本地跑没事，能出正确结果，交上去就RE后两个点，开大数组才能过 代码：

#include<bits/stdc++.h>
using namespace std;
const int N=5e5+5;
typedef long long ll;
int a[N];
ll sum[N];
ll w(int i,int j)
{
    int mid=i+j+1>>1;
    return sum[j]-sum[mid]-1ll*(j-mid)*a[mid]+1ll*(mid-i+1)*a[mid]-sum[mid]+sum[i-1];
}
ll f[N];//f(x)
int g[N];//x
ll F(int i,int j,ll k)
{
    return f[j]+w(j+1,i)-k;
}
struct qw{
    int k,l,r;
}q[N];
int n,m;
int check(ll K)
{
    int hh=1,tt=1;
    q[1]=(qw){0,1,n};
    f[0]=g[0]=0;
    for(int i=1;i<=n;i++)
    {
        if(q[hh].r==i-1) hh++;
        f[i]=F(i,q[hh].k,K);
        g[i]=g[q[hh].k]+1;
        while(hh<=tt&&(F(q[tt].l,q[tt].k,K)>F(q[tt].l,i,K)||(F(q[tt].l,q[tt].k,K)==F(q[tt].l,i,K)&&g[q[tt].k]>=g[i]))) tt--;
        if(hh<=tt)
        {
            int l=q[tt].l,r=q[tt].r;
            while(l<r)
            {
                int mid=l+r+1>>1;
                if((F(mid,q[tt].k,K)<F(mid,i,K))||(F(mid,q[tt].k,K)==F(mid,i,K)&&g[q[tt].k]<g[i])) l=mid;
                else r=mid-1;
            }
            q[tt].r=l;
            q[++tt]=(qw){i,l+1,n};
        }
        else q[++tt]=(qw){i,i+1,n};
    }
   // cout<<g[n]<<endl;
    if(g[n]<=m) return 1;
    return 0;
}
int main()
{
//	freopen("1.txt","r",stdin); 
    cin>>n>>m;
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    sort(a+1,a+n+1);
    for(int i=1;i<=n;i++) sum[i]=sum[i-1]+a[i];
    ll l=-1e11,r=1e11;
    //l,r范围
    while(l<r)
    {
        ll mid=l+r+1>>1;
        if(check(mid))
        {
            l=mid;
            if(g[n]==m)
            {
                printf("%lld\n",f[n]+mid*m);
                return 0;//
            }
        }
        else r=mid-1;
    }
    check(l);
    printf("%lld\n",f[n]+l*m);
    return 0;
}