一遍AC，爽了

/*
 * @Author: YMYS
 * @Date: 2024-12-24 16:09:54
 * @LastEditTime: 2024-12-26 09:43:25
 * @FilePath: \VScode-C&C++-Coding\Acwing\蓝桥杯辅导课\C++b组-往届省赛真题\try.cpp
 * @URL:https://www.acwing.com/problem/content/5993/
 * @Description: 5990. 拔河
 */
#include<bits/stdc++.h>

#define int long long

using namespace std;

const int N = 1010;

// int T;
// void solve(){
// }

int n;
int a[N];//前缀和数组
multiset<int> ss;

signed main()
{
#ifdef ABC
    freopen("D:/daily_Coding/VScode-C&C++-Coding/in.in", "r", stdin);
    freopen("D:/daily_Coding/VScode-C&C++-Coding/out.out", "w", stdout);
#endif
    // cin>>T;
    // while(T--){
    //     solve();
    // }
    cin>>n;

    //读取数据，并计算出前缀和数组
    for(int i=1;i<=n;i++){
        int x;
        cin>>x;
        a[i] = a[i-1] + x;
    }

    //初始化multiset【就是存入最大边界和最小边界】
    ss.insert(1e15), ss.insert(-1e15);

    int res = 1e9;//初始化答案为极大值
    for(int l=1;l<=n;l++){
        //先将左区间的每个子区间的和存入multiset数组中
        for(int i=1;i<l;i++){
            ss.insert(a[l-1] - a[i-1]);//存入(i,l-1)
        }

        //不断遍历右区间，记录sum的值，存min的左右两区间差距
        for(int r = l;r<=n;r++){
            int sum = a[r] - a[l-1];//记录(l,r)

            auto it = ss.lower_bound(sum);

            res = min(res, *it - sum);
            it--;
            res = min(res, sum - *it);
        }
    }

    cout<<res <<endl;

    return 0;
}