Boredom

标签：hat   clu   redo   lin   rom   possible   bit   pos   序列   

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.

Examples

2
1 2

2

3
1 2 3

4

9
1 2 1 3 2 2 2 2 3

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：

给定一个序列，每次从序列中选出一个数ak,获得ak的得分，同时删除序列中所有的ak−1,ak+1,

求最大得分的值。

dp[i]=max(dp[i-2]+n[i]*i，dp[i-1])
dp[i]表示的是前i个的最大值，对于第i个有取和不取的情况，对于可以取到i的情况，删掉的一定是i-1这个点，剩下的就是前i-2的情况了，由于dp[i-2]包括了取i-2的情况，于是就不用再重复考虑i-2也要加一遍的情况了，然后再考虑不取i的情况，就不用考虑i了，于是就是dp[i-1]了

#include<bits/stdc++.h>
using namespace std;
const int N = 1e6+1;
long long num[N],dp[N];
int main()
{
    int n,a,i;
    cin>>n;
    int maxa=-1;
    int mina=N;
    for(i=0;i<n;i++){
        cin>>a;
        num[a]++;
        maxa = max(a,maxa);
        mina = min(a,mina);
    }
    dp[mina] = num[mina]*mina;
    for(i=mina+1; i <= maxa; i++){
        dp[i] = max(dp[i-2]+num[i]*i,dp[i-1]);
    }
    cout<<dp[maxa]<<endl;
    return 0;
}

Boredom

标签：hat   clu   redo   lin   rom   possible   bit   pos   序列   

原文地址：https://www.cnblogs.com/clb123/p/10390233.html