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des演算法c語言

發布時間: 2021-03-06 21:36:14

1. 如何用c語言編程實現DES加密演算法

http://www.vckbase.com/sourcecode/algorithms/des.zip

2. 有關於DES演算法的C語言源代碼嘛急需，能直接運行的

//////////////////////////////////////////////////////////////////////////
/*
Provided by 王俊川, Northeastern University (www.neu.e.cn)
Email: [email protected]
This proct is free for use.
*/
//////////////////////////////////////////////////////////////////////////

#include "memory.h"
enum {ENCRYPT,DECRYPT};

//////////////////////////////////////////////////////////////////////////

// initial permutation IP
const static char IP_Table[64] = {
58, 50, 42, 34, 26, 18, 10, 2, 60, 52, 44, 36, 28, 20, 12, 4,
62, 54, 46, 38, 30, 22, 14, 6, 64, 56, 48, 40, 32, 24, 16, 8,
57, 49, 41, 33, 25, 17, 9, 1, 59, 51, 43, 35, 27, 19, 11, 3,
61, 53, 45, 37, 29, 21, 13, 5, 63, 55, 47, 39, 31, 23, 15, 7
};
// final permutation IP^-1
const static char IPR_Table[64] = {
40, 8, 48, 16, 56, 24, 64, 32, 39, 7, 47, 15, 55, 23, 63, 31,
38, 6, 46, 14, 54, 22, 62, 30, 37, 5, 45, 13, 53, 21, 61, 29,
36, 4, 44, 12, 52, 20, 60, 28, 35, 3, 43, 11, 51, 19, 59, 27,
34, 2, 42, 10, 50, 18, 58, 26, 33, 1, 41, 9, 49, 17, 57, 25
};
// expansion operation matrix
static const char E_Table[48] = {
32, 1, 2, 3, 4, 5, 4, 5, 6, 7, 8, 9,
8, 9, 10, 11, 12, 13, 12, 13, 14, 15, 16, 17,
16, 17, 18, 19, 20, 21, 20, 21, 22, 23, 24, 25,
24, 25, 26, 27, 28, 29, 28, 29, 30, 31, 32, 1
};
// 32-bit permutation function P used on the output of the S-boxes
const static char P_Table[32] = {
16, 7, 20, 21, 29, 12, 28, 17, 1, 15, 23, 26, 5, 18, 31, 10,
2, 8, 24, 14, 32, 27, 3, 9, 19, 13, 30, 6, 22, 11, 4, 25
};
// permuted choice table (key)
const static char PC1_Table[56] = {
57, 49, 41, 33, 25, 17, 9, 1, 58, 50, 42, 34, 26, 18,
10, 2, 59, 51, 43, 35, 27, 19, 11, 3, 60, 52, 44, 36,
63, 55, 47, 39, 31, 23, 15, 7, 62, 54, 46, 38, 30, 22,
14, 6, 61, 53, 45, 37, 29, 21, 13, 5, 28, 20, 12, 4
};
// permuted choice key (table)
const static char PC2_Table[48] = {
14, 17, 11, 24, 1, 5, 3, 28, 15, 6, 21, 10,
23, 19, 12, 4, 26, 8, 16, 7, 27, 20, 13, 2,
41, 52, 31, 37, 47, 55, 30, 40, 51, 45, 33, 48,
44, 49, 39, 56, 34, 53, 46, 42, 50, 36, 29, 32
};
// number left rotations of pc1
const static char LOOP_Table[16] = {
1,1,2,2,2,2,2,2,1,2,2,2,2,2,2,1
};
// The (in)famous S-boxes
const static char S_Box[8][4][16] = {
// S1
14, 4, 13, 1, 2, 15, 11, 8, 3, 10, 6, 12, 5, 9, 0, 7,
0, 15, 7, 4, 14, 2, 13, 1, 10, 6, 12, 11, 9, 5, 3, 8,
4, 1, 14, 8, 13, 6, 2, 11, 15, 12, 9, 7, 3, 10, 5, 0,
15, 12, 8, 2, 4, 9, 1, 7, 5, 11, 3, 14, 10, 0, 6, 13,
// S2
15, 1, 8, 14, 6, 11, 3, 4, 9, 7, 2, 13, 12, 0, 5, 10,
3, 13, 4, 7, 15, 2, 8, 14, 12, 0, 1, 10, 6, 9, 11, 5,
0, 14, 7, 11, 10, 4, 13, 1, 5, 8, 12, 6, 9, 3, 2, 15,
13, 8, 10, 1, 3, 15, 4, 2, 11, 6, 7, 12, 0, 5, 14, 9,
// S3
10, 0, 9, 14, 6, 3, 15, 5, 1, 13, 12, 7, 11, 4, 2, 8,
13, 7, 0, 9, 3, 4, 6, 10, 2, 8, 5, 14, 12, 11, 15, 1,
13, 6, 4, 9, 8, 15, 3, 0, 11, 1, 2, 12, 5, 10, 14, 7,
1, 10, 13, 0, 6, 9, 8, 7, 4, 15, 14, 3, 11, 5, 2, 12,
// S4
7, 13, 14, 3, 0, 6, 9, 10, 1, 2, 8, 5, 11, 12, 4, 15,
13, 8, 11, 5, 6, 15, 0, 3, 4, 7, 2, 12, 1, 10, 14, 9,
10, 6, 9, 0, 12, 11, 7, 13, 15, 1, 3, 14, 5, 2, 8, 4,
3, 15, 0, 6, 10, 1, 13, 8, 9, 4, 5, 11, 12, 7, 2, 14,
// S5
2, 12, 4, 1, 7, 10, 11, 6, 8, 5, 3, 15, 13, 0, 14, 9,
14, 11, 2, 12, 4, 7, 13, 1, 5, 0, 15, 10, 3, 9, 8, 6,
4, 2, 1, 11, 10, 13, 7, 8, 15, 9, 12, 5, 6, 3, 0, 14,
11, 8, 12, 7, 1, 14, 2, 13, 6, 15, 0, 9, 10, 4, 5, 3,
// S6
12, 1, 10, 15, 9, 2, 6, 8, 0, 13, 3, 4, 14, 7, 5, 11,
10, 15, 4, 2, 7, 12, 9, 5, 6, 1, 13, 14, 0, 11, 3, 8,
9, 14, 15, 5, 2, 8, 12, 3, 7, 0, 4, 10, 1, 13, 11, 6,
4, 3, 2, 12, 9, 5, 15, 10, 11, 14, 1, 7, 6, 0, 8, 13,
// S7
4, 11, 2, 14, 15, 0, 8, 13, 3, 12, 9, 7, 5, 10, 6, 1,
13, 0, 11, 7, 4, 9, 1, 10, 14, 3, 5, 12, 2, 15, 8, 6,
1, 4, 11, 13, 12, 3, 7, 14, 10, 15, 6, 8, 0, 5, 9, 2,
6, 11, 13, 8, 1, 4, 10, 7, 9, 5, 0, 15, 14, 2, 3, 12,
// S8
13, 2, 8, 4, 6, 15, 11, 1, 10, 9, 3, 14, 5, 0, 12, 7,
1, 15, 13, 8, 10, 3, 7, 4, 12, 5, 6, 11, 0, 14, 9, 2,
7, 11, 4, 1, 9, 12, 14, 2, 0, 6, 10, 13, 15, 3, 5, 8,
2, 1, 14, 7, 4, 10, 8, 13, 15, 12, 9, 0, 3, 5, 6, 11
};

//////////////////////////////////////////////////////////////////////////

typedef bool (*PSubKey)[16][48];

//////////////////////////////////////////////////////////////////////////

static void DES(char Out[8], char In[8], const PSubKey pSubKey, bool Type);//標准DES加/解密
static void SetKey(const char* Key, int len);// 設置密鑰
static void SetSubKey(PSubKey pSubKey, const char Key[8]);// 設置子密鑰
static void F_func(bool In[32], const bool Ki[48]);// f 函數
static void S_func(bool Out[32], const bool In[48]);// S 盒代替
static void Transform(bool *Out, bool *In, const char *Table, int len);// 變換
static void Xor(bool *InA, const bool *InB, int len);// 異或
static void RotateL(bool *In, int len, int loop);// 循環左移
static void ByteToBit(bool *Out, const char *In, int bits);// 位元組組轉換成位組
static void BitToByte(char *Out, const bool *In, int bits);// 位組轉換成位元組組

//////////////////////////////////////////////////////////////////////////

static bool SubKey[2][16][48];// 16圈子密鑰
static bool Is3DES;// 3次DES標志
static char Tmp[256], deskey[16];

//////////////////////////////////////////////////////////////////////////

//////////////////////////////////////////////////////////////////////////
// Code starts from Line 130
//////////////////////////////////////////////////////////////////////////
bool Des_Go(char *Out, char *In, long datalen, const char *Key, int keylen, bool Type)
{
if( !( Out && In && Key && (datalen=(datalen+7)&0xfffffff8) ) )
return false;
SetKey(Key, keylen);
if( !Is3DES ) { // 1次DES
for(long i=0,j=datalen>>3; i<j; ++i,Out+=8,In+=8)
DES(Out, In, &SubKey[0], Type);
} else{ // 3次DES 加密:加(key0)-解(key1)-加(key0) 解密::解(key0)-加(key1)-解(key0)
for(long i=0,j=datalen>>3; i<j; ++i,Out+=8,In+=8) {
DES(Out, In, &SubKey[0], Type);
DES(Out, Out, &SubKey[1], !Type);
DES(Out, Out, &SubKey[0], Type);
}
}
return true;
}
void SetKey(const char* Key, int len)
{
memset(deskey, 0, 16);
memcpy(deskey, Key, len>16?16:len);
SetSubKey(&SubKey[0], &deskey[0]);
Is3DES = len>8 ? (SetSubKey(&SubKey[1], &deskey[8]), true) : false;
}
void DES(char Out[8], char In[8], const PSubKey pSubKey, bool Type)
{
static bool M[64], tmp[32], *Li=&M[0], *Ri=&M[32];
ByteToBit(M, In, 64);
Transform(M, M, IP_Table, 64);
if( Type == ENCRYPT ){
for(int i=0; i<16; ++i) {
memcpy(tmp, Ri, 32);
F_func(Ri, (*pSubKey)[i]);
Xor(Ri, Li, 32);
memcpy(Li, tmp, 32);
}
}else{
for(int i=15; i>=0; --i) {
memcpy(tmp, Li, 32);
F_func(Li, (*pSubKey)[i]);
Xor(Li, Ri, 32);
memcpy(Ri, tmp, 32);
}
}
Transform(M, M, IPR_Table, 64);
BitToByte(Out, M, 64);
}
void SetSubKey(PSubKey pSubKey, const char Key[8])
{
static bool K[64], *KL=&K[0], *KR=&K[28];
ByteToBit(K, Key, 64);
Transform(K, K, PC1_Table, 56);
for(int i=0; i<16; ++i) {
RotateL(KL, 28, LOOP_Table[i]);
RotateL(KR, 28, LOOP_Table[i]);
Transform((*pSubKey)[i], K, PC2_Table, 48);
}
}
void F_func(bool In[32], const bool Ki[48])
{
static bool MR[48];
Transform(MR, In, E_Table, 48);
Xor(MR, Ki, 48);
S_func(In, MR);
Transform(In, In, P_Table, 32);
}
void S_func(bool Out[32], const bool In[48])
{
for(char i=0,j,k; i<8; ++i,In+=6,Out+=4) {
j = (In[0]<<1) + In[5];
k = (In[1]<<3) + (In[2]<<2) + (In[3]<<1) + In[4];
ByteToBit(Out, &S_Box[i][j][k], 4);
}
}
void Transform(bool *Out, bool *In, const char *Table, int len)
{
for(int i=0; i<len; ++i)
Tmp[i] = In[ Table[i]-1 ];
memcpy(Out, Tmp, len);
}
void Xor(bool *InA, const bool *InB, int len)
{
for(int i=0; i<len; ++i)
InA[i] ^= InB[i];
}
void RotateL(bool *In, int len, int loop)
{
memcpy(Tmp, In, loop);
memcpy(In, In+loop, len-loop);
memcpy(In+len-loop, Tmp, loop);
}
void ByteToBit(bool *Out, const char *In, int bits)
{
for(int i=0; i<bits; ++i)
Out[i] = (In[i>>3]>>(i&7)) & 1;
}
void BitToByte(char *Out, const bool *In, int bits)
{
memset(Out, 0, bits>>3);
for(int i=0; i<bits; ++i)
Out[i>>3] |= In[i]<<(i&7);
}
//////////////////////////////////////////////////////////////////////////
// Code ends at Line 231
//////////////////////////////////////////////////////////////////////////

3. 用C語言實現DES演算法

LZ。。。幻想啥呢。一般大家就算做過，也不會在這里因為80分就把演算法給你。

再說了，沒做過的，更不可能因為80分給你做一個。。。

還是去圖書館吧...這個。。。

4. 用C語言實現DES，RSA演算法，給100分

void initialize()
{ int i;
char c;
for (i = 11, c = 'A'; c <= 'Z'; c ++, i ++)
{ change[c] = i;
antichange[i] = c;
}
}
void changetonum(strtype str)
{ int l = strlen(str), i;
len = 0;
memset(nume, 0, sizeof(nume));
for (i = 0; i < l; i ++)
{ nume[len] = nume[len] * 100 + change[str[i]];
if (i % 2 == 1) len ++;
}
if (i % 2 != 0) len ++;
}
long binamod(long numb, long k)
{ if (k == 0) return 1;
long curr = binamod (numb, k / 2);
if (k % 2 == 0)
return curr * curr % MM;
else return (curr * curr) % MM * numb % MM;
}
long encode(long numb)
{ return binamod(numb, KK);
}
long decode(long numb)
{ return binamod(numb, PP);
}
main()
{ strtype str;
int i, a1, a2;
long curr;
initialize();
puts("Input 'Y' if encoding, otherwise input 'N':");
gets(str);
if (str[0] == 'Y')
{ gets(str);
changetonum(str);
printf("encoded: ");
for (i = 0; i < len; i ++)
{ if (i) putchar('-');
printf(" %ld ", encode(nume[i]));
}
putchar('\n');
}
else
{ scanf("%d", &len);
for (i = 0; i < len; i ++)
{ scanf("%ld", &curr);
curr = decode(curr);
a1 = curr / 100;
a2 = curr % 100;
printf("decoded: ");
if (a1 != 0) putchar(antichange[a1]);
if (a2 != 0) putchar(antichange[a2]);
}
putchar('\n');
}
putchar('\n');
system("PAUSE");
return 0;
}

5. 請問如何用C語言（或C++）實現高效率的DES演算法

開源的復有很制多啦，比如這個
http://www.ka9q.net/code/des/
匯編版和C語言版都有

6. des加密演算法(c/c++)

網上大把的源碼，別人幫你找來了，你自己就不能動一下手？

最鄙視這些要求能直接編譯運行，再要求對每一行有詳細注釋的人。

就算你懸賞200分又怎麼樣？網路的懸賞分數有什麼用？

7. 用C語言來實現DES加密演算法（很急）兩天內

DES雖然不難但是挺繁復的，代碼如下，關鍵點都有英文解釋，仔細看。各個函數的功能都可以從函數名看出來。

#include "pch.h"
#include "misc.h"
#include "des.h"

NAMESPACE_BEGIN(CryptoPP)

/* Tables defined in the Data Encryption Standard documents
* Three of these tables, the initial permutation, the final
* permutation and the expansion operator, are regular enough that
* for speed, we hard-code them. They're here for reference only.
* Also, the S and P boxes are used by a separate program, gensp.c,
* to build the combined SP box, Spbox[]. They're also here just
* for reference.
*/
#ifdef notdef
/* initial permutation IP */
static byte ip[] = {
58, 50, 42, 34, 26, 18, 10, 2,
60, 52, 44, 36, 28, 20, 12, 4,
62, 54, 46, 38, 30, 22, 14, 6,
64, 56, 48, 40, 32, 24, 16, 8,
57, 49, 41, 33, 25, 17, 9, 1,
59, 51, 43, 35, 27, 19, 11, 3,
61, 53, 45, 37, 29, 21, 13, 5,
63, 55, 47, 39, 31, 23, 15, 7
};

/* final permutation IP^-1 */
static byte fp[] = {
40, 8, 48, 16, 56, 24, 64, 32,
39, 7, 47, 15, 55, 23, 63, 31,
38, 6, 46, 14, 54, 22, 62, 30,
37, 5, 45, 13, 53, 21, 61, 29,
36, 4, 44, 12, 52, 20, 60, 28,
35, 3, 43, 11, 51, 19, 59, 27,
34, 2, 42, 10, 50, 18, 58, 26,
33, 1, 41, 9, 49, 17, 57, 25
};
/* expansion operation matrix */
static byte ei[] = {
32, 1, 2, 3, 4, 5,
4, 5, 6, 7, 8, 9,
8, 9, 10, 11, 12, 13,
12, 13, 14, 15, 16, 17,
16, 17, 18, 19, 20, 21,
20, 21, 22, 23, 24, 25,
24, 25, 26, 27, 28, 29,
28, 29, 30, 31, 32, 1
};
/* The (in)famous S-boxes */
static byte sbox[8][64] = {
/* S1 */
14, 4, 13, 1, 2, 15, 11, 8, 3, 10, 6, 12, 5, 9, 0, 7,
0, 15, 7, 4, 14, 2, 13, 1, 10, 6, 12, 11, 9, 5, 3, 8,
4, 1, 14, 8, 13, 6, 2, 11, 15, 12, 9, 7, 3, 10, 5, 0,
15, 12, 8, 2, 4, 9, 1, 7, 5, 11, 3, 14, 10, 0, 6, 13,

/* S2 */
15, 1, 8, 14, 6, 11, 3, 4, 9, 7, 2, 13, 12, 0, 5, 10,
3, 13, 4, 7, 15, 2, 8, 14, 12, 0, 1, 10, 6, 9, 11, 5,
0, 14, 7, 11, 10, 4, 13, 1, 5, 8, 12, 6, 9, 3, 2, 15,
13, 8, 10, 1, 3, 15, 4, 2, 11, 6, 7, 12, 0, 5, 14, 9,

/* S3 */
10, 0, 9, 14, 6, 3, 15, 5, 1, 13, 12, 7, 11, 4, 2, 8,
13, 7, 0, 9, 3, 4, 6, 10, 2, 8, 5, 14, 12, 11, 15, 1,
13, 6, 4, 9, 8, 15, 3, 0, 11, 1, 2, 12, 5, 10, 14, 7,
1, 10, 13, 0, 6, 9, 8, 7, 4, 15, 14, 3, 11, 5, 2, 12,

/* S4 */
7, 13, 14, 3, 0, 6, 9, 10, 1, 2, 8, 5, 11, 12, 4, 15,
13, 8, 11, 5, 6, 15, 0, 3, 4, 7, 2, 12, 1, 10, 14, 9,
10, 6, 9, 0, 12, 11, 7, 13, 15, 1, 3, 14, 5, 2, 8, 4,
3, 15, 0, 6, 10, 1, 13, 8, 9, 4, 5, 11, 12, 7, 2, 14,

/* S5 */
2, 12, 4, 1, 7, 10, 11, 6, 8, 5, 3, 15, 13, 0, 14, 9,
14, 11, 2, 12, 4, 7, 13, 1, 5, 0, 15, 10, 3, 9, 8, 6,
4, 2, 1, 11, 10, 13, 7, 8, 15, 9, 12, 5, 6, 3, 0, 14,
11, 8, 12, 7, 1, 14, 2, 13, 6, 15, 0, 9, 10, 4, 5, 3,

/* S6 */
12, 1, 10, 15, 9, 2, 6, 8, 0, 13, 3, 4, 14, 7, 5, 11,
10, 15, 4, 2, 7, 12, 9, 5, 6, 1, 13, 14, 0, 11, 3, 8,
9, 14, 15, 5, 2, 8, 12, 3, 7, 0, 4, 10, 1, 13, 11, 6,
4, 3, 2, 12, 9, 5, 15, 10, 11, 14, 1, 7, 6, 0, 8, 13,

/* S7 */
4, 11, 2, 14, 15, 0, 8, 13, 3, 12, 9, 7, 5, 10, 6, 1,
13, 0, 11, 7, 4, 9, 1, 10, 14, 3, 5, 12, 2, 15, 8, 6,
1, 4, 11, 13, 12, 3, 7, 14, 10, 15, 6, 8, 0, 5, 9, 2,
6, 11, 13, 8, 1, 4, 10, 7, 9, 5, 0, 15, 14, 2, 3, 12,

/* S8 */
13, 2, 8, 4, 6, 15, 11, 1, 10, 9, 3, 14, 5, 0, 12, 7,
1, 15, 13, 8, 10, 3, 7, 4, 12, 5, 6, 11, 0, 14, 9, 2,
7, 11, 4, 1, 9, 12, 14, 2, 0, 6, 10, 13, 15, 3, 5, 8,
2, 1, 14, 7, 4, 10, 8, 13, 15, 12, 9, 0, 3, 5, 6, 11
};

/* 32-bit permutation function P used on the output of the S-boxes */
static byte p32i[] = {
16, 7, 20, 21,
29, 12, 28, 17,
1, 15, 23, 26,
5, 18, 31, 10,
2, 8, 24, 14,
32, 27, 3, 9,
19, 13, 30, 6,
22, 11, 4, 25
};
#endif

/* permuted choice table (key) */
static const byte pc1[] = {
57, 49, 41, 33, 25, 17, 9,
1, 58, 50, 42, 34, 26, 18,
10, 2, 59, 51, 43, 35, 27,
19, 11, 3, 60, 52, 44, 36,

63, 55, 47, 39, 31, 23, 15,
7, 62, 54, 46, 38, 30, 22,
14, 6, 61, 53, 45, 37, 29,
21, 13, 5, 28, 20, 12, 4
};

/* number left rotations of pc1 */
static const byte totrot[] = {
1,2,4,6,8,10,12,14,15,17,19,21,23,25,27,28
};

/* permuted choice key (table) */
static const byte pc2[] = {
14, 17, 11, 24, 1, 5,
3, 28, 15, 6, 21, 10,
23, 19, 12, 4, 26, 8,
16, 7, 27, 20, 13, 2,
41, 52, 31, 37, 47, 55,
30, 40, 51, 45, 33, 48,
44, 49, 39, 56, 34, 53,
46, 42, 50, 36, 29, 32
};

/* End of DES-defined tables */

/* bit 0 is left-most in byte */
static const int bytebit[] = {
0200,0100,040,020,010,04,02,01
};

/* Set key (initialize key schele array) */
DES::DES(const byte *key, CipherDir dir)
: k(32)
{
SecByteBlock buffer(56+56+8);
byte *const pc1m=buffer; /* place to modify pc1 into */
byte *const pcr=pc1m+56; /* place to rotate pc1 into */
byte *const ks=pcr+56;
register int i,j,l;
int m;

for (j=0; j<56; j++) { /* convert pc1 to bits of key */
l=pc1[j]-1; /* integer bit location */
m = l & 07; /* find bit */
pc1m[j]=(key[l>>3] & /* find which key byte l is in */
bytebit[m]) /* and which bit of that byte */
? 1 : 0; /* and store 1-bit result */
}
for (i=0; i<16; i++) { /* key chunk for each iteration */
memset(ks,0,8); /* Clear key schele */
for (j=0; j<56; j++) /* rotate pc1 the right amount */
pcr[j] = pc1m[(l=j+totrot[i])<(j<28? 28 : 56) ? l: l-28];
/* rotate left and right halves independently */
for (j=0; j<48; j++){ /* select bits indivially */
/* check bit that goes to ks[j] */
if (pcr[pc2[j]-1]){
/* mask it in if it's there */
l= j % 6;
ks[j/6] |= bytebit[l] >> 2;
}
}
/* Now convert to odd/even interleaved form for use in F */
k[2*i] = ((word32)ks[0] << 24)
| ((word32)ks[2] << 16)
| ((word32)ks[4] << 8)
| ((word32)ks[6]);
k[2*i+1] = ((word32)ks[1] << 24)
| ((word32)ks[3] << 16)
| ((word32)ks[5] << 8)
| ((word32)ks[7]);
}

if (dir==DECRYPTION) // reverse key schele order
for (i=0; i<16; i+=2)
{
std::swap(k[i], k[32-2-i]);
std::swap(k[i+1], k[32-1-i]);
}
}
/* End of C code common to both versions */

/* C code only in portable version */

// Richard Outerbridge's initial permutation algorithm
/*
inline void IPERM(word32 &left, word32 &right)
{
word32 work;

work = ((left >> 4) ^ right) & 0x0f0f0f0f;
right ^= work;
left ^= work << 4;
work = ((left >> 16) ^ right) & 0xffff;
right ^= work;
left ^= work << 16;
work = ((right >> 2) ^ left) & 0x33333333;
left ^= work;
right ^= (work << 2);
work = ((right >> 8) ^ left) & 0xff00ff;
left ^= work;
right ^= (work << 8);
right = rotl(right, 1);
work = (left ^ right) & 0xaaaaaaaa;
left ^= work;
right ^= work;
left = rotl(left, 1);
}
inline void FPERM(word32 &left, word32 &right)
{
word32 work;

right = rotr(right, 1);
work = (left ^ right) & 0xaaaaaaaa;
left ^= work;
right ^= work;
left = rotr(left, 1);
work = ((left >> 8) ^ right) & 0xff00ff;
right ^= work;
left ^= work << 8;
work = ((left >> 2) ^ right) & 0x33333333;
right ^= work;
left ^= work << 2;
work = ((right >> 16) ^ left) & 0xffff;
left ^= work;
right ^= work << 16;
work = ((right >> 4) ^ left) & 0x0f0f0f0f;
left ^= work;
right ^= work << 4;
}
*/

// Wei Dai's modification to Richard Outerbridge's initial permutation
// algorithm, this one is faster if you have access to rotate instructions
// (like in MSVC)
inline void IPERM(word32 &left, word32 &right)
{
word32 work;

right = rotl(right, 4U);
work = (left ^ right) & 0xf0f0f0f0;
left ^= work;
right = rotr(right^work, 20U);
work = (left ^ right) & 0xffff0000;
left ^= work;
right = rotr(right^work, 18U);
work = (left ^ right) & 0x33333333;
left ^= work;
right = rotr(right^work, 6U);
work = (left ^ right) & 0x00ff00ff;
left ^= work;
right = rotl(right^work, 9U);
work = (left ^ right) & 0xaaaaaaaa;
left = rotl(left^work, 1U);
right ^= work;
}

inline void FPERM(word32 &left, word32 &right)
{
word32 work;

right = rotr(right, 1U);
work = (left ^ right) & 0xaaaaaaaa;
right ^= work;
left = rotr(left^work, 9U);
work = (left ^ right) & 0x00ff00ff;
right ^= work;
left = rotl(left^work, 6U);
work = (left ^ right) & 0x33333333;
right ^= work;
left = rotl(left^work, 18U);
work = (left ^ right) & 0xffff0000;
right ^= work;
left = rotl(left^work, 20U);
work = (left ^ right) & 0xf0f0f0f0;
right ^= work;
left = rotr(left^work, 4U);
}

// Encrypt or decrypt a block of data in ECB mode
void DES::ProcessBlock(const byte *inBlock, byte * outBlock) const
{
word32 l,r,work;

#ifdef IS_LITTLE_ENDIAN
l = byteReverse(*(word32 *)inBlock);
r = byteReverse(*(word32 *)(inBlock+4));
#else
l = *(word32 *)inBlock;
r = *(word32 *)(inBlock+4);
#endif

IPERM(l,r);

const word32 *kptr=k;

for (unsigned i=0; i<8; i++)
{
work = rotr(r, 4U) ^ kptr[4*i+0];
l ^= Spbox[6][(work) & 0x3f]
^ Spbox[4][(work >> 8) & 0x3f]
^ Spbox[2][(work >> 16) & 0x3f]
^ Spbox[0][(work >> 24) & 0x3f];
work = r ^ kptr[4*i+1];
l ^= Spbox[7][(work) & 0x3f]
^ Spbox[5][(work >> 8) & 0x3f]
^ Spbox[3][(work >> 16) & 0x3f]
^ Spbox[1][(work >> 24) & 0x3f];

work = rotr(l, 4U) ^ kptr[4*i+2];
r ^= Spbox[6][(work) & 0x3f]
^ Spbox[4][(work >> 8) & 0x3f]
^ Spbox[2][(work >> 16) & 0x3f]
^ Spbox[0][(work >> 24) & 0x3f];
work = l ^ kptr[4*i+3];
r ^= Spbox[7][(work) & 0x3f]
^ Spbox[5][(work >> 8) & 0x3f]
^ Spbox[3][(work >> 16) & 0x3f]
^ Spbox[1][(work >> 24) & 0x3f];
}

FPERM(l,r);

#ifdef IS_LITTLE_ENDIAN
*(word32 *)outBlock = byteReverse(r);
*(word32 *)(outBlock+4) = byteReverse(l);
#else
*(word32 *)outBlock = r;
*(word32 *)(outBlock+4) = l;
#endif
}

void DES_EDE_Encryption::ProcessBlock(byte *inoutBlock) const
{
e.ProcessBlock(inoutBlock);
d.ProcessBlock(inoutBlock);
e.ProcessBlock(inoutBlock);
}

void DES_EDE_Encryption::ProcessBlock(const byte *inBlock, byte *outBlock) const
{
e.ProcessBlock(inBlock, outBlock);
d.ProcessBlock(outBlock);
e.ProcessBlock(outBlock);
}

void DES_EDE_Decryption::ProcessBlock(byte *inoutBlock) const
{
d.ProcessBlock(inoutBlock);
e.ProcessBlock(inoutBlock);
d.ProcessBlock(inoutBlock);
}

void DES_EDE_Decryption::ProcessBlock(const byte *inBlock, byte *outBlock) const
{
d.ProcessBlock(inBlock, outBlock);
e.ProcessBlock(outBlock);
d.ProcessBlock(outBlock);
}

void TripleDES_Encryption::ProcessBlock(byte *inoutBlock) const
{
e1.ProcessBlock(inoutBlock);
d.ProcessBlock(inoutBlock);
e2.ProcessBlock(inoutBlock);
}

void TripleDES_Encryption::ProcessBlock(const byte *inBlock, byte *outBlock) const
{
e1.ProcessBlock(inBlock, outBlock);
d.ProcessBlock(outBlock);
e2.ProcessBlock(outBlock);
}

void TripleDES_Decryption::ProcessBlock(byte *inoutBlock) const
{
d1.ProcessBlock(inoutBlock);
e.ProcessBlock(inoutBlock);
d2.ProcessBlock(inoutBlock);
}

void TripleDES_Decryption::ProcessBlock(const byte *inBlock, byte *outBlock) const
{
d1.ProcessBlock(inBlock, outBlock);
e.ProcessBlock(outBlock);
d2.ProcessBlock(outBlock);
}

8. C語言實現DES演算法中各種矩陣變換

首先說下：樓主有找到DES演算法了嗎？還是打算自己寫？是這樣的？大家一般都是使用開源回的答OPNSSL，這個是好東西啊，里邊包含了全世界知名的加密演算法，而且是免費的。里邊就有你說點這些小方法。只要下載源碼，自己編譯，就能有自己的所有加密源碼了。

9. DES加密演算法C語言實現

#include<iostream.h>
class SubKey{ //定義子密鑰為一個類
public:
int key[8][6];
}subkey[16]; //定義子密鑰對象數組

class DES{
int encipher_decipher; //判斷加密還是解密
int key_in[8][8]; //用戶原始輸入的64位二進制數
int key_out[8][7]; //除去每行的最後一位校驗位
int c0_d0[8][7]; //存儲經PC-1轉換後的56位數據
int c0[4][7],d0[4][7]; //分別存儲c0,d0
int text[8][8]; //64位明文
int text_ip[8][8]; //經IP轉換過後的明文
int A[4][8],B[4][8]; //A,B分別存儲經IP轉換過後明文的兩部分,便於交換
int temp[8][6]; //存儲經擴展置換後的48位二進制值
int temp1[8][6]; //存儲和子密鑰異或後的結果
int s_result[8][4]; //存儲經S變換後的32位值
int text_p[8][4]; //經P置換後的32位結果
int secret_ip[8][8]; //經逆IP轉換後的密文
public:
void Key_Putting();
void PC_1();
int function(int,int); //異或
void SubKey_Proction();
void IP_Convert();
void f();
void _IP_Convert();
void Out_secret();
};
void DES::Key_Putting() //得到密鑰中對演算法有用的56位
{
cout<<"請輸入64位的密鑰(8行8列且每行都得有奇數個1):\n";
for(int i=0;i<8;i++)
for(int j=0;j<8;j++){
cin>>key_in[i][j];
if(j!=7) key_out[i][j]=key_in[i][j];
}
}
void DES::PC_1() //PC-1置換函數
{
int pc_1[8][7]={ //PC-1
{57, 49, 41, 33, 25, 17, 9},
{1, 58, 50, 42, 34, 26, 18},
{10, 2, 59, 51, 43, 35, 27},
{19, 11, 3, 60, 52, 44, 36},
{63, 55, 47, 39, 31, 23, 15},
{7, 62, 54, 46, 38, 30, 22},
{14, 6, 61, 53, 45, 37, 29},
{21, 13, 5, 28, 20, 12, 4}
};
int i,j;
for(i=0;i<8;i++)
for(j=0;j<7;j++)
c0_d0[i][j]=key_out[ (pc_1[i][j]-1)/8 ][ (pc_1[i][j]-1)%8 ];
}
int DES::function(int a,int b) //模擬二進制數的異或運算,a和b為整型的0和1,返回值為整型的0或1
{
if(a!=b)return 1;
else return 0;
}
void DES::SubKey_Proction() //生成子密鑰
{
int move[16][2]={ //循環左移的位數
1 , 1 , 2 , 1 ,
3 , 2 , 4 , 2 ,
5 , 2 , 6 , 2 ,
7 , 2 , 8 , 2 ,
9 , 1, 10 , 2,
11 , 2, 12 , 2,
13 , 2, 14 , 2,
15 , 2, 16 , 1
};
int pc_2[8][6]={ //PC-2
14, 17 ,11 ,24 , 1 , 5,
3 ,28 ,15 , 6 ,21 ,10,
23, 19, 12, 4, 26, 8,
16, 7, 27, 20 ,13 , 2,
41, 52, 31, 37, 47, 55,
30, 40, 51, 45, 33, 48,
44, 49, 39, 56, 34, 53,
46, 42, 50, 36, 29, 32
};
for(int i=0;i<16;i++) //生成子密鑰
{
int j,k;
int a[2],b[2];
int bb[28],cc[28];
for(j=0;j<4;j++)
for(k=0;k<7;k++)
c0[j][k]=c0_d0[j][k];
for(j=4;j<8;j++)
for(k=0;k<7;k++)
d0[j-4][k]=c0_d0[j][k];
for(j=0;j<4;j++)
for(k=0;k<7;k++){
bb[7*j+k]=c0[j][k];
cc[7*j+k]=d0[j][k];
}
for(j=0;j<move[i][1];j++){
a[j]=bb[j];
b[j]=cc[j];
}
for(j=0;j<28-move[i][1];j++){
bb[j]=bb[j+1];
cc[j]=cc[j+1];
}
for(j=0;j<move[i][1];j++){
bb[27-j]=a[j];
cc[27-j]=b[j];
}
for(j=0;j<28;j++){
c0[j/7][j%7]=bb[j];
d0[j/7][j%7]=cc[j];
}
for(j=0;j<4;j++) //L123--L128是把c0,d0合並成c0_d0
for(k=0;k<7;k++)
c0_d0[j][k]=c0[j][k];
for(j=4;j<8;j++)
for(k=0;k<7;k++)
c0_d0[j][k]=d0[j-4][k];
for(j=0;j<8;j++) //對Ci,Di進行PC-2置換
for(k=0;k<6;k++)
subkey[i].key[j][k]=c0_d0[ (pc_2[j][k]-1)/7 ][ (pc_2[j][k]-1)%7 ];
}
}
void DES::IP_Convert()
{
int IP[8][8]={ //初始置換IP矩陣
58, 50, 42, 34, 26, 18, 10, 2,
60, 52, 44, 36, 28, 20, 12, 4,
62, 54, 46, 38, 30, 22, 14, 6,
64, 56, 48, 40, 32, 24, 16, 8,
57, 49, 41, 33, 25, 17, 9, 1,
59, 51, 43, 35, 27, 19, 11, 3,
61, 53, 45, 37, 29, 21, 13, 5,
63, 55, 47, 39, 31, 23, 15, 7
};
cout<<"你好,你要加密還是解密?加密請按1號鍵(輸入1),解密請按2號鍵,並確定."<<'\n';
cin>>encipher_decipher;
char * s;
if(encipher_decipher==1) s="明文";
else s="密文";
cout<<"請輸入64位"<<s<<"(二進制):\n";
int i,j;
for(i=0;i<8;i++)
for(j=0;j<8;j++)
cin>>text[i][j];
for(i=0;i<8;i++) //進行IP變換
for(j=0;j<8;j++)
text_ip[i][j]=text[ (IP[i][j]-1)/8 ][ (IP[i][j]-1)%8 ];
}

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