Codeforces 456C - Boredom（简单DP）

Posted 2020-10-02 Farewell

Alex doesn‘t like boredom. That‘s why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.

Given a sequence a consisting of n integers. The player can make several steps. In a single step he can choose an element of the sequence (let‘s denote it ak) and delete it, at that all elements equal to ak + 1 and ak - 1 also must be deleted from the sequence. That step brings ak points to the player.

Alex is a perfectionist, so he decided to get as many points as possible. Help him.

Input

The first line contains integer n (1 ≤ n ≤ 105) that shows how many numbers are in Alex‘s sequence.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105).

Output

Print a single integer — the maximum number of points that Alex can earn.

Examples

input

2
1 2

output

2

input

3
1 2 3

output

4

input

9
1 2 1 3 2 2 2 2 3

output

10

Note

Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points.

 题意：

　　每次从n个数中选中数值为A_k将其删去，并将值为A_k-1的数和值为A_k+1的数全部删去，得到A_k的分数。问最大可获得的分数。

题解：

　　DP一下就好了。

 

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e5+5;
long long cnt[maxn],dp[maxn];
int main()
{
    int n;
    while(cin>>n)
    {
        memset(cnt,0,sizeof(cnt));
        for(int i=0;i<n;i++)
        {
            int x;
            cin>>x;
            cnt[x]++;
        }
        dp[0]=0,dp[1]=cnt[1];
        long long ans=0;
        for(int i=2;i<maxn;i++)
        {
            dp[i]=max(dp[i-1],dp[i-2]+cnt[i]*i);
            ans=max(ans,dp[i]);
        }
        cout<<ans<<endl;
    }
    return 0;
}