• 大三时候实现的,关于大整数(超过long范围)加减乘除操作的头文件,并包含了实现RSA加解密的函数


      1 #include<iostream>
      2 #include<time.h>
      3 #include<string.h>
      4 using namespace std;
      5 
      6 
      7 class BigNum{
      8 private:
      9     int *num;                                        //数值
     10     int *yu;                                        //余数
     11     int n;                                            //数值位数
     12     int YuNum;                                        //余数位数
     13     bool sym;                                        //正负,true为正
     14 public:
     15     BigNum(char *str){
     16         n = strlen(str);
     17         YuNum = 0;sym = true;
     18         num = new int[strlen(str)];
     19         for(int i = 0 ; i < strlen(str) ; i++)
     20             num[i] = str[strlen(str)-i-1]-'0';
     21     }
     22     BigNum(int A){
     23         int i,j;
     24         int a = A;
     25         for(i=0; a!=0; i++)
     26             a/=10;
     27         num = new int[i];
     28         a = A;
     29         for(j=0; j<i; j++){
     30             num[j] = a%10;
     31             a/=10;
     32         }
     33         n = i;
     34         YuNum = 0;
     35     }
     36     BigNum(bool a){
     37         YuNum = 0;sym = true;
     38         char *str = new char[256];
     39         cout<<"输入数以建立对象:"<<endl;
     40         cin>>str;
     41         n = strlen(str);
     42         num = new int[strlen(str)];
     43         for(int i = 0 ; i < strlen(str) ; i++)
     44             num[i] = str[strlen(str)-i-1]-'0';
     45     }
     46     BigNum(){
     47         sym = true;
     48         n = 0;
     49         YuNum = 0;
     50     }
     51     BigNum operator+(BigNum p){
     52         BigNum temp = BigNum();
     53         int i;
     54         if( this->sym + p.sym == 1){
     55             bool s;
     56             if(*this > p){
     57                 s = this->sym;
     58                 this->sym = p.sym;
     59                 temp = *this - p;
     60                 temp.sym = s;
     61             }
     62             else{
     63                 s = p.sym;
     64                 p.sym = this->sym;
     65                 temp = p - *this;
     66                 temp.sym = s;
     67             }
     68             return temp;
     69         }
     70         int c=0;                                                    //进位
     71         temp.n = max(this->n ,p.n );
     72         int *Num = new int[temp.n +1 ];
     73         for( i=0; i<min(this->n ,p.n); i++){
     74             Num[i]= (this->num[i] +p.num[i] +c) %10;
     75             c = (this->num[i]+p.num[i]+c)/10;
     76         }
     77         if(this->n > p.n)
     78             for( i = p.n ; i < temp.n ; i++){
     79                 Num[i] = (this->num[i]+c)%10;
     80                 c = (this->num[i]+c)/10;
     81             }
     82         else
     83             for( i=this->n;i<p.n;i++){
     84                 Num[i] = (p.num[i]+c)%10;
     85                 c = (p.num[i]+c)/10;
     86             }
     87         if(c>0){
     88             temp.n++;
     89             Num[temp.n-1]=c;
     90         }
     91         temp.num = new int[temp.n];
     92         for(int i=0;i<temp.n;i++)
     93             temp.num[i] = Num[i];
     94         return temp;
     95     }
     96     BigNum operator-(BigNum p){
     97         BigNum temp = BigNum();
     98         int i;
     99         if(this->sym + p.sym == 1){                            //异符号相减是两数绝对值相加,符号取减数
    100             bool s = this->sym;
    101             this->sym = p.sym;
    102             temp = *this + p;
    103             temp.sym = s;
    104             return temp;
    105         }
    106         if(!(*this > p)){
    107             temp = p - *this;                                //减法结果出现负数
    108             temp.sym = false;
    109             return temp;
    110         }
    111         int c=0;                                    //进位
    112         int *Num = new int[this->n];
    113         for( i=0;i<this->n;i++){
    114             if(i<p.n){
    115                 if(this->num[i]+c>=p.num[i]){
    116                     Num[i] = this->num[i]+c-p.num[i];
    117                     c=0;
    118                 }
    119                 else{
    120                     Num[i] = this->num[i]+c+10-p.num[i];
    121                     c=-1;
    122                 }
    123             }
    124             else{
    125                 if(this->num[i]+c>=0){
    126                     Num[i] = this->num[i]+c;
    127                     c = 0;
    128                 }
    129                 else{
    130                     Num[i] = this->num[i]+c+10;
    131                     c = -1;
    132                 }
    133             }
    134         }
    135         temp.n = this->n;
    136         for( i=this->n-1;Num[i]==0&&i>=0;i--)
    137             temp.n--;
    138         temp.num = new int[temp.n];
    139         for( i=0;i<temp.n;i++)
    140             temp.num[i] = Num[i];
    141         return temp;
    142     }
    143     BigNum operator*(BigNum p){
    144         int i;
    145         BigNum temp = BigNum();
    146         if(this->sym + p.sym == 1)
    147             temp.sym = false;
    148         if(this->n==0 || p.n==0)
    149             return temp;
    150         int *Num = new int[p.n+this->n+1];                        //循环中可能会用到最后一个数
    151         int c = 0;
    152         for(i=0; i < this->n+p.n; i++)
    153             Num[i] = 0;
    154         for(i=0; i < p.n+this->n; i++){
    155             for(int j=0; j < this->n; j++)
    156                 for(int k=0; k < p.n; k++){
    157                     if(i==j+k)
    158                         c += this->num[j]*p.num[k];
    159                 }
    160             Num[i]+=c%10;
    161             c=c/10;
    162         }
    163         for(i=this->n+p.n; Num[i] <= 0; )
    164             i--;
    165         temp.n=i+1;
    166         temp.num = new int[temp.n];
    167         for(i=0; i<temp.n; i++)
    168             temp.num[i] = Num[i];
    169         return temp;
    170     }
    171     bool operator> (BigNum p){                                //大于等于的效果,跟小于对立,只比较绝对值,不关心正负
    172         if(this->n > p.n)
    173             return true;
    174         else if(this->n < p.n)
    175             return false;
    176         else{
    177             for(int i=1 ; i <= p.n; i++){
    178                 if(this->num[n-i] == p.num[n-i])
    179                     continue;
    180                 else if(this->num[n-i] <= p.num[n-i])
    181                     return false;
    182                 else
    183                     return true;
    184             }
    185             return true;
    186         }
    187     }
    188     BigNum operator/(BigNum p){
    189         BigNum temp = BigNum();
    190         int i,j,k;
    191         temp.sym = this->sym + p.sym;
    192         if(!(*this > p)){
    193             temp.n = 0;
    194             temp.yu = new int[this->n];
    195             temp.YuNum = this->n;
    196             for( i=0;i<this->n;i++)
    197                 temp.yu[i] = this->num[i];
    198             return temp;
    199         }
    200         int *Num = new int[this->n-p.n+1];
    201         for( i=0;i<this->n-p.n+1;i++)                        //置零
    202             Num[i] = 0;
    203         BigNum mid = BigNum();                                    //让mid成为this的复制,在处理中要改变其中的值
    204         mid.n = this->n;
    205         mid.num = new int[mid.n];
    206         for( i = 0; i < mid.n; i++)
    207             mid.num[i] = this->num[i];
    208         int m = mid.n - p.n;                                    //用首位相减法,跟人手酸除法的过程相似,需要改变mid中的数值和位数,m存储低位没用到的位数
    209         mid.n = p.n;    mid.num += this->n - p.n;                //把mid置成跟p相同长度,开始相减,指针相应的向后移动
    210         for( i = this->n - p.n +1-1;i >= 0;  ){
    211             BigNum A = BigNum();                                //A是中间变量为了使mid的数值动态化
    212             while( mid > p ){
    213                 A = mid - p;
    214                 for( k=0; k < A.n ; k++)
    215                     mid.num[k] = A.num[k];
    216                 for( k=A.n; k < mid.n; k++)
    217                     mid.num[k] = 0;
    218                 Num[i] +=1;
    219                 for( j = mid.n-1; j >= 0 ;j--,mid.n-- )                //去掉开头的零
    220                     if(mid.num[j] != 0)
    221                         break;
    222             }
    223             if( m + mid.n<p.n || m == 0 && p >mid){                                        //剩下的数总和比P小,除法完成
    224                 temp.yu = new int[m + mid.n];
    225                 mid.num -= m;
    226                 for( k = 0; k < m + mid.n ; k++)
    227                     temp.yu[k] = mid.num[k];
    228                 temp.YuNum = m + mid.n;
    229                 //mid.n = this->n;
    230                 int j = this->n - p.n + 1;
    231                 for( ; j>0 ; j--)
    232                     if(Num[j-1] != 0)
    233                         break;
    234                 temp.num = new int[j];
    235                 for( k = 0;k < j ; k++ )
    236                     temp.num[k] = Num[k];
    237                 temp.n = j;
    238                 return temp;
    239             }
    240             else{
    241                 m -= p.n - mid.n;
    242                 mid.num -= p.n - mid.n;
    243                 i -= p.n - mid.n;
    244                 mid.n = p.n;
    245                 if( m > 0 && p > mid){                //借位
    246                     m--;
    247                     i--;
    248                     mid.num--;
    249                     mid.n++;
    250                 }
    251                 for( k = mid.n ; mid.num[k-1] == 0 && k >= 1 ; k--)
    252                     mid.n--;
    253             }
    254         }
    255         return temp;
    256     }
    257     void display(){
    258         int i;
    259         if( sym == false )
    260             cout<<'-';
    261         for( i=0 ;i < n; i++)
    262             cout<<num[n-i-1];
    263         if(YuNum > 0){
    264             cout<<endl;
    265             for( i=0; i < YuNum; i++)
    266                 cout<<yu[YuNum-i-1]<<' ';
    267         }
    268         cout<<endl;
    269     }
    270     static void swap(BigNum &a,BigNum &b){
    271         BigNum temp = a;
    272         a = b;
    273         b=temp;
    274     }
    275     static BigNum gcd(BigNum A,BigNum B){
    276         int i;
    277         BigNum temp;
    278         BigNum a;a = A;
    279         BigNum b;b = B;
    280         while(true){
    281             temp = a/b;
    282             if( a>b )
    283                 BigNum::swap( a,b );
    284             free(b.num);
    285             b.n = temp.YuNum;
    286             b.num = new int[a.n];
    287             for( i=0;i < temp.YuNum; i++){
    288                 b.num[i] = temp.yu[i];
    289             }
    290             if(  temp.YuNum==0 )
    291                 break;
    292         }
    293         return a;
    294     }
    295     void operator=(BigNum p){
    296         int i;
    297         this->~BigNum();
    298         this->sym = p.sym;
    299         this->n = p.n;
    300         this->num = new int[this->n];
    301         for( i=0; i< p.n; i++)
    302             this->num[i] = p.num[i];
    303         this->YuNum = p.YuNum;
    304         this->yu = new int[this->YuNum];
    305         for( i=0; i< p.YuNum; i++)
    306             this->yu[i] = p.yu[i];
    307     }
    308     static BigNum inv(BigNum a, BigNum b){                                            //小的放在前面
    309         BigNum a0 = a;
    310         BigNum b0 = b;
    311         BigNum t0;
    312         BigNum t("1");
    313         BigNum s0("1");
    314         BigNum s;
    315         BigNum q = a0 /b0;
    316         BigNum r = a0 - (q * b0);
    317         BigNum temp;
    318         while( r > BigNum("1") ){
    319             temp = t0 - (q * t);
    320             t0 = t;
    321             t = temp;
    322             temp = s0 - (q * s);
    323             s0 = s;
    324             s = temp;
    325             a0 = b0;
    326             b0 = r;
    327             q = a0 /b0;
    328             r = a0 - (q * b0);
    329         }
    330         r = b0;
    331         if( s.sym == false )
    332             s = s + b;
    333         return s;
    334     }
    335     bool operator==(BigNum p){
    336         if( this->n == p.n ){
    337             for(int i=0; i < p.n; i++)
    338                 if( this->num[i] != p.num[i] )
    339                     return false;
    340             return true;
    341         }
    342         return false;
    343     }
    344     char* toChar(){
    345         int i;
    346         char *str = new char[this->n+2];
    347         for(i=0; i<this->n; i++)
    348             str[this->n-i-1] = '0'+this->num[i];
    349         str[i] = '\0';
    350         return str;
    351     }
    352     BigNum mod(BigNum n){
    353         BigNum temp;
    354         if(!(*this> n)){
    355             temp.n = this->n;
    356             temp.num = new int[temp.n];
    357             for(; temp.n>0; temp.n--)
    358                 temp.num[temp.n-1] = this->num[temp.n-1];
    359             temp.n = this->n;
    360         }
    361         else{
    362             BigNum y = *this /n;
    363             temp.n = y.YuNum;
    364             temp.num = new int[temp.n];
    365             for( ; temp.n>0; temp.n--)
    366                 temp.num[temp.n-1] = y.yu[temp.n-1];
    367             temp.n = y.YuNum;
    368         }
    369         return temp;
    370     }
    371     static BigNum Square_and_Multiply(BigNum X, BigNum C, BigNum N){            //幂模运算z = x^c(mod n)
    372         int i;
    373         BigNum x = X;BigNum c = C;BigNum n = N;
    374         BigNum two("2");
    375         for(i=0; c.n>0; i++)
    376             c = c /two;
    377         bool *bol = new bool[i];
    378         c = C;
    379         for(i=0; c.n>0; i++){
    380             c = c /two;
    381             if(c.YuNum > 0)
    382                 bol[i] = 1;
    383             else
    384                 bol[i] = 0;
    385         }
    386         BigNum z("1");
    387         for( ; i >0; i--){
    388             z = (z*z).mod(n);                        //只返回正数
    389             if(bol[i-1] == 1)
    390                 z = (z*x).mod(n);
    391         }
    392         return z;
    393     }
    394     static bool Miller_Rabin(BigNum X){                //Miller Rabin算法判断大数X是否为素数,是素数返回true
    395         int i,k;
    396         BigNum x = X;
    397         BigNum two("2");
    398         x = x - BigNum("1");
    399         for(k=0; x.YuNum==0; k++){
    400             x = x /two;
    401         }
    402         BigNum m = x*two + BigNum("1");
    403         k= k- 1;
    404         BigNum b = Square_and_Multiply(BigNum("2"),m,X);
    405         x = X - BigNum("1");
    406         if( b==BigNum("1"))    return true;
    407         for(i=0; i<k; i++){
    408             if( b==x)
    409                 return true;
    410             else
    411                 b = Square_and_Multiply(b,two,X);
    412         }
    413         return false;
    414     }
    415     static BigNum create_prime(int m){
    416         int i;
    417         BigNum num("1");
    418         BigNum one("1");
    419         BigNum two("2");
    420         srand(time(0));
    421         for(i=0; i<m-1; i++){
    422             num = num* two + BigNum(rand()).mod(two);
    423         }
    424         BigNum temp;
    425         while(true){
    426             temp = num;
    427             for(i=1; temp>two; i++)
    428                 temp = temp /two;
    429             if(Miller_Rabin(num)&&i==m)
    430                 return num;
    431             else if(i>m){
    432                 num = create_prime(m);
    433                 return num;
    434             }
    435             else
    436                 num = num+ two;
    437         }
    438     }
    439 };

    以上是BigNum.h的源码,下面是应用范例:

     1 #include"BigNum.h"
     2 #include<fstream>
     3 
     4 void main(){
     5     char *str = new char[256];
     6     cout<<"输入密钥";cin>>str;
     7     BigNum c = BigNum(str);
     8     cout<<"输入n";cin>>str;
     9     BigNum n = BigNum(str);
    10     ifstream in("data.txt");
    11     ofstream out("result.txt");
    12     BigNum x;
    13     BigNum y;
    14     char g;
    15     while(!in.eof()){
    16         in.get(str,100,' ');
    17         if(!(in.eof()))
    18             in.get(g);
    19         x = BigNum( str );
    20         y = BigNum::Square_and_Multiply(x, c, n);
    21         out<<y.toChar()<<' ';
    22         y.display();
    23     }
    24     /*string ss;
    25     BigNum data;
    26     BigNum yu;                                                //余数
    27     BigNum sh;                                                //商
    28     BigNum one("1");
    29     while( !in.eof() ){
    30         in>>num;
    31         sh = BigNum("1");
    32         data = BigNum(num);
    33         for( BigNum i = BigNum("1") ; k >i; i =i +one){
    34             sh = sh * data;
    35             yu = sh / n;
    36             //free(temp->num);
    37             sh.~BigNum();
    38             sh =BigNum();
    39             sh.n =yu.YuNum;
    40             sh.num = new int[sh.n];
    41             for(int j = 0; j < yu.YuNum ; j++){
    42                 sh.num[j] = yu.yu[j];
    43             }
    44             //free(temp->yu);temp->YuNum =0;
    45         }
    46         sh.display();
    47         out<<sh.toChar()<<' ';
    48     }*/
    49     in.close();
    50     out.close();
    51     getchar();
    52     getchar();
    53 }

    其中需要文件data.txt(对其中的数据进行加密):

    599973434 2978448901 1049145034 

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  • 原文地址:https://www.cnblogs.com/zhidian314/p/2615201.html
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