我写了一个自定义加密的代码。这是一个挑战。现在它由于某些原因而无效:
加密(“abc”,“大家好”) 返回:sdfhsjfjs 解密(“abc”,“sdfhsjfjs”) Retruns:diuifidu
public int CountCharInStringAccordingToArray(string Text)
{
int Count = 0;
foreach (char x in Text)
{
Count++;
}
return Count - 1;
}
public int CountCharInString(string Text)
{
int Count = 0;
foreach (char x in Text)
{
Count++;
}
return Count;
}
public string Encrypt(string Key, string PlainText)
{
int[] TempKey = new int[CountCharInString(Key)];
int[] TempText = new int[CountCharInString(PlainText)];
int[] EncryptedInt = new int[CountCharInString(PlainText)];
char[] EncryptedChar = new char[CountCharInString(PlainText)];
for (int i = 0; i < CountCharInStringAccordingToArray(Key); i++)
{
TempKey[i] = (int)Key[i];
}
for (int i = 0; i < CountCharInStringAccordingToArray(PlainText); i++)
{
TempText[i] = (int)PlainText[i];
}
int Index = 0;
for (int i = 0; i < CountCharInStringAccordingToArray(PlainText); i++)
{
if (Index == CountCharInStringAccordingToArray(Key))
{
Index = 0;
}
EncryptedInt[i] = TempKey[Index] + TempText[i];
Index++;
EncryptedChar[i] = (char)EncryptedInt[i];
}
return new string(EncryptedChar);
}
public string Decrypt(string Key, string EncryptedText)
{
int[] TempKey = new int[CountCharInString(Key)];
int[] TempText = new int[CountCharInString(EncryptedText)];
int[] DecryptedInt = new int[CountCharInString(EncryptedText)];
char[] DecryptedChar = new char[CountCharInString(EncryptedText)];
for (int i = 0; i < CountCharInStringAccordingToArray(Key); i++)
{
TempKey[i] = (int)Key[i];
}
for (int i = 0; i < CountCharInStringAccordingToArray(EncryptedText); i++)
{
TempText[i] = (int)EncryptedText[i];
}
int Index = 0;
for (int i = 0; i < CountCharInStringAccordingToArray(EncryptedText); i++)
{
if (Index == CountCharInStringAccordingToArray(Key))
{
Index = 0;
}
DecryptedInt[i] = TempText[i] - TempKey[Index];
Index++;
DecryptedChar[i] = (char)DecryptedInt[i];
}
return new string(DecryptedChar);
}
我也知道字符串有长度属性,我忘了纠正它。
答案 0 :(得分:2)
更改行
return Count - 1;
在CountCharInStringAccordingToArray
到
return Count;
代码的表示:
String That(String key, String text, int sign) {
return new String(Enumerable.Range(0, text.Length).Select((x, i) => (char)(text.ToArray()[i]+sign*key.ToArray()[i%key.Length])).ToArray());
}
public String Encrypt(String key, String text) {
return That(key, text, 1);
}
public String Decrypt(String key, String text) {
return That(key, text, -1);
}
它的工作原理很简单。看图:
重复使用key
中的字符添加到text
,并生成一个认为加密的序列。解密只是通过减法进行的反向操作。
最大值可以存储在一个字节0x0ff
中,但可见字符的最大值为0x7e
,即~
和{{ 1}}。
所以只要您的密钥和文字都在可见字符范围内,就不会导致溢出。也就是说,您可以正确地将加密序列解密为原始序列。
答案 1 :(得分:0)
Get string length and output string
使用System;
class MainClass {
public static void Main() {
string str1 = "ABCDEabcde1234567890";
Console.WriteLine("str1: " + str1);
Console.WriteLine("Length of str1: " + str1.Length);
}
}