随机数算法加解密的程序源码

Joduska

[C] 纯文本查看 复制代码

使用方法
     加密：OneBitOneCypher plaintext cyphertext password
     解密：OneBitOneCypher cyphertext plaintext password
*/
#include <stdio.h>
#include <stdlib.h>
// struct stat 结构的头文件。
#include <sys/stat.h>
#include <fcntl.h>
#include <unistd.h>


int main(int argc, char *argv[])
{
// 取明文/密文的文件大小。
    struct stat statFileSize;
    stat(argv[1], &statFileSize);
    unsigned long ulFileSize = statFileSize.st_size;

    unsigned char *pucPlaintext_Cyphertext = malloc(ulFileSize);

// 取明文/密文的文件描述符。
    unsigned long ulPlaintext_CyphertextFD = open(argv[1], O_RDONLY, S_IRUSR | S_IWUSR);
    unsigned long ulCyphertext_PlaintextFD = open(argv[2], O_CREAT | O_WRONLY, S_IREAD | S_IWRITE);

// read API。
    read(ulPlaintext_CyphertextFD, pucPlaintext_Cyphertext, ulFileSize);

// 至少8个字符的密码（包括空格）
    unsigned char aucKeyTable[256], aucPassword[16];

    unsigned long i, j;

    for(i = 0; i < 256; ++i)
    {
Loop:
        aucKeyTable[i] = rand() & 255;

        for(j = 0; j < i; ++j)
        {
            if(aucKeyTable[j] == aucKeyTable[i])
            {
                goto Loop;
            }
        }
    }

    for(i = 0; argv[3][i]; ++i)
    {
        aucPassword[i] = argv[3][i];
    }

    srand(*((unsigned int*)aucPassword) % RAND_MAX);

    for(j = 0; j < ulFileSize; j += 256)
    {
        unsigned long k, l, ulTablePosition;

        for(k = 0; k < i; ++k)
        {
            ulTablePosition = aucPassword[k] % 256;

            aucPassword[k] = aucKeyTable[ulTablePosition];
        }

        if(!(j /256 % 2))
        {
            unsigned long *pulKeySwap1 = (unsigned long*)aucKeyTable, *pulKeySwap2 = (unsigned long*)aucKeyTable, ulKeyTemp;

            for(l = 0; l < 32; ++l)
            {
                ulTablePosition = aucPassword[l] % 32;

                if(l != ulTablePosition)
                {
                    ulKeyTemp = pulKeySwap1[l];

                    pulKeySwap1[l] = pulKeySwap2[ulTablePosition];

                    pulKeySwap2[ulTablePosition] = ulKeyTemp;
                }
            }
        }
        else
        {
            for(k = 0; k < 256; ++k)
            {
                ulTablePosition = rand() % 256;

                if(k != ulTablePosition)
                {
                    aucKeyTable[k] += aucKeyTable[ulTablePosition];

                    aucKeyTable[ulTablePosition] = aucKeyTable[k] - aucKeyTable[ulTablePosition];

                    aucKeyTable[k] -= aucKeyTable[ulTablePosition];
                }
            }
        }

        for(l = 0; l < 256; ++l)
        {
               pucPlaintext_Cyphertext[j + l] ^= aucKeyTable[l];
        }
    }

    write(ulCyphertext_PlaintextFD, pucPlaintext_Cyphertext, ulFileSize);

    return 0;
}

方法并不适用！只是分析出来给大家解密的！

461735945

谢谢楼主热心分享

shangziq

楼主写的很好，很精辟

Joduska

shangziq 发表于 2019-7-4 11:00
楼主写的很好，很精辟

看到你的头像还以为是lcg该回复了我。

94079490

学习一下加密过程

92013

看样子很6666