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 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.

 

      题意：

      给出 n（1 ~ 10 ^ 5） 个数，后给出这 n 个数，每次取一个数 k，取的同时，k - 1 与 k + 1 也会消失。每次取的数为本次取的分数，最后使求和的结果最大。

 

      思路：

      DP。dp [ i ] = MAX { dp [ i - 1 ] ， Max { dp [ 1 ] …… dp [ i - 2 ] } + num [ i ] } ，最后输出 dp [ 100000 ] 即为答案。

 

      AC：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;

ll num[100005];

int main () {
    int n;
    scanf("%d", &n);

    memset(num, 0, sizeof(num));
    for (int i = 0; i < n; ++i) {
        int ans;
        scanf("%d", &ans);
        num[ans] += (ll)ans;
    }

    ll Max = 0;
    Max = max(Max, num[1]);
    for (int i = 3; i <= 100000; ++i) {
        Max = max(num[i - 2], Max);
        num[i] = max(num[i] + Max, num[i - 1]);
    }
    printf("%I64d\n", num[100000]);

    return 0;
}