（1）有dp【i】【1/0】表示选和不选
（2）利用前缀和和map
（3关于重叠问题，定l和r）
这里

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int INF = 1e18;

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    vector<int> pre(n + 1);
    for (int i = 1; i <= n; i++) {
        pre[i] = pre[i - 1] + a[i];
    }
    vector<tuple<int, int, int>> intervals; // sum, l, r
    for (int l = 1; l <= n; l++) {
        for (int r = l; r <= n; r++) {
            intervals.emplace_back(pre[r] - pre[l - 1], l, r);
        }
    }
    sort(intervals.begin(), intervals.end());
    int min_diff = INF;
    int m = intervals.size();
    for (int i = 0; i < m; i++) {
        auto [s1, l1, r1] = intervals[i];
        for (int j = i + 1; j < m; j++) {
            auto [s2, l2, r2] = intervals[j];
            if (s2 - s1 >= min_diff) break; // 后续的差值会更大，无需继续检查
            if (r1 < l2 || r2 < l1) { // 检查区间是否不重叠
                min_diff = min(min_diff, s2 - s1);
            }
        }
    }
    cout << min_diff << "\n";
    return 0;
}