md5魔改

最近偶然看见一道魔改的md5题,

因为md5不可逆,所以也不怎么见,这个题大致就是找到魔改点,然后爆破md5(因为纯数字qwq)

主函数如下

image-20250316170001229

image-20250316170022673

MD5核心地方

image-20250316170057789

看见0x28955B88等可知这是一个类md5加密算法

MD5Encode魔改点

image-20250316170111587

很容易看出

后面的 ^0x12 ^0x34等

关键在MD5Update里面的魔改

image-20250316170123305

这里面是自定义初始话的四个幻数(别问为什么就看见三个)

这些动调都可以轻松得到

image-20250316170135475

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context->state[0] = 0x10325476;
context->state[1] = 0x12345678;
context->state[2] = 0x12345678;
context->state[3] = 0x67452301;

这个为魔改后的初始化幻数

去网上找个md5源码

MD5算法详解及实现(C语言)_c语言 md5-CSDN博客

MD5算法 —— C语言实现(字符串的加密) - mao的博客 - 博客园

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#include <stdio.h>
#include <stdlib.h>
#include <memory.h>
#include "string.h"
#include "md5.h"

unsigned char PADDING[] = {0x80,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
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};

void MD5Init(MD5_CTX* context)
{
context->count[0] = 0;
context->count[1] = 0;
// context->state[0] = 0x10325476;
// context->state[1] = 0x12345678;
// context->state[2] = 0x12345678;
// context->state[3] = 0x67452301;

context->state[0] = 0x67452301;
context->state[1] = 0xEFCDAB89;
context->state[2] = 0x98BADCFE;
context->state[3] = 0x10325476;

//0x10325476 0x12345678 0x12345678 0x67452301
}
void MD5Update(MD5_CTX* context, unsigned char* input, unsigned int inputlen)
{
unsigned int i = 0, index = 0, partlen = 0;
index = (context->count[0] >> 3) & 0x3F;
partlen = 64 - index;
context->count[0] += inputlen << 3;
if(context->count[0] < (inputlen << 3))
context->count[1]++;
context->count[1] += inputlen >> 29;

if(inputlen >= partlen)
{
memcpy(&context->buffer[index], input, partlen);
MD5Transform(context->state, context->buffer);
for(i = partlen; i + 64 <= inputlen; i += 64)
MD5Transform(context->state, &input[i]);
index = 0;
}
else
{
i = 0;
}
memcpy(&context->buffer[index], &input[i], inputlen - i);
}
void MD5Final(MD5_CTX* context, unsigned char digest[16])
{
unsigned int index = 0, padlen = 0;
unsigned char bits[8];
index = (context->count[0] >> 3) & 0x3F;
padlen = (index < 56) ? (56 - index) : (120 - index);
MD5Encode(bits, context->count, 8);
MD5Update(context, PADDING, padlen);
MD5Update(context, bits, 8);
MD5Encode(digest, context->state, 16);
}
void MD5Encode(unsigned char* output, unsigned int* input, unsigned int len)
{
unsigned int i = 0, j = 0;
while(j < len)
{
output[j] = input[i] & 0xFF;
output[j + 1] = (input[i] >> 8) & 0xFF;
output[j + 2] = (input[i] >> 16) & 0xFF;
output[j + 3] = (input[i] >> 24) & 0xFF;
i++;
j += 4;
}
}
void MD5Decode(unsigned int* output, unsigned char* input, unsigned int len)
{
unsigned int i = 0, j = 0;
while(j < len)
{
output[i] = (input[j]) |
(input[j + 1] << 8) |
(input[j + 2] << 16) |
(input[j + 3] << 24);
i++;
j += 4;
}
}
void MD5Transform(unsigned int state[4], unsigned char block[64])
{
unsigned int a = state[0];
unsigned int b = state[1];
unsigned int c = state[2];
unsigned int d = state[3];
////0x10325476 0x12345678 0x12345678 0x67452301
unsigned int x[64];
MD5Decode(x, block, 64);
FF(a, b, c, d, x[0], 7, 0xd76aa478); /* 1 */
printf("%x %x %x %x %x \n",a, b, c, d, x[0]);
FF(d, a, b, c, x[1], 12, 0xe8c7b756); /* 2 */
printf("%x %x %x %x %x \n",a, b, c, d, x[1]);
FF(c, d, a, b, x[2], 17, 0x242070db); /* 3 */
FF(b, c, d, a, x[3], 22, 0xc1bdceee); /* 4 */
FF(a, b, c, d, x[4], 7, 0xf57c0faf); /* 5 */
FF(d, a, b, c, x[5], 12, 0x4787c62a); /* 6 */
FF(c, d, a, b, x[6], 17, 0xa8304613); /* 7 */
FF(b, c, d, a, x[7], 22, 0xfd469501); /* 8 */
FF(a, b, c, d, x[8], 7, 0x698098d8); /* 9 */
FF(d, a, b, c, x[9], 12, 0x8b44f7af); /* 10 */
FF(c, d, a, b, x[10], 17, 0xffff5bb1); /* 11 */
FF(b, c, d, a, x[11], 22, 0x895cd7be); /* 12 */
FF(a, b, c, d, x[12], 7, 0x6b901122); /* 13 */
FF(d, a, b, c, x[13], 12, 0xfd987193); /* 14 */
FF(c, d, a, b, x[14], 17, 0xa679438e); /* 15 */
FF(b, c, d, a, x[15], 22, 0x49b40821); /* 16 */

/* Round 2 */
GG(a, b, c, d, x[1], 5, 0xf61e2562); /* 17 */
GG(d, a, b, c, x[6], 9, 0xc040b340); /* 18 */
GG(c, d, a, b, x[11], 14, 0x265e5a51); /* 19 */
GG(b, c, d, a, x[0], 20, 0xe9b6c7aa); /* 20 */
GG(a, b, c, d, x[5], 5, 0xd62f105d); /* 21 */
GG(d, a, b, c, x[10], 9, 0x2441453); /* 22 */
GG(c, d, a, b, x[15], 14, 0xd8a1e681); /* 23 */
GG(b, c, d, a, x[4], 20, 0xe7d3fbc8); /* 24 */
GG(a, b, c, d, x[9], 5, 0x21e1cde6); /* 25 */
GG(d, a, b, c, x[14], 9, 0xc33707d6); /* 26 */
GG(c, d, a, b, x[3], 14, 0xf4d50d87); /* 27 */
GG(b, c, d, a, x[8], 20, 0x455a14ed); /* 28 */
GG(a, b, c, d, x[13], 5, 0xa9e3e905); /* 29 */
GG(d, a, b, c, x[2], 9, 0xfcefa3f8); /* 30 */
GG(c, d, a, b, x[7], 14, 0x676f02d9); /* 31 */
GG(b, c, d, a, x[12], 20, 0x8d2a4c8a); /* 32 */

/* Round 3 */
HH(a, b, c, d, x[5], 4, 0xfffa3942); /* 33 */
HH(d, a, b, c, x[8], 11, 0x8771f681); /* 34 */
HH(c, d, a, b, x[11], 16, 0x6d9d6122); /* 35 */
HH(b, c, d, a, x[14], 23, 0xfde5380c); /* 36 */
HH(a, b, c, d, x[1], 4, 0xa4beea44); /* 37 */
HH(d, a, b, c, x[4], 11, 0x4bdecfa9); /* 38 */
HH(c, d, a, b, x[7], 16, 0xf6bb4b60); /* 39 */
HH(b, c, d, a, x[10], 23, 0xbebfbc70); /* 40 */
HH(a, b, c, d, x[13], 4, 0x289b7ec6); /* 41 */
HH(d, a, b, c, x[0], 11, 0xeaa127fa); /* 42 */
HH(c, d, a, b, x[3], 16, 0xd4ef3085); /* 43 */
HH(b, c, d, a, x[6], 23, 0x4881d05); /* 44 */
HH(a, b, c, d, x[9], 4, 0xd9d4d039); /* 45 */
HH(d, a, b, c, x[12], 11, 0xe6db99e5); /* 46 */
HH(c, d, a, b, x[15], 16, 0x1fa27cf8); /* 47 */
HH(b, c, d, a, x[2], 23, 0xc4ac5665); /* 48 */

/* Round 4 */
II(a, b, c, d, x[0], 6, 0xf4292244); /* 49 */
II(d, a, b, c, x[7], 10, 0x432aff97); /* 50 */
II(c, d, a, b, x[14], 15, 0xab9423a7); /* 51 */
II(b, c, d, a, x[5], 21, 0xfc93a039); /* 52 */
II(a, b, c, d, x[12], 6, 0x655b59c3); /* 53 */
II(d, a, b, c, x[3], 10, 0x8f0ccc92); /* 54 */
II(c, d, a, b, x[10], 15, 0xffeff47d); /* 55 */
II(b, c, d, a, x[1], 21, 0x85845dd1); /* 56 */
II(a, b, c, d, x[8], 6, 0x6fa87e4f); /* 57 */
II(d, a, b, c, x[15], 10, 0xfe2ce6e0); /* 58 */
II(c, d, a, b, x[6], 15, 0xa3014314); /* 59 */
II(b, c, d, a, x[13], 21, 0x4e0811a1); /* 60 */
II(a, b, c, d, x[4], 6, 0xf7537e82); /* 61 */
II(d, a, b, c, x[11], 10, 0xbd3af235); /* 62 */
II(c, d, a, b, x[2], 15, 0x2ad7d2bb); /* 63 */
II(b, c, d, a, x[9], 21, 0xeb86d391); /* 64 */
state[0] += a;
state[1] += b;
state[2] += c;
state[3] += d;
}

int main(int argc, char* argv[])
{
int i;
unsigned char encrypt[] = "87654321";
unsigned char decrypt[16];
MD5_CTX md5;
MD5Init(&md5);
MD5Update(&md5, encrypt, strlen((char*) encrypt));
MD5Final(&md5, decrypt);
printf("加密前:%s\n加密后:", encrypt);
for(i = 0; i < 16; i++)
{
printf("%02x", decrypt[i]);
}

getchar();

return 0;
//0x10325476 0x12345678 0x12345678 0x67452301
}

md5.h

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#ifndef MD5_H
#define MD5_H

typedef struct
{
unsigned int count[2];
unsigned int state[4];
unsigned char buffer[64];
}MD5_CTX;


#define F(x,y,z) ((x & y) | (~x & z))
#define G(x,y,z) ((x & z) | (y & ~z))
#define H(x,y,z) (x^y^z)
#define I(x,y,z) (y ^ (x | ~z))
#define ROTATE_LEFT(x,n) ((x << n) | (x >> (32-n)))
#define FF(a,b,c,d,x,s,ac) \
{ \
a += F(b,c,d) + x + ac; \
a = ROTATE_LEFT(a,s); \
a += b; \
}
#define GG(a,b,c,d,x,s,ac) \
{ \
a += G(b,c,d) + x + ac; \
a = ROTATE_LEFT(a,s); \
a += b; \
}
#define HH(a,b,c,d,x,s,ac) \
{ \
a += H(b,c,d) + x + ac; \
a = ROTATE_LEFT(a,s); \
a += b; \
}
#define II(a,b,c,d,x,s,ac) \
{ \
a += I(b,c,d) + x + ac; \
a = ROTATE_LEFT(a,s); \
a += b; \
}
void MD5Init(MD5_CTX *context);
void MD5Update(MD5_CTX *context,unsigned char *input,unsigned int inputlen);
void MD5Final(MD5_CTX *context,unsigned char digest[16]);
void MD5Transform(unsigned int state[4],unsigned char block[64]);
void MD5Encode(unsigned char *output,unsigned int *input,unsigned int len);
void MD5Decode(unsigned int *output,unsigned char *input,unsigned int len);

#endif

直接把幻数一改发现还是不对

(我测试时用的87654321去测试的)

可以去让程序跑下看看加密的md5值为多少

image-20250316170201357

这个地方下断点

image-20250316170214564

因为到这个地方还没进行后面的异或所以这里是只有前面魔改而生成的md5(正常md5到这里就结束了,后面只是个Encode,换句话说就是如果前面没有魔改到这里就是正常的md5值)

所以关键就变成还原算法了

如何自己加密生成这个31133B716DBC9AF54A27BDA3D1CD29A0

因为我们改幻数后还是不对,应该还有别的魔改

自己编译一份正常的md5加密对照

魔改的

image-20250316170226708

自己编译得

image-20250316170239148

不难看出这里是不一样得

源码中对应得是这里

image-20250316170251814

动调时看参数修改就行

确定下x y z的位置

image-20250316170303253

改完就是这样(这里面的苦只有自己弄了才知道qwq)

然后测试还是不对

然后就这样输出看看从哪里开始错了

image-20250316170317321

发现到FF最后两个那里开始错了

image-20250316170332027

打印出来的x[14]为0x40,x[15]为0

然而实际上他们的值为

image-20250316170343171

在初始话后修改下

image-20250316170353774

然后就得到

image-20250316170404045

加上后面的异或,动调拿到密文让ai写个爆破

image-20250316170415209

得到

image-20250316170424851

附上脚本

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#include <stdio.h>
#include <stdlib.h>
#include <memory.h>
#include "string.h"
#include "md5.h"

unsigned char PADDING[] = {0x80,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
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};

void MD5Init(MD5_CTX* context)
{
context->count[0] = 0;
context->count[1] = 0;
context->state[0] = 0x10325476;
context->state[1] = 0x12345678;
context->state[2] = 0x12345678;
context->state[3] = 0x67452301;

// context->state[0] = 0x67452301;
// context->state[1] = 0xEFCDAB89;
// context->state[2] = 0x98BADCFE;
// context->state[3] = 0x10325476;

//0x10325476 0x12345678 0x12345678 0x67452301
}
void MD5Update(MD5_CTX* context, unsigned char* input, unsigned int inputlen)
{
unsigned int i = 0, index = 0, partlen = 0;
index = (context->count[0] >> 3) & 0x3F;
partlen = 64 - index;
context->count[0] += inputlen << 3;
if(context->count[0] < (inputlen << 3))
context->count[1]++;
context->count[1] += inputlen >> 29;

if(inputlen >= partlen)
{
memcpy(&context->buffer[index], input, partlen);
MD5Transform(context->state, context->buffer);
for(i = partlen; i + 64 <= inputlen; i += 64)
MD5Transform(context->state, &input[i]);
index = 0;
}
else
{
i = 0;
}
memcpy(&context->buffer[index], &input[i], inputlen - i);
}
void MD5Final(MD5_CTX* context, unsigned char digest[16])
{
unsigned int index = 0, padlen = 0;
unsigned char bits[8];
index = (context->count[0] >> 3) & 0x3F;
padlen = (index < 56) ? (56 - index) : (120 - index);
MD5Encode(bits, context->count, 8);
MD5Update(context, PADDING, padlen);
MD5Update(context, bits, 8);
MD5Encode(digest, context->state, 16);
}
void MD5Encode(unsigned char* output, unsigned int* input, unsigned int len)
{
unsigned int i = 0, j = 0;
while(j < len)
{
output[j] = (input[i] & 0xFF)^0x12;
output[j + 1] = ((input[i] >> 8) & 0xFF)^0x34;
output[j + 2] = ((input[i] >> 16) & 0xFF)^0x56;
output[j + 3] = ((input[i] >> 24) & 0xFF)^0x78;
i++;
j += 4;
}
}
void MD5Decode(unsigned int* output, unsigned char* input, unsigned int len)
{
unsigned int i = 0, j = 0;
while(j < len)
{
output[i] = (input[j]) |
(input[j + 1] << 8) |
(input[j + 2] << 16) |
(input[j + 3] << 24);
i++;
j += 4;
}
}
void MD5Transform(unsigned int state[4], unsigned char block[64])
{
unsigned int a = state[0];//0x10325476
unsigned int b = state[1];//0x12345678
unsigned int c = state[2];//0x12345678
unsigned int d = state[3];//0x67452301


unsigned int x[64];



MD5Decode(x, block, 64);

x[14]=0x78563452;
x[15]=0x78563412;

FF(a, b, c, d, x[0], 7, 0xd76aa478); /* 1 */
FF(d, a, b, c, x[1], 12, 0xe8c7b756); /* 2 */
FF(c, d, a, b, x[2], 17, 0x242070db); /* 3 */
FF(b, c, d, a, x[3], 22, 0xc1bdceee); /* 4 */
FF(a, b, c, d, x[4], 7, 0xf57c0faf); /* 5 */
FF(d, a, b, c, x[5], 12, 0x4787c62a); /* 6 */
FF(c, d, a, b, x[6], 17, 0xa8304613); /* 7 */
FF(b, c, d, a, x[7], 22, 0xfd469501); /* 8 */
FF(a, b, c, d, x[8], 7, 0x698098d8); /* 9 */
FF(d, a, b, c, x[9], 12, 0x8b44f7af); /* 10 */
FF(c, d, a, b, x[10], 17, 0xffff5bb1); /* 11 */
FF(b, c, d, a, x[11], 22, 0x895cd7be); /* 12 */
FF(a, b, c, d, x[12], 7, 0x6b901122); /* 13 */
FF(d, a, b, c, x[13], 12, 0xfd987193); /* 14 */
FF(c, d, a, b, x[14], 17, 0xa679438e); /* 15 */
FF(b, c, d, a, x[15], 22, 0x49b40821); /* 16 */


/* Round 2 */
GG(a, b, c, d, x[1], 5, 0xf61e2562); /* 17 */
GG(d, a, b, c, x[6], 9, 0xc040b340); /* 18 */
GG(c, d, a, b, x[11], 14, 0x265e5a51); /* 19 */
GG(b, c, d, a, x[0], 20, 0xe9b6c7aa); /* 20 */
GG(a, b, c, d, x[5], 5, 0xd62f105d); /* 21 */
GG(d, a, b, c, x[10], 9, 0x2441453); /* 22 */
GG(c, d, a, b, x[15], 14, 0xd8a1e681); /* 23 */
GG(b, c, d, a, x[4], 20, 0xe7d3fbc8); /* 24 */
GG(a, b, c, d, x[9], 5, 0x21e1cde6); /* 25 */
GG(d, a, b, c, x[14], 9, 0xc33707d6); /* 26 */
GG(c, d, a, b, x[3], 14, 0xf4d50d87); /* 27 */
GG(b, c, d, a, x[8], 20, 0x455a14ed); /* 28 */
GG(a, b, c, d, x[13], 5, 0xa9e3e905); /* 29 */
GG(d, a, b, c, x[2], 9, 0xfcefa3f8); /* 30 */
GG(c, d, a, b, x[7], 14, 0x676f02d9); /* 31 */
GG(b, c, d, a, x[12], 20, 0x8d2a4c8a); /* 32 */


/* Round 3 */
HH(a, b, c, d, x[5], 4, 0xfffa3942); /* 33 */
HH(d, a, b, c, x[8], 11, 0x8771f681); /* 34 */
HH(c, d, a, b, x[11], 16, 0x6d9d6122); /* 35 */
HH(b, c, d, a, x[14], 23, 0xfde5380c); /* 36 */
HH(a, b, c, d, x[1], 4, 0xa4beea44); /* 37 */
HH(d, a, b, c, x[4], 11, 0x4bdecfa9); /* 38 */
HH(c, d, a, b, x[7], 16, 0xf6bb4b60); /* 39 */
HH(b, c, d, a, x[10], 23, 0xbebfbc70); /* 40 */
HH(a, b, c, d, x[13], 4, 0x289b7ec6); /* 41 */
HH(d, a, b, c, x[0], 11, 0xeaa127fa); /* 42 */
HH(c, d, a, b, x[3], 16, 0xd4ef3085); /* 43 */
HH(b, c, d, a, x[6], 23, 0x4881d05); /* 44 */
HH(a, b, c, d, x[9], 4, 0xd9d4d039); /* 45 */
HH(d, a, b, c, x[12], 11, 0xe6db99e5); /* 46 */
HH(c, d, a, b, x[15], 16, 0x1fa27cf8); /* 47 */
HH(b, c, d, a, x[2], 23, 0xc4ac5665); /* 48 */

/* Round 4 */
II(a, b, c, d, x[0], 6, 0xf4292244); /* 49 */
II(d, a, b, c, x[7], 10, 0x432aff97); /* 50 */
II(c, d, a, b, x[14], 15, 0xab9423a7); /* 51 */
II(b, c, d, a, x[5], 21, 0xfc93a039); /* 52 */
II(a, b, c, d, x[12], 6, 0x655b59c3); /* 53 */
II(d, a, b, c, x[3], 10, 0x8f0ccc92); /* 54 */
II(c, d, a, b, x[10], 15, 0xffeff47d); /* 55 */
II(b, c, d, a, x[1], 21, 0x85845dd1); /* 56 */
II(a, b, c, d, x[8], 6, 0x6fa87e4f); /* 57 */
II(d, a, b, c, x[15], 10, 0xfe2ce6e0); /* 58 */
II(c, d, a, b, x[6], 15, 0xa3014314); /* 59 */
II(b, c, d, a, x[13], 21, 0x4e0811a1); /* 60 */
II(a, b, c, d, x[4], 6, 0xf7537e82); /* 61 */
II(d, a, b, c, x[11], 10, 0xbd3af235); /* 62 */
II(c, d, a, b, x[2], 15, 0x2ad7d2bb); /* 63 */
II(b, c, d, a, x[9], 21, 0xeb86d391); /* 64 */
state[0] += a;
state[1] += b;
state[2] += c;
state[3] += d;
}

const char* target_md5 = "abaf09d8d1c3b31414443875ffc3b766"; // "abaf09d8d1c3b31414443875ffc3b766";

// 生成所有8位数字组合并检查MD5
void brute_force_md5() {
unsigned char encrypt[9]; // 8位数字 + 1个空字符
unsigned char decrypt[16];
MD5_CTX md5;

// 遍历所有8位数字组合
for (long long i = 0; i <= 99999999; i++) {
// 将数字格式化为8位字符串
snprintf((char*)encrypt, sizeof(encrypt), "%08lld", i);

// 计算MD5
MD5Init(&md5);
MD5Update(&md5, encrypt, strlen((char*)encrypt));
MD5Final(&md5, decrypt);

// 将MD5结果转换为十六进制字符串
char md5_str[33];
for (int j = 0; j < 16; j++) {
sprintf(&md5_str[j * 2], "%02x", decrypt[j]);
}

// 检查是否匹配目标MD5
if (strcmp(md5_str, target_md5) == 0) {
printf("找到匹配的数字组合: %s\n", encrypt);
printf("对应的MD5值: %s\n", md5_str);
return;
}

// 打印进度(可选)
if (i % 1000000 == 0) {
printf("当前进度: %lld/99999999\n", i);
}
}

printf("未找到匹配的数字组合。\n");
}

int main(int argc, char* argv[])
{
// int i;
// unsigned char encrypt[] = "87654321";//21232f297a57a5a743894a0e4a801fc3
// unsigned char decrypt[16];
// MD5_CTX md5;
// MD5Init(&md5);
// MD5Update(&md5, encrypt, strlen((char*) encrypt));
// MD5Final(&md5, decrypt);
// printf("加密前:%s\n加密后:", encrypt);
// for(i = 0; i < 16; i++)
// {
// printf("%02x", decrypt[i]);
// }


brute_force_md5();
return 0;

}

md5.h

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#ifndef MD5_H
#define MD5_H

typedef struct
{
unsigned int count[2];
unsigned int state[4];
unsigned char buffer[64];
}MD5_CTX;

// b c d
#define F(x,y,z) (z^x&(y^z))
#define G(x,y,z) (y^z&(x^y))
#define H(x,y,z) (x^y^z)
#define I(x,y,z) (y ^ (x | ~z))
#define ROTATE_LEFT(x,n) ((x << n) | (x >> (32-n)))
#define FF(a,b,c,d,x,s,ac) \
{ \
a += F(b,c,d) + x + ac; \
a = ROTATE_LEFT(a,s); \
a += b; \
}
#define GG(a,b,c,d,x,s,ac) \
{ \
a += G(b,c,d) + x + ac; \
a = ROTATE_LEFT(a,s); \
a += b; \
}
#define HH(a,b,c,d,x,s,ac) \
{ \
a += H(b,c,d) + x + ac; \
a = ROTATE_LEFT(a,s); \
a += b; \
}
#define II(a,b,c,d,x,s,ac) \
{ \
a += I(b,c,d) + x + ac; \
a = ROTATE_LEFT(a,s); \
a += b; \
}

void MD5Init(MD5_CTX *context);
void MD5Update(MD5_CTX *context,unsigned char *input,unsigned int inputlen);
void MD5Final(MD5_CTX *context,unsigned char digest[16]);
void MD5Transform(unsigned int state[4],unsigned char block[64]);
void MD5Encode(unsigned char *output,unsigned int *input,unsigned int len);
void MD5Decode(unsigned int *output,unsigned char *input,unsigned int len);

#endif

拿到这题到做出来用了两天,虽然玩了一天半,qwq,我是小废物