des算法的实现c++
时间: 2024-05-06 11:15:57 浏览: 16
DES(Data Encryption Standard)是一种对称加密算法,下面是一个简单的DES算法的实现(只用于学习和了解,不适用于实际加密):
```c++
#include <iostream>
#include <bitset>
#include <cstring>
using namespace std;
// 初始置换表
const int IP[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
};
// 逆初始置换表
const int IP_INV[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
};
// 扩展置换表
const int E[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
};
// S盒
const int S_BOX[8][4][16] = {
{
{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}
},
{
{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}
},
{
{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}
},
{
{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}
},
{
{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}
},
{
{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}
},
{
{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}
},
{
{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}
}
};
// P盒
const int P[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
};
// 密钥置换表
const int KEY_PC1[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
};
// 子密钥置换表
const int KEY_PC2[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
};
// 循环左移位数表
const int KEY_SHIFT[16] = {
1, 1, 2, 2,
2, 2, 2, 2,
1, 2, 2, 2,
2, 2, 2, 1
};
// 将64位明文/密文分为左右两部分(各32位)
void split(string msg, bitset<32>& left, bitset<32>& right) {
for (int i = 0; i < 32; i++) {
left[i] = msg[i];
right[i] = msg[i + 32];
}
}
// 将左右两部分合并为64位明文/密文
string merge(bitset<32> left, bitset<32> right) {
string res;
for (int i = 0; i < 32; i++) {
res += left[i] ? '1' : '0';
}
for (int i = 0; i < 32; i++) {
res += right[i] ? '1' : '0';
}
return res;
}
// 初始置换
void IP_transform(string& msg) {
string res(64, '0');
for (int i = 0; i < 64; i++) {
res[i] = msg[IP[i] - 1];
}
msg = res;
}
// 逆初始置换
void IP_INV_transform(string& msg) {
string res(64, '0');
for (int i = 0; i < 64; i++) {
res[i] = msg[IP_INV[i] - 1];
}
msg = res;
}
// 扩展置换
bitset<48> E_transform(bitset<32> right) {
bitset<48> res;
for (int i = 0; i < 48; i++) {
res[47 - i] = right[32 - E[i]];
}
return res;
}
// S盒变换
bitset<32> S_transform(bitset<48> right) {
bitset<32> res;
for (int i = 0; i < 8; i++) {
int row = right[47 - i * 6] * 2 + right[47 - i * 6 - 5];
int col = right[47 - i * 6 - 1] * 8 + right[47 - i * 6 - 2] * 4 + right[47 - i * 6 - 3] * 2 + right[47 - i * 6 - 4];
int num = S_BOX[i][row][col];
for (int j = 3; j >= 0; j--) {
res[31 - i * 4 - j] = num % 2;
num /= 2;
}
}
return res;
}
// P盒变换
bitset<32> P_transform(bitset<32> right) {
bitset<32> res;
for (int i = 0; i < 32; i++) {
res[31 - i] = right[32 - P[i]];
}
return res;
}
// 生成子密钥
void generate_sub_key(string key, string sub_key[16]) {
bitset<56> key_bitset;
bitset<28> left, right;
for (int i = 0; i < 56; i++) {
key_bitset[55 - i] = key[KEY_PC1[i] - 1] == '1';
}
for (int i = 0; i < 28; i++) {
left[27 - i] = key_bitset[55 - i];
right[27 - i] = key_bitset[27 - i];
}
for (int i = 0; i < 16; i++) {
left = ((left << KEY_SHIFT[i]) & 0xfffffff) | (left >> (28 - KEY_SHIFT[i]));
right = ((right << KEY_SHIFT[i]) & 0xfffffff) | (right >> (28 - KEY_SHIFT[i]));
bitset<56> merge_key(left.to_ulong());
merge_key <<= 28;
merge_key |= bitset<56>(right.to_ulong());
for (int j = 0; j < 48; j++) {
sub_key[i][j] = merge_key[55 - KEY_PC2[j]] ? '1' : '0';
}
}
}
// DES加密
string des_encrypt(string plaintext, string key) {
string sub_key[16];
generate_sub_key(key, sub_key);
IP_transform(plaintext);
bitset<32> left, right, new_left;
split(plaintext, left, right);
for (int i = 0; i < 16; i++) {
new_left = right;
right = P_transform(S_transform(E_transform(right) ^ bitset<48>(sub_key[i])));
right ^= left;
left = new_left;
}
string ciphertext = merge(right, left);
IP_INV_transform(ciphertext);
return ciphertext;
}
// DES解密
string des_decrypt(string ciphertext, string key) {
string sub_key[16];
generate_sub_key(key, sub_key);
IP_transform(ciphertext);
bitset<32> left, right, new_right;
split(ciphertext, left, right);
for (int i = 15; i >= 0; i--) {
new_right = left;
left = P_transform(S_transform(E_transform(left) ^ bit
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