CF 455A(Boredom-dp)

A. Boredom

time limit per test

memory limit per test

input

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.

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 and ak also must be deleted from the sequence. That step brings ak

Alex is a perfectionist, so he decided to get as many points as possible. Help him.

Input

n (1 ≤ n ≤ 105) that shows how many numbers are in Alex's sequence.

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

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

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;
}