Report problem solving: luogu P1115 (the largest sub-segment and template)

Topic link: P1115 largest sub-segment and
tell you, this day I tune title is orange title ......
linear readily available, put the papers Solution:

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n[200001],p,ans[200001]={0};
    int sum=-9999999;//|x|<=10000   QWQ
    cin>>p;
    for(int i=1;i<=p;i++)
    {
        cin>>n[i];//输入
        ans[i]=max(ans[i-1]+n[i],n[i]);//DP
        sum=max(sum,ans[i]);//取最大值也同时进行，节约时间
    }
    cout<<sum;//直接输出
    return 0;
}

I can think of a black problem, doing so become \ (O (n ^ 2) \) , how do we maintain any section of the largest sub-segment and can be used to maintain the tree line, I began to think of a \ (O (n ^ 2logn) \) , is clearly false, then learn the next, but also out of some \ (SB \) error, finally \ (AC \) a.

\(Code\):

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const int MAXN=200005;
int t[MAXN],n;
struct seg_t
{
    int l,r,sum,pre,suf,maxn;
    seg_t()
    {
        l=r=sum=pre=suf=maxn=0;
    }
}a[MAXN<<2];
void update(int k)
{
    a[k].pre=max(a[k<<1].pre,a[k<<1].sum+a[k<<1|1].pre);
    a[k].suf=max(a[k<<1|1].suf,a[k<<1|1].sum+a[k<<1].suf);
    a[k].maxn=max(max(a[k<<1].maxn,a[k<<1|1].maxn),a[k<<1].suf+a[k<<1|1].pre);
    a[k].sum=a[k<<1].sum+a[k<<1|1].sum;
}
void build(int k,int l,int r)
{
    a[k].l=l,a[k].r=r;
    if(l==r)
    {
        a[k].maxn=a[k].sum=a[k].suf=a[k].pre=t[l];
        return;
    }
    int mid=(l+r)>>1;
    build(k<<1,l,mid),build(k<<1|1,mid+1,r); 
    update(k);
}
seg_t query(int k,int l,int r)
{
    int mid=(a[k].l+a[k].r)>>1;
    if(a[k].l==l&&a[k].r==r) return a[k];
    if(r>=mid) return query(k<<1,l,r);
    else if(l>=mid+1) return query(k<<1|1,l,r);
    else
    {
        seg_t ll,rr,ans;
        ll=query(k<<1,l,mid),rr=query(k<<1|1,mid+1,r);
        ans.sum=ll.sum+rr.sum;
        ans.maxn=max(ll.maxn,rr.maxn);
        ans.pre=max(ll.pre,ll.sum+rr.pre);
        ans.suf=max(rr.suf,rr.sum+ll.suf);
        ans.maxn=max(ans.maxn,ll.suf+rr.pre);
        return ans;
    }
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&t[i]);
    build(1,1,n);
    printf("%d\n",query(1,1,n).maxn);
    return 0;
}

This segment tree we really have not seen