首先根據題目中給的要求，推一下方差的柿子。

\[v\times m^2 = m\times \sum x^2 - 2 \times sum \times sum +sum*sum\]

所以\(ans = v*m^2 = m\times \sum x^2 - sum*sum\)

那我們實際上就是最大化平方和。

由於題目限制了要分\(m\)段。所以我們的\(dp\)狀態就是\(f[i][j]\)表示前\(i\)個數分了\(j\)段。
那麼一個比較顯然的轉移
\(dp[i][p]=min(dp[j][p-1]+(s[i]-s[j]^2))\)

然後直接套斜率優化就好了！

但是要注意的是，因為題目中對第二維有點限制，所以我們要開\(m\)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#define mk make_pair
#define ll long long
using namespace std;
inline int read()
{
  int x=0,f=1;char ch=getchar();
  while (!isdigit(ch)) {if (ch=='-') f=-1;ch=getchar();}
  while (isdigit(ch)) {x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
  return x*f;
}
const int maxn = 3010;
struct Point{
    int x,y;
};
ll chacheng(Point x,Point y)
{
    ll xx = x.x;
    ll yy = x.y;
    ll xxx = y.x;
    ll yyy = y.y;
    return xx*yyy-xxx*yy;
}
struct Node{
    Point q[maxn];
    int head=1,tail=0;
    bool count(Point i,Point j,Point k)
    {
        Point x,y;
        x.x=k.x-i.x;
        x.y=k.y-i.y;
        y.x=k.x-j.x;
        y.y=k.y-j.y;
        if (chacheng(x,y)<=0ll) return true;
        return false;
    }
    void push(Point x)
    {
       while (tail>=head+1 && count(q[tail-1],q[tail],x)) tail--;
       q[++tail]=x;
    }
    void pop(int lim)
    {
        while (tail>=head+1 && q[head+1].y-q[head].y<lim*(q[head+1].x-q[head].x)) head++;
    }
};
Node q[maxn];
int sum[maxn];
int n,m;
int dp[maxn][maxn];
int main()
{
  n=read();m=read();
  for (int i=1;i<=n;i++) sum[i]=read();
  for (int i=1;i<=n;i++) sum[i]+=sum[i-1];
  for (int i=0;i<=n;i++) q[i].push((Point){0,0});
  for (int i=1;i<=n;i++)
  {
    for (int j=1;j<=min(i,m);j++)
    {
        q[j-1].pop(2*sum[i]);
        Point now = q[j-1].q[q[j-1].head];
        dp[i][j]=now.y-now.x*now.x + (sum[i]-now.x)*(sum[i]-now.x);
        q[j].push((Point){sum[i],dp[i][j]+sum[i]*sum[i]});
    }
  }
  //cout<<dp[n][m]<<" "<<m<<" "<<sum[n]*sum[n]<<endl;
  cout<<dp[n][m]*m-sum[n]*sum[n]<<endl;
  return 0;
}