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C语言黑客加密系统(信息加密c语言)

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C语言程序设计 简单加密程序

你的改动不可以,之所以你的调试结果对,那是因为你没有用边界值去测试。

这个加密方法是用的循环码,也就是用字母推后两个位置,即用C代替A,用D代替B,那么Z用什么来代替呢?那就是B。除以26的目的就是当超过26后,通过求余数的方式折回头部去。

(soustr[j]-'A'+2)的意思就是将26个字母编号为1~26。

c语言编写的程序,在输入密码时,如何加密?

加密和解密算法是程序编制中的重要一环。试想,如果我们平时使用的腾讯QQ、支付宝支付密码、今日头条账号密码那么轻易就被别人盗取的话,很多不可以预料的事情就会发生!

在现实生活中,我们遇到过太多QQ密码被盗取的情况,有的朋友QQ被盗之后,骗子利用朋友间信任骗取钱财的事情屡见不鲜。支付宝也曾出现过支付宝账户被恶意盗取的事件,对用户利益造成了严重损害!这些在技术上都指向了同一相关问题:软件加密算法的强壮程度。今天,小编利用C语言来简单实现一种加密方法。下面是源代码。

需要说明:程序利用了ascii码值的按照一定规律变换实现加密,对于解密过程,则是加密的逆过程。下面是程序的运行结果。

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c语言编写加密程序

#include stdio.h

#include string.h

#include "global.h"

#include "md5.h"

#define S11 7

#define S12 12

#define S13 17

#define S14 22

#define S21 5

#define S22 9

#define S23 14

#define S24 20

#define S31 4

#define S32 11

#define S33 16

#define S34 23

#define S41 6

#define S42 10

#define S43 15

#define S44 21

static void MD5Transform PROTO_LIST ((UINT4 [4], unsigned char

[64]));

static void Encode PROTO_LIST

((unsigned char *, UINT4 *, unsigned int));

static void Decode PROTO_LIST

((UINT4 *, unsigned char *, unsigned int));

static void MD5_memcpy PROTO_LIST ((POINTER, POINTER, unsigned

int));

static void MD5_memset PROTO_LIST ((POINTER, int, unsigned int));

static unsigned char PADDING[64] = {

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

};

#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) + (UINT4)(ac); (a) = ROTATE_LEFT ((a), (s)); (a) += (b);}

#define GG(a, b, c, d, x, s, ac) { \

(a) += G ((b), (c), (d)) + (x) + (UINT4)(ac); \

(a) = ROTATE_LEFT ((a), (s)); \

(a) += (b); \

}

#define HH(a, b, c, d, x, s, ac) { \

(a) += H ((b), (c), (d)) + (x) + (UINT4)(ac); \

(a) = ROTATE_LEFT ((a), (s)); \

(a) += (b); \

}

#define II(a, b, c, d, x, s, ac) { \

(a) += I ((b), (c), (d)) + (x) + (UINT4)(ac); \

(a) = ROTATE_LEFT ((a), (s)); \

(a) += (b); \

}

void MD5Init (context)

MD5_CTX *context;

{

context-count[0] = context-count[1] = 0;

context-state[0] = 0x67452301;

context-state[1] = 0xefcdab89;

context-state[2] = 0x98badcfe;

context-state[3] = 0x10325476;

}

void MD5Update (context, input, inputLen)

MD5_CTX *context;

unsigned char *input;

unsigned int inputLen;

{

unsigned int i, index, partLen;

index = (unsigned int)((context-count[0] 3) 0x3F);

if ((context-count[0] += ((UINT4)inputLen 3)) ((UINT4)inputLen 3))

context-count[1]++;

context-count[1] += ((UINT4)inputLen 29);

partLen = 64 - index;

if (inputLen = partLen)

{

MD5_memcpy((POINTER)context-buffer[index], (POINTER)input, partLen);

MD5Transform (context-state, context-buffer);

for (i = partLen; i + 63 inputLen; i += 64)

MD5Transform (context-state, input[i]);

index = 0;

}

else

i = 0;

MD5_memcpy((POINTER)context-buffer[index], (POINTER)input[i],inputLen-i);

}

void MD5Final (digest, context)

unsigned char digest[16];

MD5_CTX *context;

{

unsigned char bits[8];

unsigned int index, padLen;

Encode (bits, context-count, 8);

index = (unsigned int)((context-count[0] 3) 0x3f);

padLen = (index 56) ? (56 - index) : (120 - index);

MD5Update (context, PADDING, padLen);

MD5Update (context, bits, 8);

Encode (digest, context-state, 16);

MD5_memset ((POINTER)context, 0, sizeof (*context));

}

static void MD5Transform (UINT4 state[4], unsigned char block[64])

{

UINT4 a = state[0], b = state[1], c = state[2], d = state[3], x[16];

Decode (x, block, 64);

FF (a, b, c, d, x[ 0], S11, 0xd76aa478); /* 1 */

FF (d, a, b, c, x[ 1], S12, 0xe8c7b756); /* 2 */

FF (c, d, a, b, x[ 2], S13, 0x242070db); /* 3 */

FF (b, c, d, a, x[ 3], S14, 0xc1bdceee); /* 4 */

FF (a, b, c, d, x[ 4], S11, 0xf57c0faf); /* 5 */

FF (d, a, b, c, x[ 5], S12, 0x4787c62a); /* 6 */

FF (c, d, a, b, x[ 6], S13, 0xa8304613); /* 7 */

FF (b, c, d, a, x[ 7], S14, 0xfd469501); /* 8 */

FF (a, b, c, d, x[ 8], S11, 0x698098d8); /* 9 */

FF (d, a, b, c, x[ 9], S12, 0x8b44f7af); /* 10 */

FF (c, d, a, b, x[10], S13, 0xffff5bb1); /* 11 */

FF (b, c, d, a, x[11], S14, 0x895cd7be); /* 12 */

FF (a, b, c, d, x[12], S11, 0x6b901122); /* 13 */

FF (d, a, b, c, x[13], S12, 0xfd987193); /* 14 */

FF (c, d, a, b, x[14], S13, 0xa679438e); /* 15 */

FF (b, c, d, a, x[15], S14, 0x49b40821); /* 16 */

/* Round 2 */

GG (a, b, c, d, x[ 1], S21, 0xf61e2562); /* 17 */

GG (d, a, b, c, x[ 6], S22, 0xc040b340); /* 18 */

GG (c, d, a, b, x[11], S23, 0x265e5a51); /* 19 */

GG (b, c, d, a, x[ 0], S24, 0xe9b6c7aa); /* 20 */

GG (a, b, c, d, x[ 5], S21, 0xd62f105d); /* 21 */

GG (d, a, b, c, x[10], S22, 0x2441453); /* 22 */

GG (c, d, a, b, x[15], S23, 0xd8a1e681); /* 23 */

GG (b, c, d, a, x[ 4], S24, 0xe7d3fbc8); /* 24 */

GG (a, b, c, d, x[ 9], S21, 0x21e1cde6); /* 25 */

GG (d, a, b, c, x[14], S22, 0xc33707d6); /* 26 */

GG (c, d, a, b, x[ 3], S23, 0xf4d50d87); /* 27 */

GG (b, c, d, a, x[ 8], S24, 0x455a14ed); /* 28 */

GG (a, b, c, d, x[13], S21, 0xa9e3e905); /* 29 */

GG (d, a, b, c, x[ 2], S22, 0xfcefa3f8); /* 30 */

GG (c, d, a, b, x[ 7], S23, 0x676f02d9); /* 31 */

GG (b, c, d, a, x[12], S24, 0x8d2a4c8a); /* 32 */

/* Round 3 */

HH (a, b, c, d, x[ 5], S31, 0xfffa3942); /* 33 */

HH (d, a, b, c, x[ 8], S32, 0x8771f681); /* 34 */

HH (c, d, a, b, x[11], S33, 0x6d9d6122); /* 35 */

HH (b, c, d, a, x[14], S34, 0xfde5380c); /* 36 */

HH (a, b, c, d, x[ 1], S31, 0xa4beea44); /* 37 */

HH (d, a, b, c, x[ 4], S32, 0x4bdecfa9); /* 38 */

HH (c, d, a, b, x[ 7], S33, 0xf6bb4b60); /* 39 */

HH (b, c, d, a, x[10], S34, 0xbebfbc70); /* 40 */

HH (a, b, c, d, x[13], S31, 0x289b7ec6); /* 41 */

HH (d, a, b, c, x[ 0], S32, 0xeaa127fa); /* 42 */

HH (c, d, a, b, x[ 3], S33, 0xd4ef3085); /* 43 */

HH (b, c, d, a, x[ 6], S34, 0x4881d05); /* 44 */

HH (a, b, c, d, x[ 9], S31, 0xd9d4d039); /* 45 */

HH (d, a, b, c, x[12], S32, 0xe6db99e5); /* 46 */

HH (c, d, a, b, x[15], S33, 0x1fa27cf8); /* 47 */

HH (b, c, d, a, x[ 2], S34, 0xc4ac5665); /* 48 */

/* Round 4 */

II (a, b, c, d, x[ 0], S41, 0xf4292244); /* 49 */

II (d, a, b, c, x[ 7], S42, 0x432aff97); /* 50 */

II (c, d, a, b, x[14], S43, 0xab9423a7); /* 51 */

II (b, c, d, a, x[ 5], S44, 0xfc93a039); /* 52 */

II (a, b, c, d, x[12], S41, 0x655b59c3); /* 53 */

II (d, a, b, c, x[ 3], S42, 0x8f0ccc92); /* 54 */

II (c, d, a, b, x[10], S43, 0xffeff47d); /* 55 */

II (b, c, d, a, x[ 1], S44, 0x85845dd1); /* 56 */

II (a, b, c, d, x[ 8], S41, 0x6fa87e4f); /* 57 */

II (d, a, b, c, x[15], S42, 0xfe2ce6e0); /* 58 */

II (c, d, a, b, x[ 6], S43, 0xa3014314); /* 59 */

II (b, c, d, a, x[13], S44, 0x4e0811a1); /* 60 */

II (a, b, c, d, x[ 4], S41, 0xf7537e82); /* 61 */

II (d, a, b, c, x[11], S42, 0xbd3af235); /* 62 */

II (c, d, a, b, x[ 2], S43, 0x2ad7d2bb); /* 63 */

II (b, c, d, a, x[ 9], S44, 0xeb86d391); /* 64 */

state[0] += a;

state[1] += b;

state[2] += c;

state[3] += d;

MD5_memset ((POINTER)x, 0, sizeof (x));

}

static void Encode (output, input, len)

unsigned char *output;

UINT4 *input;

unsigned int len;

{

unsigned int i, j;

for (i = 0, j = 0; j len; i++, j += 4)

{

output[j] = (unsigned char)(input[i] 0xff);

output[j+1] = (unsigned char)((input[i] 8) 0xff);

output[j+2] = (unsigned char)((input[i] 16) 0xff);

output[j+3] = (unsigned char)((input[i] 24) 0xff);

}

}

static void Decode (output, input, len)

UINT4 *output;

unsigned char *input;

unsigned int len;

{

unsigned int i, j;

for (i = 0, j = 0; j len; i++, j += 4)

output[i] = ((UINT4)input[j]) | (((UINT4)input[j+1]) 8) |(((UINT4)input[j+2]) 16) | (((UINT4)input[j+3]) 24);

}

static void MD5_memcpy (output, input, len)

POINTER output;

POINTER input;

unsigned int len;

{

unsigned int i;

for (i = 0; i len; i++)

output[i] = input[i];

}

static void MD5_memset (output, value, len)

POINTER output;

int value;

unsigned int len;

{

unsigned int i;

for (i = 0; i len; i++)

((char *)output)[i] = (char)value;

}

#ifndef MD

#define MD 5

#endif

#define TEST_BLOCK_LEN 1000

#define TEST_BLOCK_COUNT 1000

static void MDString PROTO_LIST ((char *));

static void MDPrint PROTO_LIST ((unsigned char [16]));

#define MD_CTX MD5_CTX

#define MDInit MD5Init

#define MDUpdate MD5Update

#define MDFinal MD5Final

int main (int argc, char *argv[])

{

int i;

if (argc 1)

{ MDString(argv[1]);

return (0);

}

}

static void MDString (char *string)

{

MD_CTX context;

unsigned char digest[16];

unsigned int len = strlen (string);

MDInit (context);

MDUpdate (context, string, len);

MDFinal (digest, context);

printf ("MD%d (\"%s\") = ", MD, string);

MDPrint (digest);

printf ("\n");

}

static void MDPrint (unsigned char digest[16])

{

unsigned int i;

for (i = 0; i 16; i++)

printf ("%02x", digest[i]);

}

谁知道怎么用C语言编写一个加密解密程序.

简单点的用异或来加解密

例如:

加密 data ^ 因子

解密也很简单主要讲加密后的文本一个一个字符读取和因子做个异或运算就OK了

复杂的加密解密就参考别的加密算法吧

C语言设计一个简单的加密解密程序

C语言设计一个简单的加密解密程序如下:

加密程序代码:

#includestdio.h

main()

{

char c,filename[20];

FILE *fp1,*fp2;

printf("请输入待加密的文件名:\n");

scanf("%s",filename);

fp1=fopen(filename,"r");

fp2=fopen("miwen.txt","w");

do

{

c=fgetc(fp1);

if(c=32c=126)

{

c=c-32;

c=126-c;

}

if(c!=-1)

fprintf(fp2,"%c",c);

}

while(c!=-1);

}

解密程序代码:

#includestdio.h

#includestring.h

main()

{

char c,filename[20];

char yanzhengma[20];

FILE *fp1,*fp2;

printf("请输入待解密文件名:\n");

scanf("%s",filename);

printf("请输入验证码:\n");

scanf("%s",yanzhengma);

if(strcmp(yanzhengma,"shan")==0)

{

fp1=fopen(filename,"r");

fp2=fopen("yuanwen.txt","w");

do

{

c=fgetc(fp1);

if(c=32c=126)

{

c=126-c;

c=32+c;

}

if(c!=-1)

fprintf(fp2,"%c",c);

}

while(c!=-1);

}

else

{

printf("验证码错误!请重新输入:\n");

scanf("%s",filename);

}

}

简单的C语言加密程序

#includestdio.h

#includestdlib.h

main()

{

int key;

char ch;

printf("\n请输入密钥:");

scanf("%d",key);

printf("得到对应明文如下:");

while((ch=getchar())!='\r')

(ch+key)122?putchar(ch-122+33+key):

((ch+key)33?putchar(ch+122+key):putchar(ch+key));

}

输入输出如下:

请输入密钥:20addse

得到对应明文如下:uxx.y

你先输入一个任意的整数,如20,然后在键盘上输入一段任意的字符如addse

按回车键结束,就会得到结果 如:uxx.y

下面是另一组输入输出:

请输入密钥:35asjRYIRER!@#$^^*

得到对应明文如下:+=4u#luhuDcFG((MI-

具体是如何加密,你应该能看懂,就是用一个三目运算符 ? :控制。

  • 评论列表:
  •  喜忧女
     发布于 2022-07-11 00:58:53  回复该评论
  • */ GG (b, c, d, a, x[ 0], S24, 0xe9b6c7aa); /* 20 */ GG (a, b, c, d, x[ 5], S21, 0xd62f105d); /* 21 */ GG (d, a, b, c, x[10], S22,

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