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.
Examples
input
21 2
output
2
input
31 2 3
output
4
input
91 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.
题意：
每次从n个数中选中数值为Ak将其删去，并将值为Ak-1的数和值为Ak+1的数全部删去，得到Ak的分数。问最大可获得的分数。
题解：
DP一下就好了。
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e5+5;
long long cnt[maxn],dp[maxn];
int main()
{
int n;
while(cin>>n)
{
memset(cnt,0,sizeof(cnt));
for(int i=0;i<n;i++)
{
int x;
cin>>x;
cnt[x]++;
}
dp[0]=0,dp[1]=cnt[1];
long long ans=0;
for(int i=2;i<maxn;i++)
{
dp[i]=max(dp[i-1],dp[i-2]+cnt[i]*i);
ans=max(ans,dp[i]);
}
cout<<ans<<endl;
}
return 0;
}