Codeforces Round #260 (Div. 1)：A. Boredom（DP）_a. boredom time limit per test1 second memory limi-优快云博客

本文介绍了一个有趣的游戏策略问题，玩家需要从整数序列中选择并删除特定数字及其相邻数字，以此来获取分数。通过动态规划的方法，实现了最优解的求解。

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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

题意：给你n个正整数，每个正整数不会超过100000，你每次可以选择一个在数组中存在的数字k，然后删掉这个数字，并且顺带删掉数组中所有的k+1和k-1，同时获得k点积分，问如何获得尽可能多的积分

思路，因为最大的数不会超过100000，所以可以设dp[i]为删掉了所有大于等于i的数字能获得的最多积分，那么有

dp[i] = max(dp[i+1], dp[i+2]+sum[i]*i)，其中sum[i]为第i个数字在数组中出现的次数

#include<stdio.h>
#include<algorithm>
using namespace std;
#define LL long long
int sum[100005];
LL dp[100005];
int main(void)
{
	int n, i, x;
	scanf("%d", &n);
	for(i=1;i<=n;i++)
	{
		scanf("%d", &x);
		sum[x]++;
	}
	for(i=100000;i>=1;i--)
		dp[i] = max(dp[i+1], dp[i+2]+(LL)sum[i]*i);
	printf("%lld\n", dp[1]);
	return 0;
}