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 ≤ 10^5) that shows how many numbers are in Alex’s sequence. The second line contains n integers a1, a2, …, an (1 ≤ ai ≤ 10^5)

Output

Print a single integer — the maximum number of points that Alex can earn.
Sample Input
91 2 1 3 2 2 2 2 3
Sample Output
10

Hint

Consider the 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.

問題面の大意

一連の数列では,akサイズの要素を1つ削除するたびにスコアakが得られるが,ak−1サイズでak+1の要素も削除される.取得できる最大点数を求める.

構想

これは簡単なdp問題で、ブラシメーターで解決します.まず、値が等しい要素については、配列cnt[]を宣言できることは明らかです.cnt[i]は、要素iが数列に現れる回数を意味します.動的計画セクションでは、配列dp[]を宣言し、dp[i]は、要素を小さいものから大きいものに、その要素に取ったときに得られる最大スコアを表します.明らかに、2つの状況があります.

値i-1の要素を選択すると、値iの要素は無視され、dp[i]=dp[i-1].

値i-2の要素と値iの要素を選択し、値i-1の要素を無視し、dp[i]=dp[i-2]+cnt[i]*i;

また,問題は点数最大値をとることを要求するので,状態遷移方程式:dp[i]=max(dp[i-1],dp[i-2]+cnt[i]*i)を得る.
ACコード:

#include
using namespace std;
long long cnt[100000+5];
long long dp[100000+5];
int main()
{
 long long a,n,maxx=-1;
 scanf("%lld",&n);
 while(n--)
 {
  scanf("%lld",&a);
  maxx=max(a,maxx);// dp 
  cnt[a]++;
 }
 dp[1]=cnt[1];
 for(long long i=2;i<=maxx;i++)
  dp[i]=max(dp[i-1],dp[i-2]+cnt[i]*i);
 printf("%lld",dp[maxx]);
 return 0;
}