原始题目

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.

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.

题目大意

给出一个数列a1,a2,a3...ai...an。每次选定一个数ai，并且所有与ai-1，ai+1相等的数全部消去，这样消去一次得分为ai。最终把数列全部消去完，问最终最高得分是多少。

解题思路

dp，数组A[ai]表示ai出现的次数，并且记录最大值mmax，从0到mmax走一遍dp。

状态转移方程为dp[i]=max(dp[i-1],dp[i-2]+A[i]*i)，即从小到大消去到ai=i时的最高得分为：消去到i-1的最高得分（因为消去i-1时所有i=i-1+1全部被消去）与消去到i-2的最高得分加上消去A[i]个i的得分（即跳过i-1不消去）。

由于涉及求和，ai乘上出现次数会超int范围，故使用long long。

解题代码

#include <queue>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
//dp[i]=max(dp[i-1],dp[i-2]+a[i]*i)
using namespace std;
const int maxn=100100;
long long dp[maxn],a[maxn];
long long n,t,mmax;
void solve()
{
	//0不出现，初始化 1 
	dp[1]=a[1];
	for(long long i=2;i<=mmax;i++)
	{
		dp[i]=max(dp[i-1],dp[i-2]+a[i]*i);
	}
	printf("%lld\n",dp[mmax]);
}
int main()
{
	while(~scanf("%lld",&n))
	{
		memset(dp,0,sizeof(dp));
		memset(a,0,sizeof(a));
		long long maxx=0;
		for(long long i=1;i<=n;i++)
		{
			scanf("%lld",&t);
			if(t>mmax) mmax=t;
			a[t]++;
		}
//		for(long long i=1;i<=n;i++)
//			printf("a[%d]=%d\n",i,a[i]);
		solve();
	}
	return 0;
}

收获与反思

建立好状态转移方程就好，注意long long。