破解vigenere密码算法C++版.....

cpp 代码

// P40第9题
// 破解vigenere密码课本P15---2.3.3所用的算法
//
#include
#include
//寻找密钥长度
int find_key_lenth(char*pass,int len)
{
    //pass密文,len 密文长度；移位统计相等的密文，取其最大的步数为d;
    int d=0,count,MaxCount=0;
    for(int step=1;step<10;step++)      //移动步数从1-10;
    {
        count=0;
        for(int j=0;j
        {
            if(pass[j]==pass[j+step])
                count++;
        }
        if(count>MaxCount)
        {
            MaxCount=count;
            d=step;
        }
    }
    return d;
}
// 发现密钥并解密。。。
void decode(char*pass,char*ming,int d,int len)
{
    float v[26]={0};               //V或W向量组；
    int per_len=len/d;             //每组长度；
    double A[26]={0.082,0.015,0.028,0.043,0.127,              //英文字母频率表A
              0.022,0.02,0.061,0.07,0.002,0.008,
              0.04,0.024,0.067,0.075,0.019,0.001,
              0.06,0.063,0.091,0.028,0.01,0.023,0.001,0.02,0.001};

    double B[26]={0};              //存储W*A值
    char*key;                      //密钥
    key=new char[d];
    for(int i=0;i
    {
        int j=0;
        while(true)                     //统计每组a--z出现的频率存在V[26]中
        {
            if((i+d*j)>=len) break;
            v[pass[i+d*j]-'a']+=1;
            j++;
        }
        for(int k=0;k<26;k++)          //计算W
            v[k]=v[k]/per_len;

        for(k=0;k<26;k++)              //计算B[i]=Ai*V;
        {
            for(int l=0;l<26;l++)
                B[k]+=A[l]*v[(l+k)%26];
        }
        //找出B中的与0.065最接近的值其的下标即为密钥
        double max=1;
        int c;
        for(k=0;k<26;k++)
        {
            if(fabs(B[k]-0.065)
            {
            max=fabs(B[k]-0.065);c=k;
            }
        }
        key[i]=c;
        //清空B，V；
        for(k=0;k<26;k++)
        {
            B[k]=0;
            v[k]=0;
        }
        printf("%c",'a'+key[i]);
    }
    //解密并显示
    printf("\n\n明文:\n");
    for(i=0;i
    {
      int tmp;
      tmp=pass[i]-'a';
      ming[i]=(tmp-key[i%d]+26)%26+'a';
      printf("%c",ming[i]);
    }
    printf("\n\n");
    return ;
}
int main()
{
    char password[1000]={0}; //密文
    char mingwen[1000]={0};  //明文
    FILE* fp;
    if((fp=fopen("xkju.txt","r+"))==NULL)
    {
        printf("xkju.txt open error!\n");
        return 1;
    }
    int i=0,d;
    printf("密文:");
    while((password[i++]=fgetc(fp))!=EOF);
    for(int j=0;j
    {
        if(j%59==0) printf("\n");
        printf("%c",password[j]);
    }
    d=find_key_lenth(password,i-1);
    printf("\n\nd==%d\n",d);
    printf("key=");
    decode(password,mingwen,d,i-1);
    fclose(fp);
    return 0;
}

发现指针有指针的好处,有好多问题都是JAVA不太方便处理的