我真的觉得发布这个很愚蠢,但因为我的问题没有答案,而且我仍然是初级程序员,所以我会发布这个:
//List of keys
byte[] Key0 = { 10, 20, 30, 40, 50, 60, 70, 80, 90, 100 }; //mode 10 = 0
byte[] Key1 = { 1, 3, 5, 7, 9, 11, 13, 15, 17, 19 }; // i + 2
byte[] Key2 = { 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 }; // i mode 2 = 0
byte[] Key3 = { 66, 77, 88, 99, 111, 222, 110, 112, 114, 115 }; // mode 11 = 0
byte[] Key4 = { 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121 }; //x^2
byte[] Key5 = { 6,17,34,57,86,121,162,209 }; //3x^2+2x+1
byte[] Key6 = { 77,78,79,80,81,82,83,84,85,86,87 }; // only in range
//Add all keys to the list
List<byte[]> oKeysList = new List<byte[]>();
oKeysList.Add(Key0);
oKeysList.Add(Key1);
oKeysList.Add(Key2);
oKeysList.Add(Key3);
oKeysList.Add(Key4);
oKeysList.Add(Key5);
oKeysList.Add(Key6);
Random oRandom = new Random();
//Generate random key index to be used in the encryption
int ListSelectedIndex = oRandom.Next(0, oKeysList.Count);
byte[] GeneratedKey = oKeysList[ListSelectedIndex];
//Generate 3 random number from the selected key and concate the key index to it
byte[] GeneratedBytes = new byte[4];
for (int i = 0; i < 3; i++)
{
GeneratedBytes[i] = GeneratedKey[oRandom.Next(0,GeneratedKey.Length)];
}
//Add the list of key index
GeneratedBytes[3] = (byte)ListSelectedIndex;
//Return the genreated bytes
return GeneratedBytes;
如您所见,我生成了这个 4 字节数组以及使用 RNG Cryptography 生成的 8 字节,当我想检查我的序列时,我取最后 4 个字节并使用它们之间的数学关系,我想生成许多序列并成为能够检查它们是否有效。我知道它可能是非常古老的方法,安全性非常差,所以如果有人可以帮助我或在我的代码中添加任何内容或提出任何新的建议,我将非常感激。