Codeforces Round #260 (Div. 1) 455 A. Boredom （DP）

题目链接：http://codeforces.com/problemset/problem/455/A

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

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.

题意：

给定一个序列。每次从序列中选出一个数ak,获得ak的得分。同一时候删除序列中全部的ak−1,ak+1,

求最大得分的值。

思路：

存下每一个数的个数放在c中。消除一个数i,会获得c[i]*i的值（由于能够消除c[i]次），

假设从0的位置開始向右消去，那么。消除数i时。i-1可能选择了消除。也可能没有，

假设消除了i-1，那么i值就已经不存在，dp[i] = dp[i-1]。

假设没有被消除，那么dp[i] = dp[i-2]+ c[i]*i。

代码例如以下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define LL __int64
const 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;
}