我正在寫一個類的程序首先用預定的密鑰加密字符串。那部分完成了。下一部分是我遇到問題或本身不存在問題的地方。它是一個冗餘問題。在此之後,我應該對字符串和加密字符串進行KPA查找密鑰。 這是工作,但我使用像15嵌套循環的蠻力。有沒有另一種方法來做到這一點?而不是遞歸地做!蠻力在Java自己的加密
static String Key = null;
public static void main(String[] args) {
long startTime = System.nanoTime();
long startTime1 = System.currentTimeMillis();
int cntr = 0;
String key = "AAAAAAAAAAADDDAM";
String plaintext = "Secretfoemotherd";
StringBuilder cipher = new StringBuilder();
StringBuilder brutus = new StringBuilder();
byte[] ciphertext = encrypt(byteT(key), byteT(plaintext));
for (int i = 0; i < ciphertext.length; i++) {
cipher.append(ciphertext[i]);
}
while (true) {
char[] nkey = new char[16];
for (int i1 = 65; i1 < 122; i1++) {
nkey[0] = (char) i1;
for (int i2 = 65; i2 < 122; i2++) {
nkey[1] = (char) i2;
for (int i3 = 65; i3 < 122; i3++) {
nkey[2] = (char) i3;
for (int i4 = 65; i4 < 122; i4++) {
nkey[3] = (char) i4;
for (int i5 = 65; i5 < 122; i5++) {
nkey[4] = (char) i5;
for (int i6 = 65; i6 < 122; i6++) {
nkey[5] = (char) i6;
for (int i7 = 65; i7 < 122; i7++) {
nkey[6] = (char) i7;
for (int i8 = 65; i8 < 122; i8++) {
nkey[7] = (char) i8;
for (int i9 = 65; i9 < 122; i9++) {
nkey[8] = (char) i9;
for (int i10 = 65; i10 < 122; i10++) {
nkey[9] = (char) i10;
for (int i11 = 65; i11 < 122; i11++) {
nkey[10] = (char) i11;
for (int i12 = 65; i12 < 122; i12++) {
nkey[11] = (char) i12;
for (int i13 = 65; i13 < 122; i13++) {
nkey[12] = (char) i13;
for (int i14 = 65; i14 < 122; i14++) {
nkey[13] = (char) i14;
for (int i15 = 65; i15 < 122; i15++) {
nkey[14] = (char) i15;
for (int i16 = 65; i16 < 122; i16++) {
nkey[15] = (char) i16;
cntr++;
byte[] brutusCipher = Crack(
byteC(nkey),
byteT(plaintext));
for (int k = 0; k < brutusCipher.length; k++) {
brutus.append(brutusCipher[k]);
}
if (brutus
.toString()
.equals(cipher
.toString())) {
System.out
.println("found it");
System.out
.println("Key: "
+ Key);
System.out
.println("Brutus: "
+ brutus);
System.out
.println("i ran: "
+ cntr
+ "times");
long endTime = System
.nanoTime();
System.out
.println("time:"
+ (endTime - startTime)
+ " ns");
long endTime1 = System
.currentTimeMillis();
System.out
.println("Took "
+ (endTime1 - startTime1)
+ " ms");
return;
}
brutus.setLength(0);
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
public static byte[] byteT(String s) {
return s.getBytes();
}
public static byte[] byteC(char[] s) {
StringBuilder temp = new StringBuilder();
for (int i = 0; i < s.length; i++) {
temp.append(s[i]);
}
Key = temp.toString();
return temp.toString().getBytes();
}
public static byte[] encrypt(byte[] key, byte[] plaintext) {
byte[] d = new byte[key.length];
System.out.println(key.length);
for (int i = 0; i < key.length; i++) {
d[i] = (byte) (key[i]^plaintext[i]);
}
return d;
}
public static byte[] Crack(byte[] key, byte[] plaintext) {
byte[] n = new byte[key.length];
for (int i = 0; i < key.length; i++) {
n[i] = (byte) (key[i]^plaintext[i]);
}
return n;
}
}
這是迄今爲止最嵌套循環結構,我已經看到了:) –
Yeeeeeah這就是爲什麼我在這裏:) – Becktor