Boredom
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 a k) and delete it, at that all elements equal to a k + 1 and a k - 1 also must be deleted from the sequence. That step brings a k points to the player.
Alex is a perfectionist, so he decided to get as many points as possible. Help him.
題目大意：給你一個長度爲n的序列 a，你每次能取序列裏面的一個數 a[i]（得到a[i]分數），並且序列裏面 a[i]-1 和 a[i]+1 的個數都要減一（如果個數爲0，那就不減），問最後能得到的最大分數；
首先要明白，一個數 a[i] 要取，一定是把這個序列的所有 a[i] 都取掉，要不就不取 a[i] ；
現在要判斷的就是存不存在 a[i] 和 a[i]-1 或者 a[i]+1 一起取？
不存在，可以發現如果:
- a[i] 的個數比 a[i]-1 大，取了 a[i]，a[i]-1 的個數就沒有了；
- a[i] 的個數比 a[i]-1 小，取了 a[i]，a[i]-1 的個數有，但是之前在計算 a[i]-1 時如果取了a[i]-1，那麼 a[i] 的個數就沒有了；
- a[i] 的個數和 a[i]-1 一樣，那和第一種情況一樣；
所以如果 a[i] 取，那麼 a[i]-1 就一定不能取，狀態要從 a[i]-2 轉化過來；
代碼：
#include<bits/stdc++.h>
#define LL long long
#define pa pair<int,int>
#define ls k<<1
#define rs k<<1|1
#define inf 0x3f3f3f3f
using namespace std;
const int N=100100;
const int M=2000100;
const LL mod=1e9+7;
int a[N],sum[N],mx;
LL dp[N],ans;
int main(){
int n;cin>>n;
for(int i=1;i<=n;i++){
cin>>a[i];
sum[a[i]]++;
mx=max(mx,a[i]);
}
ans=dp[1]=sum[1];
for(int i=2;i<=mx;i++){
dp[i]=max(dp[i-1],dp[i-2]+(LL)i*(LL)sum[i]);
ans=max(dp[i],ans);
}
cout<<ans<<endl;
return 0;
}