//---------------------------------------------------------------------------

#include <vcl.h>
#pragma hdrstop

#include "slnum_p.h"
//---------------------------------------------------------------------------


#pragma package(smart_init)
#pragma resource "*.dfm"
TForm1 *Form1;

slnum   a,b,c;
//---------------------------------------------------------------------------
__fastcall TForm1::TForm1(TComponent* Owner)
        : TForm(Owner)
{
}
//---------------------------------------------------------------------------
//---------------從string類中取出字符串轉(zhuǎn)換為char * -------------------------------
char * strtoa(String ss)
{
     char *sa;
     sa=new char[strlen(ss.c_str())+1];
     strcpy(sa,ss.c_str() );
     return sa;
}
/*-----------------------------------------------------------------------
版權(quán)所有：貴州省天柱縣二中  羅國(guó)文，郵編：556600，Email:xuefu998@yahoo.com.cn
本程序提供了一種關(guān)于素?cái)?shù)判定的新算法，以這些基本函數(shù)為基礎(chǔ)，可幫助我們
處理關(guān)于素?cái)?shù)的許多命題。
版權(quán)聲明：在您使用和推廣本程序時(shí)尊重作者版權(quán)，不準(zhǔn)在不經(jīng)許可的情況下發(fā)表
-------------------------------------------------------------------------*/

slnum Test_1(slnum x)            //求模2的倍數(shù)
{
     slnum n;
     n=x/2;
     return  n;
}

int Test_2(slnum x)             //判斷x的余數(shù)，返回值{0，1}
{
     int k;
     k=x.sl[0]%2;
     return k;
}

int Test_P(slnum n)        //P素性檢驗(yàn)
{
    slnum m("1");
    slnum M;
    M=(n-1)/3;
    for(m=1;(m<=M);m=m+1) {
          if(n%(m*2+1)==m) return 0;
         }
    return 1;
}

//數(shù)x的素性檢驗(yàn)，x為素?cái)?shù)返回1，否則為0
int Test_X(slnum x)
{
   if (Test_2(x)==0) return 0;
   if (x.arr_n ==1&&atol(x.soure )==1) return 0;
   slnum n;
   n=Test_1(x);
   return (Test_P(n));
}


/*
void Out_X(slnum a,slnum b)    //輸出[a,b]間的素?cái)?shù)
{ slnum x;
  Form1->Memo1->Lines->Clear();
  if (a.arr_n ==1&&a.sl[0]<3)  Form1->Memo1->Lines->Append('2');
  if (a.arr_n ==1&&a.sl[0]<5)  Form1->Memo1->Lines->Append('3');
  for(x=a;x<=b;x=x+1)
  {  if(Test_X(x)==1)  Form1->Memo1->Lines->Append(x.soure );}
}

void Out_N1(slnum a,slnum b)    //輸出[a,b]間使6n+1為素的n值
{  slnum n;
   for(n=a;n<=b;n=n+1)
   { if(Test_P(n,1)==1) cout<<n.soure <<endl;}
}

void Out_N_1(slnum a,slnum b)    //輸出[a,b]間使6n-1為素的n值
{  slnum n;
   for(n=a;n<=b;n=n+1)
   { if(Test_P(n,-1)==1) cout<<n.soure<<endl;}
}
//輸出[a,b]間的孿生素?cái)?shù)
void Out_2P(slnum a,slnum b)
{
        slnum n,n1,n2;
        n1=Test_1(a);
        n2=Test_1(b);
        for  (n=n1;n<=n2;n=n+1)
            if(Test_P(n,-1)*Test_P(n,1)==1)
               cout<<"["<<n.soure<<":("<<n*6-1<<" , "<<n*6+1<<")]"<<endl;
}


//求出任意的大于4的整數(shù)m為中心的第一對(duì)素?cái)?shù)
int Out_Q(slnum m)
{
   slnum S;int s;
   slnum n, k;
   n=Test_1(m);
   S=(m%6);
   s=S.sl[0];
   if(n==0)return  0 ;
   switch (s) {
     case 5 :
                for(k=0;k<n;k=k+1)
                {if(Test_P(n+k,-1)*Test_P(n-k,-1)==1)
                   { cout<<"["<<m*2<<":("<<(n+k)*6-1<<" , "<<(n-k)*6-1<<")]"<<endl;
                     return 0;}
                 }
                   cout<<"Not find this number!!"<<endl;
     case 4 :
              for(k=0;k<n;k=k+1)
                {if(Test_P(n+1+k,1)*Test_P(n-k,1)==1)
                   { cout<<"["<<m*2<<":("<<(n+1+k)*6+1<<" , "<<(n-k)*6+1<<")]"<<endl;
                     return 0;}
                 if(Test_P(n+k,1)*Test_P(n+1-k,1)==1)
                   { cout<<"["<<m*2<<":("<<(n+k)*6+1<<" , "<<(n+1-k)*6+1<<")]"<<endl;
                     return 0;}
                 }
                   cout<<"Not find this number!!"<<endl;
     case 3 :
              for(k=0;k<n;k=k+1)
                {if(Test_P(n+1+k,1)*Test_P(n-k,-1)==1)
                   { cout<<"["<<m*2<<":("<<(n+1+k)*6+1<<" , "<<(n-k)*6-1<<")]"<<endl;
                     return 0;}
                 if(Test_P(n+k,1)*Test_P(n+1-k,-1)==1)
                   { cout<<"["<<m*2<<":("<<(n+k)*6+1<<" , "<<(n+1-k)*6-1<<")]"<<endl;
                     return 0;}
                 if(Test_P(n+1+k,-1)*Test_P(n-k,1)==1)
                   { cout<<"["<<m*2<<":("<<(n+1+k)*6-1<<" , "<<(n-k)*6+1<<")]"<<endl;
                     return 0;}
                 if(Test_P(n+k,-1)*Test_P(n+1-k,1)==1)
                   { cout<<"["<<m*2<<":("<<(n+k)*6-1<<" , "<<(n+1-k)*6+1<<")]"<<endl;
                     return 0;}
                 }
                  cout<<"Not find this number!!"<<endl;
     case 2 :
              for(k=0  ; k < n ; k=k+1)
                {if(Test_P(n+1+k,-1)*Test_P(n-k,-1)==1)
                   { cout<<"["<<m*2<<":("<<(n+1+k)*6-1<<" , " <<(n-k)*6-1<<")]"<<endl;
                     return 0;}
                 if(Test_P(n+k,-1)*Test_P(n+1-k,-1)==1)
                   { cout<<"["<<m*2<<":("<<(n+k)*6-1<<" , "<<(n+1-k)*6-1<<")]"<<endl;
                     return 0;}
                 }
                 cout<<"Not find this number!!"<<endl;
     case 1 :
                for(k=0;k<n;k=k+1)
                {if(Test_P(n+k,1)*Test_P(n-k,1)==1)
                    {cout<<"["<<m*2<<":("<<(n+k)*6+1<<" , "<<(n-k)*6+1<<")]"<<endl;
                    return 0; }
                }
                cout<<"Not find this number!!"<<endl;
     case 0 :
                for(k=0;k<n;k=k+1)
                {if(Test_P(n+k,1)*Test_P(n-k,-1)==1)
                   { cout<<"["<<m*2<<":("<<(n+k)*6+1<<" , "<<(n-k)*6-1<<")]"<<endl;
                     return 0;}
                 if(Test_P(n+k,-1)*Test_P(n-k,1)==1)
                   { cout<<"["<<m*2<<":("<<(n+k)*6-1<<" , "<<(n-k)*6+1<<")]"<<endl;
                     return 0;}
                 }
                 cout<<"Not find this number!!"<<endl;
     }
     return 1;
}


*/



void __fastcall TForm1::Button1Click(TObject *Sender)
{
  slnum x; int i;
  Form1->Memo1->Lines->Clear();
  if (a.arr_n ==1&&a.sl[0]<=2) { Form1->Memo1->Lines->Append('2'),i=1;}
  for(x=a;x<=b;x=x+1)
          if(Test_X(x)==1){++i;  Form1->Memo1->Lines->Append(x.soure );}
  char *s,*ss;
  s=new char[65536];
  ss=new char[65536];
  strcpy(ss,"這個(gè)范圍內(nèi)的素?cái)?shù)有：");
  ltoa(i,s,10);
  strcat(ss,s);
  strcat(ss,"個(gè)");
  Form1->Label7->Caption =ss;
}
//---------------------------------------------------------------------------



void __fastcall TForm1::Edit2Exit(TObject *Sender)
{
      b.setval(strtoa(Form1->Edit2->Text));
}
//---------------------------------------------------------------------------

void __fastcall TForm1::Edit1Exit(TObject *Sender)
{
      a.setval(strtoa(Form1->Edit1->Text));
}
//---------------------------------------------------------------------------


void __fastcall TForm1::Button6Click(TObject *Sender)
{
  slnum x; int i=0;
  Form1->Memo1->Lines->Clear();
  for(x=a;x<=b;x=x+1)
          if(Test_P(x)==1){++i;  Form1->Memo1->Lines->Append(x.soure );}
  char *s,*ss;
  s=new char[65536];
  ss=new char[65536];
  strcpy(ss,"這個(gè)范圍內(nèi)的Q數(shù)有：");
  ltoa(i,s,10);
  strcat(ss,s);
  strcat(ss,"個(gè)");
  Form1->Label7->Caption =ss;
}
//---------------------------------------------------------------------------