CF 455A(Boredom-dp)_限累天0f455b-优快云博客

本文链接：https://blog.youkuaiyun.com/nike0good/article/details/38457917

介绍了一个消除无聊的游戏算法，玩家通过删除序列中的元素获得分数，目标是在遵循特定规则的情况下获得最高分。使用动态规划解决该问题，确保每一步选择都能带来最大化的收益。

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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.

Dp，设数i出现a[i]次

很容易发现，从n取到i（i全取）的最优值f{i}只与f(i+1)和f(i+2)相关

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100000+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int n,a[MAXN]={0};
ll f[MAXN]={0};
int main()
{
//	freopen("seq.in","r",stdin);
//	freopen(".out","w",stdout);
	MEM(a)
	cin>>n;
	int p,m=0;
	For(i,n)
	{
		scanf("%d",&p);
		a[p]++;
		m=max(m,p);
	}
	ll ans=0;
	ForD(i,m)
	{
		f[i]=max(f[i+1],f[i+2]+(ll)i*a[i]);
		ans=max(ans,f[i]);
	}
	cout<<ans<<endl;
	
	
	return 0;
}