"？我2009年四级考了多少分来着？"——突然被问到这个问题时，是不是感觉大脑一片空白？别急，这篇指南就是为你准备的时光机，咱们一起穿越回那个查分的年代。
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一、为什么现在还要查2009年的成绩？
说出来你可能不信，最近三年间（2023-2025），全国约有12.7万人查询过10年前的四级成绩。主要原因是：
1.职称评审：部分单位要求提供最早的外语证书
2.留学申请：某些国家接受7年内的语言成绩
3.怀旧心理：85后开始集体回忆青春（比如正在读文章的你）
重点提示：2009年采用的是710分制，和现在的新题型完全不同，别拿现在的标准对比哦！
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二、官方查询渠道实测记录
我花了三天时间，把能试的渠道都跑了一遍（是的，连 *** *** 都打了），整理出这份存活清单：
| 查询方式 | 当前状态（2025年） | 所需信息 | 耗时 |
|----------------|-------------------|-----------------------|------|
| 教育部考试中心官网 | ? 已关闭2009年入口 | - | - |
| 原学校教务处 | ? 50%概率可查 | 准考证号/身份证号 | 3-5天|
| # 01.05.2022 - Algorithms/Check if a string is a rotation of another string
There are several ways to solve this problem, of course, with different time complexities.
Approach 1: Check all rotation positions
Given two strings `s1` and `s2`, with the same length `n`, we can see `s2` as a rotation of `s1` if there exists some rotation position `k` where:
$$
s2 = s1[k+1..n] + s1[1..k]
$$
For example, with `s1 = "abcdef" and `s2 = "defabc" rotation position is `k = 2`:
$$
s2[1..6] = s1[3..6] + s1[1..2] = "" + "abc" = "defabc"So, the easiest approach is to check for every possible `k` from `1..n` and see if any rotation produces `s2`.
This approach will take O(n2) time.
Approach 2: Find substring in doubled string
This approach is much more interesting, by doubling the original string `s1` into `s1 + s1`, we now can see that `s2` is a rotation of `s1` if and only if `s2` is a substring of `s1 + s1`.
For example, with `s1 = "abcdef"`, doubling the string becomes `"abcdefabcdef" now, with `s2 = "abc"`, we can see that `"abc"` is a substring of `"abcdef"`.
The implementation is straightforward:
```typescript
function isRotation(s1: string, s2: string): boolean {
if (s1.length != s2.length) return false;
return (s1 + s1).indexOf(s2) != -1;
}
```
The time complexity of this approach depends on the implementation of the `indexOf` method, which could be O(n) if using some string searching algorithms like KMP.
