cf 155A Boredom dp

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A. Boredom

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 test(s)

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.

思路：
DP[i]:代表删除第I个被删除时获得的最大值状态转移方程：
DP[i] = max(DP[i -1], DP[i – 2] + num[i] * i) // num[I]代表i这个数的个数
// ↑ 删除i-1的时候附带这删除i ↑ 直接删除i

代码：

#define _CRT_SBCURE_MO_DEPRECATE  #include<iostream>  #include<stdlib.h>  #include<stdio.h>  #include<cmath>  #include<algorithm>  #include<string>  #include<string.h>  #include<set>  #include<queue>  #include<stack>  #include<functional>   using namespace std;const int maxn = 100000 + 10;typedef __int64 ll;const int INF = 1e9;ll s[maxn];ll dp[maxn];ll n;int main(){while (scanf("%I64d", &n)!=EOF){memset(dp, 0, sizeof(dp));memset(s, 0, sizeof(s));ll m = 0;//zui dall t;for (int i = 1; i <= n; i++){scanf("%I64d", &t);s[t]++;if (t > m) m = t;}dp[0] = 0;dp[1] = s[1];for (int i = 2; i <= m; i++){dp[i] = max(dp[i - 1], dp[i - 2] + s[i] * i);}printf("%I64d\n", dp[m]);}//system("pause");return 0;}