Codeforces Round #260 (Div. 1) A

A. Boredom

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/455/problem/A

Description

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 Input

2
1 2

Sample Output

2

HINT

题意

给你n个数，你每次可以选择删除去一个数x，但是等于x+1和等于x-1的数都得删去

你每一次操作可以得x分

题解：

dp，dp[i]表示到i后能够得到的最大分数

dp[i]=max(dp[i-1],dp[i-2]+a[i-1]*(i-1));

代码：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)  
#define maxn 2000001
#define mod 10007
#define eps 1e-5
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//**************************************************************************************

ll a[100010],dp[100010];
int main()
{
    int n=read();
    for(int i=0;i<n;i++)
    {
        int x=read();
        a[x]++;
    }
    for(int i=2;i<100010;i++)
        dp[i]=max(dp[i-1],dp[i-2]+a[i-1]*(i-1));
    cout<<dp[100009]<<endl;
}