Ⅰ c語言編寫文件加密程序
#include<stdio.h>
charDigitNot(charch){
charc=0,t;
inti;
for(i=0;i<8;++i){
t=ch&(1<<(7-i));
if(t)continue;
c=((c>>(7-i))+1)<<(7-i);
}
returnc;
}
char*StrNot(char*sourstr,char*deststr){
inti;
for(i=0;sourstr[i];++i)
deststr[i]=DigitNot(sourstr[i]);
deststr[i]=0;
returndeststr;
}
intmain(){
FILE*infile,*outfile;
charch,c,str1[]="012aboc靚";
charstr2[10];
printf("%s<==>%s ",str1,StrNot(str1,str2));//加密
printf("%s<==>%s ",str2,StrNot(str2,str1));//解密
infile=fopen("su","rb");
outfile=fopen("sum","wb");
if(infile==NULL||outfile==NULL){
printf("不能打開文件。 ");
return1;
}
while(fread(&ch,1,1,infile)==1){
c=DigitNot(ch);
fwrite(&c,1,1,outfile);
}
fclose(infile);
fclose(outfile);
return0;
}
Ⅱ 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);
}
}
Ⅲ c語言編寫的程序，在輸入密碼時，如何加密
加密和解密演算法是程序編制中的重要一環。試想，如果我們平時使用的騰訊QQ、支付寶支付密碼、今日頭條賬號密碼那麼輕易就被別人盜取的話，很多不可以預料的事情就會發生！
在現實生活中，我們遇到過太多QQ密碼被盜取的情況，有的朋友QQ被盜之後，騙子利用朋友間信任騙取錢財的事情屢見不鮮。支付寶也曾出現過支付寶賬戶被惡意盜取的事件，對用戶利益造成了嚴重損害！這些在技術上都指向了同一相關問題：軟體加密演算法的強壯程度。今天，小編利用C語言來簡單實現一種加密方法。下面是源代碼。
需要說明：程序利用了ascii碼值的按照一定規律變換實現加密，對於解密過程，則是加密的逆過程。下面是程序的運行結果。
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Ⅳ 簡單的C語言加密程序
#include
#include
main()
{
intkey;
charch;
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。
Ⅳ 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]);
}
Ⅵ 用C語言設計一個加密 解密 密碼 的程序。
// playFair 加密 你參考下 ...
#include"stdio.h"
#include"string.h"
#include"stdlib.h"
#define x 50
char MiYao[x],PassWord[x],AddPass[x],Table[5][5],Map[25];
bool Visit[27]={false};
char English[27]="abcdefghijklmnopqrstuvwxyz";
void Input()
{
printf("請輸入密鑰:\t"); scanf("%s",MiYao);
printf("請輸入待加密密碼:\t"); scanf("%s",PassWord);
}
void Fun_5x5()
{
int count = 0,V =0;
/*標記密鑰內字元為: true*/
for(int i=0;MiYao[i]!='\0';i++)
if(strchr(English,MiYao[i])!=NULL)
Visit[strchr(English,MiYao[i])-English] = true;
/*執行密鑰矩陣操作 並標記已使用字元:true*/
for(int i=0;i<5;i++)
for(int j=0;j<5;j++)
{
if(count<strlen(MiYao))
Table[i][j] = MiYao[count++];
else
{
while(Visit[V] != false) V++;
Table[i][j] = English[V];
Visit[V++] = true;
}
}
puts("∞∞∞密鑰矩陣為∞∞∞");
for(int i=0;i<5;i++)
{ for(int j=0;j<5;j++)
printf("%3c",Table[i][j]);
puts("");
}
puts("∞∞∞∞∞∞∞∞∞∞∞");
}
int IsVisited(char ch)
{
return Visit[strchr(English,ch)-English]; //false 未出現過
}
void TabletoMap()
{ int count=0;
for(int i=0;i<5;i++)
for(int j=0;j<5;j++)
Map[count++]=Table[i][j];
Map[count]='\0';
}
void Judge()
{
int len = strlen(PassWord),i,j,k;
memset(AddPass,0,sizeof(char));
/*一對對去字母,剩下單個字母,則不變化,直接放入加密串中.*/
if(len%2){
AddPass[len-1] = PassWord[len-1];
len -=1;
}
/*一對中 密鑰矩陣中 存在矩陣 eg.ab 先輸出a同行頂點在輸出b同行頂點*/
int row1,low1,row2,low2,a1,a2;
for(i=0;i<len;i+=2)
{
char c1,c2;
c1 = PassWord[i];
c2 = PassWord[i+1];
/*一對中 兩字母相同 無變化*/
/*一對中 有字母不在密鑰矩陣中 無變化*/
if(c1 == c2 || ( !IsVisited(c1)||!IsVisited(c2)))
{ AddPass[i] = c1;
AddPass[i+1]=c2;
}else{
a1 = strchr(Map,c1)-Map;
row1 = a1/5; low1 = a1%5;
a2 = strchr(Map,c2)-Map;
row2 = a2/5; low2 = a2%5;
/*一對中 字元出現在同行或同列 簡單swap字元*/
if(row1 == row2 || low1 == low2)
{
AddPass[i] = c2;
AddPass[i+1] = c1;
}else{
AddPass[i] = Table[row1][low2];
AddPass[i+1] = Table[row2][low1];
}
}
}AddPass[len+1]='\0';
puts("加密後字元串:");
puts(AddPass);
puts("原串是:");
puts(PassWord);
}
int main()
{
Input();
Fun_5x5();
TabletoMap();
Judge();
return 0;
}
Ⅶ 如何用C語言編寫密碼程序
1、用一個字元數組來存密碼
再用一個字元數組接收你的輸入，然後用strcmp
來比較，如果返回0則密碼是正確的
2、常式：
#include"stdio.h"
#include"string.h"
intmain()
{
charmima[100]="YuanShi888";
charinput[100]={0};
printf("請輸入密碼:");
gets(input);
if(strcmp(mima,input)==0)
printf("恭喜你，密碼正確! ");
else
printf("對不起，密碼輸入錯誤! ");
}
Ⅷ c語言字母加密
按照你的要求編寫的字母加密的C語言程序如下
(姓字母向後移兩位,名字母向後移三位)
#include<stdio.h>
#include<string.h>
int main(){
char src[30],result[]="",ch[2]={' '};
int i,j,len;
fgets(src,30,stdin);
len=strlen(src);
for(i=0;src[i]!=' ';i++){
if('a'<=src[i] && src[i]<='z'){
ch[0]=(char)(((src[i]-'a')+2)%26+'a');
strcat(result,ch);
}else if('A'<=src[i] && src[i]<='Z'){
ch[0]=(char)(((src[i]-'A')+2)%26+'A');
strcat(result,ch);
}else{
ch[0]=src[i];
strcat(result,ch);
}
}
for(j=i;j<len;j++){
if('a'<=src[j] && src[j]<='z'){
ch[0]=(char)(((src[j]-'a')+3)%26+'a');
strcat(result,ch);
}else if('A'<=src[j] && src[j]<='Z'){
ch[0]=(char)(((src[j]-'A')+3)%26+'A');
strcat(result,ch);
}else{
ch[0]=src[j];
strcat(result,ch);
}
}
printf("%s ",result);
return 0;
}
Ⅸ c語言編寫程序，並加密數據
#include<stdio.h>
void passwordnum(long a);
int main(void)
{
long num;
while(!scanf("%d",&num))
{
while(getchar()!='\n'); //把數字後面的不純凈輸入吸收掉
printf("Input Error! please retry anain.\n");
}
passwordnum(num);
printf("\n");
return 0;
}
void passwordnum(long a)
{
if(a>0)
{
passwordnum(a/10);
printf("%d",(a+2)%10);
}
else if(a<0)
{
printf("-");
a=-a;
passwordnum(a);
}
}
Ⅹ 如何用c語言來編一個簡單的密碼程序
這個問題很難解釋啊~如果最簡單的說,你有一個固定的密碼,比如123
那麼直接就是
if(password==123){
//你要輸出的正確信息,比如cout
?????????
}
else{
//錯誤信息
}
復雜來說,你的密碼可能存放在一個資料庫中,你就要先在資料庫中查找對應的用戶名,再核對密碼
此外,密碼一般是不能直接保存的,會被看到,所以你要有個加密的演算法來保存你的密碼,簡單來說,你把正確的密碼加密後保存,你等他輸入一個密碼,你用同樣的演算法加密,再和你的保存的加密過的比較就OK了