题解：dp

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
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.

 1 #include<stdio.h> 2 #include<string.h> 3 #include<algorithm> 4 using namespace std;  5 typedef long long LL; 6 LL dp[100000],num[100000]; 7 int main() 8 { 9     int n,i;10     while(~scanf("%d",&n))11     {12     for(i=1;i<=n;i++)13     {14         int x;15         scanf("%d",&x);16         num[x]++;17     }18     for(i=2;i<=n;i++)19     {20         dp[i]=max(dp[i-1],dp[i-2]+(LL)num[i]*i);21     }22     printf("%lld\n",dp[n]);23     }24     return 0;25 }

（寒假CF3）B - 坑坑