Description
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.
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).
Print a single integer — the maximum number of points that Alex can earn.
2
1 2
2
3
1 2 3
4
9
1 2 1 3 2 2 2 2 3
10
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：我们每次都删除一个数字n，并且n-1，n+1也删除，我们要获得n的总和要最大
解法：DP，我们记录每个数字出现次数，这样删除之后还能乘以i得到一个和，我们还要把最大的数字找出来
这样我们就从2—>max 有dp[i]=dp[i-1]，因为i-1也要删除的，dp[i]=dp[i-2]+i*出现次数，求他们之间的最大值
注意起始dp[0],dp[1]
#include<bits/stdc++.h>
using namespace std;
map<long long,long long>q;
long long dp[100010];
int main()
{
long long n;
long long m;
long long MAX=-1;
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>m;
q[m]++;
MAX=max(MAX,m);
}
dp[0]=0;
dp[1]=q[1];
for(int i=2;i<=MAX;i++)
{
dp[i]=max(dp[i-1],(long long)(dp[i-2]+i*q[i]));
}
cout<<dp[MAX]<<endl;
return 0;
}