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Codeforces 455A - Boredom

时间:2017-07-26 19:12:23      阅读:264      评论:0      收藏:0      [点我收藏+]

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
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.

 

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<algorithm>
 5 #include<cstdlib>
 6 #include<cstring>
 7 using namespace std;
 8 long long a[100010],ans[100010];
 9 int main()
10 {
11     int n,i,k;
12     cin>>n;
13     for(i=0;i<n;i++){
14         scanf("%d",&k);
15         a[k]++;
16     }
17     ans[1]=a[1];
18     for(i=2;i<=100002;i++){
19         ans[i]=max(ans[i-1],ans[i-2]+i*a[i]);
20     }
21     cout<<ans[100002]<<endl;
22     return 0;
23 }

 

Codeforces 455A - Boredom

原文:http://www.cnblogs.com/shixinzei/p/7241322.html

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