CodeForces 455A Boredom （DP）

Boredom

题目链接：

http://acm.hust.edu.cn/vjudge/contest/121334#problem/G

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.

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

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

Hint

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：

给出n个数，每次任意选择其中一个数Ai：
删除Ai(一个)以及所有的Ai-1 Ai+1; 此次删除操作得分为Ai;
问删除所有元素最多可以获得多少分.


##题解：

由于数组元素的范围是10^5，故可以排序后直接DP：
dp[i][0/1]分别表示删除i或不删除(由其他数删掉)所获得的最大分数.
转移方程：
dp[i][0] = max(dp[i-1][0], dp[i-1][1]);
dp[i][1] = dp[i-1][0] + cnt[i]*i;


##代码：
``` cpp
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LL long long
#define eps 1e-8
#define maxn 101000
#define mod 100000007
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;

int n;

int cnt[maxn];

LL dp[maxn][2];

int main(int argc, char const *argv[])

{

//IN;

while(scanf("%d", &n) != EOF)
{
    memset(cnt, 0, sizeof(cnt));
    for(int i=1; i<=n; i++) {
        int x; scanf("%d", &x);
        cnt[x]++;
    }
    memset(dp, 0, sizeof(dp));
    for(int i=1; i<=100000; i++) {
        dp[i][0] = max(dp[i-1][0], dp[i-1][1]);
        dp[i][1] = dp[i-1][0] + (LL)(cnt[i])*(LL)(i);
    }
    LL ans = max(dp[100000][0], dp[100000][1]);
    printf("%I64d\n", ans);
}
return 0;

}