=max(ans,f[i+dis][j]);}
				if (i+dis+1<=M) {f[i+dis+1][j+1]=max(f[i+dis+1][j+1],f[i][j]+a[i+dis+1]);ans=max(ans,f[i+dis+1][j+1]);}
				if (dis-1>0&&i+dis-1<=M) {f[i+dis-1][j-1]=max(f[i+dis-1][j-1],f[i][j]+a[i+dis-1]);ans=max(ans,f[i+dis-1][j-1]);}
			}
	}
	
	
	cout<


















						
		
	
							
						
```