codeforces-455A

题目连接：http://codeforces.com/contest/455/problem/A

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 a1, a2, ..., an (1 ≤ ai ≤ 105).

Output

Print a single integer — the maximum number of points that Alex can earn.

题目大意：有n个数，选择一个（设为ak）删除可以得ak分，但是数列中所有等于ak+1或ak-1的数必须删去，求最多可以得多少分。

很典型的dp题目，我比较菜所以状态设置为比较好理解的  dp[i][0]代表不选i的最大值，dp[i][1]代表选i得到的最大值。

转移方程就很好写了，边界也很清晰，详见代码

#include<bits/stdc++.h>

using namespace std;
typedef long long LL;

LL dp[][];
LL c[]={};
int main()
{
    int n;
    int in;
    ;
    cin>>n;
    ;i<n;i++)
    {
        scanf("%d",&in);
        c[in]++;
        z=max(z,in);
    }
    memset(dp,,sizeof(dp));
    ;i<=z;i++)
    {
        dp[i][]=dp[i-][]+c[i]*LL(i);
        dp[i][]=max(dp[i-][],dp[i-][]);
    }
    cout<<max(dp[z][],dp[z][]);
}