#include

#include

long code(long m,long e,long n)//-----------加解密函数:用c=m^e(modn)加密，用m=c^d(modn)解密

{ long c=1;

e=e+1;

while(e!=1)

{

    c=c*m;

    c=c%n;

    e--;

}

return c;

}

int (long a,long b)//--------------------判断两个数是否互素

{

    long r0=a;

    long r1=b;

    int q,temp;

    while(r1!=0)

    {

        q=r0/r1;

        temp=r0;

        r0=r1;

        r1=temp-q*r1;

    }

    if(r0==1)//------------------------------最大公因数为1时两数互素

        return 1;             

    else return 0;        

}

long privatekey(long a,long b)//-------------用欧几里德算法求私钥d

{

    long r,r1=1;

    long r0=b;

    long s0,t0,e,s1,q,temprivatekey;

    r1=a;

    t0=0;

    e=1;

    s0=1;

    s1=0;

    q=r0/r1;

    r=r0-q*r1;

while(r>0)

    {

        temprivatekey=t0-q*e;

        t0=e;

        e=temprivatekey;

        temprivatekey=s0-q*s1;

        s0=s1;

        s1=temprivatekey;

        r0=r1;

        r1=r;

        q=r0/r1;

        r=r0-q*r1;

    }

while(e<0)

    {

        e=e+b;

    }

    return e;    

}

void main()//----------------------------------------主函数

{

    long p,q,e,d,m,n,t;

cout<<"请输入两个素数 p,q: ";

cin>>p>>q;

    n=p*q;

cout<<"n为 :"<    t=(p-1)*(q-1);//------------------------------------欧拉函数值

cout<<"n的欧拉函数值 :"<cout<<"请输入公钥e: ";

cin>>e;//-------------------------------------------输入公钥e,满足t>e>1

    int u;

    u=(e,t);//----------------------------------------判断e是否与n的欧拉函数值互素

if(1    {

     cout<<"请重新输入: ";

     cin>>e;

    }

    long e2=0;

    for(int i=0;;i++)

    {

        if((e2*e)%t==1)

            break;

        else e2++;

    }

cout<<"e2为："<    int w;

cout<<"输入1加密,输入2解密:";

cin>>w;

    switch(w)

    {

    case 1 ://----------------------------------------------加密

        {

         cout<<"请输入明文:";

         cin>>m;//---------------------------------------输入明文

         if(m<=n)

             cout<< "密文为:"<            else

            {

                long i=0,g,h;

                g=n;

                h=m;

                while(g!=0)//----------------------------------------求n的长度

                {

                    g/=10;

                    i++;

                }

                long j=0;

                while(h!=0)//----------------------------------------求明文长度

                {

                    h/=10;

                    j++;

                } 

                long x=0,long c[1000];//-----------------------------数组c用来存放加密后的各个分组

                int k=0;//-------------------------------------------k将记录明文分的组数

                while(j!=0)

                {

                    x=m/pow(10,j-i);

                 if(x>n)

                    { 

                        x=x/10;

                        c[k]=code(x,e,n);

                        m=m-x*pow(10,j-i+1);

                        j=j-1;

                        k++;

                    } 

                    else

                    {

                        c[k]=code(x,e,n);

                        m=m-x*pow(10,j-i);

                        j=j-2;

                        k++;

                    }

                 if(i>j)//----------------------------------------------最后一个分组的长度小于n的长度时直接加密

                        c[k]=code(m,e,n);

                }

             cout<<"密文输出:";

             for(long l=0;l                 cout<             cout<            }

            break;

        }

    case 2 ://------------------------------------------------解密

        {

         cout<<"请输入密文:"<         cin>>m;

            long d;

            d=privatekey(e,t);

         if(m<=n)

             cout<< "密文为:"<            else

            {

                long i=0,g,h;

                g=n;

                h=m;

                while(g!=0)

                {

                    g/=10;

                    i++;

                }

                long j=0;

                while(h!=0)

                {

                    h/=10;

                    j++;

                } 

                long x=0,long c[1000];

                int k=0;

                while(j!=0)

                {

                    x=m/pow(10,j-i);

                 if(x>n)

                    { 

                        x=x/10;

                        c[k]=code(x,d,n);

                        m=m-x*pow(10,j-i+1);

                        j=j-1;

                        k++;

                    } 

                    else

                    {

                        c[k]=code(x,d,n);

                        m=m-x*pow(10,j-i);

                        j=j-2;

                        k++;

                    }

                 if(i>j)

                        c[k]=code(m,d,n);

                }

             for(long l=0;l                    

                 cout<             cout<            }

            break;

        }

}

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