高手帮忙解几个MD5加密的数据

有没有搞错，md5是单向散列解不开的

穷举破解吧
虽然麻烦点

这个是不可能解开的，MD5 是单向散列函数，只能从明文计算出散列值，很难从散列值推算出原文，也很难重新给出一个明文使得他们的散列值相等。

MD5 在目前来说是安全的签名算法。
穷举法

暴力破解吧.

MD5不可逆的.
不过楼主还是别费劲了

无法还原,不过可以给你一个MD5的加密算法,我改写在JAVA了,你要是能还原,你就牛了,呵呵
__________________________________________________________________
public class MD5Bean {
/* 下面?些S11-S44??上是一个4*4的矩?，在原始的C??中是用#define ??的，
?里把它???成?static final是表示了只?，切能在同一个?程空?内的多个
Instance?共享*/

static final int S12 = 12;
static final int S11 = 7;
static final int S13 = 17;
static final int S14 = 22;
static final int S21 = 5;
static final int S22 = 9;
static final int S23 = 14;
static final int S24 = 20;
static final int S31 = 4;
static final int S32 = 11;
static final int S33 = 16;
static final int S34 = 23;
static final int S41 = 6;
static final int S42 = 10;
static final int S43 = 15;
static final int S44 = 21;
static final byte[] PADDING = { -128, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 };

/* 下面的三个成?是MD5?算?程中用到的3个核心数据，在原始的C??中
被定?到MD5_CTX??中

*/

private long[] state = new long[4]; // state (ABCD)
private long[] count = new long[2]; // number of bits, modulo 2^64 (lsb first)
private byte[] buffer = new byte[64]; // input buffer

/* digestHexStr是MD5的唯一一个公共成?，是最新一次?算?果的
16?制ASCII表示.
*/

public String digestHexStr;

/* digest,是最新一次?算?果的2?制内部表示，表示128bit的MD5?.
*/

private byte[] digest = new byte[16];

/*
getMD5ofStr是?MD5最主要的公共方法，入口参数是?想要?行MD5??的字符串
返回的是??完的?果，?个?果是从公共成?digestHexStr取得的．
*/

public String getMD5ofStr(String inbuf) {
md5Init();
md5Update(inbuf.getBytes(), inbuf.length());
md5Final();
digestHexStr = """";
for (int i = 0; i < 16; i++) {
digestHexStr += byteHEX(digest[i]);
}
return digestHexStr;
}

// ?是MD5?个?的?准?造函数，JavaBean要求有一个public的并且没有参数的?造函数
public MD5Bean() {
md5Init();
return;
}

/* md5Init是一个初始化函数，初始化核心?量，装入?准的幻数 */

private void md5Init() {
count[0] = 0L;
count[1] = 0L;
///* Load magic initialization constants.
state[0] = 0x67452301L;
state[1] = 0xefcdab89L;
state[2] = 0x98badcfeL;
state[3] = 0x10325476L;
return;
}

/* F, G, H ,I 是4个基本的MD5函数，在原始的MD5的C??中，由于它?是
??的位?算，可能出于效率的考?把它???成了宏，在java中，我?把它?
??成了private方法，名字保持了原来C中的。 */

private long F(long x, long y, long z) {
return (x & y) | ((~x) & z);
}

private long G(long x, long y, long z) {
return (x & z) | (y & (~z));
}

private long H(long x, long y, long z) {
return x ^ y ^ z;
}

private long I(long x, long y, long z) {
return y ^ (x | (~z));
}

/*
FF,GG,HH和II将?用F,G,H,I?行近一???
FF, GG, HH, and II transformations for rounds 1, 2, 3, and 4.
Rotation is separate from addition to prevent recomputation.
*/

private long FF(long a, long b, long c, long d, long x, long s, long ac) {
a += F(b, c, d) + x + ac;
a = ((int)a << s) | ((int)a >>> (32 - s));
a += b;
return a;
}

private long GG(long a, long b, long c, long d, long x, long s, long ac) {
a += G(b, c, d) + x + ac;
a = ((int)a << s) | ((int)a >>> (32 - s));
a += b;
return a;
}

private long HH(long a, long b, long c, long d, long x, long s, long ac) {
a += H(b, c, d) + x + ac;
a = ((int)a << s) | ((int)a >>> (32 - s));
a += b;
return a;
}

private long II(long a, long b, long c, long d, long x, long s, long ac) {
a += I(b, c, d) + x + ac;
a = ((int)a << s) | ((int)a >>> (32 - s));
a += b;
return a;
}

/*
md5Update是MD5的主?算?程，inbuf是要??的字?串，inputlen是?度，?个
函数由getMD5ofStr?用，?用之前需要?用md5init，因此把它??成private的
*/

private void md5Update(byte[] inbuf, int inputLen) {
int i, index, partLen;
byte[] block = new byte[64];
index = (int)(count[0] >>> 3) & 0x3F;
// /* Update number of bits */
if ((count[0] += (inputLen << 3)) < (inputLen << 3))
count[1] ++;
count[1] += (inputLen >>> 29);
partLen = 64 - index;
// Transform as many times as possible.
if (inputLen >= partLen) {
md5Memcpy(buffer, inbuf, index, 0, partLen);
md5Transform(buffer);
for (i = partLen; i + 63 < inputLen; i += 64) {
md5Memcpy(block, inbuf, 0, i, 64);
md5Transform(block);
}
index = 0;
} else
i = 0;
///* Buffer remaining input */
md5Memcpy(buffer, inbuf, index, i, inputLen - i);
}

/*
md5Final整理和填写?出?果
*/

private void md5Final() {
byte[] bits = new byte[8];
int index, padLen;
///* Save number of bits */
Encode(bits, count, 8);
///* Pad out to 56 mod 64.
index = (int)(count[0] >>> 3) & 0x3f;
padLen = (index < 56) ? (56 - index) : (120 - index);
md5Update(PADDING, padLen);
///* Append length (before padding) */
md5Update(bits, 8);
///* Store state in digest */
Encode(digest, state, 16);
}

/* md5Memcpy是一个内部使用的byte数?的?拷?函数，从input的inpos?始把len?度的
字?拷?到output的outpos位置?始
*/

private void md5Memcpy(byte[] output, byte[] input, int outpos, int inpos, int len) {
int i;
for (i = 0; i < len; i++)
output[outpos + i] = input[inpos + i];
}

/*
md5Transform是MD5核心??程序，有md5Update?用，block是分?的原始字?
*/

private void md5Transform(byte[] block) {
long a = state[0], b = state[1], c = state[2], d = state[3];
long[] x = new long[16];
Decode(x, block, 64);

/* Round 1 */

a = FF(a, b, c, d, x[0], S11, 0xd76aa478L);

/* 1 */

d = FF(d, a, b, c, x[1], S12, 0xe8c7b756L);

/* 2 */

c = FF(c, d, a, b, x[2], S13, 0x242070dbL);

/* 3 */

b = FF(b, c, d, a, x[3], S14, 0xc1bdceeeL);

/* 4 */

a = FF(a, b, c, d, x[4], S11, 0xf57c0fafL);

/* 5 */

d = FF(d, a, b, c, x[5], S12, 0x4787c62aL);

/* 6 */

c = FF(c, d, a, b, x[6], S13, 0xa8304613L);

/* 7 */

b = FF(b, c, d, a, x[7], S14, 0xfd469501L);

/* 8 */

a = FF(a, b, c, d, x[8], S11, 0x698098d8L);

/* 9 */

d = FF(d, a, b, c, x[9], S12, 0x8b44f7afL);

/* 10 */

c = FF(c, d, a, b, x[10], S13, 0xffff5bb1L);

/* 11 */

b = FF(b, c, d, a, x[11], S14, 0x895cd7beL);

/* 12 */

a = FF(a, b, c, d, x[12], S11, 0x6b901122L);

/* 13 */

d = FF(d, a, b, c, x[13], S12, 0xfd987193L);

/* 14 */

c = FF(c, d, a, b, x[14], S13, 0xa679438eL);

/* 15 */

b = FF(b, c, d, a, x[15], S14, 0x49b40821L);

/* 16 */

/* Round 2 */

a = GG(a, b, c, d, x[1], S21, 0xf61e2562L);

/* 17 */

d = GG(d, a, b, c, x[6], S22, 0xc040b340L);

/* 18 */

c = GG(c, d, a, b, x[11], S23, 0x265e5a51L);

/* 19 */

b = GG(b, c, d, a, x[0], S24, 0xe9b6c7aaL);

/* 20 */

a = GG(a, b, c, d, x[5], S21, 0xd62f105dL);

/* 21 */

d = GG(d, a, b, c, x[10], S22, 0x2441453L);

/* 22 */

c = GG(c, d, a, b, x[15], S23, 0xd8a1e681L);

/* 23 */

b = GG(b, c, d, a, x[4], S24, 0xe7d3fbc8L);

/* 24 */

a = GG(a, b, c, d, x[9], S21, 0x21e1cde6L);

/* 25 */

d = GG(d, a, b, c, x[14], S22, 0xc33707d6L);

/* 26 */

c = GG(c, d, a, b, x[3], S23, 0xf4d50d87L);

/* 27 */

b = GG(b, c, d, a, x[8], S24, 0x455a14edL);

/* 28 */

a = GG(a, b, c, d, x[13], S21, 0xa9e3e905L);

/* 29 */

d = GG(d, a, b, c, x[2], S22, 0xfcefa3f8L);

/* 30 */

c = GG(c, d, a, b, x[7], S23, 0x676f02d9L);

/* 31 */

b = GG(b, c, d, a, x[12], S24, 0x8d2a4c8aL);

/* 32 */

/* Round 3 */

a = HH(a, b, c, d, x[5], S31, 0xfffa3942L);

/* 33 */

d = HH(d, a, b, c, x[8], S32, 0x8771f681L);

/* 34 */

c = HH(c, d, a, b, x[11], S33, 0x6d9d6122L);

/* 35 */

b = HH(b, c, d, a, x[14], S34, 0xfde5380cL);

/* 36 */

a = HH(a, b, c, d, x[1], S31, 0xa4beea44L);

/* 37 */

d = HH(d, a, b, c, x[4], S32, 0x4bdecfa9L);

/* 38 */

c = HH(c, d, a, b, x[7], S33, 0xf6bb4b60L);

/* 39 */

b = HH(b, c, d, a, x[10], S34, 0xbebfbc70L);

/* 40 */

a = HH(a, b, c, d, x[13], S31, 0x289b7ec6L);

/* 41 */

d = HH(d, a, b, c, x[0], S32, 0xeaa127faL);

/* 42 */

c = HH(c, d, a, b, x[3], S33, 0xd4ef3085L);

/* 43 */

b = HH(b, c, d, a, x[6], S34, 0x4881d05L);

/* 44 */

a = HH(a, b, c, d, x[9], S31, 0xd9d4d039L);

/* 45 */

d = HH(d, a, b, c, x[12], S32, 0xe6db99e5L);

/* 46 */

c = HH(c, d, a, b, x[15], S33, 0x1fa27cf8L);

/* 47 */

b = HH(b, c, d, a, x[2], S34, 0xc4ac5665L);

/* 48 */

/* Round 4 */

a = II(a, b, c, d, x[0], S41, 0xf4292244L);

/* 49 */

d = II(d, a, b, c, x[7], S42, 0x432aff97L);

/* 50 */

c = II(c, d, a, b, x[14], S43, 0xab9423a7L);

/* 51 */

b = II(b, c, d, a, x[5], S44, 0xfc93a039L);

/* 52 */

a = II(a, b, c, d, x[12], S41, 0x655b59c3L);

/* 53 */

d = II(d, a, b, c, x[3], S42, 0x8f0ccc92L);

/* 54 */

c = II(c, d, a, b, x[10], S43, 0xffeff47dL);

/* 55 */

b = II(b, c, d, a, x[1], S44, 0x85845dd1L);

/* 56 */

a = II(a, b, c, d, x[8], S41, 0x6fa87e4fL);

/* 57 */

d = II(d, a, b, c, x[15], S42, 0xfe2ce6e0L);

/* 58 */

c = II(c, d, a, b, x[6], S43, 0xa3014314L);

/* 59 */

b = II(b, c, d, a, x[13], S44, 0x4e0811a1L);

/* 60 */

a = II(a, b, c, d, x[4], S41, 0xf7537e82L);

/* 61 */

d = II(d, a, b, c, x[11], S42, 0xbd3af235L);

/* 62 */

c = II(c, d, a, b, x[2], S43, 0x2ad7d2bbL);

/* 63 */

b = II(b, c, d, a, x[9], S44, 0xeb86d391L);

/* 64 */

state[0] += a;
state[1] += b;
state[2] += c;
state[3] += d;
}

/*Encode把long数?按?序拆成byte数?，因?java的long?型是64bit的，
只拆低32bit，以??原始C??的用途
*/

private void Encode(byte[] output, long[] input, int len) {
int i, j;
for (i = 0, j = 0; j < len; i++, j += 4) {
output[j] = (byte)(input[i] & 0xffL);
output[j + 1] = (byte)((input[i] >>> 8) & 0xffL);
output[j + 2] = (byte)((input[i] >>> 16) & 0xffL);
output[j + 3] = (byte)((input[i] >>> 24) & 0xffL);
}
}

/*Decode把byte数?按?序合成成long数?，因?java的long?型是64bit的，
只合成低32bit，高32bit清零，以??原始C??的用途
*/

private void Decode(long[] output, byte[] input, int len) {
int i = 0, j = 0;
for (i = 0, j = 0; j < len; i++, j += 4)
output[i] = b2iu(input[j]) | (b2iu(input[j + 1]) << 8) | (b2iu(input[j + 2]) << 16) | (b2iu(input[j + 3]) << 24);
return;
}

/*
b2iu是我写的一个把byte按照不考?正?号的原?的”升位”程序，因?java没有unsigned?算
*/

public static long b2iu(byte b) {
return b < 0 ? b & 0x7F + 128 : b;
}

/*byteHEX()，用来把一个byte?型的数??成十六?制的ASCII表示，
因?java中的byte的toString无法???一点，我?又没有C?言中的
sprintf(outbuf,""%02X"",ib)
*/

public static String byteHEX(byte ib) {
char[] Digit = { '0','1','2','3','4','5','6','7','8','9',
'A','B','C','D','E','F' };
char[] ob = new char[2];
ob[0] = Digit[(ib >>> 4) & 0X0F];
ob[1] = Digit[ib & 0X0F];
String s = new String(ob);
return s;
}

public String encode(String s) {
return s;
}

public String decode(String s) {
return getMD5ofStr(s);
}

public static void main(String args[]) {

MD5Bean m = new MD5Bean();

String aaa = m.getMD5ofStr(""genehxm"");
String bbb = m.getMD5ofStr(""genehxm"");

System.out.println(""aaa===========""+aaa);
System.out.println(""BBB===========""+bbb);
}
}
"