merge: Merge pull request #8 from BetterECNU/organize_files
feat: 整理现有数论试卷
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title: 2024Spring_期末
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category:
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- 软件工程学院
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- 课程资料
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tag:
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- 试卷
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author:
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- タクヤマ
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- KirisameVanilla
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- zeyi2
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---
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## 华东师范大学期末试卷（A）
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### 一、选择题（共 10 小题，每小题 2 分，共 20 分）
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> **1. 设 $m > 1$ 是整数，$(a, m) = 1$，则下列选项中不正确的是**
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A. 若 $b \equiv a \pmod{m}$，则 $\mathrm{ord}_m(b) = \mathrm{ord}_m(a)$.
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B. $a^d \equiv a^k \pmod{m}$ 成立的充要条件是 $d \equiv k \pmod{m}$.
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C. 若 $a' \cdot a \equiv 1 \pmod{m}$，则 $\mathrm{ord}_m(a') = \mathrm{ord}_m(a)$.
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D. 若 $\mathrm{ord}_m(a) = st$，则 $\mathrm{ord}_m(a^s) = t$.
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> **2. 下列哪个数不是模 11 的原根？**
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A. 7 B. 6 C. 4 D. 2
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> **3. 9 模 14 的指数 $\mathrm{ord}_{14}(9)$ 是**
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A. 6 B. 3 C. 2 D. 1
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> **4. 设 $a, b, c \in \mathbb{Z}, c \ne 0$，下列结论不正确的是**
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A. 若 $c \mid a, c \mid b$，则 $c \mid (a + b)$.
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B. 若 $c \mid a$，则 $c \mid ab$.
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C. 若 $bc \mid ac$，则 $b \mid a$.
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D. 若 $c \mid (a^2 - b^2)$，则 $c \mid (a - b)$ 或 $c \mid (a + b)$.
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> **5. 模 40 的简化剩余系中元素的个数为**
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A. 16 B. 28 C. 39 D. 40
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> **6. 已知 $\mathrm{ord}_{137}(47) = 136$, $\mathrm{ord}_{739}(47) = 82$，则 $\mathrm{ord}_{101243}(47) =$**
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A. 136 B. 82 C. 5576 D. 11152
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> **7. 设 $n$ 为整数，则下列选项中一定可以被 6 整除的是**
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A. $n(n^2 + 1)$
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B. $n(n^2 - 1)$
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C. $n(n + 1)$
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D. $n(n - 1)$
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> **8. 下列选项中正确的是**
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A. 若 $m = 1458$，则模 $m$ 的原根不存在。
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B. 1275 是 Carmichael 数。
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C. 2047 是对于基 2 的拟素数。
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D. 给定整数 $m > 1$，$(a,m) = (b,m) = 1$，则 $\mathrm{ord}_m(a \cdot b) = \mathrm{ord}_m(a)\ \cdot \mathrm{ord}_m(b)$.
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> **9. 模 24 的一个简化剩余系为**
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A. $\{-1, 2, 3, 5, 7, 9, 19, 20\}$
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B. $\{-7, -1, 9, 13, 17, 2, 23\}$
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C. $\{1, 5, 7, 11, 13, 17, 19, 23\}$
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D. $\{3, 7, 11, 13, 17, 19, 23\}$
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> **10. 以下哪个数不是模 71 的二次剩余？**
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A. 35 B. 36 C. 37 D. 38
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---
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### 二、填空题（共 10 小题，每小题 3 分，共 30 分）
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1. 13 模 21 的指数 $\mathrm{ord}_{21}(13) =$ ________.
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2. $3^{865749} \mod 11 =$ ________.
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3. 同余方程 $17x \equiv 14 \pmod{21}$ 的解为 ________.
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4. 已知 $a = 123, b = 321$，则有 $s =$ ________, $t =$ ________，使得 $sa + tb = (a, b) =$ ________.
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5. 下面的方程组的解为 ________.
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$$ \begin{cases}
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3x + 5y \equiv 38 \pmod{47} \\
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x - y \equiv 10 \pmod{47}
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\end{cases} $$
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6. $\left( \frac{65}{103} \right) =$ ________.
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7. 同余方程 $6x \equiv 3 \pmod{9}$ 的解为 ________.
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8. 模 29 的最小正原根为 ________.
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9. $2^{2002}$ 被 7 除所得的余数为 ________.
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10. 已知 443 是素数，同余方程 $x^2 \equiv 26 \pmod{443}$ 有 ________ 个解。
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---
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### 三、计算题（共 25 分）
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> **1. 判断方程 $x^{15} \equiv 14 \pmod{41}$ 解的个数，并求出所有解（15 分）**
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模 41 以 6 为原根的指数表如下，其中第一列表示十位数，第一行表示个位数，交叉位置表示该数的指数：
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| | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
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|-----|---|---|---|---|---|---|---|---|---|---|
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| 0 | | 40| 26| 15| 12| 22| 1 | 39| 38| 30|
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| 1 | 8 | 3 | 27| 31| 25| 37| 24| 33| 16| 9 |
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| 2 | 34| 14| 29| 36| 13| 4 | 17| 5 | 11| 7 |
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| 3 | 23| 28| 10| 18| 19| 21| 2 | 32| 35| 6 |
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| 4 | 20| | | | | | | | | |
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---
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> **2. 计算 Legendre 符号（10 分）**
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1) $\left( \frac{33}{317} \right)$
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2) $\left( \frac{286}{563} \right)$
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---
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### 四、证明题（25 分）
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> **1. 证明：121 是对基 3 的拟素数。（10 分）**
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> **2. 设 $n$ 为偶数, $p$ 为素数, 且 $p \mid n^{4} + 1$, 证明 $p \equiv 1 \pmod 8$ (15 分)**
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@ -9,8 +9,14 @@ tag:
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- 数论
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---
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# 课程评价
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## 资源列表
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- [课件 2022级更新 (3-1 到 5)](https://drive.vanillaaaa.org/SharedCourses/%E8%BD%AF%E4%BB%B6%E5%B7%A5%E7%A8%8B%E5%AD%A6%E9%99%A2/%E4%BF%A1%E6%81%AF%E5%AE%89%E5%85%A8%E6%95%B0%E5%AD%A6%E5%9F%BA%E7%A1%80/2022%E7%BA%A7)
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- [2024春学期期末试卷](https://courses.ecnu.vanillaaaa.org/courses/软件工程学院/信息安全数学基础（一）/2024Spring_期末.html)
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## 2024Spring
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### 沈家辰
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- 初等数论, 涉及知识较为广阔, 但是并不要求高妙的数学技巧, 也不需要惊人的注意力, 认真学习完全可以对付
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- 考试不难, 老师会捞
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---
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title: 2024Fall_期末_含答案
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category:
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- 软件工程学院
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- 课程资料
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tag:
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- 试卷
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author:
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- タクヤマ
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---
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## 说明
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本学期（2024年秋季学期）的期末试卷难度**相当**低，本人评价为有手就行，故没有特意记忆所有题目，而是挑出一些有代表性的题目以供参考
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|
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我们这里再提供一道填空题，一道选择题，以显示这张卷子的**诚意**
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<ol>
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<li>
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设 $G$ 是一个群，对于 $\forall a,b \in G$， $a*b=b*a$， 则 $G$ 是 ____ 群
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</li>
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<li>
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设 $R$ 是一个环，对于 $\forall a,b \in R$， $a+b=b+a$， 则 $R$ 是？
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A. 整环 B. 交换环 C. 含幺环 D. 环
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</li>
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</ol>
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很大程度上，老师的出题和给分是十分宽容的，请大家好好学习这门课程，**数学并不是妖魔鬼怪！**
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## 题目
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> **1. 设 $G=\{a,b,c,e\}$ 是一个群， $H=\{a,e\}$ 是 $G$ 的子群，写出 $H$ 的所有左陪集**
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<details>
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<summary>解：</summary>
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将 $G$ 中元素 $g$ 各个代入，计算 $gH$
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- $g = e$：
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$eH = H$
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- $g = a$：
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$aH = \{ a \cdot a, a \cdot e \} = \{ a^2, a \}$，由于 $G$ 是群，且 $H$ 是子群，所以 $a^2$ 必须是 $G$ 中元素。
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故 $a^2 = e$，则： $aH = \{ e, a \} = H$
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- $g = b$：
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$bH = \{ b \cdot a, b \cdot e \} = \{ ba, b \}$，
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假设 $ba = c$ ，则：
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$bH = \{ c, b \}$
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- $g = c$：
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$cH = \{ c \cdot a, c \cdot e \} = \{ ca, c \}$，
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假设 $ca = b$ ，则：
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$cH = \{ b, c \} = bH$
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综上所述，H的所有子陪集是 $\{ e, a \}, \{ b, c \}$
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</details>
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***
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> **2. 写出 $R=Z/6Z$ 的所有零因子**
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<details>
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<summary>解：</summary>
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$R=Z/6Z \cong Z_6$，我们只要考虑 $Z_6$ 上的性质：
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显然有 $2\*3 \equiv 0 \pmod{6}, 4*3 \equiv 0 \pmod{6}$，所以零因子是2，3，4
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</details>
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***
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> **3. 定义在有限域 $F_{17}$ 上的椭圆曲线 $E: x^3 + 2x + 3 = y^2$ 上有点 $P(2, 7), Q(11, 8)$ , 计算 $P+Q$ , $2P$**
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<details>
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<summary>解：</summary>
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直接计算即可，我们这里直接给出答案： $P+Q = (8, 15)$ , $2P = (14, 15)$
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</details>
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***
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> **4. 求域 $F_{16}=F_2[x]/(x^4+x^3+1)$的一个生成元 $g(x)$，并用 $g(x)$ 的幂表示 $F_{16}$ 中的所有非零元**
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<details>
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<summary>解：</summary>
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这时候有同学要问了，生成元怎么找啊？其实很简单，直接验证就可以了，（出于强大的直觉和观察力）我们在这里直接验证 $x$ :
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| $x^n$ | $x^n\pmod{x^4+x^3+1}$ |
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| -------- | --------------------- |
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| $x^0$ | $1$ |
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| $x^1$ | $x$ |
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| $x^2$ | $x^2$ |
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| $x^3$ | $x^3$ |
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| $x^4$ | $x^3 + 1$ |
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| $x^5$ | $x^3 + x + 1$ |
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| $x^6$ | $x^3 + x^2 + x + 1$ |
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| $x^7$ | $1 + x^2 + x$ |
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| $x^8$ | $x^2 + x + 1$ |
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| $x^9$ | $x^3 + x^2 + x$ |
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| $x^{10}$ | $x^3 + x^2 + x + 1$ |
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| $x^{11}$ | $x^3 + x + 1$ |
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| $x^{12}$ | $x^3 + 1$ |
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| $x^{13}$ | $x^3$ |
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| $x^{14}$ | $x^2$ |
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如上表，由 $x$ 生成的15个非零元素互不相等，所以 $x$ 确实是生成元，非零元的表示如表中所示
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</details>
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***
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> **5. 证明：有限环的特征一定不为0**
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证明：请参考课件Chap7.pdf中27-28页
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@ -1,67 +0,0 @@
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# 答案
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##
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### 1.设 $G=\{a,b,c,e\}$ 是一个群， $H=\{a,e\}$ 是 $G$ 的子群，写出 $H$ 的所有左陪集
|
||||
|
||||
解：
|
||||
将 $G$ 中元素 $g$ 各个代入，计算 $gH$
|
||||
|
||||
- $g = e$：
|
||||
$eH = H$
|
||||
|
||||
- $g = a$：
|
||||
$aH = \{ a \cdot a, a \cdot e \} = \{ a^2, a \}$，由于 $G$ 是群，且 $H$ 是子群，所以 $a^2$ 必须是 $G$ 中元素。
|
||||
故 $a^2 = e$，则： $aH = \{ e, a \} = H$
|
||||
|
||||
- $g = b$：
|
||||
$bH = \{ b \cdot a, b \cdot e \} = \{ ba, b \}$，
|
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假设 $ba = c$ ，则：
|
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$bH = \{ c, b \}$
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|
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- $g = c$：
|
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$cH = \{ c \cdot a, c \cdot e \} = \{ ca, c \}$，
|
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假设 $ca = b$ ，则：
|
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$cH = \{ b, c \} = bH$
|
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|
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综上所述，H的所有子陪集是 $\{ e, a \}, \{ b, c \}$
|
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|
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**2.写出 $R=Z/6Z$ 的所有零因子**
|
||||
|
||||
解：
|
||||
$R=Z/6Z \cong Z_6$，我们只要考虑 $Z_6$ 上的性质：
|
||||
显然有 $2\*3 \equiv 0 \pmod{6}, 4*3 \equiv 0 \pmod{6}$，所以零因子是2，3，4
|
||||
|
||||
**3.定义在有限域 $F_{17}$ 上的椭圆曲线 $E: x^3 + 2x + 3 = y^2$ 上有点 $P(2, 7), Q(11, 8)$ , 计算 $P+Q$ , $2P$**
|
||||
|
||||
解：
|
||||
直接计算即可，我们这里直接给出答案： $P+Q = (8, 15)$ , $2P = (14, 15)$
|
||||
|
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**4.求域 $F_{16}=F_2[x]/(x^4+x^3+1)$的一个生成元 $g(x)$，并用 $g(x)$ 的幂表示 $F_{16}$ 中的所有非零元**
|
||||
|
||||
解：
|
||||
这时候有同学要问了，生成元怎么找啊？其实很简单，直接验证就可以了，（出于强大的直觉和观察力）我们在这里直接验证 $x$ :
|
||||
|
||||
| $x^n$ | $x^n\pmod{x^4+x^3+1}$ |
|
||||
|------|----------|
|
||||
| $x^0$ | $1$ |
|
||||
| $x^1$ | $x$ |
|
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| $x^2$ | $x^2$ |
|
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| $x^3$ | $x^3$ |
|
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| $x^4$ | $x^3 + 1$ |
|
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| $x^5$ | $x^3 + x + 1$ |
|
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| $x^6$ | $x^3 + x^2 + x + 1$ |
|
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| $x^7$ | $1 + x^2 + x$ |
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| $x^8$ | $x^2 + x + 1$ |
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| $x^9$ | $x^3 + x^2 + x$ |
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| $x^{10}$ | $x^3 + x^2 + x + 1$ |
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| $x^{11}$ | $x^3 + x + 1$ |
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| $x^{12}$ | $x^3 + 1$ |
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| $x^{13}$ | $x^3$ |
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| $x^{14}$ | $x^2$ |
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如上表，由 $x$ 生成的15个非零元素互不相等，所以 $x$ 确实是生成元，非零元的表示如表中所示
|
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|
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### 5.证明：有限环的特征一定不为0
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|
||||
证明：请参考课件Chap7.pdf中27-28页
|
||||
|
|
@ -1,19 +0,0 @@
|
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## 说明：
|
||||
本学期（2024年秋季学期）的期末试卷难度**相当**低，本人评价为有手就行，故没有特意记忆所有题目，而是挑出一些有代表性的题目以供参考
|
||||
|
||||
<ol>
|
||||
<li>设 $G=\{a,b,c,e\}$ 是一个群， $H=\{a,e\}$ 是 $G$ 的子群，写出 $H$ 的所有左陪集</li>
|
||||
<li>写出 $R=Z/6Z$ 的所有零因子</li>
|
||||
<li>定义在有限域 $F_{17}$ 上的椭圆曲线 $E: x^3 + 2x + 3 = y^2$ 上有点 $P(2, 7), Q(11, 8)$ , 计算 $P+Q$ , $2P$</li>
|
||||
<li>求域 $F_{16}=F_2[x]/(x^4+x^3+1)$的一个生成元 $g(x)$，并用 $g(x)$ 的幂表示 $F_{16}$ 中的所有非零元</li>
|
||||
<li>证明：有限环的特征一定不为0</li>
|
||||
</ol>
|
||||
|
||||
我们这里再提供一道填空题，一道选择题，以显示这张卷子的**诚意**
|
||||
<ol>
|
||||
<li>设 $G$ 是一个群，对于 $\forall a,b \in G$， $a*b=b*a$， 则 $G$ 是 ____ 群</li>
|
||||
<li>设 $R$ 是一个环，对于 $\forall a,b \in R$， $a+b=b+a$， 则 $R$ 是？
|
||||
A. 整环 B.交换环 C.含幺环 D.环</li>
|
||||
</ol>
|
||||
|
||||
很大程度上，老师的出题和给分是十分宽容的，请大家好好学习这门课程，**数学并不是妖魔鬼怪！**
|
||||
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|
|
@ -8,6 +8,14 @@ tag:
|
|||
- 数学基础
|
||||
- 抽象代数
|
||||
---
|
||||
## 资源列表
|
||||
|
||||
### 这门课是抽象代数，但是并没有涉及到抽象代数真正复杂的部分，较为浅尝辄止，所以课程本身的学习难度并不高。
|
||||
### 同样地，这门课实际考试不难，老师会捞，请大家不要害怕。
|
||||
- [课件 2022级更新 (6-1 到 9, 商密介绍)](https://drive.vanillaaaa.org/SharedCourses/%E8%BD%AF%E4%BB%B6%E5%B7%A5%E7%A8%8B%E5%AD%A6%E9%99%A2/%E4%BF%A1%E6%81%AF%E5%AE%89%E5%85%A8%E6%95%B0%E5%AD%A6%E5%9F%BA%E7%A1%80/2022%E7%BA%A7)
|
||||
- [2024秋学期期末试卷](https://courses.ecnu.vanillaaaa.org/courses/软件工程学院/信息安全数学基础（二）/2024Fall_期末_含答案.html)
|
||||
|
||||
## 2024Spring
|
||||
|
||||
### 沈家辰
|
||||
|
||||
- 这门课是抽象代数，但是并没有涉及到抽象代数真正复杂的部分，较为浅尝辄止，所以课程本身的学习难度并不高。
|
||||
- 同样地，这门课实际考试不难，老师会捞，请大家不要害怕。
|
||||
|
|
|
|||
|
|
@ -1,5 +1,5 @@
|
|||
---
|
||||
title: 信息安全数学基础（二）
|
||||
title: 形式语言与自动机理论
|
||||
category:
|
||||
- 软件工程学院
|
||||
- 课程评价
|
||||
|
|
@ -9,5 +9,4 @@ tag:
|
|||
- 抽象代数
|
||||
---
|
||||
|
||||
### 这门课是抽象代数，但是并没有涉及到抽象代数真正复杂的部分，较为浅尝辄止，所以课程本身的学习难度并不高。
|
||||
### 同样地，这门课实际考试不难，老师会捞，请大家不要害怕。
|
||||
|
||||
|
|
|
|||
|
|
@ -1846,76 +1846,76 @@
|
|||
]
|
||||
},
|
||||
"node_modules/@shikijs/core": {
|
||||
"version": "3.12.0",
|
||||
"resolved": "https://registry.npmjs.org/@shikijs/core/-/core-3.12.0.tgz",
|
||||
"integrity": "sha512-rPfCBd6gHIKBPpf2hKKWn2ISPSrmRKAFi+bYDjvZHpzs3zlksWvEwaF3Z4jnvW+xHxSRef7qDooIJkY0RpA9EA==",
|
||||
"version": "3.12.1",
|
||||
"resolved": "https://registry.npmjs.org/@shikijs/core/-/core-3.12.1.tgz",
|
||||
"integrity": "sha512-j9+UDQ6M50xvaSR/e9lg212H0Fqxy3lYd39Q6YITYQxfrb5VYNUKPLZp4PN9f+YmRcdpyNAm3obn/tIZ2WkUWg==",
|
||||
"dev": true,
|
||||
"license": "MIT",
|
||||
"dependencies": {
|
||||
"@shikijs/types": "3.12.0",
|
||||
"@shikijs/types": "3.12.1",
|
||||
"@shikijs/vscode-textmate": "^10.0.2",
|
||||
"@types/hast": "^3.0.4",
|
||||
"hast-util-to-html": "^9.0.5"
|
||||
}
|
||||
},
|
||||
"node_modules/@shikijs/engine-javascript": {
|
||||
"version": "3.12.0",
|
||||
"resolved": "https://registry.npmjs.org/@shikijs/engine-javascript/-/engine-javascript-3.12.0.tgz",
|
||||
"integrity": "sha512-Ni3nm4lnKxyKaDoXQQJYEayX052BL7D0ikU5laHp+ynxPpIF1WIwyhzrMU6WDN7AoAfggVR4Xqx3WN+JTS+BvA==",
|
||||
"version": "3.12.1",
|
||||
"resolved": "https://registry.npmjs.org/@shikijs/engine-javascript/-/engine-javascript-3.12.1.tgz",
|
||||
"integrity": "sha512-mwif5T3rEBSMn/1m9dNi4WmB4dxH4VfYqreQMLpbFYov8MM3Gus98I549amFMjtEmYDAkTKGP7bmsv1n9t9I+A==",
|
||||
"dev": true,
|
||||
"license": "MIT",
|
||||
"dependencies": {
|
||||
"@shikijs/types": "3.12.0",
|
||||
"@shikijs/types": "3.12.1",
|
||||
"@shikijs/vscode-textmate": "^10.0.2",
|
||||
"oniguruma-to-es": "^4.3.3"
|
||||
}
|
||||
},
|
||||
"node_modules/@shikijs/engine-oniguruma": {
|
||||
"version": "3.12.0",
|
||||
"resolved": "https://registry.npmjs.org/@shikijs/engine-oniguruma/-/engine-oniguruma-3.12.0.tgz",
|
||||
"integrity": "sha512-IfDl3oXPbJ/Jr2K8mLeQVpnF+FxjAc7ZPDkgr38uEw/Bg3u638neSrpwqOTnTHXt1aU0Fk1/J+/RBdst1kVqLg==",
|
||||
"version": "3.12.1",
|
||||
"resolved": "https://registry.npmjs.org/@shikijs/engine-oniguruma/-/engine-oniguruma-3.12.1.tgz",
|
||||
"integrity": "sha512-hbYq+XOc55CU7Irkhsgwh8WgQbx2W5IVzHV4l+wZ874olMLSNg5o3F73vo9m4SAhimFyqq/86xnx9h+T30HhhQ==",
|
||||
"dev": true,
|
||||
"license": "MIT",
|
||||
"dependencies": {
|
||||
"@shikijs/types": "3.12.0",
|
||||
"@shikijs/types": "3.12.1",
|
||||
"@shikijs/vscode-textmate": "^10.0.2"
|
||||
}
|
||||
},
|
||||
"node_modules/@shikijs/langs": {
|
||||
"version": "3.12.0",
|
||||
"resolved": "https://registry.npmjs.org/@shikijs/langs/-/langs-3.12.0.tgz",
|
||||
"integrity": "sha512-HIca0daEySJ8zuy9bdrtcBPhcYBo8wR1dyHk1vKrOuwDsITtZuQeGhEkcEfWc6IDyTcom7LRFCH6P7ljGSCEiQ==",
|
||||
"version": "3.12.1",
|
||||
"resolved": "https://registry.npmjs.org/@shikijs/langs/-/langs-3.12.1.tgz",
|
||||
"integrity": "sha512-Y1MbMfVO5baRz7Boo7EoD36TmzfUx/I5n8e+wZumx6SlUA81Zj1ZwNJL871iIuSHrdsheV4AxJtHQ9mlooklmg==",
|
||||
"dev": true,
|
||||
"license": "MIT",
|
||||
"dependencies": {
|
||||
"@shikijs/types": "3.12.0"
|
||||
"@shikijs/types": "3.12.1"
|
||||
}
|
||||
},
|
||||
"node_modules/@shikijs/themes": {
|
||||
"version": "3.12.0",
|
||||
"resolved": "https://registry.npmjs.org/@shikijs/themes/-/themes-3.12.0.tgz",
|
||||
"integrity": "sha512-/lxvQxSI5s4qZLV/AuFaA4Wt61t/0Oka/P9Lmpr1UV+HydNCczO3DMHOC/CsXCCpbv4Zq8sMD0cDa7mvaVoj0Q==",
|
||||
"version": "3.12.1",
|
||||
"resolved": "https://registry.npmjs.org/@shikijs/themes/-/themes-3.12.1.tgz",
|
||||
"integrity": "sha512-9JrAm9cA5hqM/YXymA3oAAZdnCgQf1zyrNDtsnM105nNEoEpux4dyzdoOjc2KawEKj1iUs/WH2ota6Atp7GYkQ==",
|
||||
"dev": true,
|
||||
"license": "MIT",
|
||||
"dependencies": {
|
||||
"@shikijs/types": "3.12.0"
|
||||
"@shikijs/types": "3.12.1"
|
||||
}
|
||||
},
|
||||
"node_modules/@shikijs/transformers": {
|
||||
"version": "3.12.0",
|
||||
"resolved": "https://registry.npmjs.org/@shikijs/transformers/-/transformers-3.12.0.tgz",
|
||||
"integrity": "sha512-HcJwlvMAyZzOY+ayEAGE891BdJ7Vtio+qdWUTF9ki4d0LIkDb6DBz8ynOWGAEglHv6eQs/WcAWf/h6ina6IgCw==",
|
||||
"version": "3.12.1",
|
||||
"resolved": "https://registry.npmjs.org/@shikijs/transformers/-/transformers-3.12.1.tgz",
|
||||
"integrity": "sha512-crGh3cSZf6mwg3K2W8i79Ja+q4tVClRHdHLnUGi5arS58+cqdzsbkrEZBDMyevf9ehmjFUWDTEwCMEyp9I3z0g==",
|
||||
"dev": true,
|
||||
"license": "MIT",
|
||||
"dependencies": {
|
||||
"@shikijs/core": "3.12.0",
|
||||
"@shikijs/types": "3.12.0"
|
||||
"@shikijs/core": "3.12.1",
|
||||
"@shikijs/types": "3.12.1"
|
||||
}
|
||||
},
|
||||
"node_modules/@shikijs/types": {
|
||||
"version": "3.12.0",
|
||||
"resolved": "https://registry.npmjs.org/@shikijs/types/-/types-3.12.0.tgz",
|
||||
"integrity": "sha512-jsFzm8hCeTINC3OCmTZdhR9DOl/foJWplH2Px0bTi4m8z59fnsueLsweX82oGcjRQ7mfQAluQYKGoH2VzsWY4A==",
|
||||
"version": "3.12.1",
|
||||
"resolved": "https://registry.npmjs.org/@shikijs/types/-/types-3.12.1.tgz",
|
||||
"integrity": "sha512-Is/p+1vTss22LIsGCJTmGrxu7ZC1iBL9doJFYLaZ4aI8d0VDXb7Mn0kBzhkc7pdsRpmUbQLQ5HXwNpa3H6F8og==",
|
||||
"dev": true,
|
||||
"license": "MIT",
|
||||
"dependencies": {
|
||||
|
|
@ -2912,6 +2912,16 @@
|
|||
"vuepress": "2.0.0-rc.24"
|
||||
}
|
||||
},
|
||||
"node_modules/@vuepress/plugin-redirect/node_modules/commander": {
|
||||
"version": "14.0.0",
|
||||
"resolved": "https://registry.npmjs.org/commander/-/commander-14.0.0.tgz",
|
||||
"integrity": "sha512-2uM9rYjPvyq39NwLRqaiLtWHyDC1FvryJDa2ATTVims5YAS4PupsEQsDvP14FqhFr0P49CYDugi59xaxJlTXRA==",
|
||||
"dev": true,
|
||||
"license": "MIT",
|
||||
"engines": {
|
||||
"node": ">=20"
|
||||
}
|
||||
},
|
||||
"node_modules/@vuepress/plugin-rtl": {
|
||||
"version": "2.0.0-rc.112",
|
||||
"resolved": "https://registry.npmjs.org/@vuepress/plugin-rtl/-/plugin-rtl-2.0.0-rc.112.tgz",
|
||||
|
|
@ -3070,15 +3080,15 @@
|
|||
}
|
||||
},
|
||||
"node_modules/@vueuse/core": {
|
||||
"version": "13.8.0",
|
||||
"resolved": "https://registry.npmjs.org/@vueuse/core/-/core-13.8.0.tgz",
|
||||
"integrity": "sha512-rmBcgpEpxY0ZmyQQR94q1qkUcHREiLxQwNyWrtjMDipD0WTH/JBcAt0gdcn2PsH0SA76ec291cHFngmyaBhlxA==",
|
||||
"version": "13.9.0",
|
||||
"resolved": "https://registry.npmjs.org/@vueuse/core/-/core-13.9.0.tgz",
|
||||
"integrity": "sha512-ts3regBQyURfCE2BcytLqzm8+MmLlo5Ln/KLoxDVcsZ2gzIwVNnQpQOL/UKV8alUqjSZOlpFZcRNsLRqj+OzyA==",
|
||||
"dev": true,
|
||||
"license": "MIT",
|
||||
"dependencies": {
|
||||
"@types/web-bluetooth": "^0.0.21",
|
||||
"@vueuse/metadata": "13.8.0",
|
||||
"@vueuse/shared": "13.8.0"
|
||||
"@vueuse/metadata": "13.9.0",
|
||||
"@vueuse/shared": "13.9.0"
|
||||
},
|
||||
"funding": {
|
||||
"url": "https://github.com/sponsors/antfu"
|
||||
|
|
@ -3088,9 +3098,9 @@
|
|||
}
|
||||
},
|
||||
"node_modules/@vueuse/metadata": {
|
||||
"version": "13.8.0",
|
||||
"resolved": "https://registry.npmjs.org/@vueuse/metadata/-/metadata-13.8.0.tgz",
|
||||
"integrity": "sha512-BYMp3Gp1kBUPv7AfQnJYP96mkX7g7cKdTIgwv/Jgd+pfQhz678naoZOAcknRtPLP4cFblDDW7rF4e3KFa+PfIA==",
|
||||
"version": "13.9.0",
|
||||
"resolved": "https://registry.npmjs.org/@vueuse/metadata/-/metadata-13.9.0.tgz",
|
||||
"integrity": "sha512-1AFRvuiGphfF7yWixZa0KwjYH8ulyjDCC0aFgrGRz8+P4kvDFSdXLVfTk5xAN9wEuD1J6z4/myMoYbnHoX07zg==",
|
||||
"dev": true,
|
||||
"license": "MIT",
|
||||
"funding": {
|
||||
|
|
@ -3098,9 +3108,9 @@
|
|||
}
|
||||
},
|
||||
"node_modules/@vueuse/shared": {
|
||||
"version": "13.8.0",
|
||||
"resolved": "https://registry.npmjs.org/@vueuse/shared/-/shared-13.8.0.tgz",
|
||||
"integrity": "sha512-x4nfM0ykW+RmNJ4/1IzZsuLuWWrNTxlTWUiehTGI54wnOxIgI9EDdu/O5S77ac6hvQ3hk2KpOVFHaM0M796Kbw==",
|
||||
"version": "13.9.0",
|
||||
"resolved": "https://registry.npmjs.org/@vueuse/shared/-/shared-13.9.0.tgz",
|
||||
"integrity": "sha512-e89uuTLMh0U5cZ9iDpEI2senqPGfbPRTHM/0AaQkcxnpqjkZqDYP8rpfm7edOz8s+pOCOROEy1PIveSW8+fL5g==",
|
||||
"dev": true,
|
||||
"license": "MIT",
|
||||
"funding": {
|
||||
|
|
@ -3591,13 +3601,13 @@
|
|||
}
|
||||
},
|
||||
"node_modules/commander": {
|
||||
"version": "14.0.0",
|
||||
"resolved": "https://registry.npmjs.org/commander/-/commander-14.0.0.tgz",
|
||||
"integrity": "sha512-2uM9rYjPvyq39NwLRqaiLtWHyDC1FvryJDa2ATTVims5YAS4PupsEQsDvP14FqhFr0P49CYDugi59xaxJlTXRA==",
|
||||
"version": "8.3.0",
|
||||
"resolved": "https://registry.npmjs.org/commander/-/commander-8.3.0.tgz",
|
||||
"integrity": "sha512-OkTL9umf+He2DZkUq8f8J9of7yL6RJKI24dVITBmNfZBmri9zYZQrKkuXiKhyfPSu8tUhnVBB1iKXevvnlR4Ww==",
|
||||
"dev": true,
|
||||
"license": "MIT",
|
||||
"engines": {
|
||||
"node": ">=20"
|
||||
"node": ">= 12"
|
||||
}
|
||||
},
|
||||
"node_modules/connect-history-api-fallback": {
|
||||
|
|
@ -4533,16 +4543,6 @@
|
|||
"katex": "cli.js"
|
||||
}
|
||||
},
|
||||
"node_modules/katex/node_modules/commander": {
|
||||
"version": "8.3.0",
|
||||
"resolved": "https://registry.npmjs.org/commander/-/commander-8.3.0.tgz",
|
||||
"integrity": "sha512-OkTL9umf+He2DZkUq8f8J9of7yL6RJKI24dVITBmNfZBmri9zYZQrKkuXiKhyfPSu8tUhnVBB1iKXevvnlR4Ww==",
|
||||
"dev": true,
|
||||
"license": "MIT",
|
||||
"engines": {
|
||||
"node": ">= 12"
|
||||
}
|
||||
},
|
||||
"node_modules/kind-of": {
|
||||
"version": "6.0.3",
|
||||
"resolved": "https://registry.npmjs.org/kind-of/-/kind-of-6.0.3.tgz",
|
||||
|
|
@ -5929,18 +5929,18 @@
|
|||
"license": "ISC"
|
||||
},
|
||||
"node_modules/shiki": {
|
||||
"version": "3.12.0",
|
||||
"resolved": "https://registry.npmjs.org/shiki/-/shiki-3.12.0.tgz",
|
||||
"integrity": "sha512-E+ke51tciraTHpaXYXfqnPZFSViKHhSQ3fiugThlfs/om/EonlQ0hSldcqgzOWWqX6PcjkKKzFgrjIaiPAXoaA==",
|
||||
"version": "3.12.1",
|
||||
"resolved": "https://registry.npmjs.org/shiki/-/shiki-3.12.1.tgz",
|
||||
"integrity": "sha512-eMlxVaXyuNQAQCaMtDKQjKv0eVm+kA6fsZtv9UqKgspP+7lWCVi7SoN+cJq1dawvIDQY7TI3SixamztotM6R6Q==",
|
||||
"dev": true,
|
||||
"license": "MIT",
|
||||
"dependencies": {
|
||||
"@shikijs/core": "3.12.0",
|
||||
"@shikijs/engine-javascript": "3.12.0",
|
||||
"@shikijs/engine-oniguruma": "3.12.0",
|
||||
"@shikijs/langs": "3.12.0",
|
||||
"@shikijs/themes": "3.12.0",
|
||||
"@shikijs/types": "3.12.0",
|
||||
"@shikijs/core": "3.12.1",
|
||||
"@shikijs/engine-javascript": "3.12.1",
|
||||
"@shikijs/engine-oniguruma": "3.12.1",
|
||||
"@shikijs/langs": "3.12.1",
|
||||
"@shikijs/themes": "3.12.1",
|
||||
"@shikijs/types": "3.12.1",
|
||||
"@shikijs/vscode-textmate": "^10.0.2",
|
||||
"@types/hast": "^3.0.4"
|
||||
}
|
||||
|
|
|
|||
Loading…
Reference in New Issue