做法就是先枚举右区间的区间和，循环从下标1开始，然后从2开始 全部插入到set里面，
然后枚举左区间，先把从下标1开始的组合删掉，再把下标2的值的组合删掉，然后枚举左区间的组合，用二分来查找差距最小的值
cpp
#include <iostream>
#include <set>
using namespace std;
typedef long long LL;
const int N = 1e3+10;
LL s[N],a[N];
int n,sum;
LL retleft,retright;
multiset <LL> mp;
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cin >> n;
LL ret = 1e9;
mp.insert(-1e9);
mp.insert(1e9);
for(int i = 1;i<=n;i++)
{
cin >> a[i];
s[i] = s[i-1]+a[i];
}
for(int i = 1;i<=n;i++)
{
for(int j = i;j<=n;j++)
{
mp.insert(s[j]-s[i-1]);
}
}
for(int i = 1;i<=n;i++)
{
for(int j = i;j<=n;j++)
{
auto p = mp.find(s[j]-s[i-1]);
mp.erase(p);
}
for(int j = i;j>=1;j--)
{
LL k = s[i]-s[j-1];//s[j]
auto it = mp.lower_bound(k);
LL t1 = *it;
it--;
LL t2 = *it;
LL dis = min(abs(t1-k),abs(t2-k));
ret = min(ret,dis);
}
}
cout << ret << endl;
}