CF 455A(Boredom-dp)

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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A. 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 a1a2, ..., 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;
}




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