The 2022 ICPC Asia Regionals Online Contest (II) B

B ： Non-decreasing Array

You are given a non-decreasing array of integers a1,a2,…,an. In one operation, when the current length of the array is m:

Firstly you can choose an index i(1<i<m) and delete ai (m decrease 1) or you can do nothing,

Secondly you can choose an index i(1<i<m) and change ai to any integer.

You should ensure that the array is non-decreasing after every delete or change.

Now you want to know that after operating k(1≤k≤n) times, when the current length of the array is m, what is the maximum value of ∑(i=2-m)(ai−ai_1)^2.

You need to answer for each k(1≤k≤n), different queries are independent of each other.

输入格式:

The first line contains one integer n(3≤n≤100).

The second line contains n integers a1,a2,...,an(−10^9≤ai≤10^9).

输出格式:

Output n lines, each of which contains a single integer—the i-th number is for the answer of k=i.

输入样例:
5
1 2 3 4 5
输出样例:
10
16
16
16
16
代码长度限制

16 KB

时间限制

400 ms

内存限制

dp[i][k] 为前[1-i]中删除k个数，且a[1],a[i]不删除的最大权值

1次操作对应两次删除

#include<bits/stdc++.h>
using namespace std;
typedef double db;
#define int long long
const int N=110;
int a[N],n;
int dp[N][N];
int dfs(int id,int k)
{if (dp[id][k]!=-1)return dp[id][k];int mx=0;for(int i=1;i<id;i++){//del[i+1,id-1]int cnt=(id-1)-(i+1)+1;//删除的个数if (cnt>k)continue;mx=max(mx,dfs(i,k-cnt)+(a[id]-a[i])*(a[id]-a[i]));}return dp[id][k]=mx;
}
signed main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cin>>n;for(int i=1; i<=n; i++){cin>>a[i];}memset(dp,-1,sizeof dp);for(int i=1; i<=n; i++){cout<<dfs(n,min(n-2,i*2))<<"\n";}return 0;
}

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输入格式:

输出格式:

输入样例:

输出样例:

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