# Codeforces round #260 (Div. 1) A. Boredom (DP)

Source: Internet
Author: User

A. boredomtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard 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 sequenceAConsistingNIntegers. the player can make several steps. In a single step he can choose an element of the sequence (Let's denote itAK) And delete it, at that all elements equalAK? +? 1 andAK? -? 1 also must be deleted from the sequence. That step bringsAKPoints to the player.

Alex is a perfectionist, so he decided to get as your points as possible. Help him.

Input

The first line contains integerN(1? ≤?N? ≤? 105) that shows how should numbers are in Alex's sequence.

The second line containsNIntegersA1,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.

Ideas:

Place the number of each number in C. If I is eliminated, the value of C [I] * I will be obtained (because C [I] times can be eliminated ),
If the elimination starts from the position 0 to the right, then the I-1 may have chosen to remove the number I, or not,
If the I-1 is removed, the I value does not exist, DP [I] = DP [I-1],
If not eliminated, DP [I] = DP [I-2] + C [I] * I.

The Code is as follows:

`#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define LL __int64const int MAXN = 100017;LL c[MAXN], dp[MAXN];void init(){memset(dp,0,sizeof(dp));memset(c,0,sizeof(c));}int main(){LL n;LL tt;int i, j;while(~scanf("%I64d",&n)){init();int maxx = 0;for(i = 1; i <= n; i++){scanf("%I64d",&tt);if(tt > maxx)maxx = tt;c[tt]++;}dp[0] = 0, dp[1] = c[1];for(i = 2; i <= maxx; i++){dp[i] = max(dp[i-1],dp[i-2]+c[i]*i);}printf("%I64d\n",dp[maxx]);}return 0;}`

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