具体原理和证明过程参见ElGamal加密体制,看个例子:
假设发送方为A,接收方为B,B选择素数p=13171,生成元g=2,私钥x=23。A欲用EIGamal算法将消息m=nupt加密为密文C后传送给B。消息m按英文字母表a=00,b=01,…,z=25编码,求加解密过程。
原题是bupt,我的代码中用的是nuptlovebupt(南邮)。
#include<bits/stdc++.h>
#include<time.h>
using namespace std;
string e_to_binary(int e)//模指数转二进制数
{
string str="";
int tail=0;
for(int i=0;i<32;i++)
{
int t=(e&1);
e=e>>1;
if(t==1)
str+='1';
else
str+='0';
}
for(int i=str.size()-1;i>=0;i--)
{
if(str[i]!='0')
{
tail=i;
break;
}
}
for(int i=0;i<(1+tail/2);i++)
{
char ch=str[i];
str[i]=str[tail-i];
str[tail-i]=ch;
}
return str.substr(0,tail+1);
}
int computation(string e_binary,int m,int n)//加密解密的数学运算
{
int c=1;
for(int i=0;i<e_binary.size();i++)
{
c=(c*c)%n;
if(e_binary[i]=='1')
{
c=(c*m)%n;
}
}
return c;
}
vector<string> encrypt_message(string message,int g,int y,int p)
{
unordered_map<char,string> mapp;
vector<string> en_message;
for(char ch='a';ch<='z';ch++)
{
mapp[ch]=to_string(ch-'a');
if(mapp[ch].size()<2)
mapp[ch]='0'+mapp[ch];
}
srand((int)time(NULL));
for(int i=0;i<message.size();i+=2)
{
int r= rand()%(p-1);
string tmp_str=e_to_binary(r);
int tmp_num=computation(tmp_str,g,p);
string tmp_str2=to_string(tmp_num);
while(tmp_str2.size()<4)
tmp_str2='0'+tmp_str2;
en_message.emplace_back(tmp_str2);//en_message+=tmp_str2;
string tmp_str_m=mapp[message[i]]+mapp[message[i+1]];
int tmp_num_m=stoi(tmp_str_m);
tmp_num_m=(tmp_num_m*computation(tmp_str,y,p))%p;
tmp_str_m=to_string(tmp_num_m);
while(tmp_str_m.size()<4)
tmp_str_m='0'+tmp_str_m;
en_message.emplace_back(tmp_str_m);//en_message+=tmp_str_m;
}
return en_message;
}
string decrypt_message(vector<string> en_message,int private_key,int p)
{
unordered_map<char,string> mapp;
unordered_map<string,char> mp;
string de_message="";
for(char ch='a';ch<='z';ch++)
{
mapp[ch]=to_string(ch-'a');
if(mapp[ch].size()<2)
mapp[ch]='0'+mapp[ch];
mp[mapp[ch]]=ch;
}
string e_binary=e_to_binary(private_key);
for(int i=0;i<en_message.size();i+=2)
{
string str_c1=en_message[i],str_c2=en_message[i+1];
int num_c1=stoi(str_c1),num_c2=stoi(str_c2);
int tmp_num=num_c2%p;
int tmp_num2=computation(e_binary,num_c1,p);
for(int i=1;;i++)
{
if(i*tmp_num2%p==1)
{
tmp_num2=i;
break;
}
}
tmp_num=tmp_num*tmp_num2%p;
string tmp_str=to_string(tmp_num);
while(tmp_str.size()<4)
tmp_str='0'+tmp_str;
de_message+=mp[tmp_str.substr(0,2)];
de_message+=mp[tmp_str.substr(2,2)];
}
return de_message;
}
int main()
{
int p=13171,g=2,private_key=23;
string e_binary=e_to_binary(private_key);
int y=computation(e_binary,g,p);
string message="nuptlovebupt";
cout<<"message:"<<message<<endl;
vector<string> en_message=encrypt_message(message,g,y,p);
cout<<"en_message:";
for(int i=0;i<en_message.size();i+=2)
cout<<"("<<en_message[i]<<","<<en_message[i+1]<<")";
cout<<endl;
string de_message=decrypt_message(en_message,private_key,p);
cout<<"de_message:"<<de_message<<endl;
return 0;
}