在昨晚 CF Round 640 div4 E 题的官方题解中，它是这么说的：

The intended solution for this problem uses $O(n^2)$ time and $O(n)$ memory. Firstly, let's calculate $cnt_i$ for each $i$ from $1$ to $n$, where $cnt_i$ is the number of occurrences of $i$ in $a$. This part can be done in $O(n)$.

Then let's iterate over all segments of a of length at least $2$ maintaining the sum of the current segment $sum$. We can notice that we don't need sums greater than $n$ because all elements do not exceed $n$. So if the current sum does not exceed $n$ then add $cnt_{sum}$ to the answer and set $cnt_{sum}:=0$ to prevent counting the same elements several times. This part can be done in $O(n^2)$.

官方代码也是按这个写的：

#include <bits/stdc++.h>

using namespace std;

int main() {
	int t;
	cin >> t;
	while (t--) {
		int n;
		cin >> n;
		vector<int> a(n);
		vector<int> cnt(n + 1);
		int ans = 0;
		for (auto &it : a) {
			cin >> it;
			++cnt[it];
		}
		for (int l = 0; l < n; ++l) {
			int sum = 0;
			for (int r = l; r < n; ++r) {
				sum += a[r];
				if (l == r) continue;
				if (sum <= n) {
					ans += cnt[sum];
					cnt[sum] = 0;
				}
			}
		}
		cout << ans << endl;
	}
}

但是我当时没敢这么写，因为这道题有多组数据，我认为时间复杂度是 $O(n^2t)$。

（$t\leq1000,n\leq8000$）

那么请问为什么能过呢/kk