Codeforces Round #286 C. Mr. Kitayuta, the Treasure Hunter dp

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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 1010212727

output

3

input

8 8919283645556678

output

6

input

13 788916171718212324242630

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)  → ...

一共有30001个岛屿，下标为0~30000，有些岛屿藏有宝藏。你从0往下标大的方向跳，第一步跳的距离为d。如果上一步跳的距离为l，这一步就可以跳l-1或l或l+1（距离必须大于0）。问最多拿到多少宝藏。

设dp[i][j]表示到达第i个岛屿，跳了j步得到的最大宝藏数，那么动态转移方程：

dp[to][j]=max(dp[to][j],dp[i][j]+treasure[to])

dp[to+1][j+1]=max(dp[to+1][j+1],dp[i][j]+treasure[to+1])

dp[to-1][j-1]=max(dp[to-1][j-1],dp[i][j]+treasure[to-1])

如果开一个dp[30000][30000]的数组的话会MLE，那么在到达30000之前，最大步长是1+2+3+...+250>30000。所以第二维开到500就够了。

//93 ms 59700 KB#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int dp[30007][507],trea[30007];int main(){    int n,k;    while(scanf("%d%d",&n,&k)!=EOF)    {        int a;        memset(dp,-1,sizeof(dp));        memset(trea,0,sizeof(trea));        for(int i=0; i<n; i++)        {            scanf("%d",&a);            trea[a]++;        }        dp[k][250]=trea[k];        int ans=dp[k][250];        for(int i=k; i<=30000; i++)            for(int j=1; j<=500; j++)            {                if(dp[i][j]==-1)continue;                int to=i+k+j-250;                if(to<=30000)                {                    dp[to][j]=max(dp[to][j],dp[i][j]+trea[to]);                    ans=max(ans,dp[to][j]);                }                if(to+1<=30000)                {                    dp[to+1][j+1]=max(dp[to+1][j+1],dp[i][j]+trea[to+1]);                    ans=max(ans,dp[to+1][j+1]);                }                if(to-i>1&&to<=30000)                {                    dp[to-1][j-1]=max(dp[to-1][j-1],dp[i][j]+trea[to-1]);                    ans=max(ans,dp[to-1][j-1]);                }            }        printf("%d\n",ans);    }    return 0;}