（寒假CF3）B

题解：dp

Description
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 Input
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
Hint
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.

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std; 
typedef long long LL;
LL dp[100000],num[100000];
int main()
{
    int n,i;
    while(~scanf("%d",&n))
    {
    for(i=1;i<=n;i++)
    {
        int x;
        scanf("%d",&x);
        num[x]++;
    }
    for(i=2;i<=n;i++)
    {
        dp[i]=max(dp[i-1],dp[i-2]+(LL)num[i]*i);
    }
    printf("%lld
",dp[n]);
    }
    return 0;
}