CodeForces 455A：Boredom （动态规划）

Boredom

Time limit:1000 ms Memory limit:262144 kB

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Problem 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.

Example

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.

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题意：

给出n个元素，让我们来挑选，如果选了 ak，获得ak点数，同时与ak+1和ak−1相等的元素都要被删除，问选完所有元素后的所能获得的最大点数

解题思路：

首先输入的时候用桶排序记录好。如果我们要选择其中一个元素才能获得最大价值，那么我们肯定要把与这个元素相等的都选了。所以选择一种元素的价值为 a[i]*i，接着就是dp了，用p记录前考虑i-1一个能获得的最大价值，q记录前i个能获得的最大价值，就有p=max(p,q);q=max(q,p+a[i])，当然右边的p和q都是上一层的。

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Code：

#include <iostream>#include <string>#include <cstring>#include <algorithm>#include <cstdio>#define mem(a,b) memset(a,b,sizeof(a))using namespace std;typedef long long LL;const int maxn=1e5;LL a[maxn+5];int main(){    int n;    scanf("%d",&n);    for(int i=0;i<n;i++)    {        int num;        scanf("%d",&num);        a[num]++;    }    for(int i=1;i<=maxn;i++)        a[i]=a[i]*i;    LL p=0,q=0;    for(int i=1;i<=maxn;i++)    {        //p代表不选第i个        LL temp=p;        p=max(p,q);        //q代表选第i个        if(temp+a[i]>q)            q=temp+a[i];    }    if(p>q)        cout<<p<<endl;    else        cout<<q<<endl;    return 0;}