请按以下步骤设计一个c语言程序:1. 输入64bit的明文进行IP置换后,分成左右两个分支各为32bit, 左边L0:32bit ,右边R0:32bit 2. 右分支:L1 = R0 左分支:引入48bit的密钥,R1=L0 异或 f(R0,K1) 3. 相同的操作进行16次的运算循环,算出相应的,R1~R16,L0~L16 4. 最后在进行IP的逆序置换,将左右两个分支再次合并为64bit密文 本次实验实现IP置换和IP逆置换算法。
时间: 2024-02-09 20:13:17 浏览: 69
好的,以下是一个简单的C语言程序,实现了您所要求的功能:
```
#include <stdio.h>
// IP置换
int IP_table[] = {
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
};
// IP逆置换
int IP_inv_table[] = {
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
};
// f函数,输入R0和K1,输出48bit结果
int f(int R0, int K1) {
// TODO: 实现f函数
return 0;
}
int main() {
// 输入64bit明文
long long plaintext;
printf("请输入64bit明文:");
scanf("%lld", &plaintext);
// IP置换
long long permuted_plaintext = 0;
for (int i = 0; i < 64; i++) {
int bit = (plaintext >> (64 - IP_table[i])) & 1;
permuted_plaintext |= (long long)bit << (63 - i);
}
// 分成左右两个32bit分支
int L0 = permuted_plaintext >> 32;
int R0 = permuted_plaintext & 0xffffffff;
// 进行16次的运算循环
int L = L0, R = R0;
for (int i = 1; i <= 16; i++) {
int tmp = R;
R = L ^ f(R, K[i]);
L = tmp;
}
// 合并左右两个分支
long long permuted_ciphertext = ((long long)R << 32) | L;
// IP逆置换
long long ciphertext = 0;
for (int i = 0; i < 64; i++) {
int bit = (permuted_ciphertext >> (64 - IP_inv_table[i])) & 1;
ciphertext |= (long long)bit << (63 - i);
}
// 输出64bit密文
printf("64bit密文为:%lld\n", ciphertext);
return 0;
}
```
请注意,上述代码中的f函数尚未实现,您需要根据具体需求自行实现。此外,K1和K数组也尚未定义,您需要根据具体需求定义并初始化这些变量。
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