题面描述
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.
亚历克斯不喜欢无聊。这就是为什么每当他感到无聊时，他就会想出一些游戏。在一个漫长的冬日傍晚，他想出了一个游戏并决定玩它。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.
给定一个由n个整数组成的序列。玩家可以做几个步骤。在单个步骤中，他可以选择序列的元素（假设为\(a_k\)）并删除它，此时，所有等于\(a_k+1和a_k-1\)的元素也必须从序列中删除。这个步骤给玩家带来\(a_k\)点数。Alex is a perfectionist, so he decided to get as many points as possible. Help him.
亚历克斯是个完美主义者，所以他决定得到尽可能多的分数。帮助他。输入格式
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). 第一行包含一个整数n（\(1≤n≤105\)），表示Alex序列中有多少个数字。 第二行包含n个整数\(a1,a2,…,an(1≤105)\)输出格式
输出一个整数——Alex可以获得的最大点数
样例
样例输入
9
1 2 1 3 2 2 2 2 3样例输出
10
题解
先求出数列中每一个数字k的出现次数num[k]
考虑取任意一个数\(x\)时只会影响到\(x+1\)和\(x-1\),我们可以先设dp[i]表示选取num后可以取得的最大值。因为任意取两个数\(a和b\),若选取\(a\)后可以选取\(b\),则选取\(b\)后可以选取\(a\),因此我们只考虑\(x与x-1\)之间的关系。这样我们就很容易得到递推式：\[ dp[i]=\begin{cases} 0 && i=0\\ num[1]*1 && i=1\\ max(dp[i-1],dp[i-2]+num[i]*i) && else \end{cases} \] 注意，最后一重for循环要从2循环至已知的maxn#include#define maxn 1000050using namespace std;inline char get(){ static char buf[3000],*p1=buf,*p2=buf; return p1==p2 && (p2=(p1=buf)+fread(buf,1,3000,stdin),p1==p2)?EOF:*p1++;}inline long long read(){ register char c=getchar();register long long f=1,_=0; while(c>'9' || c<'0')f=(c=='-')?-1:1,c=getchar(); while(c<='9' && c>='0')_=(_<<3)+(_<<1)+(c^48),c=getchar(); return _*f;}long long note,n,a[maxn],dp[maxn];long long op=0;int main(){ //freopen("1.txt","r",stdin); n=read(); for(register long long i=1;i<=n;i++)a[i]=read(),dp[a[i]]+=a[i],note=max(note,a[i]); for(register long long i=2;i<=note;i++)dp[i]=max(dp[(i)-1],dp[(i)-2]+dp[i]),op=max(dp[i],op); cout< <