From 06e8343031c77b39b97206542b9556d44ea64014 Mon Sep 17 00:00:00 2001
From: zine_yu <46991452+zine0@users.noreply.github.com>
Date: Mon, 15 May 2023 02:25:58 +0800
Subject: [PATCH] fix(docs/other/bvid_desc.md) (#663)

---
 docs/other/bvid_desc.md | 97 ++++++++++++++++++++++++++---------------
 1 file changed, 61 insertions(+), 36 deletions(-)

diff --git a/docs/other/bvid_desc.md b/docs/other/bvid_desc.md
index 25bb0e1..d3da777 100644
--- a/docs/other/bvid_desc.md
+++ b/docs/other/bvid_desc.md
@@ -25,13 +25,15 @@
 
 ## 算法概述
 
-算法以及程序主要参考[知乎@mcfx的回答](https://www.zhihu.com/question/381784377/answer/1099438784)
+~~算法以及程序主要参考[知乎@mcfx的回答](https://www.zhihu.com/question/381784377/answer/1099438784)~~
+实际上该算法并不完整,新的算法参考自[【揭秘】av号转bv号的过程](https://www.bilibili.com/video/BV1N741127Tj)
 
 ### av->bv算法
 
 注：本算法及示例程序仅能编解码`avid < 29460791296`，且暂无法验证`avid >= 29460791296`的正确性
+再注：本人不清楚新算法能否编解码`avid >= 29460791296`
 
-1. a = (avid ⊕ 177451812) + 8728348608
+1. a = (avid ⊕ 177451812) + 100618342136696320
 2. 以 i 为循环变量循环 6 次 b[i] = (a / 58 ^ i) % 58
 3. 将 b[i] 中各个数字转换为以下码表中的字符
 
@@ -39,23 +41,31 @@
 
 > fZodR9XQDSUm21yCkr6zBqiveYah8bt4xsWpHnJE7jL5VG3guMTKNPAwcF
 
-4. 初始化字符串 b[i]=`BV1 4 1 7 `
+4. 初始化字符串 b[i]=` `
 
 5. 按照以下字符顺序编码表编码并填充至 b[i]
 
 字符顺序编码表：
 
-> 0 -> 11
+> 0 -> 9
 >
-> 1 -> 10
+> 1 -> 8
 >
-> 2 -> 3
+> 2 -> 1
 >
-> 3 -> 8
+> 3 -> 6
 >
-> 4 -> 4
+> 4 -> 2
 >
-> 5 -> 6
+> 5 -> 4
+>
+> 6 -> 0
+> 
+> 7 -> 7
+>
+> 8 -> 3
+>
+> 9 -> 5
 
 
 ### bv->av算法
@@ -66,41 +76,56 @@
 
 使用 Python、C、TypeScript、Java、Kotlin 以及 Golang 等语言作为示例，欢迎社区提交更多例程
 
+注: 新算法只提供了Python版本
 ### Python
 
 ```python
-table = 'fZodR9XQDSUm21yCkr6zBqiveYah8bt4xsWpHnJE7jL5VG3guMTKNPAwcF' # 码表
-tr = {} # 反查码表
-# 初始化反查码表
-for i in range(58):
-    tr[table[i]] = i
-s = [11, 10, 3, 8, 4, 6] # 位置编码表
-XOR = 177451812 # 固定异或值
-ADD = 8728348608 # 固定加法值
+XOR = 177451812
+ADD = 100618342136696320
+TABLE = "fZodR9XQDSUm21yCkr6zBqiveYah8bt4xsWpHnJE7jL5VG3guMTKNPAwcF"
+MAP = {
+    0:9,
+    1:8,
+    2:1,
+    3:6,
+    4:2,
+    5:4,
+    6:0,
+    7:7,
+    8:3,
+    9:5
+}
+def av2bv(av):
+    av = (av ^ XOR) + ADD
+    bv = [""]*10
+    for i in range(10):
+        bv[MAP[i]] = TABLE[(av//58**i)%58]
+    return "".join(bv)
 
-def bv2av(x):
-    r = 0
-    for i in range(6):
-        r += tr[x[s[i]]] * 58 ** i
-    return (r - ADD) ^ XOR
+def bv2av(bv):
+    av = [""]*10
+    s = 0
+    for i in range(10):
+        s += TABLE.find(bv[MAP[i]])*58**i
+    av=(s-ADD)^XOR
 
-def av2bv(x):
-    x = (x ^ XOR) + ADD
-    r = list('BV1  4 1 7  ')
-    for i in range(6):
-        r[s[i]] = table[x // 58 ** i % 58]
-    return ''. join(r)
+    return(av)
 
-print(av2bv(170001))
-print(bv2av('BV17x411w7KC'))
+def main():
+    while True:
+        mod = input("1.AV2BV\n2.BV2AV\n3.Exit\n你的选择:")
+        if mod == "1":
+            print("BV号是: BV"+av2bv(int(input("AV号是:"))))
+        elif mod == "2":
+            print("AV号是 AV"+str(bv2av(input("BV号是"))))
+        elif mod == "3":
+            break
+        else:
+            print("输入错误请重新输入")
+
+main()
 ```
 
-输出为：
-
-```
-BV17x411w7KC
-170001
-```
 
 ### C