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

C. Boredom

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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\n", dp[maxn]);
    return 0;
}