#include<stdio.h>
void main()
{
    int a[6]={79,59,12,2,79,35};
    int i,j,k,l,s;
    int count=0;
    for(i=97;i<=122;i++)
        for(j=97;j<=122;j++)
            for(k=97;k<=122;k++)
            {
            //    s=i*10000+j*100+k;
                count=0;
            //    for(l=0;l<6;l++)
            //    {
                    /*
                    if(s^a[l]<=90&&s^a[l]>=65||s^a[l]<=122&&s^a[l]>=97)
                        count++;
                    */
                    if((i^a[0]<=90&&i^a[0]>=65)||(i^a[0]<=122&&i^a[0]>=97))
                    {
                        count++;
                        printf("%d\n",i^a[0]);
                    }
                    if((j^a[1]<=90&&j^a[1]>=65)||(j^a[1]<=122&&j^a[1]>=97))
                    {
                        count++;
                        printf("%d\n",j^a[1]);
                    }
                    if((k^a[2]<=90&&k^a[2]>=65)||(k^a[2]<=122&&k^a[2]>=97))
                    {
                        count++;
                        printf("%d\n",k^a[2]);
                    }
                    if((i^a[3]<=90&&i^a[3]>=65)||(i^a[3]<=122&&i^a[3]>=97))
                    {
                        count++;
                        printf("%d\n",i^a[3]);
                    }
                    if((j^a[4]<=90&&j^a[4]>=65)||(j^a[4]<=122&&j^a[4]>=97))
                    {
                        count++;
                        printf("%d\n",j^a[4]);
                    }
                    if((k^a[5]<=90&&k^a[5]>=65||k^a[5]<=122&&k^a[5]>=97))
                    {
                        count++;
                        printf("%d\n",k^a[5]);
                    }
                    if(count==6)
                        printf("%c,%c,%c,%c,%c,%c\n",i^a[0],j^a[1],k^a[2],i^a[3],j^a[4],k^a[5]);
            //    }
            }
}

不知道我的思路对了吗？虽然，结果不是我想要的，我只想知道思路对了没有。谢谢

原以为会计算好长时间，没想到一敲回车就出来了，好没成就感

[root@Dora6 crack]#./crack
data length is 1201
Congratulations, the key is: god
(The Gospel of John, chapter 1) 1 In the beginning the Word already existed. He was with God, and he was God. 2 He was in the beginning with God. 3 He created everything there is. Nothing exists that he didn't make. 4 Life itself was in him, and this life gives light to everyone. 5 The light shines through the darkness, and the darkness can never extinguish it. 6 God sent John the Baptist 7 to tell everyone about the light so that everyone might believe because of his testimony. 8 John himself was not the light; he was only a witness to the light. 9 The one who is the true light, who gives light to everyone, was going to come into the world. 10 But although the world was made through him, the world didn't recognize him when he came. 11 Even in his own land and among his own people, he was not accepted. 12 But to all who believed him and accepted him, he gave the right to become children of God. 13 They are reborn! This is not a physical birth resulting from human passion or plan, this rebirth comes from God.14 So the Word became human and lived here on earth among us. He was full of unfailing love and faithfulness. And we have seen his glory, the glory of the only Son of the Father.
Cost of time: 0sec 2341usec

是不是题目太简单了？

ps: VxWorks 肯定以为我是在回复他，

哈哈，算法不难写，但是结果是要人工一个个找的，找得我眼睛都花了，不过还好昨晚半夜把它找了出来
有那位大侠知道自动检测结果的代码吗，比如在程序中循环找到要找的译文（这样的译文应该只有英文、标点、括号等等，而不该有$#@这些符号，检测代码就检测每次循环解密的译文是否满足这些条件）时，就停止循环，并保存这个译文，程序结束拜个先

求指教~自动检测的代码~~

[ 本帖最后由 stophin 于 2011-8-28 10:49 编辑 ]

求指教，自动检测的代码~~~~

[ 本帖最后由 stophin 于 2011-8-28 10:49 编辑 ]

字频，或者找一个一定会有的英文单词，比如the，有这个词就算是成功解密。

MD5加密算法是怎样的呢？

这个问题的难点就是明文的识别，你都用生物智能了还写什么程序

字频比较靠谱，至于找一定会有的单词就不怎么严谨了
不要光说不练，写个代码试试，又不复杂