字符翻翻看

感谢1407刘可盈同学提供

uses crt;

var

  p:array[1..4,1..5] of byte;

  q:array[1..4,1..5] of char;

  a,b,c,d,e,bs,g,i,j,f,s,x1,y1,x2,y2:byte;

begin

  clrscr;

  cursoroff;

  gotoxy(33,12);  //封面页

  writeln('字 符 翻 翻 看');

  gotoxy(35,14);

  writeln('By  刘可盈');

  delay(5000);

  clrscr;

 

  gotoxy(19,10);

  writeln('规则：每次可以翻两个数，如果这两个数对应的');

  gotoxy(19,12);

  writeln('字符相同便消除，请在20步内消除尽量多的字符');

  gotoxy(32,15);

  writeln('按回车键开始游戏');

  readln;

  clrscr;

 

  randomize;  //随机生成字符

  repeat

    a:=0;b:=0;c:=0;d:=0;e:=0;

    for  i:=1 to 4 do

      for j:=1 to 5 do begin

        p[i,j]:=random(5);

        case p[i,j] of

          0:begin q[i,j]:='@'; a:=a+1; end;

          1:begin q[i,j]:='#'; b:=b+1; end;

          2:begin q[i,j]:='$'; c:=c+1; end;

          3:begin q[i,j]:='&'; d:=d+1; end;

          4:begin q[i,j]:='%'; e:=e+1; end;

        end;

      end;

  until (a mod 2=0) and (b mod 2=0) and (c mod 2=0) and (d mod 2=0) and (e mod 2=0);  //检查字符是否成对

 

  bs:=21;

  gotoxy(12,20);

  write('the first: ');

  gotoxy(12,21);

  write('the second: ');

  gotoxy(12,5);

  write('step: ');

  cursoron;

 

  while (bs>1) and (g<20) do begin  //当步数为0或全部消完时结束游戏

    textcolor(white);

    bs:=bs-1;

    gotoxy(18,5);

    write(bs:2);

    for i:=1 to 4 do begin

      gotoxy(32,7+i*2);

      for j:=1 to 5 do if p[i,j]=6 then begin textcolor(darkgray);write(q[i,j]:3);end  //已消除的变成灰色字符

                                   else begin textcolor(yellow);write((i-1)*5+j:3);end;  //输出编号

      writeln;

    end;

 

    gotoxy(24,21);

    clreol;

    repeat  //读入第一个数

      gotoxy(23,20);

      clreol;

      textcolor(yellow);

      readln(f);

      x1:=(f-1) div 5+1;

      y1:=(f-1) mod 5+1;

    until (f>0) and (f<21) and (p[x1,y1]<6);  //防止输入的数超出范围或已消除

    gotoxy(30+y1*3,x1*2+7);

    write(q[x1,y1]:2);

    repeat  //读入第二个数

      gotoxy(24,21);

      clreol;

      textcolor(yellow);

      readln(s);

      x2:=(s-1) div 5+1;

      y2:=(s-1) mod 5+1;

    until (s>0) and (s<21) and (p[x2,y2]<6) and (f<>s);

    gotoxy(30+y2*3,x2*2+7);

    write(q[x2,y2]:2);

    delay(2000);

 

    if (p[x1,y1]=p[x2,y2]) then begin  //标记消除

      g:=g+2;

      p[x1,y1]:=6;

      p[x2,y2]:=6;

    end;

  end;

 

  clrscr;

  gotoxy(38,12);

  textcolor(lightred);

  if g=20 then write('Win!')

          else write('Fail');

  delay(5000);

end.