Codeforces Round #260 (Div. 2)C. Boredom（dp）

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 a_k) and delete it, at that all elements equal to a_k + 1 and a_k - 1 also must be deleted from the sequence. That step brings a_k 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, ..., a_n (1 ≤ a_i ≤ 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.

给出n个数，每次能够选择消除一个值为ai的数，那么全部ai-1。ai+1的数也会被消掉，同一时候会获得值ai，问最多能够获得多少？

看完题就可想到这应该是一个dp问题，首先，哈希一下，存下每一个数的个数放在p中，消除一个数i,会获得p[i]*i的值（由于能够消除p[i]次），假设从0的位置開始向右消去，那么，消除数i时，i-1可能选择了消除，也可能没有。假设消除了i-1，那么i值就已经不存在，dp[i] = dp[i-1]。假设没有被消除。那么dp[i] = dp[i-2]+ p[i]*i。

那么初始的dp[0] = 0 ; dp[1] = p[1] ;得到了公式 dp[i] = max（dp[i-1],dp[i-2]+p[i]*i）.

#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL __int64
using namespace std;
LL p[110000] , dp[110000] ;
int main()
{
    LL i , n , x , maxn = -1;
    memset(p,0,sizeof(p));
    memset(dp,0,sizeof(dp));
    scanf("%I64d", &n);
    for(i = 1 ; i <= n ; i++)
    {
        scanf("%I64d", &x);
        if(x > maxn)
            maxn = x ;
        p[x]++ ;
    }
    dp[1] = p[1] ;
    for(i = 2 ; i <= maxn ; i++)
    {
        dp[i] = max( dp[i-1],dp[i-2]+p[i]*i );
    }
    printf("%I64d
", dp[maxn]);
    return 0;
}