skip32

时间:2022-09-29 03:51:20

在某些应用场景下,需要对数字进行加密,skip32无疑是一种很实用的算法,网上有python的源码(https://bitbucket.org/anuraguniyal/skip32.py/src/403577f22bc2300b30b746526f52b6cf3c8cad51/skip32.py?at=master),可惜没有java的版本。下面送上自己写的java版本:

import java.nio.ByteBuffer;
import java.nio.ByteOrder; public class Skip32 {
public static void main(String[] args) {
long number = 4294967295L;
String key = "djwiji6789";
String enumber = skip32encrypt(number, key);
Long dnumber = skip32decrypt(enumber, key);
System.out.println(enumber);
System.out.println(dnumber);
} public static String skip32encrypt(long number, String key) {
return Long.toString(skip32(number, key, true), 36);
} public static long skip32decrypt(String base36text, String key) {
return skip32(Long.parseLong(base36text, 36), key, false);
} private static long skip32(long number, String key, boolean encrypt) {
ByteBuffer buffer = ByteBuffer.allocate(8);
buffer.order(ByteOrder.LITTLE_ENDIAN);
buffer.putLong(number);
byte[] b = buffer.array();
int[] a = new int[b.length];
for (int i = 0; i < b.length; i++) {
a[i] = b[i] & 0xFF;
} int buf[] = new int[4];
for (int i = 0; i < 4; i++) {
buf[i] = a[i];
} int[] k = new int[10];
for (int i = 0; i < k.length; i++) {
k[i] = key.codePointAt(i);
} _skip32(k, buf, encrypt); int index = 0;
int firstByte = (0x000000FF & buf[index]);
int secondByte = (0x000000FF & buf[index + 1]);
int thirdByte = (0x000000FF & buf[index + 2]);
int fourthByte = (0x000000FF & buf[index + 3]); long anUnsignedInt = ((long) (fourthByte << 24 | thirdByte << 16
| secondByte << 8 | firstByte)) & 0xFFFFFFFFL;
return anUnsignedInt;
} private static void _skip32(int[] key, int[] buf, boolean encrypt) { int k;
int i;
int kstep;
int wl, wr; if (encrypt) {
kstep = 1;
k = 0;
} else {
kstep = -1;
k = 23;
} wl = ((buf[0] << 8) + buf[1]);
wr = ((buf[2] << 8) + buf[3]); for (i = 0; i < 24 / 2; ++i) {
wr ^= g(key, k, wl) ^ k;
k += kstep;
wl ^= g(key, k, wr) ^ k;
k += kstep;
} buf[0] = (wr >> 8);
buf[1] = (wr & 0xFF);
buf[2] = (wl >> 8);
buf[3] = (wl & 0xFF);
} private static int g(int[] key, int k, int w) {
int g1, g2, g3, g4, g5, g6; g1 = ((w >> 8) & 0xff);
g2 = (w & 0xff); g3 = (F[g2 ^ key[(4 * k) % 10]] ^ g1);
g4 = (F[g3 ^ key[(4 * k + 1) % 10]] ^ g2);
g5 = (F[g4 ^ key[(4 * k + 2) % 10]] ^ g3);
g6 = (F[g5 ^ key[(4 * k + 3) % 10]] ^ g4); return ((g5 << 8) + g6);
} private static final int[] F = new int[] { 0xa3, 0xd7, 0x09, 0x83, 0xf8,
0x48, 0xf6, 0xf4, 0xb3, 0x21, 0x15, 0x78, 0x99, 0xb1, 0xaf, 0xf9,
0xe7, 0x2d, 0x4d, 0x8a, 0xce, 0x4c, 0xca, 0x2e, 0x52, 0x95, 0xd9,
0x1e, 0x4e, 0x38, 0x44, 0x28, 0x0a, 0xdf, 0x02, 0xa0, 0x17, 0xf1,
0x60, 0x68, 0x12, 0xb7, 0x7a, 0xc3, 0xe9, 0xfa, 0x3d, 0x53, 0x96,
0x84, 0x6b, 0xba, 0xf2, 0x63, 0x9a, 0x19, 0x7c, 0xae, 0xe5, 0xf5,
0xf7, 0x16, 0x6a, 0xa2, 0x39, 0xb6, 0x7b, 0x0f, 0xc1, 0x93, 0x81,
0x1b, 0xee, 0xb4, 0x1a, 0xea, 0xd0, 0x91, 0x2f, 0xb8, 0x55, 0xb9,
0xda, 0x85, 0x3f, 0x41, 0xbf, 0xe0, 0x5a, 0x58, 0x80, 0x5f, 0x66,
0x0b, 0xd8, 0x90, 0x35, 0xd5, 0xc0, 0xa7, 0x33, 0x06, 0x65, 0x69,
0x45, 0x00, 0x94, 0x56, 0x6d, 0x98, 0x9b, 0x76, 0x97, 0xfc, 0xb2,
0xc2, 0xb0, 0xfe, 0xdb, 0x20, 0xe1, 0xeb, 0xd6, 0xe4, 0xdd, 0x47,
0x4a, 0x1d, 0x42, 0xed, 0x9e, 0x6e, 0x49, 0x3c, 0xcd, 0x43, 0x27,
0xd2, 0x07, 0xd4, 0xde, 0xc7, 0x67, 0x18, 0x89, 0xcb, 0x30, 0x1f,
0x8d, 0xc6, 0x8f, 0xaa, 0xc8, 0x74, 0xdc, 0xc9, 0x5d, 0x5c, 0x31,
0xa4, 0x70, 0x88, 0x61, 0x2c, 0x9f, 0x0d, 0x2b, 0x87, 0x50, 0x82,
0x54, 0x64, 0x26, 0x7d, 0x03, 0x40, 0x34, 0x4b, 0x1c, 0x73, 0xd1,
0xc4, 0xfd, 0x3b, 0xcc, 0xfb, 0x7f, 0xab, 0xe6, 0x3e, 0x5b, 0xa5,
0xad, 0x04, 0x23, 0x9c, 0x14, 0x51, 0x22, 0xf0, 0x29, 0x79, 0x71,
0x7e, 0xff, 0x8c, 0x0e, 0xe2, 0x0c, 0xef, 0xbc, 0x72, 0x75, 0x6f,
0x37, 0xa1, 0xec, 0xd3, 0x8e, 0x62, 0x8b, 0x86, 0x10, 0xe8, 0x08,
0x77, 0x11, 0xbe, 0x92, 0x4f, 0x24, 0xc5, 0x32, 0x36, 0x9d, 0xcf,
0xf3, 0xa6, 0xbb, 0xac, 0x5e, 0x6c, 0xa9, 0x13, 0x57, 0x25, 0xb5,
0xe3, 0xbd, 0xa8, 0x3a, 0x01, 0x05, 0x59, 0x2a, 0x46 };
}

运行结果:

1ev336c
4294967295

使用注意:以上算法只对无符号32位数字有效。超过32位,可以参考https://github.com/dstar4138/jskipjack/blob/master/src/cipher/SkipJack.java

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