CF 455A Boredom

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
input
Copy
2
1 2
output
Copy
2
input
Copy
3
1 2 3
output
Copy
4
input
Copy
9
1 2 1 3 2 2 2 2 3
output
Copy
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】

给你一个数组a，里面有n个整数。你每次可以选择数组中的一个元素a_k，从数组中删掉它，再删掉所有值等于a_k + 1 或者 a_k - 1的元素，这样你可以得到 a_k 分。你可以重复进行多次该操作，请问你最后最多能得多少分？

【分析】

先求出数列中每一个数字k的出现次数num[k]

状态转移方程：如果取得第i-1个数，那么第i-2和第i个数均不可取；反之可取得第i和第i-2个数

第i个状态的值为 (取得第i-1个数的得分) 与 (取得第i-2个数得分和取得当前数的得分之和) 的最大值

注意，最后一重for循环要从2循环至已知的maxn

【代码】

#include<cstdio>
#include<iostream>
using namespace std;
const int N=1e5+5;
inline int read(){
	register int x=0;register char ch=getchar();
	for(;ch<'0'||ch>'9';ch=getchar());
	for(;ch>='0'&&ch<='9';ch=getchar()) x=(x<<3)+(x<<1)+ch-'0';
	return x;
}
int n,mx;long long cnt[N],f[N];
int main(){
	n=read();
	for(int i=1,x;i<=n;i++) mx=max(mx,x=read()),cnt[x]+=x;
	f[1]=cnt[1];
	for(int i=2;i<=mx;i++) f[i]=max(f[i-1],f[i-2]+cnt[i]);
	printf("%I64d
",f[mx]);
	return 0;
}