『壹』 C語言設計一個簡單的加密解密程序
C語言設計一個簡單的加密解密程序如下：
加密程序代碼：
#include<stdio.h>
main()
{
char c,filename[20];
FILE *fp1,*fp2;
printf("請輸入待加密的文件名:\n");
scanf("%s",filename);
fp1=fopen(filename,"r");
fp2=fopen("miwen.txt","w");
do
{
c=fgetc(fp1);
if(c>=32&&c<=126)
{
c=c-32;
c=126-c;
}
if(c!=-1)
fprintf(fp2,"%c",c);
}
while(c!=-1);
}
解密程序代碼：
#include<stdio.h>
#include<string.h>
main()
{
char c,filename[20];
char yanzhengma[20];
FILE *fp1,*fp2;
printf("請輸入待解密文件名:\n");
scanf("%s",filename);
printf("請輸入驗證碼:\n");
scanf("%s",yanzhengma);
if(strcmp(yanzhengma,"shan")==0)
{
fp1=fopen(filename,"r");
fp2=fopen("yuanwen.txt","w");
do
{
c=fgetc(fp1);
if(c>=32&&c<=126)
{
c=126-c;
c=32+c;
}
if(c!=-1)
fprintf(fp2,"%c",c);
}
while(c!=-1);
}
else
{
printf("驗證碼錯誤！請重新輸入:\n");
scanf("%s",filename);
}
}
『貳』 VB中如何編寫一個加密程序
編寫一個加密軟體，要求將源文件按位元組逐位倒排序加密法加密。
位元組逐位倒排序加密法是以比特為單位的換位加密方法，用vb實現的具體演算法是：
(1) 以二進制模式打開源文件；
(2) 從源文件第i位讀取一個位元組，假設為字母「a」，得到「a」的ascii值為65；
(3) 將65轉換成八位二進制串為「01000001」；
(4) 將「01000001」按位元組逐位倒排序得另一個八位二進制串「10000010」；
(5) 將「10000010」轉換成十進制再寫回源文件第i位置，完成一個位元組的加密；
(6) 重復(2)、(3)、(4)和(5)，直到所有位元組加密結束。
為了使程序模塊化，我們用函數過程bytetobin完成將位元組型數據轉換成二進制串（其實質就是將十進制數轉換成八位二進制串）；用函數過程bintobyte將二進制串轉換成位元組型數據（實質是將八位二進制串轉換成十進制數）：用函數過程reverse將八位二進制串逐位倒排序。具體程序如下：
function bytetobin(m as byte) as string ' 將位元組型數據轉換成八位二進制字元串
dim c$
c$ = ""
do while m <> 0
r = m mod 2
m = m \ 2
c$ = r & c$
loop
c$ = right("00000000" & c$, 8)
bytetobin = c$
end function
function reverse(m as string) as string ' 將八位二進制字元串顛倒順序
dim i%, x$
x = ""
for i = 1 to 8
x = mid(m, i, 1) & x
next i
reverse = x
end function
function bintobyte(m as string) as byte ' 將八位二進制串轉換成十進制
dim x as string * 1, y%, z%
z = 0
for i = 1 to 8
x = mid(m, i, 1)
y = x * 2 ^ (8 - i)
z = z + y
next i
bintobyte = z
end function
private sub command1_click()
dim x as byte, i%, fname$
fname = inputbox("請輸入要加密的文件名！注意加上路徑名：")
if dir(fname) = "" then
msgbox "文件不存在！"
exit sub
end if
open fname for binary as #1 ' 以二進制訪問模式打開待加密文件
for i = 1 to lof(1) ' lof函數是求文件長度的內部函數
get #1, i, x ' 取出第i個位元組
x = bintobyte(reverse(bytetobin(x))) ' 這里調用了三個自定義函數
put #1, i, x ' 將加密後的這個位元組寫回到文件原位置
next i
close
msgbox "任務完成！"
end sub
『叄』 寫一個java加密程序
publicclass_Test2
{
publicstaticvoidmain(String[]args)throwsFileNotFoundException,IOException
{
//為了便於理解,所以有的部分為了通俗寫得不夠好
Scannersc=newScanner(System.in);
Stringline=sc.nextLine();
line=encrypt(line);
System.out.println(line);
newFileOutputStream("d:\word.txt").write((line+" ").getBytes());
}
//該方法是將一行數據簡單加密(愷撒加密)
privatestaticStringencrypt(Stringline){
char[]chars=line.toCharArray();
for(inti=0;i<chars.length;i++)
{
//由ASCII碼表得知大寫字母是65-90,小寫字母是97-122
if((chars[i]>=65&&chars[i]<90)||(chars[i]>=97&&chars[i]<122))
{
chars[i]=(char)(chars[i]+1);
}
if(chars[i]==90)
{
chars[i]='a';
}
if(chars[i]==122)
{
chars[i]='A';
}
}
returnnewString(chars);
}
}
『肆』 c語言編寫加密程序
#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]);
}
『伍』 高分求用C++寫一道完整的加密程序
#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
#include <string.h>
void encrypt(char *in, char *pwd, char *out)
{
FILE *infile, *outfile;
char ch;
int i=0, PwdLen=0;
if ((infile=fopen(in, "rb"))==NULL)
{
printf("無法打開 %s\n"亂姿, in);
getch();
exit(1);
}
if ((outfile=fopen(out, "wb"))==NULL)
{
printf("無法打開或創建 %s\n", out);
getch();
exit(1);
}
while (pwd[PwdLen++]); //這里用來應付密碼為空擾戚的情況
PwdLen--;
ch=fgetc(infile);
while (!feof(infile))
{
fputc(ch^pwd[i>=PwdLen?i=0:i++], outfile);
ch=fgetc(infile);
}
fclose(infile);
fclose(outfile);
}
int main(int argc, char *argv[])
{
char in[256];
char out[256];
char pwd[256];
if (argc!=4)
{
printf("要加／解緩陪陵密的文件路徑和文件名：\n");
gets(in);
printf("密碼：\n");
gets(pwd);
printf("輸出路徑和文件名：\n");
gets(out);
encrypt(in, pwd, out);
printf("\n\n任務完成!\n\n");
getch();
}
else encrypt(argv[1], argv[2], argv[3]);
return 0;
『陸』 C#里寫出一段程序對字元串進行加密。
string pText = "fdaei"; //明文
char[] pTextChar = pText.ToCharArray();
string cText = (Convert.ToChar(pTextChar[pTextChar.Length - 1] + 3)).ToString(); /閉坦/最後一個字元世態棗置搜拆於第一位，同時ASCII + 3
for (int i = 0; i < pTextChar.Length - 1; i++)
{
cText += (Convert.ToChar(pTextChar[i] + 3)).ToString(); //除最後一個字元外所有字元後移一位，同時ASCII + 3
}
MessageBox.Show(cText); //密文
『柒』 簡單的C語言加密程序
#include<stdio.h>
#include<stdlib.h>
main()
{
int key;
char ch;
printf("\n請輸入密鑰：");
scanf("%d",&key);
printf("得到對應明文如下：");
while((ch=getchar())!='\r')
(ch+key)>122?putchar(ch-122+33+key):
((ch+key)<33?putchar(ch+122+key):putchar(ch+key));
}
輸入輸出如下：
請輸入密鑰：20addse
得到對應明文如下：uxx.y
你先輸入一個任意的整數，如20，然後在鍵盤上輸入一段任意的字元如addse
按回車鍵結束，就會得到結果 如：uxx.y
下面是另一組輸入輸出：
請輸入密鑰：35asjRYIRER!@#$^^*&
得到對應明文如下：+=4u#luhuDcFG((MI-
具體是如何加密，你應該能看懂，就是用一個三目運算符 ? :控制。