CF#260 C.Boredom

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.

Sample test(s)

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.

渣渣被虐的体无完肤，闭关修行去了，附上渣代码。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
#include<cmath>
using namespace std;
const int maxn=1e5+10;
int hash[maxn];
long long dp[maxn];
int main()
{
    int n,x;
    while(~scanf("%d",&n))
    {
        int minn=0;
        memset(hash,0,sizeof(hash));
        for(int i=0;i<n;i++)
        {
            scanf("%d",&x);
            if(x>minn)
                minn=x;
            hash[x]++;
        }
        dp[0]=0;
        dp[1]=hash[1];
        for(int i=2;i<=minn;i++)
        {
            dp[i]=dp[i-1];
            long long temp=(long long)i*hash[i];
            if(dp[i-2]+temp>dp[i-1])
                dp[i]=dp[i-2]+temp;
        }
        printf("%I64d\n",dp[minn]);
    }
    return 0;
}

CF#260 C.Boredom