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fix(docs/other/bvid_desc.md) (#663)
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@ -25,13 +25,15 @@
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## 算法概述
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算法以及程序主要参考[知乎@mcfx的回答](https://www.zhihu.com/question/381784377/answer/1099438784)
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~~算法以及程序主要参考[知乎@mcfx的回答](https://www.zhihu.com/question/381784377/answer/1099438784)~~
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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)
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### av->bv算法
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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
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再注:本人不清楚新算法能否编解码`avid >= 29460791296`
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1. a = (avid ⊕ 177451812) + 8728348608
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1. a = (avid ⊕ 177451812) + 100618342136696320
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2. 以 i 为循环变量循环 6 次 b[i] = (a / 58 ^ i) % 58
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3. 将 b[i] 中各个数字转换为以下码表中的字符
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@ -39,23 +41,31 @@
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> fZodR9XQDSUm21yCkr6zBqiveYah8bt4xsWpHnJE7jL5VG3guMTKNPAwcF
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4. 初始化字符串 b[i]=`BV1 4 1 7 `
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4. 初始化字符串 b[i]=` `
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5. 按照以下字符顺序编码表编码并填充至 b[i]
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字符顺序编码表:
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> 0 -> 11
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> 0 -> 9
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>
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> 1 -> 10
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> 1 -> 8
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>
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> 2 -> 3
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> 2 -> 1
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>
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> 3 -> 8
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> 3 -> 6
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>
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> 4 -> 4
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> 4 -> 2
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>
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> 5 -> 6
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> 5 -> 4
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>
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> 6 -> 0
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>
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> 7 -> 7
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>
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> 8 -> 3
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>
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> 9 -> 5
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### bv->av算法
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@ -66,41 +76,56 @@
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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
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注: 新算法只提供了Python版本
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### Python
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```python
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table = 'fZodR9XQDSUm21yCkr6zBqiveYah8bt4xsWpHnJE7jL5VG3guMTKNPAwcF' # 码表
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tr = {} # 反查码表
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# 初始化反查码表
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for i in range(58):
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tr[table[i]] = i
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s = [11, 10, 3, 8, 4, 6] # 位置编码表
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XOR = 177451812 # 固定异或值
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ADD = 8728348608 # 固定加法值
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XOR = 177451812
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ADD = 100618342136696320
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TABLE = "fZodR9XQDSUm21yCkr6zBqiveYah8bt4xsWpHnJE7jL5VG3guMTKNPAwcF"
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MAP = {
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0:9,
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1:8,
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2:1,
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3:6,
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4:2,
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5:4,
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6:0,
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7:7,
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8:3,
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9:5
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}
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def av2bv(av):
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av = (av ^ XOR) + ADD
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bv = [""]*10
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for i in range(10):
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bv[MAP[i]] = TABLE[(av//58**i)%58]
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return "".join(bv)
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def bv2av(x):
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r = 0
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for i in range(6):
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r += tr[x[s[i]]] * 58 ** i
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return (r - ADD) ^ XOR
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def bv2av(bv):
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av = [""]*10
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s = 0
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for i in range(10):
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s += TABLE.find(bv[MAP[i]])*58**i
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av=(s-ADD)^XOR
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def av2bv(x):
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x = (x ^ XOR) + ADD
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r = list('BV1 4 1 7 ')
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for i in range(6):
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r[s[i]] = table[x // 58 ** i % 58]
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return ''. join(r)
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return(av)
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print(av2bv(170001))
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print(bv2av('BV17x411w7KC'))
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def main():
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while True:
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mod = input("1.AV2BV\n2.BV2AV\n3.Exit\n你的选择:")
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if mod == "1":
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print("BV号是: BV"+av2bv(int(input("AV号是:"))))
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elif mod == "2":
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print("AV号是 AV"+str(bv2av(input("BV号是"))))
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elif mod == "3":
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break
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else:
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print("输入错误请重新输入")
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main()
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```
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输出为:
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```
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BV17x411w7KC
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170001
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```
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### C
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